Produkte endlicher nilpotenter

Gruppen

Diplomarbeit

vorgelegt dem Fachbereich Mathematik an der Johannes-Gutenberg-Universit^{}ät Mainz

im Mai 1993

von Burkhard H^{}öfling

Boppstraße 1 6500 Mainz 1

Das Thema der Arbeit wurde von Prof. Dr. Bernhard Amberg gestellt.

Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1 Basic properties of products of groups . . . . . . . . . . . . . . . . . 6 1.1 Subgroups of products of groups . . . . . . . . . . . . . . . . . . . . . . 6 1.2 Factorizers, prefactorizers, and quotient groups . . . . . . . . . . . . . . . 11 1.3 Classes of groups and factorizations . . . . . . . . . . . . . . . . . . . . 14

2 Factorizations of finite soluble groups . . . . . . . . . . . . . . . . . 18 2.1 Hall subgroups and Hall systems . . . . . . . . . . . . . . . . . . . . . 18 2.2 Products of subgroups of coprime order . . . . . . . . . . . . . . . . . . 20 2.3 Groups with many factorized subgroups . . . . . . . . . . . . . . . . . . 24

3 Subgroups of products of two finite nilpotent groups . . . . . . . . . 27 3.1 First results about products of nilpotent subgroups . . . . . . . . . . . . . 27 3.2 Primitive groups that are the product of two nilpotent subgroups . . . . . . 31 3.3 Pronormal and abnormal subgroups . . . . . . . . . . . . . . . . . . . . 35 3.4 Projectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.5 ℌ-maximal subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 3.6 Injectors and radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4 Structural properties of a product of two finite nilpotent groups . . 50 4.1 Fundamental results about the p-length of a p-soluble group . . . . . . . . . 50 4.2 Bounds on the π-lengths of products of two finite nilpotent subgroups …. 52 4.3 Connections between Fitting length and π-lengths . . . . . . . . . . . . . 57 4.4 Bounds on Fitting length and derived length . . . . . . . . . . . . . . . . 60 4.5 Properties of the Fitting quotient group . . . . . . . . . . . . . . . . . . 73

List of symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

Introduction

A group G is said to be the product of its subgroups G_{i} for i ∈ I if G is generated by the G_{i} and the set G_{i}G_{j} is a subgroup of G for all i, j ∈ I. In this terminology, a famous result of P. Hall [24] published in 1937 can be formulated as follows: The finite soluble groups G are precisely the finite groups which are the product of certain of their Sylow subgroups. This shows in particular that the finite soluble groups are groups which are the product of certain of their nilpotent subgroups.

For about 20 years, it remained an open question whether the converse of the last statement is also true, namely whether a finite group that is the product of certain nilpotent subgroups is soluble. In 1955, Itô [32] obtained a first result in this direction, namely that a product of two (possibly infinite) abelian subgroups is metabelian.

The original question was answered positively by Wielandt [45] and Kegel [33] in 1958 and 1961 respectively. Their result, stating that a finite group is soluble if and only if it is the product of finitely many nilpotent subgroups, has become known as the Kegel-Wielandt theorem.

This motivates the following general question: If the group G is the product of its subgroups G_{i} with i ∈ I, and certain group-theoretical properties of the G_{i} are known, what group-theoretical properties does the group G have?

One problem in answering this question is that in general, the group G is not uniquely determined by the G_{i} since the concept of a product of subgroups includes e.g. direct and semidirect products. Observe however that these kinds of products are not typical because the factors are not necessarily normal in G (for a finite group that is the product of two nonnormal subgroups see Example 3.6.8 below).

A second question, often encountered when dealing with questions of the first type, is the following: Which subgroups of G inherit the product structure of G, or, more concretely, which subgroups of G are conjugate to (isomorphic to) a prefactorized subgroup S of G, i.e. a subgroup that is the product of its subgroups S ∩ G_{i}?

In this Diplomarbeit, we will be concerned mainly with groups G that are the product of two subgroups A and B. Then a subgroup S of G is prefactorized if S = (S∩A)(S∩B).

A prefactorized subgroup that also contains A ∩ B will be termed factorized. Chapter 1 will be dedicated to the study of factorized and in particular prefactorized subgroups in the subgroup lattice of the group G = AB and in quotient groups.

Since products of finite nilpotent subgroups are soluble by the Kegel-Wielandt theorem, we study in Chapter 2 finite groups G which are the product of two subgroups and which satisfy some properties related to solubility. Our results are based on a result of Wielandt [44] who proved that for every prime p, the finite group G has a prefactorized Sylow p-subgroup. We will show in particular that if G is soluble, then it possesses a Hall system that consists entirely of prefactorized Hall subgroups: a Hall system of this type will play an important role in Section 3.3.

Furthermore, we will show in Section 2.2 and Section 2.3 that a relatively large number of subgroups of G is prefactorized or factorized if A and B have coprime indices or orders respectively: if G is finite and soluble, then it is possible to find a prefactorized or factorized subgroup among the conjugates of any subgroup.

In Chapter 3, we specialize to groups G that are the products of two nilpotent subgroups A and B. One of the main tools of this and the following chapter will be a theorem due to Gross which gives important information about the structure of G when G is primitive. This theorem will be proved in Section 3.2 in a somewhat more general form, based on the proof of Gross’ result given in [4].

The rest of Chapter 3 is dedicated to finding prefactorized or factorized conjugates for certain subgroups of G: we will show that for every abnormal subgroup of G, there is exactly one factorized conjugate (Proposition 3.3.5). This result will then be used to reduce the question whether certain pronormal subgroups of G are prefactorized or factorized to questions about normal subgroups. As a first application, we will be able to improve a result of Heineken [28], showing that for every Schunck class ℌ containing all finite nilpotent groups, G has a unique factorized ℌ-projector. More generally, we will show in Theorem 3.5.1 that every ℌ-maximal subgroup has a factorized conjugate. However, these factorized subgroups are not necessarily isomorphic.

A corresponding result about 𝔉-injectors for arbitrary Fitting classes 𝔉 does not hold (see Example 3.6.8). Nevertheless, using our results about pronormal subgroups, we will prove in Section 3.6 that a product of two finite nilpotent subgroups has a prefactorized or factorized 𝔉-injector if the 𝔉-radical of every product of two finite nilpotent subgroups is prefactorized or factorized respectively and that in this case, every finite group that is the product of two nilpotent subgroups has exactly one prefactorized 𝔉-injector. By Proposition 3.6.9, this result applies in particular to all saturated Fitting-formations and thus to all subgroup-closed Fitting classes. This result generalizes a well-known result of Amberg [1] and Pennington [40] stating that the Fitting subgroup of a product of two finite nilpotent subgroups is always factorized.

We do not know wheter our results about injectors and radicals can be extended to certain other classes of subgroups. In this context, it is of interest to ask whether the hypercentre of a product G of two finite nilpotent subgroups is prefactorized, since such a group has a factorized system normalizer (see Corollary 3.3.13). On the other hand, an example due to Gillam [19] shows that Z(G) is not necessarily prefactorized.

Also, a number of other characteristic subgroups of the group G = AB need not be prefactorized (or factorized), among them the derived subgroup G′, the nilpotent residual G^{𝔑} and the subgroups O^{π}(G) (see Example 3.4.4).

If π is an arbitrary set of primes, we show in Chapter 4 how to obtain upper bounds on the π-length of a group that is the product of two finite nilpotent subgroups A and B in terms of certain invariants of A and B from bounds on the p-length of a finite soluble group in terms of a Sylow p-subgroup. We use the bounds found by Hall and Higman [26], supplemented by some additional results of Gross [21] and Berger and Gross [9] when p = 2, to show e.g. that for every set π of odd primes,

l_{π}(G) ≤ max{d(A), d(B)}

and thus for any set π of primes,

l_{π}(G) ≤ max{d(A), d(B)} + 1.

Our bounds for the p-lengths are best-possible in the sense of [26] whenever the bounds in terms of the Sylow p-subgroups are best-possible.

As a further result, it will turn out that nilpotent length and the p-lengths of a group that is the product of two finite nilpotent subgroups are not too far removed from each other: We will show in Theorem 4.3.1 that

n(G) ≤ 2 max_{p∈±⊬ℙ}{l_{p}(G)},

n(G) ≤ 2 max_{p∈±⊬ℙ}{l_{p′}(G)} + 1,

and

n(G) ≤ max_{p∈±⊬ℙ}{l_{p}(G) + l_{p′}(G)}.

Using essentially the above results, the nilpotent length of a product G of two finite nilpotent subgroups A and B can also be bounded in terms of invariants of A and B. In particular, we show that

n(G) ≤ 2d(A) + max{0, 2d_{2}(A) − 4} + 1

and

n(G) ≤ d(A) + d(B) + max{0, d_{2}(A) − 2, d_{2}(B) − 2}.

These bounds are best-possible at least for groups of odd order.

Although it seems to be an open question whether the derived length of G can be bounded by the derived lengths of A and B, we can still obtain some information about the derived length of G: if π is the set of common prime divisors of the orders of A and B, then we obtain in Section 4.4:

\begin{matrix}d(G/Φ(G) ∩ O_{π}(G)) ≤ & max\{c_{2}(A), \frac{1}{2}d_{2′}(A)(d_{2′}(A) + 1)\} \\ + & max\{c_{2}(B), \frac{1}{2}d_{2′}(B)(d_{2′}(B) + 1)\}, \end{matrix}

\begin{matrix}d(G/Φ(G) ∩ O_{π}(G)) = & \frac{1}{2}d(A)(d(A) + 1) + \frac{1}{2}d(B)(d(B) + 1) \\ + & max\{\frac{1}{2}d_{2}(A)(d_{2}(A) + 1), \frac{1}{2}d_{2}(B)(d_{2}(B) + 1)\} \end{matrix}

and also

d(G/F (G)) ≤ (d(A) + max{0, d_{2}(A) − 1})(d(B) + max{0, d_{2}(B) − 1}).

Finally, in Section 4.5, further structural information about the Fitting quotient group G/F (G) will be obtained; this will give rise to relate the class of products of finite nilpotent groups to other classes of finite soluble groups and in particular to the class of groups ‘with many Sylow bases’ introduced by Huppert in [30].

The notation used is standard and follows Doerk and Hawkes [13], Robinson [43] and Amberg, Franciosi and de Giovanni [4]. For details, the reader is referred to the list of symbols at the end of the text.

Chapter 1

Basic properties of products of groups

1.1 Subgroups of products of groups

Let G be a group. If X and Y are subsets of G, define

XY = {xy | x ∈ X, y ∈ Y }.

We first recall the following elementary lemma which allows to calculate the cardinality of the set AB; for the proof, see e.g. Robinson [43], 1.3.11 or Doerk and Hawkes [13], A.1.5.

1.1.1 Lemma. Let A and B be subgroups of the group G, then |AB|·|A∩B| = |A|·|B|. In particular if A and B are finite, then

|AB| = \frac{|A| · |B|}{|A ∩ B|}.

If X and Y are subgroups of G, one is particularly interested in whether XY is again a subgroup of G. In this case, X and Y are said to be permutable because of the well-known equivalence of (i) and (ii) in the following

1.1.2 Lemma. Let G be a group and let A and B be subgroups of G. Then the following statements about the set AB are equivalent:

(i) AB is a subgroup of G; (ii) AB = BA (i.e. A and B are permutable); (iii) [a, b] ∈ AB for all a ∈ A and b ∈ B; (iv) a^{b} ∈ AB for all a ∈ A and b ∈ B.

Proof. (i) ⇒ (ii). Let g ∈ AB. Since AB is a group, also g^{−1} ∈ AB: write g^{−1} = ab with a ∈ A and b ∈ B. Then g = b^{−1}a^{−1} ∈ BA and therefore AB ⊆ BA. Similarly, BA ⊆ AB so that AB = BA.

(ii) ⇒ (iii). Let a ∈ A and b ∈ B, then we have [a, b] = a^{−1}b^{−1}ab ∈ ABAB = A(AB)B = AB as required.

(iii) ⇒ (iv). Since a^{b} = a[a, b] for all a ∈ A and b ∈ B, we have a^{b} ∈ A(AB) = AB.

(iv) ⇒ (i). Let a_{1}, a_{2} ∈ A and b_{1}, b_{2} ∈ B. We have to show that (a_{1}b_{1})^{−1}(a_{2}b_{2}) ∈ AB. But this is true since

(a_{1}b_{1})^{−1}(a_{1}b_{1}) = b_{1}^{−1}a_{1}^{−1}a_{2}b_{2} = (a_{1}^{−1}a_{2})^{b_{1}}b_{1}^{−1}b_{2} ∈ (AB)BB = AB.

The following lemma will be needed later.

1.1.3 Lemma. . Let G be a group with subgroups A, B and C such that AC = CA and BC = CB. Then also ⟨A, B⟩C = C⟨A, B⟩.

Proof. Let g ∈ ⟨A, B⟩ and c ∈ C. By Lemma 1.1.2, it suffices to show that g^{c} ∈ ⟨A, B⟩. Since g ∈ ⟨A, B⟩, g can be written g = a_{1}b_{1} … a_{r}b_{r} with a_{i} ∈ A and b_{i} ∈ B for all b = 1 … r. Since by Lemma 1.1.2, a_{i}^{c} ∈ AC and b_{i}^{c} ∈ BC for all i, we obtain

g^{c} = a_{1}^{c}b_{1}^{c} … a_{r}^{c}b_{r}^{c} ∈ (ACBC) … (ACBC)_{⟵r times⟶} = (AB) … (AB)_{⟵r times⟶}C

because C permutes with both A and B. Since clearly

(AB) … (AB)_{⟵r times ⟶} ⊆ ⟨A, B⟩,

we have g^{c} ∈ ⟨A, B⟩C.

A group G is said to be the product of its subgroups A and B if G = AB. Sometimes such a group is also called factorized by A and B^{1} or simply factorized. A subgroup S of G = AB is called prefactorized if S = (S ∩ A)(S ∩ B), or equivalently, if every s ∈ S can be written s = ab with a ∈ A ∩ S and b ∈ B ∩ S. Following Wielandt [45], a subgroup S of G is called factorized if whenever s = ab with a ∈ A and b ∈ B, then a ∈ S (and b ∈ S). Since every g ∈ G, and thus every element g of S, can be written g = ab with a ∈ A and b ∈ B, every factorized subgroup of G is prefactorized. It is also clear that every subgroup of G containing A or B is factorized; in particular G is factorized.

For some statements concerning groups which are the product of two of their subgroups, the following generalization of Dedekind’s modular law is useful:

1 We will avoid this terminology wherever possible since this might lead to confusion when dealing with

prefactorized subgroups S which are — as groups — factorized by A ∩ S and B ∩ S. In the opinion of

the author, it might have been preferable to use the terms factorized and strongly factorized instead

of prefactorized and factorized respectively because of this problem of terminology. As we will see

later, many results about factorized subgroups also hold for the subgroups now called prefactorized,

which could be an indication that the concept of being prefactorized might be the more natural one

when dealing with subgroups of factorized groups.

1.1.4 Lemma. Let G be a group. If X, Y and U are subsets of G such that U^{−1} = {u^{−1} | u ∈ U} ⊆ U, then

(i) if XU ⊆ X, then (X ∩ Y)U = X ∩ Y U and (ii) if UX ⊆ X, then U(X ∩ Y) = X ∩ UY. Proof. (i) The proof follows that usually given for Dedekind’s modular law: Clearly,

(X ∩ Y)U ⊆ XU ∩ Y U ⊆ X ∩ Y U. Now let x ∈ X ∩ Y U and write x = yu with y ∈ Y and u ∈ U. Then xu^{−1} ∈ XU ∩ Y ⊆ X ∩ Y and therefore x = xu^{−1}u ∈ (X ∩ Y)U. The proof of (ii) is similar.

The following lemma shows in particular that prefactorized subgroups are factorized if and only if they contain A ∩ B:

1.1.5 Lemma. Suppose that the group G is the product of its subgroups A and B. Then the following statements are equivalent:

(i) S is factorized; (ii) S = (S ∩ A)(S ∩ B) and A ∩ B ≤ S; (iii) A ∩ SB = A ∩ S; (iv) A∩SB ⊆ S; (v) AS∩B = S∩B; (vi) AS∩B ⊆ S; (vii) S = (AS∩B)(A∩SB); (viii) S = (AS ∩ B)S; (ix) S = S(A ∩ SB): (x) S = AS ∩ BS; (xi) S = SA ∩ SB.

Proof. (i) ⇒ (ii). Since every factorized subgroup is prefactorized, it remains to show that A ∩ B ≤ S. For every x ∈ A ∩ B, we have 1 = xx^{−1} ∈ S and, of course, x ∈ A and x^{−1} ∈ B. So by the definition of a factorized subgroup, we have x ∈ S and hence A ∩ B ≤ S.

(ii) ⇒ (iii). Since S = (A ∩ S)(B ∩ S), we have A ∩ SB = A ∩ (S ∩ A)(S ∩ B)B = A∩(S ∩A)B. By the modular law (or by Lemma 1.1.4), A∩(S ∩A)B = (S ∩A)(A∩B) = S ∩ A since also A ∩ B ⊆ S.

(iii) ⇒ (iv). This is trivial.

(iv) ⇒ (i). If s = ab ∈ S with a ∈ A and b ∈ B, then a = sb^{−1} ∈ A ∩ SB ⊆ S whence S is factorized.

The implications (i) ⇒ (ii) ⇒ (v) ⇒ (vi) ⇒ (i) can be proved similarly.

To prove that (vii), (viii), (ix), (x) and (xi) are equivalent to S being factorized, suppose first that S is factorized. Then S = (S ∩ B)(A ∩ S) ⊆ (AS ∩ B)(A ∩ BS) = S, so (vii) holds. Also, (viii) and (ix) hold since AS ∩ B ⊆ S and A ∩ SB ⊆ S by (iv) and (vi) respectively. (x) and (xi) follow from (viii) and (ix) by Lemma 1.1.4.

Conversely, if one of the equations (vii), (viii), (ix), (x) and (xi) hold, then it is evident that at least one of the sets AS ∩ B or A ∩ SB is contained in S proving statements (iv) or (vi) respectively, which are both equivalent to S being factorized.

Note. It is easy to see that for any subgroup S of G = AB, the set A∩SB equals the set {a ∈ A | ab ∈ S for some b ∈ B} and AS ∩ B = {b ∈ A | ab ∈ S for some a ∈ A}.

The next lemma studies the behaviour of factorized (prefactorized) subgroups in the subgroup lattice of a factorized group G. The statements about factorized subgroups can also be found in Chapter 1 of [4] from which also most proofs are derived. To see that the intersection of two prefactorized subgroups is not necessarily prefactorized, see Example 1.2.3.

1.1.6 Lemma. Let the group G be the product of its subgroups A and B.

(i) If S is prefactorized (factorized) in G, then T ≤ S is prefactorized (factorized) with respect to the factorization S = (S ∩ A)(S ∩ B) of S if and only if T is prefactorized (factorized) in G.

(ii) The intersection of any family of factorized subgroups of G is factorized.

(iii) The intersection of a prefactorized subgroup and a factorized subgroup is prefac-

torized. (iv) If the subgroup S of G is the union of the prefactorized subgroups S_{i} where i ∈ I

of G, then S is prefactorized. It is factorized, provided one of the subgroups S_{i} is factorized.

(v) The product of two prefactorized subgroups one of which is normal in G is prefactorized. It is factorized, provided that either of the subgroups is factorized.

(vi) The product of any number of prefactorized normal subgroups is prefactorized. It is factorized if one of the normal subgroups is factorized.

Proof. (i) The statement concerning prefactorized subgroups follows from the fact that by definition,

T = (S ∩ A ∩ T)(S ∩ B ∩ T) = (A ∩ T)(B ∩ T)

so that T is a prefactorized subgroup of G. In case S is a factorized subgroup of G, we have A ∩ B = (A ∩ S) ∩ (B ∩ S); thus if T is factorized in S, then T contains (A ∩ S) ∩ (B ∩ S) = A ∩ B so that by Lemma 1.1.5, T is a factorized subgroup of G.

(ii) Let {S_{i} | i ∈ I} be a family of factorized subgroups of G. If s = ab ∈ ⋂ S_{i}, then s ∈ S_{i} for all i, and hence a ∈ S_{i} for all i by the definition of a factorized subgroup. Hence a is contained in A ∩ (⋂ S_{i}), which shows that ⋂ S_{i} is factorized.

(iii) Let S and P be a factorized and a prefactorized subgroup of G. Every g ∈ P ∩ S can be written g = ab with a ∈ A ∩ P and b ∈ P ∩ B since P is prefactorized. Now S is factorized and also g ∈ S, so we have a ∈ S and b ∈ S, that is, a ∈ A ∩ P ∩ S and b ∈ B ∩ P ∩ S. This shows that S ∩ P is prefactorized.

(iv) Let s ∈ S, then s ∈ S_{i} for some i ∈ I. Since S_{i} is prefactorized, we have s = ab with a ∈ A ∩ S_{i} and b ∈ B ∩ S_{i}, so in particular a ∈ A ∩ S and b ∈ B ∩ S. This shows that S is prefactorized. If one of the S_{i} is factorized, it contains A ∩ B and so A ∩ B ≤ S whence S is factorized.

(v) Let N and P be prefactorized subgroups of G with N ⊴ N. Then

P N = (P ∩ A)(P ∩ B)N
= (P ∩ A)N(P ∩ B)
= (P ∩ A)(N ∩ A)(N ∩ B)(P ∩ B)
≤ (P N ∩ A)(P N ∩ B)
≤ P N,

which shows that P N is prefactorized. The statement about factorized subgroups follows as in (iv).

(vi) Let {N_{i}}_{i∈I} be a family of prefactorized normal subgroups of G. By (v), the product of two prefactorized normal subgroups is prefactorized and since it is clearly normal, the statement is true for every finite index set I. For arbitrary index sets, the product of the N_{i} is the union of all products of a finite number of the N_{i}, so the full result follows from (iv).

The following lemma shows in particular that if G is the product of its subgroups A and B, then also G = A^{x}B^{y} for all x, y ∈ G.

1.1.7 Lemma (Wielandt [44]). Let the group G be the product of its subgroups A and B. If A_{0} and B_{0} are normal subgroups of A and B respectively such that A_{0}B_{0} = B_{0}A_{0}, then for every x and y ∈ G, there is a z ∈ G such that A_{0}^{x} = A_{0}^{z} and B_{0}^{y} = B_{0}^{z} and thus B_{0}^{x}B_{0}^{y} = B_{0}^{y}A_{0}^{x} = (A_{0}B_{0})^{z}.

Proof. Let x, y ∈ G and write xy^{−1} = a^{−1}b with a ∈ A and b ∈ B. Let z = ax = by. Then A_{0}^{z} = A_{0}^{ax} = A_{0}^{x} since A_{0} ⊴ A; similarly B_{0}^{Z} = B_{0}^{y}. So A_{0}^{x}B_{0}^{y}) = (A_{o}B_{o})^{z} is a subgroup of G which is equivalent to A_{0}^{x}B_{0}^{y} = B_{0}^{y}A_{0}^{x} by Lemma 1.1.2.

This has the following consequence:

1.1.8 Lemma. If G = AB = AC = BC where A, B and C are subgroups of G and C^{g} is factorized with respect to G = AB for some g ∈ G, i.e. C^{g} = (A ∩ C^{g})(B ∩ C^{g}) and C ≥ A ∩ B, then G = C.

Proof. Since G = AC = CA, we deduce from Lemma 1.1.7 that also AC^{g} = C^{g}A which is conjugate to G, i.e. G = AC^{g}. Similarly, we obtain that G = BC^{g} so that we may assume w.l.o.g. that C is factorized. Now by the modular law, G = A(B ∩ C) = (A ∩ C)B, therefore, using the modular law again, G = A(B ∩ C) ∩ (A ∩ C)B = (A ∩ (A ∩ C)B)(B ∩ C) = (A ∩ C)(A ∩ B)(B ∩ C) ≤ C.

Note that the preceding lemma becomes false if we only have C = (A ∩ C)(B ∩ C), that is, if C is only a prefactorized subgroup of G = AB: take any group G ≠ 1 and put A = B = G and C = 1 ≠ G.

The following lemma will be needed later.

1.1.9 Lemma. Let the group G be the product of its subgroups A and B and suppose that {N_{i} | i ∈ I} is a set of factorized normal subgroups of G. Then

⋂_{i∈I} AN_{i} = A(⋂_{i∈I} N_{i}).

Proof. Clearly, ⋂_{i∈I} AN_{i} contains A(⋂_{i∈I} N_{i}). To prove the other inclusion, observe first that, since every subgroup of G that contains A or B is factorized, hence AN_{i} is factorized, AN_{i} = A(B ∩ N_{i}) whence B ∩ AN_{i} = (B ∩ N_{i})(A ∩ B) = B ∩ N_{i}.

Now also ⋂_{i∈I} AN_{i} is factorized by Lemma 1.1.6, hence

⋂_{i∈I} AN_{i} = A(B ∩ (⋂_{i∈I} AN_{i})) = A(⋂_{i∈I}(B ∩ AN_{i})) = A(⋂_{i∈I}(B ∩ N_{i})) ≤ A(⋂_{i∈I} N_{i}).

1.2 Factorizers, prefactorizers, and quotient groups

For every subgroup S of the factorized group G = AB, the intersection of all factorized subgroups of G which contain S is factorized by Lemma 1.1.6. Clearly, this is the unique minimal factorized group that contains S. This subgroup is called the factorizer of S; we denote it with X_{G}(S). If S is normal in G, its factorizer can be described as follows:

1.2.1 Lemma. Let the group G be the product of its subgroups A and B. If N is a normal subgroup of G and X = X_{G}(N) is the factorizer of N in G, then AX = AN and BX = BN, hence A ∩ X = A ∩ BN = A ∩ BX and B ∩ X = B ∩ AN = B ∩ AX Thus X = AN ∩ BN has a triple factorization

X = (AN ∩ B)N = N(A ∩ BN) = (AN ∩ B)(A ∩ BN).

Proof. Obviously, AN and BN are factorized subgroups of G containing N, therefore X ≤ AN and X ≤ BN, from which it follows that AX = AN and BX = BN. The remaining statements follow directly from this and from Lemma 1.1.5

The next lemma shows that homomorphic images of factorized groups inherit the factorization of G.

1.2.2 Lemma. If the group G is the product of its subgroups A and B and N is a normal subgroup of G, then

(i) G/N is the product of its subgroups AN/N and BN/N.

(ii) If S is prefactorized in G, then SN/N is prefactorized in G/N.

(iii) If N is prefactorized (with respect to G = AB) and N ≤ S, then S/N is a

prefactorized subgroup of G/N if and only if S is a prefactorized subgroup of G. (iv) If N ≤ S, then S is a factorized subgroup of G if and only if S/N is factorized

in G/N. In particular, AN/N ∩ BN/N = X_{G}(N)/N.

Proof. (i) This is trivial.

(ii) Clearly, SN/N = (S ∩A)N/N ·(S ∩B)N/N ≤ (SN/N ∩AN/N)·(SN/N ∩BN/N) which is contained in SN/N. This shows that SN/N is prefactorized.

(iii) If S is prefactorized, it is clear by the preceding statement that S/N is prefactorized. Conversely, suppose that S/N is prefactorized, or equivalently, that S = (S∩AN)(S∩BN). Then by the modular law, S = (S∩AN)(S∩BN) = (S∩A)N(S∩B). Since N is prefactorized and N ≤ S, we have S = (S ∩ A)(N ∩ A)(N ∩ B)(S ∩ B) = (S ∩ A)(S ∩ B) which shows that S is prefactorized.

(iv) Suppose first that S is factorized. We already know that S/N is prefactorized. Moreover, AN ∩ BN ≤ AS ∩ SB = S by Lemma 1.1.5 and so AN/N ∩ BN/N ≤ S/N; thus by the same lemma, S/N is factorized.

Conversely, suppose that S/N is factorized, then by Lemma 1.1.5, S/N contains AN/N∩BN/N whence S contains the subgroup AN∩BN. By Lemma 1.2.1, AN∩BN = X_{G}(N) = (AN ∩ B)(A ∩ NB) and so by the modular law,

S = (S ∩ AN)(S ∩ BN)
= (S ∩ A)N(S ∩ B)
≤ (S ∩ A)(A ∩ NB)(AN ∩ B)(S ∩ B)
= (S ∩ A)(S ∩ B),

observing that A ∩ BN ≤ A ∩ S and B ∩ AN ≤ B ∩ S. This also shows that A ∩ B ≤ S and so by Lemma 1.1.5, S is factorized.

The definition of a factorizer cannot be extended to prefactorized subgroups, since the intersection of two prefactorized subgroups is not necessarily prefactorized, not even if the subgroups are normal, as the following simple example shows:

1.2.3 Example. Let V be a vector space of dimension 3 over some field F and let {x_{1}, x_{2}, x_{3}} be a base of V. Let A = ⟨x_{1}, x_{2}⟩ and B = ⟨x_{2}, x_{3}⟩, then the subgroups H_{1} = ⟨x_{1}, x_{2} + x_{3}⟩ = ⟨x_{1}⟩ ⊕ ⟨x_{2} + x_{3}⟩ and H_{2} = ⟨x_{1} + x_{2}, x_{3}⟩ = ⟨x_{1} + x_{2}⟩ ⊕ ⟨x_{3}⟩ are prefactorized but their intersection H = ⟨x_{1} +x_{2} +x_{3}⟩ intersects A and B trivially, hence H cannot be prefactorized. If we choose F to be a field of prime order, H_{1} and H_{2} are minimal subject to containing H and being prefactorized while the factorizer of H is G by Lemma 1.2.1.

However if the group G is the product of its subgroups A and B and H ≤ G, let

S = {S ≤ G | H ≤ S and S is prefactorized}.

We call the minimal elements of S prefactorizers of H in G. Note, however, that unlike factorizers, prefactorizers need not exist if the group G is infinite.

1.2.4 Lemma. Let the group G be the product of its subgroups A and B. for two subgroups A and B.

(i) If H ≤ G and S is a prefactorizer of H in G, then S = X_{S}(H) ≤ X_{G}(H).

(ii) If N is a normal subgroup of G and S is a prefactorizer of N in G, then S has a triple factorization S = (A ∩ S)(B ∩ S) = (A ∩ S)N = (B ∩ S)N.

Proof. (i) If S is a prefactorizer of H in G, then X_{S}(H), the factorizer of H with respect to the factorization S = (S ∩ A)(S ∩ B) of S, is a prefactorized subgroup of G contained in S. Therefore we have X_{S}(H) = S by the minimality of S. Since X_{G}(H) is factorized, X_{G}(H)∩S is prefactorized by Lemma 1.1.6 and H ≤ X_{G}(H)∩S ≤ S whence S = X_{G}(H) ∩ S by the definition of S.

(ii) This follows directly from the fact that S = X_{S}(N) and the characterization of X_{N}(S) given in Lemma 1.2.1.

1.3 Classes of groups and factorizations

A class ℭ of groups is a class in the set-theoretical sense whose elements are groups and that satisfies the condition: if G ∈ ℭ and H ≅ G, then H ∈ ℭ. In other words, a class of groups is the union of isomorphy classes of groups.^{1} Following Doerk and Hawkes [13], Chapter II, a closure operation C is a map

C: {group classes} → {group classes}

such that C∅ = ∅ and for every class ℭ of groups, ℭ ⊆ Cℭ and Cℭ = C(Cℭ) and if 𝔇 is a class of groups with ℭ ⊆ 𝔇, then Cℭ ⊆ C𝔇.

We introduce the following closure operations on classes of groups

Qℭ = {G/N | G ∈ ℭ, N ⊴ G}
Sℭ = {H | ∃ G ∈ ℭ such that H ≤ G}
S_{n}ℭ = {S | ∃ G ∈ ℭ such that S ◁◁ G}
R_{0}ℭ = {G | ∃ N_{1}, …, N_{r} ⊴ H with G/N_{i} ∈ ℭ and ⋂_{i=1}^{r} N_{i} = 1}
D_{0}ℭ = {G | ∃ G_{1}, …, G_{r} ∈ ℭ with G = /\_{i=1}^{r} G_{i}}
N_{0}ℭ = {G | ∃ S_{1}, …, S_{r} ◁◁ H: S_{i} ∈ ℭ, G = ⟨S_{1}, …, S_{r}⟩}
E_{Φ}ℭ = {G | G/N ∈ ℭ for some N ⊴ G with N ≤ Φ(G)}

If C is a closure operation and ℭ is a class of groups, we say that ℭ is C-closed if Cℭ = ℭ; if D is another closure operation, we define CDℭ = C(Dℭ); observe that the latter class need not be D-closed. Therefore we define ⟨C,D⟩ℭ to be the smallest class of groups that is C-closed and D-closed. We also recall that SD_{0} = ⟨S,D_{0}⟩ and that a SD_{0}-closed class is R_{0}-closed.

A formation 𝔉 is a class of groups that is Q-closed and R_{0}-closed. If ℭ is any class of groups, QR_{0}ℭ is the smallest formation containing ℭ and QSD_{0}ℭ is the smallest subgroupclosed formation containing ℭ. A class ℭ of groups is said to be saturated if it is E_{Φ}-closed.

If ℭ is a class of groups, define

Pℭ = {G | G/N ∈ ℭ whenever G/N is primitive for an N ⊴ G}

(recall that a group is said to be primitive if it has a maximal subgroup whose core is trivial; see also Section 3.2). A class ℌ of finite groups is called a Schunck class if it is Pℌ = ℌ, i.e. the finite group G belongs to ℌ if every primitive epimorphic image of G lies in ℌ. Schunck classes are Q-closed and D_{0}-closed, and every saturated formation of finite groups is a Schunck class. It should be observed that P is not a closure operation, but PQ is; moreover for any class ℭ of groups, PQℭ is the smallest Schunck class containing ℭ.

1 Some authors impose the additional condition on a class of groups that it must contain the class of

groups with one element. We do not make such an assumption.

The following simple lemma is basic when dealing with Schunck classes.

1.3.1 Lemma. Let ℨ be a Q-closed class of finite groups. If ℌ is a Schunck class, then a group of minimal order in ℨ \ ℌ is primitive.

Proof. Let G be a group of minimal order that belongs to ℨ \ ℌ. Since ℨ is Q-closed, G/N ∈ ℨ for every normal subgroup and so for every N ≠ 1, we have G/N ∈ ℌ by the minimality of G. Since the group G does not belong to ℌ, it must be primitive by the very definition of a Schunck class.

Let G be a finite group and ℭ a class of groups. If ℭ is R_{0}-closed, let G^{ℭ} denote the intersection of all normal subgroups N of G such that G/N ∈ ℭ. Then G/G^{ℭ} ∈ R_{0}ℭ = ℭ and we call G^{ℭ} the ℭ-residual of G; G^{ℭ} is obviously a characteristic subgroup of G.

Similarly, if ℭ is N_{0}-closed, then the subgroup generated by all subnormal ℭ-subgroups of G is a (characteristic) ℭ-subgroup of G which is called the ℭ-radical of G and is denoted by G_{ℭ}.

For any class of groups ℭ, we define the characteristic char(ℭ) of ℭ to be the set of primes p such that ℭ contains a cyclic p-group, and for every set π of primes, we define

ℭ_{π} = {G ∈ ℭ | σ(G) ⊆ π}.

The following lemma shows in particular that the class of finite groups that are the product of (two) nilpotent subgroups is Q-closed and D_{0}-closed.

1.3.2 Lemma. Suppose that 𝔛 and 𝔜 are classes of groups and let

ℨ = {G | ∃ A, B ≤ G with A ∈ 𝔛 and B ∈ 𝔜 such that G = AB}.

(i) If 𝔛 and 𝔜 are Q-closed, then ℨ is Q-closed.

(ii) If 𝔛 and 𝔜 are D_{0}-closed, then ℨ is D_{0}-closed.

Proof. (i) Let G ∈ ℨ, then G is the product of two subgroups A and B with A ∈ 𝔛 and B ∈ 𝔜. Now if N ⊴ G, then G/N is the product of its subgroups AN/N and BN/N by Lemma 1.2.2. Now by an isomorphism theorem, AN/N ≅ A/A ∩ N ∈ Q𝔛 = 𝔛 and similarly BN/N ∈ 𝔜, proving that G/N ∈ ℨ.

(ii) Let G = /\_{i=1}^{r} G_{i} and suppose that G_{i} ∈ ℨ for i = 1, …, r. Then each G_{i} is the product of two subgroups A_{i} and B_{i} with A_{i} ∈ 𝔛 and B_{i} ∈ 𝔜. Now A = /\_{i=1}^{r} A_{i} ∈ D_{0}𝔛 = 𝔛 and B = /\_{i=1}^{r} B_{i} ∈ D_{0}𝔜 = 𝔜. Since the subset AB of G = /\_{i=1}^{r} G_{i} clearly contains all G_{i}, G is the product of the 𝔛-group A and the 𝔜-group B.

Combining Lemma 1.3.1 and Lemma 1.3.2, we obtain

1.3.3 Lemma. Let 𝔛 and 𝔜 be Q-closed classes of groups and let ℨ be the class of finite groups that are the product of an 𝔛-group with a 𝔜-group. If ℌ is a Schunck class, then a group of minimal order in ℨ \ ℌ is primitive.

1.3.4 Lemma. Let G be a finite group and suppose that ℭ is an R_{0}-closed class of groups. If X is a subgroup of the group G such that XN_{1}/N_{1} and XN_{2}/N_{2} ∈ ℭ for two normal subgroups N_{1} and N_{2} of G with N_{1} ∩ N_{2} = 1, then X ∈ ℭ.

Proof. We have XN_{i}/N_{i} ≅ X/X ∩ N_{i} ∈ 𝔉 and (X ∩ N_{1}) ∩ (X ∩ N_{2}) = 1, therefore X ∈ ℭ.

Remark. This can be used to show that there is even some weak form of R_{0}-closure of the product of an 𝔛-group and a 𝔜-group in case the classes 𝔛 and 𝔜 are R_{0}-closed: if the group G is the product of its subgroups A and B and there are normal subgroups N_{1} and N_{2} with N_{1} ∩ N_{2} = 1 and such that G/N_{i} is the product of the 𝔛-subgroup AN_{i}/N_{i} and the 𝔜-subgroup BN_{i}/N_{i} for i = 1, 2, then G is the product of the 𝔛-subgroup A and the 𝔜-subgroup B.

A dual of the following statement due to Lockett can be found in [13], II.2.12:

1.3.5 Lemma. Let G be a finite group and suppose that the class 𝔉 is a Fitting formation, i.e. a ⟨Q,R_{0},S_{n},N_{0}⟩-closed class of groups.

(i) If N_{1} and N_{2} are two normal subgroups of G with N_{1} ∩ N_{2} = 1 and R_{i}/N_{i} ∈ 𝔉 for i = 1, 2, then R_{1} ∩ R_{2} ∈ 𝔉.

(ii) If R_{1}/N_{1} and R_{2}/N_{2} are the 𝔉-radicals of G/N_{1} and G/N_{2} respectively, then

R_{1} ∩ R_{2} is the 𝔉-radical of G. Proof. (i) We have

(R_{1} ∩ R_{2})N_{1}/N_{1} = (R_{1} ∩ R_{2}N_{1})/N_{1} ⊴ R_{1}/N_{1}

so that

(R_{1} ∩ R_{2})/(R_{1} ∩ R_{2} ∩ N_{1}) ≅ (R_{1} ∩ R_{2})N_{1}/N_{1} ∈ 𝔉.

Similarly,

(R_{1} ∩ R_{2})/(R_{1} ∩ R_{2} ∩ N_{2}) ∈ 𝔉

so that also R_{1} ∩ R_{2} ∈ 𝔉 because (R_{1} ∩ R_{2} ∩ N_{1}) ∩ (R_{1} ∩ R_{2} ∩ N_{2}) = 1.

(ii) Let R denote the 𝔉-radical of G. Then RN_{i}/N_{i} is a normal 𝔉-subgroup of G/N_{i} for i = 1, 2, therefore R ≤ R_{1} ∩ R_{2}. On the other hand, the normal subgroup R_{1} ∩ R_{2} is contained in 𝔉 by (i), proving the other inclusion.

Factorizations of finite soluble groups

2.1 Hall subgroups and Hall systems

Finite soluble groups can be characterized by the fact that they possess Hall π-subgroups for every set π of primes (Hall [24]; see also Doerk and Hawkes [13], Sections I.3 and I.4). Note that we also consider G itself and the unit subgroup as Hall subgroups of G for the sets ±⊬ℙ and ∅ of primes respectively.

If G is a finite soluble group, a set Σ of Hall subgroups of G is called a Hall system of G if Σ contains a Hall π-subgroup of G for every set of primes π and HK = KH for all H, K ∈ Σ, we have HK = KH (observe that this implies that Σ contains exactly one Hall π-subgroup for every set of primes π by Lemma 1.1.1).

Given a Hall p′-subgroup G_{p′} of G for every prime p, it is easy to see that

G_{π} = ⋂_{π⊆p′} G_{p′}

is a Hall π-subgroup of G and the set

Σ = {G_{π} | π a set of primes}

is a Hall system of G. The set {G_{p′} | p a prime} is called a complement basis of Σ. Consequently, Hall systems always exist in finite soluble groups. Moreover, any two Hall systems are conjugate, i.e. if Σ and T are Hall systems, then there is g ∈ G such that T = Σ^{g} = {H^{g} | H ∈ Σ}.

Also, the set {G_{p} | p a prime} containing exactly one Sylow p-subgroup G_{p} of G for each prime is called a Sylow basis if G_{p}G_{q} = G_{q}G_{p} whenever p and q are distinct primes. In this case, a Hall system of G can be obtained by defining the G_{π}’s to be the product of all G_{p} where p ∈ π.

If S is a subgroup of the finite soluble group G, then a Hall system Σ is said to reduce into S if for every set of primes π, the intersection of the Hall π-subgroup G_{π} ∈ Σ with S yields a Hall π-subgroup of S; in this case, the set

Σ ∩ S = {G_{π} ∩ S | G_{π} ∈ Σ}

is a Hall system of S. Conversely, it can be shown that every Hall system Σ^{∗} of S can be extended to a Hall system Σ of G, i.e. every subgroup contained in Σ^{∗} is contained in some subgroup contained in Σ.

Following Hall [25], a group is said to satisfy the property E_{π} (existence) if it possesses a Hall π-subgroup; it satisfies C_{π} (conjugacy) if it satisfies E_{π} and all its Hall πsubgroups are conjugate; finally, it satisfies D_{π} (dominance) if it has C_{π} and moreover every π-subgroup is contained in some Hall π-subgroup. Thus the finite soluble groups are precisely the finite groups that satisfy D_{π} for all sets π of primes.

The following lemma which relates the Sylow structure of a group G to a given factorization of G was first proved by Wielandt [44] for Sylow subgroups and Hall p′-groups of soluble groups:

2.1.1 Lemma. Let G = AB be a finite group satisfying D_{π}. If A_{π} and B_{π} are Hall πsubgroups of A and B respectively, then there are a ∈ A and b ∈ B such that A_{π}^{a}B_{π}^{b} is a Hall π-subgroup of G. Furthermore, A ∩ B possesses Hall π-subgroups one of which is A_{π}^{a} ∩ B_{π}^{b}.

Proof. By the property D_{π}, the subgroup A_{π} is contained in a Hall π-subgroup G_{π} and also B_{π} is contained in a Hall π-subgroup which is conjugate to G_{π} by the property D_{π}, so we have B_{π} ≤ G_{π}^{g} for a suitable g ∈ G. Write g = ab^{−1}, then A_{π}^{a} ≤ G_{π}^{a} and B_{π}^{b} ≤ G_{π}^{gb} = G_{π}^{a}. Therefore, replacing A_{π}, B_{π} and G_{π} by suitable conjugates, we may suppose that A_{π} and B_{π} are contained in G_{π}.

By Lemma 1.1.1, the order of the group G is

|G| = \frac{|A||B|}{|A ∩ B|};

thus the order of a Hall π-subgroup (which equals the π-part of |G|) is

|G_{π}| = \frac{|A_{π}||B_{π}|}{|A ∩ B|_{π}}

where |A∩B|_{π} is the π-part of |A∩B|. Now the order of A_{π} ∩B_{π} is a π-number dividing |A ∩ B|_{π}; hence

|G_{π}| ≤ \frac{|A_{π}||B_{π}|}{|A_{π} ∩ B_{π}|} = |A_{π}B_{π}|.

Since we also have A_{π}B_{π} ⊆ G_{π}, we must have A_{π}B_{π} = G_{π} as required. So |G_{π}| = |A_{π}B_{π}|, from which we deduce that |A ∩ B|_{π} = |A_{π} ∩ B_{π}|; therefore the subgroup A_{π} ∩ B_{π} is a Hall subgroup of A ∩ B.

Recall that as a consequence of the Feit-Thompson theorem [14], every π-separable group satisfies D_{π} and that the class of π-separable groups is subgroup-closed so that every π-separable group satisfies the above lemma.

In the case of a soluble group, one is interested in Hall systems rather than single Hall subgroups. The following proposition states the existence of Hall systems in which every Hall subgroup is of the form just described.

2.1.2 Proposition. If the soluble group G is the product of its subgroups A and B, then there is a Hall system of G of the form {A_{π}B_{π} | π a set of primes} which reduces into A and B.

Proof. G is soluble, hence satisfies D_{π} for every set π of primes. By the preceding proposition, for every prime p there is a Hall p′-subgroup G_{p′} of G of the form G_{p′} = A_{p′}B_{p′} for suitable Hall p′-subgroups A_{p′} and B_{p′} of A and B respectively.

Extend the complement bases {G_{p′}}, {A_{p′}} and {B_{p′}} to Hall systems Σ = {G_{π}}, {A_{π}} and {B_{π}} of G, A and B respectively, then it is clear that A_{π} ≤ A ∩ G_{π} and B_{π} ≤ B ∩ G_{π} for every set π of primes; since A_{π} and B_{π} are Hall subgroups of A and B respectively, we have A_{π} = A∩G_{π} and B_{π} = B ∩G_{π} and as in the proof of Lemma 2.1.1, G_{π} = A_{π}B_{π}. So Σ is the required Hall system.

Remark. By induction on the number of factors, the last proposition as well as Lemma 2.1.1 can be extended to finite soluble groups that are the product of any (finite) number of subgroups.

2.2 Products of subgroups of coprime order

Let the group G be the product of its subgroups A and B. We establish some criteria for certain subgroups of G to be factorized or prefactorized, based on the orders or indices of A and B.

The next lemma is well-known:

2.2.1 Lemma. If A and B are subgroups of the finite group G such that |G : A| and |G : B| are coprime, then G = AB.

Proof. Under the hypothesis of the lemma, we have |G : A ∩ B| = |G : A| · |G : B|; together with the formula from Lemma 1.1.1, we obtain that |AB| = |G| and thus we have G = AB.

Recall that a subnormal subgroup S of a group G has subnormal defect n if n is the least integer such that there is a subnormal series of G of the form

G = S_{0} ▷ S_{1} ▷ ⋯ ▷ S_{n} = S.

Such a chain can be obtained by recursively defining S_{0} = G and S_{i+1} = S^{S_{i}} for all positive integers i (see Doerk and Hawkes [13], A.14.7 or Robinson [43], Chapter 13.1).

2.2.2 Lemma. Let the group G be the product of its subgroups A and B. and suppose that |G : A| and |G : B| are finite.

(i) if N ⊴ G and X = X_{G}(N) denotes its factorizer, then |N : N ∩ A| = |X : X ∩ A| divides |G : A| and |N : N ∩ B| = |X : X ∩ B| divides |G : B|.

(ii) Every subgroup S of G contains a prefactorized subgroup N ⊴ S of finite index;

moreover N ≤ A ∩ B. (iii) If the indices of A and B are coprime, then every subnormal subgroup of G is

prefactorized.

Proof. (i) By Lemma 1.2.1, we have AX = AN which shows that AX is a subgroup of G. Therefore

|X : X ∩ A| = |AX : A| = |AN : A| = |N : N ∩ A|

which divides |G : A|, similarly for B.

(ii) Let S be a subgroup of G. Since the indices of A and B are finite, so is |G : A ∩ B| and thus also |G : (A ∩ B)_{G}|. Then also |S : S ∩ (A ∩ B)_{G}| is finite; moreover N = S ∩ (A ∩ B)_{G} is trivially prefactorized and so N is the required normal subgroup of S.

(iii) As a first case, assume that S ⊴ G; then S/N is finite where N = S ∩ (A ∩ B)_{G}; moreover since N is prefactorized, by Lemma 1.2.2, S is a prefactorized subgroup of G/N if (and only if) S/N is prefactorized in G/N. Thus we may assume w.l.o.g. that N = 1 and that S is finite.

Now by part (i), |S : S ∩ A| divides |G : A| and |S : S ∩ B| divides |G : B|. Therefore the indices of A ∩ S and B ∩ S are coprime and we have S = (A ∩ S)(B ∩ S) by Lemma 2.2.1.

There remains the case when S is a nonnormal subnormal subgroup of defect n > 1 in G: we have already proved that K = S^{G} ⊴ G is prefactorized and that the indices of K ∩ A and K ∩ B are coprime. Since the subnormal defect of S in K is n − 1, by induction on the subnormal defect of S, the subgroup S is a prefactorized subgroup of N, hence is prefactorized in G by Lemma 1.1.6.

For a periodic group G, i.e. a group in which every element has finite order, we define σ(G) to be the set of primes that divide the order of an element of G. If G is finite, then clearly σ(G) is the set of prime divisors of |G|. If π is a set of primes, the group G is called a π-group if σ(G) ⊆ π, that is, if the order of every element is divisible only by primes in π.

2.2.3 Lemma. Let G = AB be a (possibly infinite) group and let N be a normal subgroup of G. If X denotes the factorizer of N in G, then σ(X/N) ⊆ σ(AN/N)∩σ(BN/N).

Proof. By Lemma 1.2.1, we have X = (A ∩ BN)N = (AN ∩ B)N and so X/N ≤ AN/N and X/N ≤ BN/N. Therefore σ(X/N) ⊆ σ(AN/N) ∩ σ(BN/N).

2.2.4 Corollary. Let the group G be the product of its subgroups A and B. If A and B have coprime orders, then every subnormal subgroup of G is factorized.

Proof. If N ⊴ G, this follows from lemma since we must have X/N = 1 and therefore N = X_{G}(N) is factorized. If S is subnormal in G, the result follows by induction on the defect of S in G like in the proof of Lemma 2.2.2.

Using transfinite induction, the last result can be extended from subnormal subgroups to descendant subgroups, observing that the intersection of arbitrarily many factorized subgroups is factorized by Lemma 1.1.6.

Under certain additional hypotheses, and in particular when G is finite and soluble, it is possible to extend Lemma 2.2.2, (iii) and Corollary 2.2.4 to a statement about all subgroups of the group G: there are factorized (prefactorized) conjugates for all subgroups of the finite group.

2.2.5 Lemma. Let G be a finite group which is the product of its subgroups A and B with (|A|, |B|) = 1 and such that G satisfies D_{π} and D_{π′} where π and π′ denote the sets of prime divisors of |A| and |B| respectively. If S is a subgroup of G which satisfies E_{π} and E_{π′}, then S possesses a factorized conjugate S^{g}, i. e. S^{g} = (S^{g} ∩ A)(S^{g} ∩ B) for a g ∈ G.

More generally, if H is a factorized subgroup of G that satisfies D_{π} and D_{π′} and contains S, then there is a h ∈ H such that S^{h} ≤ H is factorized.

Proof. Clearly, A and B are Hall π and π′-subgroups of G. If S_{π} and S_{π′} denote the Hall π- and π′-subgroups of S, then S = S_{π}S_{π′}. Now S_{π} and S_{π′} are contained in Hall subgroups A^{x} and B^{y} of G where x, y ∈ G. By Lemma 1.1.7, there is a g ∈ G with A^{x} = A^{g^{−1}} and B^{y} = B^{g^{−1}}. Then we also have S^{g} = S_{π}^{g}S_{π′}^{g} with S_{π}^{g} ≤ A and S_{π′}^{g} ≤ B, hence S^{g} = (S^{g} ∩ A)(S^{g} ∩ B). Note also that S^{g} ≥ A ∩ B = 1.

The second statement follows directly from the first, observing that A ∩ H and B ∩ H are π- and π′-groups respectively.

Note that Lemma 2.2.5 applies in particular to all subgroups of π-separable groups G when π is the set of prime divisors of |A|. To extend Lemma 2.2.5 to the prefactorized case, we need the following

2.2.6 Lemma. Let the group G be the product of its subgroups A and B. If A ≤ A^{∗} and B ≤ B^{∗} for two subgroups A^{∗} and B^{∗} of G, then every subgroup that is prefactorized with respect to the factorization G = AB is also prefactorized with respect to the factorization G = A^{∗}B^{∗} of G.

Proof. Let S be a subgroup of G that is prefactorized with respect to the factorization G = AB, i.e. S = (S ∩ A)(S ∩ B). Clearly, S = (S ∩ A)(S ∩ B) ⊆ (S ∩ A^{∗})(S ∩ B^{∗}) ⊆ S, proving that S = (S ∩ A^{∗})(S ∩ B^{∗}). Application to the case S = G shows in particular that G is the product of A^{∗} and B^{∗}.

It is easy to see that the preceding lemma becomes false if we substitute ‘factorized’ for ‘prefactorized’: choose any group G ≠ 1 and let A = 1 and A^{∗} = B = B^{∗} = G, then the unit subgroup is factorized with respect to G = AB but not with respect to G = A^{∗}B^{∗}.

We can now extend Lemma 2.2.5 to a similar result for prefactorized subgroups; observe that the next lemma is true for every soluble group G that is the product of two subgroups which have coprime indices.

2.2.7 Lemma. Let G be a finite soluble group which is the product of its subgroups A and B and such that (|G : A|, |G : B|) = 1. Let π be a set of primes such that A contains a Sylow p-subgroup of G for all p ∈ π and B contains a Sylow p-subgroup of G for all p ∈ π′. If G satisfies D_{π} and D_{π′}, then for every subgroup S of G there is a prefactorized conjugate S^{g}, i. e. S^{g} = (S^{g} ∩ A)(S^{g} ∩ B) for a g ∈ G.

More generally, let H be a prefactorized subgroup of G such that A ∩ H and B ∩ H have coprime indices. If H satisfies D_{π} and D_{π′} and contains S, then there is a h ∈ H such that S^{h} ≤ H is prefactorized.

Proof. Since (|G : A|, |G : B|) = 1, the subgroups A and B contain Hall π- and π′-subgroups A_{π} and B_{π′} respectively of G. Since the indices of A_{π} and B_{π′} are coprime, we have G = A_{π}B_{π′} by Lemma 2.2.1 and so every subgroup of S possesses a conjugate which is factorized with respect to the factorization G = A_{π}B′_{π} by Lemma 2.2.7. So by Lemma 2.2.6, this conjugate is prefactorized with respect to the factorization G = AB.

The second statement follows from the first by considering H instead of G.

2.3 Groups with many factorized subgroups

Let the finite soluble group G be the product of its subgroups A and B. We have seen in Lemma 2.2.5 and Lemma 2.2.7 that every subgroup S of G has a conjugate which is factorized (prefactorized) if the orders (indices) of A and B are coprime; more precisely, this conjugate can already be found in certain factorized (prefactorized) subgroups containing S.

Conversely, if G is a soluble group which is the product of two subgroups A and B in which every subgroup has a factorized (prefactorized) conjugate in every factorized (prefactorized) subgroup of G, then we will show that the orders (the indices) of A and B are coprime. Observe that for finite soluble groups, the condition that A and B have coprime indices is equivalent to the condition that A and B contain Hall π and π′-subgroups of G for a suitable set π of primes.

First, we deal with the nilpotent case:

2.3.1 Proposition. Suppose that the finite nilpotent group G is the product of its subgroups A and B and that every subgroup of G has a prefactorized conjugate. Then A and B contain Hall π-and π′-subgroups of G for some set of primes π. In particular, if G is a p-group for some prime p, then G = A or G = B.

Proof. As a first step, we prove the proposition for p-groups G. Suppose that the proposition is false and let G be a counterexample. Clearly, if G = A, the proposition is fulfilled with π = ±⊬ℙ; similarly if G = B (let π = ∅). So we must have A < G and B < G. Since G is nilpotent, A and B are contained in maximal normal subgroups A^{∗} and B^{∗} of G. By Lemma 2.2.6, every subgroup that is prefactorized with respect to the factorization G = AB of G is also prefactorized with respect to the factorization G = A^{∗}B^{∗} so that G is also a counterexample with respect to the latter factorization. Therefore we may suppose w.l.o.g. that A and B are maximal normal subgroups of G.

Let N = A ∩ B, then G/N is the direct product of the two cyclic groups AN/N and BN/N, both of order p: A/N = ⟨a⟩N/N and B/N = ⟨b⟩N/N for suitable a ∈ A and b ∈ B. Now the diagonal subgroup D/N = ⟨ab⟩N/N is also a maximal normal subgroup of G/N which is not prefactorized in G/N since it intersects AN/N and BN/N trivially. Since N is prefactorized, by Lemma 1.2.2, D cannot be a prefactorized subgroup of G. So D, being a normal subgroup of G, does not have a prefactorized conjugate. This contradiction shows that we must have G = A or G = B.

Now let G be any finite nilpotent group and suppose that P is a Sylow p-subgroup of G for some prime p. If S is a subgroup of P, then every conjugate of S in G is already conjugate to S in P since P is centralized by the Hall p′-subgroup of G. On the other hand, we know by Proposition 2.1.2 that P = A_{p}B_{p} where A_{p} and B_{p} are the Hall psubgroups of A and B respectively. So we may apply the first part to P and obtain that P = A_{p} or P = B_{p}. Therefore every Sylow subgroup of G is contained in A or B. This shows that if π is the set of primes p such that A contains a Sylow p-subgroup of G, then A contains a Hall π-subgroup and B contains a Hall π′-subgroup of G.

To analyse the case when every subgroup has a factorized conjugate, we need the following simple

2.3.2 Lemma. Suppose that the finite group G is the product of its subgroups A and B. If A and B have coprime indices and A ∩ B = 1, then A and B have coprime orders.

Proof. This follows at once from Lemma 1.1.1.

Then we obtain

2.3.3 Corollary. Suppose that the finite nilpotent group G is the product of its subgroups A and B and that every subgroup of G has a factorized conjugate. Then A and B have coprime orders, hence are Hall subgroups of G.

Proof. Since the unit subgroup is normal and thus factorized, by Lemma 1.1.5, we must have A ∩ B = 1. So the result follows from Proposition 2.3.1 and Lemma 2.3.2.

We formulate the result of this section for prefactorized subgroups first.

2.3.4 Proposition. Suppose that the finite group G is the product of its subgroups A and B.

(i) If for every prefactorized subgroup H of G, every subgroup S ≤ H has a prefactorized conjugate in H, then A and B have coprime indices; moreover if H is a prefactorized subgroup of G, then H ∩ A and H ∩ B have coprime indices in H.

(ii) Suppose that the indices of A and B in G are coprime and that every subgroup of G satisfies D_{π} and D_{π′} for a set π of primes such that A contains a Sylow psubgroup of G for every p ∈ π and B contain Sylow p-subgroups of G for all p ∈ π′. Let H be a prefactorized subgroup of G satisfying D_{π} and D′_{π}. If |H : H ∩ A| and |H : H ∩ B| are coprime and S ≤ H, then there is a h ∈ H such that S^{h} ≤ H is prefactorized.

Proof. (i) Let p be a prime, then since every finite group satisfies D_{p} by Sylow’s theorem, by Lemma 2.1.1, G possesses a prefactorized Sylow p-subgroup P = (A∩P)(B∩P). Now by hypothesis, every subgroup of P has a prefactorized conjugate in P, and so by Proposition 2.3.1, P is contained in A or B. If π is the set of primes for which A contains a Sylow p-subgroup of G, then B contains a Sylow p-subgroup for all remaining primes p ∈ π′. So the index of A is a π′-number while that of B is a π-number.

Since every prefactorized subgroup H of G likewise satisfies the hypotheses of (i), it is clear that also A ∩ H and B ∩ H have coprime indices in H.

(ii) This has already been proved in Lemma 2.2.7.

Unfortunately, the statements (i) and (ii) are not equivalent. However, it should be noted that (i) implies (ii) if the group G is finite and soluble since in this case, G satisfies D_{π} for every set of primes π.

In the case when A and B have coprime orders, a much more satisfactory result can be proved, which even holds for arbitrary finite groups.

2.3.5 Theorem. Let the finite soluble group G be the product of its subgroups A and B. Then the following statements are equivalent:

(i) For every factorized subgroup H of G and every subgroup S ≤ H, there is an h ∈ H such that S^{h} is factorized.

(ii) The subgroups A and B of G have coprime orders and every subgroup of G sat-

isfies D_{π} and D_{π′}, where π is the set of prime divisors of A. Proof. (i) ⇒ (ii). From Proposition 2.3.4 and Lemma 2.3.2, we obtain that A and B

have coprime orders. Since the property (i) is inherited by subgroups and the situation is completely symmetrical for the sets π and π′, it remains to show that G itself satisfies D_{π}: it is clear that G possesses a Hall π-subgroup, namely A. If P is a π-subgroup of G, there is a conjugate P^{g} of P for some g ∈ G, with P^{g} = (P^{g} ∩ A)(P^{g} ∩ B) and since B is a π′-group, we have P^{g} ∩ B = 1 and so P^{g} is contained in A. So the conjugates of A are the maximal π-subgroups of G and every π-subgroup is contained in a conjugate of A; hence G satisfies D_{π}.

(ii) ⇒ (i). This is the result of Lemma 2.2.5.

For an additional result in the case when the finite soluble group G is the product of two nilpotent subgroups, see Section 3.3.

Subgroups of products of two finite nilpotent groups

3.1 First results about products of nilpotent subgroups

The following theorem, known as the Kegel-Wielandt theorem, is probably the most important theorem about finite groups that are the product of two finite nilpotent subgroups; note that this theorem remains true for products of any finite number of nilpotent subgroups.

3.1.1 Theorem (Wielandt [45] and Kegel [33]). Let the finite group G be the product of its nilpotent subgroups A and B. Then G is soluble.

A proof of the Kegel-Wielandt theorem can also be found in [4], Section 2.4 and in [31], VI, § 4.

3.1.2 Proposition. Let G be a finite group. If A, B ≤ G such that AB^{g} = B^{g}A for all g ∈ G, then:

(i) (Kegel [33]) if AB < G, then AB^{G} < G or A^{G}B < G; in particular, A or B is contained in a proper normal subgroup of G.

(ii) (Wielandt [46]) [A, B] ⊴ A^{AB} ∩ B^{AB} ◁◁ G.

Proof. (i) Suppose that AB^{G} = A^{G}B = G. We have to show that G = AB. We proceed by induction on |G : A| + |G : B|, observing that the statement is trivial if G = A = B. Also, (i) is true if A is normal in G, we may assume that A^{x} ≠ A for some x ∈ G.

Since every conjugate of B permutes with both A and A^{x}, by Lemma 1.1.3, B also permutes with A_{1} = ⟨A, A^{x}⟩. Then we have G = A_{1}B by induction hypothesis. Since x ∈ G = AB^{G}, we can write x = ab^{y} with a ∈ A, b ∈ B and y ∈ G. and w.l.o.g., we may assume that a = 1.

By Lemma 1.1.7, also G = A_{1}B^{y} = ⟨A, A^{x}, B^{y}⟩ = AB^{y} and thus also G = AB.

(ii) If G = AB, then it is clear that [A, B] and A^{AB} ∩ B^{AB} are normal subgroups of G and that [A, B] ≤ A^{AB} ∩ B^{AB}. Therefore we may suppose that AB < G.

From (i), we infer that A^{G}B < G or AB^{G} < G; w.l.o.g. suppose the former. Then by finite induction, [A, B] ⊴ A^{AB} ∩B^{AB} ⊴ A^{G}B and so also [A, B] ⊴ A^{AB} ∩B^{AB} ⊴ A^{G} ⊴ G which shows that the subgroups in question are subnormal in G.

We mention an important special case of the previous proposition:

3.1.3 Corollary. Let G = AB be a finite group that satisfies D_{π} for a set π of primes. If A and B have normal Hall π-subgroups A_{π} and B_{π}, then:

(i) If G is not a π-group, then A_{π} or B_{π} is contained in a proper normal subgroup of G.

(ii) The subgroups [A_{π}, B_{π}] and A_{π}^{B_{π}} ∩ B_{π}^{A_{π}} are subnormal π-subgroups of G.

(iii) If A_{π} and B_{π} are nilpotent, then the normal Hall π-subgroup A_{π} ∩ B_{π} of A ∩ B, and thus every π-subgroup of A ∩ B, is a subnormal subgroup of G.

Proof. By Lemma 2.1.1, the group G possesses a Hall π-subgroup of the form A_{π}^{a}B_{π}^{b} for some a ∈ A and b ∈ B. Since the Hall π-subgroups of A and B are normal, we have A_{π}^{a}B_{π}^{b} = A_{π}B_{π}. Thus by Lemma 1.1.7, every conjugate of A_{π} permutes with every conjugate of B_{π}, and consequently [A_{π}, B_{π}] and A_{π}^{B_{π}} ∩ B_{π}^{A_{π}} are subnormal π-subgroups of G by Proposition 3.1.2.

3.1.4 Proposition (Pennington [40]). Let G = AB be a finite group satisfying D_{π}. If A and B have normal Hall π-subgroups, then O_{π}(G) = (A ∩ O_{π}(G))(B ∩ O_{π}(G)) and A_{π} ∩ B_{π} ≤ O_{π}(G).

Proof. Write O = O_{π}(G). Since G satisfies D_{π}, by Lemma 2.1.1, there is a Hall πsubgroup of G of the form H = A_{π}B_{π}. We show that O is a factorized subgroup of H = A_{π}B_{π} (which does not, however, imply that O is factorized in G because H is not necessarily factorized).

Consider the group G/O: since the normal Hall π-subgroups A_{π}O/O and B_{π}O/O are normal in AO/O and BO/O respectively, every conjugate of A_{π}O/O permutes with B_{π}O/O and so by Proposition 3.1.2, A_{π}O/O ∩ B_{π}O/O is a subnormal π-subgroup of G. Therefore A_{π}O ∩ B_{π}O ≤ O, showing that O = A_{π}O ∩ B_{π}O. Thus by Lemma 1.2.1, O equals its factorizer in H, i.e. O is factorized in H as required.

This leads to the following

3.1.5 Theorem (Amberg [1], Pennington [40]). If G is the product of two finite nilpotent subgroups A and B, then F (G), the Fitting subgroup of G, is factorized. Therefore the subgroup A ∩ B is subnormal in G.

Proof. Since F (G) is the product of all O_{p}(G) and by Sylow’s theorem, G satisfies D_{{p}} for all primes p dividing the order of G, it is clear from the preceding Proposition 3.1.4 that F (G) is prefactorized. Also, A∩B is nilpotent, hence is the direct product of all A_{p} ∩ B_{p} and so F (G) also contains A ∩ B. This also shows that A ∩ B ◁◁ F (G) is subnormal in G.

We will see in Section 3.6 that the last two results can be generalized to radicals with respect to arbitrary Fitting formations.

Also, the fact that A∩B, and hence every subgroup of A∩B, is a subnormal subgroup of G has lead to the following result of Maier [36] and Wielandt [47]: they show that if the finite group G is the product of two subgroups A and B and S ≤ A ∩ B such that S ◁◁ A and S ◁◁ B, then S ◁◁ G. Further results in this direction can also be found in [38] and [11].

3.1.6 Corollary. If the finite group G is the product of its nilpotent subgroups A and B and N ⊴ G, then the factorizer and every prefactorizer of N is subnormal in G.

Proof. Consider the factor group G/N. Then the subgroup AN/N ∩BN/N is subnormal in G/N the preceding theorem. Therefore the subgroup X = AN ∩BN is subnormal in G. On the other hand, by Lemma 1.2.1, X equals the factorizer of N in G.

For prefactorizers S, the result follows from the fact that N ≤ S ≤ X and X/N is nilpotent.

3.1.7 Corollary. Let G be a finite group which is the product of its nilpotent subgroups A and B. If G possesses a nilpotent normal subgroup N such that G = AN = BN, then G is nilpotent.

Proof. If N is a nilpotent normal subgroup of G, then its factorizer (with respect to the factorization G = AB) is contained in the factorized subgroup F (G) of G. But by Lemma 1.2.1, the factorizer of N equals AN ∩ BN = G and so G is nilpotent.

For generalizations of the last result to certain classes of infinite groups, we refer the reader to Section 6.3 of Amberg, Franciosi and de Giovanni [4]. We will also see in Corollary 3.5.2 that in fact N does not necessarily have to be normal in G.

The following proposition shows that every finite group G that is the product of two proper nilpotent subgroups has a proper factorized normal subgroup.

3.1.8 Proposition (Kegel [33]). If the finite group G is the product of its nilpotent subgroups A and B and A ≠ B, then A or B is contained in a proper normal subgroup of G.

Proof. If G = 1, there is nothing to prove. Therefore suppose by finite induction that the proposition is true for all groups of smaller order than |G| and let N be a minimal normal subgroup of G. If AN ≠ BN, then by induction hypothesis, AN/N or BN/N is contained in a proper normal subgroup of G/N and therefore also A or B is contained in a proper normal subgroup of G.

Therefore we may assume that AN = BN = G, and since A, B and N are nilpotent, G is nilpotent by Corollary 3.1.7. Then, however, at least one of A and B is contained in a maximal subgroup of G which is normal in G.

Kegel’s result has been extended by Amberg, Franciosi and de Giovanni [3] to certain classes of (possibly infinite) groups G which are the product of their subgroups A_{1}, …, A_{r} satisfying some nilpotency condition: if at least one of the subgroups A_{i} is properly contained in G, then some A_{i} is contained in a proper normal subgroup of G. This result holds in particular if the subgroups A_{i} are finite and nilpotent.

On the other hand, a similar result about minimal normal subgroups of a product of finite nilpotent subgroups does not hold: If p is any prime, J. D. Gillam [19] gives an example of a p-group P of order p^{6} which is the product of two subgroups A and B but A and B do not contain normal subgroups of P. This also shows that the centre of P cannot be prefactorized: since A ∩ Z(P) and B ∩ Z(P) are normal subgroup of P, we must have Z(P) ≠ 1 = (A ∩ Z(P))(B ∩ Z(P))

We mention a consequence Proposition 3.1.8 that might be of interest.

3.1.9 Proposition. Let G be a finite group. If A and B are nilpotent subgroups of G such that AB^{g} = B^{g}A for all g ∈ G, then A ∩ B is subnormal in G.

Proof. If A = B (= G), the group G is nilpotent and hence every subgroup of G is subnormal. Therefore we may suppose that A ≠ B and by Proposition 3.1.8, one of the factors, say A, is contained in a normal subgroup N of G. Now by the modular law, for all g ∈ G,

A(N ∩ B^{g}) = N ∩ AB^{g} = N ∩ B^{g}A(N ∩ B^{g})A

for all g ∈ G which shows that A permutes with every conjugate of B ∩ N. Therefore by induction on the order of G, A ∩ B = A ∩ B ∩ N is subnormal in N, hence in G.

3.2 Primitive groups that are the product of two nilpotent subgroups

Let G be group that possesses a faithful representation as a a transitive permutation group on the set X. Recall that G is called primitive if we have |Y | = 1 for every subset Y of X such that Y g = Y or Y g ∩ Y = ∅ for all g ∈ G. By [43], 7.2.3, G is primitive if and only if for every x ∈ X, the stabilizer M = {g ∈ G | x^{g} = x} of x is a maximal subgroup of G. Since G is transitive, the conjugates of M in G are precisely the stabilizers of the elements of X, so it is clear that M_{G} = 1. On the other hand, if an arbitrary group G possesses a maximal subgroup M with M_{G} = 1, then the permutation representation of G on the cosets of M is clearly faithful and has M as a stabilizer, hence G is primitive by the theorem stated above.

The next simple lemma gives a sufficient condition for a finite group to be primitive. Theorem 3.2.2 below will show that this condition is also necessary:

3.2.1 Lemma. Let G be a finite group. If G possesses a maximal subgroup M < G that supplements every minimal normal subgroup of G, then G is primitive and M is a stabilizer of G.

Proof. Suppose that M_{G} ≠ 1, then M_{G} contains a minimal normal subgroup N of G. But then N ≤ M = MN = G, contradicting the maximality of M.

The class of finite primitive groups can be divided into three disjoint subclasses as the following theorem (Baer [6], see also Doerk and Hawkes [13], A.15.2) shows:

3.2.2 Theorem. Let G be a finite group. Then the following statements are equivalent:

(i) G is primitive with stabilizer M; (ii) G satisfies one of the following statements:

(1) G has a unique minimal normal subgroup N; moreover N is abelian, N = C_{G}(N) and N is complemented by M.

(2) G has a unique minimal normal subgroup N; N is non-abelian, C_{G}(N) = 1 and N is supplemented by M. Furthermore, if V is a minimal supplement to N, then N ∩ V ≤ Φ(V),

(3) G has exactly two isomorphic non-abelian minimal normal subgroups N and N^{∗}. Moreover C_{G}(N) = N^{∗} and C_{G}(N^{∗}) = N and N ≅ N^{∗} ≅ NN^{∗} ∩ M. M complements N and N^{∗}, and if V < G supplements N and N^{∗}, then V complements N and N^{∗}. Also, M ∩ NN^{∗} is a (normal) subgroup of M isomorphic with N (and N^{∗}).

If G is a finite soluble group, clearly every minimal normal subgroup of G is abelian, so a finite primitive soluble group must belong to the class of groups described in (1) above.

Primitive finite soluble groups can be characterized in several ways:

3.2.3 Lemma. Let G be a finite soluble group and suppose that N is a minimal normal subgroup of G whose order is divisible by the prime p. Then the following statements are equivalent:

(i) G is primitive; (ii) N = C_{G}(N); (iii) N = O_{p}(G) for the prime p and O_{p′}(G) = 1; (iv) N = F (G);

(v) Φ(G) = 1 and N is the only minimal normal subgroup of G.

Proof. Note first that N is elementary abelian of exponent p since G is soluble.

(i) ⇒ (ii): This follows directly from Theorem 3.2.2 since N is abelian.

(ii) ⇒ (iii): If N = C_{G}(N) and N^{∗} ≠ N is a minimal normal subgroup, then [N.N^{∗}] ≤ N ∩ N^{∗} = 1, which shows that N^{∗} ≤ C_{G}(N) = N, a contradiction. Therefore N is the unique minimal normal subgroup of G. So if O_{p′}(G) ≠ 1, we must have N ≤ O_{p′}(G) which is, of course, impossible, N being a p-group. Therefore O_{p′}(G) = 1. Let O = O_{p}(G), then N ≤ O; moreover O is nilpotent and so Z(O) ≠ 1. Z(O) is characteristic in O, hence normal in G so that N ≤ Z(O) and so O ≤ C_{G}(N) = N, proving that N = O = O_{p}(G).

(iii) ⇒ (iv): Clearly, the Hall p′-subgroup of F (G) is characteristic in F (G), so it is normal in G and hence contained in O_{p′}(G) = 1. Therefore F (G) is a p-group and F (G) = O_{p}(G) = N.

(iv) ⇒ (v): Φ(G) is a characteristic subgroup of G properly contained in F (G) = N. Therefore we must have Φ(G) = 1. Clearly, every minimal normal subgroup of the soluble group G is abelian, therefore contained in the Fitting subgroup of G. Since F (G) = N is itself a minimal normal subgroup, it follows that N is the unique minimal normal subgroup of G.

(v) ⇒ (i): Since Φ(G) = 1, there is a maximal subgroup M of G which does not contain N. If M_{G} ≠ 1, the normal subgroup M_{G} must contain the unique minimal normal subgroup N of G. But then N ≤ M, a contradiction. So we must have M_{G} = 1 and G is primitive.

We will now analyse primitive groups G which are the product of two finite nilpotent subgroups. The main result, namely that if G is non-nilpotent, then one of these subgroups is a Sylow p-subgroup and the other a Hall p′-subgroup, is due to Gross [22].

First, we prove a generalization of this result which is based on the proof of Gross’ statement that can be found in [4], Lemma 2.5.2. For the definition of a π-separable group, we refer the reader to Section 4.1.

3.2.4 Lemma. Let the π-separable group G be the product of its subgroups A and B and suppose that A and B have normal Hall π′-subgroups. Furthermore, suppose that G does not contain nontrivial normal π′-subgroups and let O = O_{π}(G). Then

(i) C_{G}(O) ≤ O; (ii) if O is the unique minimal normal subgroup of G, then A or B is a π-group; (iii) if in addition A and B have normal Hall π-subgroups, then AO/O and BO/O

are Hall π- and Hall π′-subgroups of G/O and O is factorized; (iv) if in addition, O is abelian and O < G, then A and B are Hall π- and Hall π′-

subgroups of G. Thus every subgroup of G has a factorized conjugate (in particular, every normal subgroup of G is factorized), and if G = AC = BC for some C ≤ G, then G = C.

Proof. (i) This is a result of Hall and Higman [26], Lemma 1.2.3.

(ii) Let A_{π′} and B_{π′} be the Hall π′-subgroups of A and B respectively, then A_{π′}B_{π′} is a Hall π′-subgroup of G by Lemma 2.1.1, because every π-separable group satisfies D_{π} and D_{π′}, and since A_{π′} and B_{π′} are normal subgroups of A and B respectively, we have A_{π′}B_{π′}^{g} = B_{π′}^{g}A_{π′} for all g ∈ G by Lemma 1.1.7. Now by Proposition 3.1.2, [A_{π′}^{g}, B_{π′}^{g}] is a subnormal π′-subgroup of G and as such it is contained in O_{π′}(G) = 1. Therefore also [A_{π′}^{G}, B_{π′}^{G}] = 1. If, say, B is not a π-group, then B_{π′}^{G} ≠ 1, thus O ≤ B_{π′}^{G} and [A_{π′}, O] ≤ [A_{π′}^{G}, B_{π′}^{G}] = 1. Therefore, A_{π′} ≤ C_{G}(O) ≤ O and A must be a πgroup.

(iii) Suppose w.l.o.g. that A is a π-group. As above, G has a Hall π-subgroup of the form A_{π}B_{π} = AB_{π} by Lemma 2.1.1. Now B_{π} ⊴ B and therefore B_{π}^{G} = B_{π}^{A} which is contained in the π-group AB_{π}. So B_{π}^{G} is a normal π-subgroup of G, and B_{π} must be contained in O. Hence AO/O is a π-group and BO/O is a π′-group. Moreover by Corollary 2.2.4, 1 = O/O is a factorized subgroup of G/O and so by Lemma 1.2.2, O is factorized.

(iv) Since O_{ππ′}(G)/O is a π′-group, it is contained in BO/O = B_{π′}O/O. Now if O is abelian, then both O and B_{π′} centralize B_{π} ≤ O and so O_{ππ′}(G) centralizes B_{π}: Therefore B_{π} ≤ Z(O_{ππ′}(G)) ≤ C_{G}(O) = O. By the minimality of O, we have either Z(O_{ππ′}(G)) = 1 or Z(O_{ππ′}(G)) = C_{G}(O). From the latter, it follows directly that O = O_{ππ′}(G) = G, contrary to our assumption that O < G. So we must have Z(O_{ππ′}(G)) = 1 and hence B_{π} = 1. So by Lemma 1.1.1, A and B must be Hall subgroups of G. The remaining statements follow from Lemma 2.2.5 and Lemma 1.1.8.

We summarize the results of Lemma 3.2.4 for primitive groups that are the product of two nilpotent subgroups A and B

3.2.5 Lemma. Let the finite soluble group G = AB be the product of its nilpotent subgroups A and B. Suppose that G is primitive and denote with N the unique minimal normal subgroup of G. Then

(i) N = F (G) = O_{p}(G) is an elementary abelian p-group for some prime p and O_{p′}(G) = 1;

(ii) A or B is a Sylow p-subgroup of G containing N; if A ≠ B, then the other is a

Hall p′-subgroup. In particular, A ∩ B = 1. (iii) If F_{2}/N = F (G/N), then F_{2}/N is a p′-group and every prime divisor of |B|

divides already |F_{2}|. (iv) If A and B are proper subgroups of G, then A and B are maximal nilpotent

subgroups of G.

(v) If A ≠ B, then every subgroup of G possesses a factorized conjugate.

(vi) If A ≠ B, then G = BC = CA for a subgroup C, then G = C.

Proof. First, we deal first with the case when G is nilpotent. Since then a maximal subgroup of G is normal, a primitive nilpotent group is cyclic of prime order. So either G = A = B or A = 1 and B = G or A = G and B = 1. In all three cases, it is easy to see that the lemma holds. Therefore we may suppose from now on that A, B and N are properly contained in G.

(i) follows directly from Lemma 3.2.3.

(ii), (v) and (vi) follow directly from Lemma 3.2.4 with π = {p}, since we have already excluded the case when G = N.

(iii) If P/N is a Sylow p-subgroup of F_{2}/N, then P is a normal p-subgroup of G since F_{2}/N is nilpotent. This shows that P = N and F_{2}/N is a p′-group. So F_{2}/N is contained in the Hall p′-subgroup BN/N of G/N. Therefore Z(BN/N) ≤ Z(F_{2}/N) ≤ C_{G/N}(F_{2}/N) ≤ F_{2}/N. Since Z(BN/N) ≅ Z(B) is the direct product of the centres of the primary components of BN/N, this shows that every prime divisor of B divides F_{2}/N.

(iv) Suppose that A is contained in a proper nilpotent subgroup H of G, then G = HB and, applying the same arguments to G = HB, we obtain H∩B = 1. So by Lemma 1.1.1, we must have |A| = |H| and thus A = H, similarly for B. Therefore A and B are maximal nilpotent subgroups of G.

For further results about the structure of G, see also Lemma 4.2.1.

3.3 Pronormal and abnormal subgroups

Recall that a subgroup P of the finite group G is called pronormal if P and P^{g} are conjugate in the subgroup ⟨P, P^{g}⟩ for every g ∈ G. If P ≤ H ≤ G, then, of course, P is also a pronormal subgroup of H.

Pronormal subgroups of finite soluble groups can also be characterized in the following way:

3.3.1 Theorem (Mann [39]). A subgroup P of a finite soluble group G is pronormal if and only if each Hall system of G reduces into exactly one conjugate of P.

A proof of this can also be found in [13], I.6.6.

3.3.2 Proposition. Let G be a finite soluble group which is the product of its nilpotent subgroups A and B. If S is a prefactorized subgroup of G, then the Hall system

Σ = {A_{π}B_{π} | π ⊆ ±⊬ℙ}

reduces into S. Moreover, Σ is the only Hall system of G that reduces into every prefactorized subgroup of G. In particular Σ is the only Hall system of G consisting entirely of prefactorized Hall subgroups.

Proof. By Proposition 2.1.2, Σ is a Hall system of G. Since S is the product of its nilpotent subgroups S ∩ A and S ∩ B, also the subgroup S possesses a Hall system of the form

{S_{π} = (S ∩ A)_{π}(S ∩ B)_{π} | π ⊆ ±⊬ℙ}

where (S ∩ A)_{π} and (S ∩ B)_{π} are the Hall π-subgroups of S ∩ A and S ∩ B respectively. Since A_{π} and B_{π} are the unique Hall π-subgroups of A and B respectively, we must have (S ∩ A)_{π} ≤ A_{π} and (S ∩ B)_{π} ≤ B_{π}. This shows that for every set π of primes, the Hall π-subgroup S_{π} = (S ∩ A)_{π}(S ∩ B)_{π} of S is contained in the π-subgroup S ∩ A_{π}B_{π}, and therefore that S_{π} = S ∩ A_{π}B_{π}. Since this is true for every set π of primes, we have shown that Σ reduces into S.

Now suppose that Σ^{∗} is another Hall system of G reducing into every prefactorized subgroup of G. If π is a set of primes, then Σ^{∗} reduces into the Hall subgroup A_{π}B_{π} of G. Therefore A_{π}B_{π} must be contained in a Hall π-subgroup H ∈ Σ^{∗} and so A_{π}B_{π} ∈ Σ^{∗}. Continuing like this for every set π of primes, we have Σ = Σ^{∗} as required.

The next proposition is a direct consequence of Proposition 3.3.2 and Theorem 3.3.1.

3.3.3 Proposition. Let the finite group G be the product of its nilpotent subgroups A and B. If P is a pronormal subgroup of G, then P has a unique prefactorized conjugate.

Proof. If P^{g} and P^{h} are prefactorized conjugates of P for g, h ∈ G, then by Proposition 3.3.2, the Hall system

Σ = {(A_{π}B_{π} | π ⊆ ±⊬ℙ}

reduces into both P^{g} and P^{h}. Since P is pronormal, we must have P^{g} = P^{h} by Theorem 3.3.1.

A subgroup S of G is called abnormal if g ∈ ⟨S, S^{g}⟩ for every g ∈ G. Of course, every abnormal subgroup of G is pronormal in G. Also, if S ≤ H ≤ G, then, S is abnormal in H. The following lemma shows in particular that an abnormal subgroup cannot be contained in a proper normal subgroup of G.

Remark. By induction on the number of factors, it can be proved that Proposition 3.3.2 and Proposition 3.3.3 even holds for groups G that are the product of finitely many finite nilpotent subgroups.

3.3.4 Lemma. Let S be an abnormal subgroup of the (possibly infinite) group G. If S ≤ K ⊴ H ≤ G, then H = K. Thus S = N_{G}(S).

Proof. For all h ∈ H, we have S^{h} ≤ K^{h} = K and therefore h ∈ ⟨S, S^{h}⟩ ≤ K. So we have H = K. The second statement follows from the fact that S ⊴ N_{G}(S).

3.3.5 Proposition. Let G = AB be the product of the finite nilpotent subgroups A and B. Then every abnormal subgroup of G possesses exactly one factorized conjugate.

Proof. Let S be an abnormal subgroup of G. If S = G, then there is nothing to prove, so suppose that S < G and let M be a maximal subgroup of G which contains S. By Lemma 3.3.4, M cannot be normal in G and therefore M_{G} < M < G and G/M_{G} cannot be nilpotent. Hence we may apply Lemma 3.2.5 to show that M/M_{G} possesses a factorized conjugate, say, M^{g}/M_{G}. So by Lemma 1.2.2, also M^{g} is factorized in G. Now S^{g} is clearly an abnormal subgroup of M^{g} and by induction on the order of G, the subgroup S^{g} possesses a factorized conjugate in M^{g} which is also factorized in G by Lemma 1.1.6 and is clearly conjugate in G to S as required. The uniqueness of this conjugate follows from Proposition 3.3.3.

Since every maximal nonnormal subgroup of a group G is abnormal, we also have

3.3.6 Corollary. Suppose that the group G is the product of its finite nilpotent subgroups A and B. Then every maximal nonnormal subgroup of G has a unique factorized conjugate.

The following results shows that in general, one cannot expect every conjugate of a subgroup to be prefactorized or factorized.

3.3.7 Corollary. If the finite group G is the product of its subgroups A and B and every maximal subgroup of G is prefactorized, then G is nilpotent.

Proof. Let M be a maximal subgroup of G. Since M is a pronormal subgroup of G, M has a unique prefactorized conjugate by Proposition 3.3.3. Since on the other hand, every conjugate of G is a maximal subgroup, hence prefactorized, it follows that M ⊴ G. So every maximal subgroup of G is normal and G must be nilpotent.

This result can be used to characterize those products of two finite nilpotent subgroups in which every subgroup is factorized or prefactorized.

3.3.8 Proposition. Let G be a finite soluble group which is the product of its nilpotent subgroups A and B. Then the following statements are equivalent:

(i) Every subgroup of the nilpotent group G has a factorized (prefactorized) conjugate. (ii) G is nilpotent; its subgroups A and B have coprime orders (coprime indices). (iii) Every subgroup of G is factorized (prefactorized).

Proof. (i) ⇒ (ii): This follows immediately from Proposition 2.3.1 and Corollary 2.3.3 respectively.

(ii) ⇒ (iii). Obviously, every subgroup of G is subnormal, hence this follows from Lemma 2.2.2 and Corollary 2.2.4 respectively.

(iii) ⇒ (i): We have to show that G is nilpotent. But this follows at once from Corollary 3.3.7

The next lemma shows that abnormal subgroups can be characterized as the normalizers of the pronormal subgroups:

3.3.9 Lemma. Let G be a (possibly infinite) group. Then the normalizer of a pronormal subgroup of G is abnormal in G. Therefore a subgroup of G is abnormal if and only if it is pronormal and self-normalized.

Proof. Suppose that P is a pronormal subgroup of G and let g ∈ G. We have to show that g ∈ H = ⟨N_{G}(P), N_{G}(P)^{g}⟩. Since ⟨P, P^{g}⟩ ≤ H, there is an h ∈ H such that P^{h} = P^{g}. Hence gh^{−1} ∈ N_{G}(P) and consequently g ∈ ⟨N_{G}(P), H⟩ = H, proving that N_{G}(P) is abnormal.

From this, it follows that a self-normalized pronormal subgroup is abnormal. On the other hand, we have already proved in Lemma 3.3.4 that abnormal subgroups are selfnormalized.

A factorized (abnormal) subgroup clearly contains A ∩ B, hence we obtain

3.3.10 Corollary. Let the group G be the product of its nilpotent subgroups A and B. If P is any pronormal subgroup of G, then it is normalized by some conjugate of A ∩ B.

Suppose that the group G is the product of its nilpotent subgroups A and B and that P is a prefactorized pronormal subgroup of G. Then N_{G}(P) is abnormal in G, and therefore we know that some conjugate of N_{G}(P) is factorized. The next proposition shows that N_{G}(P) itself is factorized (and therefore no other conjugate of N_{G}(P) can be factorized). This means that the investigation whether a pronormal subgroup P of G has a prefactorized (factorized) conjugate may be reduced to finding the factorized conjugate N_{G}(P)^{g} of its normalizer and to checking whether the conjugate P^{g} of P that lies in N_{G}(P)^{g} = N_{G}(P^{g}) is a prefactorized (factorized) normal subgroup of N_{G}(P)^{g}.

3.3.11 Proposition. Let the group G be the product of its subgroups A and B and let P be a pronormal subgroup of G. Then the following statements are equivalent:

(i) P is prefactorized; (ii) N_{G}(P) is factorized and P is a prefactorized (normal) subgroup of N_{G}(P). Proof. Suppose that P is a prefactorized pronormal subgroup of G. Since N_{G}(P) is

abnormal in G, by Proposition 3.3.5, there is a g ∈ G such that N_{G}(P)^{g} = N_{G}(P^{g}) is factorized. Now by Proposition 3.3.2, the Hall system Σ defined there reduces into N_{G}(P^{g}) and since P^{g} ⊴ N_{G}(P^{g}), Σ reduces into P^{g} as well.

On the other hand, Σ also reduces into P by Proposition 3.3.2 since P is prefactorized, and so we must have P^{g} = P by Proposition 3.3.3, showing that the normalizer N_{G}(P) = N_{G}(P^{g}) of P is factorized.

The other implication follows directly from Lemma 1.1.6.

3.3.12 Corollary. Let the group G be the product of its subgroups A and B. If P is a pronormal subgroup of G such that N_{G}(P) is factorized, then the factorizer X_{G}(P) of P can be written X_{G}(P) = N_{A}(P)P ∩ N_{B}(P)P and it possesses a triple factorization

X_{G}(P) = A^{∗}P = B^{∗}P = A^{∗}B^{∗}

where P ⊴ X, A^{∗} = N_{A}(P) ∩ N_{B}(P)P ≤ A and B^{∗} = N_{A}(P)P ∩ N_{B}(P) ≤ B.

Proof. N_{G}(P) is factorized, so we have X_{G}(P) ≤ N_{G}(P) which shows that X_{G}(P) is also the factorizer of P in N_{G}(P) = N_{A}(P)N_{B}(P). Since P ⊴ N_{G}(P), the corollary now follows directly from Lemma 1.2.1.

It also follows from Proposition 3.3.11 that the system normalizer of the Hall system Σ is factorized:

3.3.13 Corollary. Suppose that the group G is the product of its finite nilpotent subgroups and let Σ be the Hall system defined in Proposition 3.3.2. Then N_{G}(H) is factorized for all H ∈ Σ and also the system normalizer

N_{G}(Σ) = ⋂_{H∈Σ} N_{G}(H)

of Σ is factorized.

Proof. Since every Hall subgroup H ∈ Σ is prefactorized, it follows from Proposition 3.3.11 that N_{G}(H) is factorized for all H ∈ Σ. The system normalizer N_{G}(Σ) of Σ is factorized since by Lemma 1.1.6, the intersection of any number of factorized subgroups is factorized.

Question. Recall that the hypercentre of a finite soluble group equals the intersection of its system normalizers. Is the hypercentre of a finite group G that is the product of two nilpotent subgroups prefactorized? (This question is motivated by the fact that a corresponding result holds for F-injectors: if an F-injector of a group G that is the product of two nilpotent subgroups is factorized (prefactorized), then also its core, the F-radical of G, is factorized (prefactorized); see Proposition 3.6.4 for details.)

On the other hand, by an example of Heineken [28], the hypercentre does not necessarily contain A ∩ B and hence it is not factorized. Also, an example of Gillam [19] of a finite p-group G which is the product of two subgroups A and B which do not contain normal subgroups of G shows that it is possible to have A∩Z(G) = 1 and B ∩Z(G) = 1 whence the centre Z(G) of G is not necessarily prefactorized.

3.4 Projectors

Let 𝔛 be a class of groups. An 𝔛-subgroup X of a group G is 𝔛-maximal in G if for every 𝔛-subgroup Y of G with X ≤ Y, it follows that X = Y. A subgroup P of G is called an 𝔛-projector if P N/N is an 𝔛-maximal subgroup of G/N for every normal subgroup N of G.

Recall from Section 1.3 that a class ℌ of finite groups is called a Schunck class if G ∈ ℌ whenever all primitive homomorphic images of G lie in ℌ. The Schunck classes of finite soluble groups are precisely the classes for which every finite soluble group G possesses an ℌ-projector. In this case, the ℌ-projectors of G are conjugate; see e.g. [13], III.3.10, 3.21, and by [13], III.3.22, if P is an ℌ-projector of G and P ≤ H ≤ G, then P is an ℌ-projector of H as well. This also shows that ℌ-projectors are pronormal.

In view of Proposition 3.3.11, we examine first the case when a projector is normal in G; cf. Blessenhohl and Gasch^{}ütz [10].

3.4.1 Lemma. Let G be a finite soluble group and suppose that ℌ is a Schunck class such that an ℌ-projector P is normal in G. Then P = O^{π′}(G) where π is the characteristic of ℌ.

Proof. Suppose first that a prime p ∈ π divides the order of G/P. Then G/P has a subgroup of order p which belongs to ℌ and so 1 = P/P is not ℌ-maximal in G/P, contradicting the fact that P is an ℌ-projector. Therefore G/P is a π′-group and O^{π′}(G) ≤ P.

Now suppose that O^{π′}(G) < P and let M be a maximal normal subgroup of P containing O^{π}(G). Then P/M is a π′-group and since G (and hence P) is soluble, P/M is cyclic of prime order p ∈ π′. But P/M ∈ Qℌ = ℌ whose characteristic is π, hence does not contain cyclic p-groups. This contradiction shows that P = O^{π′}(G).

We will also make use of the following property of saturated formations.

3.4.2 Proposition. If 𝔉 is a saturated formation of characteristic π, then every group G ∈ 𝔉 is a π-group.

Proof. See e.g. Doerk and Hawkes [13], IV.4.3.

This result is false for arbitrary Schunck classes, e.g. for the classes of finite π-perfect (soluble) groups, i.e. the class of finite (soluble) groups G for which O^{π}(G) = G, if ∅ ⫋ π ⫋ ±⊬ℙ.

For saturated formations, the next result has been proved by Heineken [28]:

3.4.3 Theorem. Let ℌ be a Schunck class and suppose that the group G is the product of its finite nilpotent subgroups A and B.

(i) If char(ℌ) contains σ(A) ∩ σ(B), then G has a unique factorized ℌ-projector. (ii) If ℌ is a saturated formation, then G has a unique prefactorized ℌ-projector.

Proof. Note first that in both cases, the uniqueness of the factorized or prefactorized projector follows from Proposition 3.3.3 since an ℌ-projector is pronormal. Therefore it remains to prove the existence of such projectors in G.

Let P be an ℌ-projector of G. Since N_{G}(P) is an abnormal subgroup of G, by Proposition 3.3.5, it has a factorized conjugate N_{G}(P)^{g} = N_{G}(P^{g}) where g ∈ G. Therefore by Proposition 3.3.11, it is enough to show that P^{g} is a prefactorized subgroup of N_{G}(P^{g}), hence we may assume that G = N_{G}(P^{g}) and P ⊴ G. Then we have P = O^{π′}(G) by the preceding Lemma 3.4.1, where π = char(ℌ).

Now under the hypothesis of (i), the groups AP/P and BP/P have coprime orders since π contains all common prime divisors of A and B. Therefore by Corollary 2.2.4, 1 = P/P is a factorized subgroup of G/P and so by Lemma 1.2.2, P is factorized in G.

In the case of (ii), P is a π-group by Proposition 3.4.2 since ℌ is a saturated formation. On the other hand, since G/P is a π′-group, P is the unique Hall π-subgroup of G, which must be prefactorized by Lemma 2.1.1.

Let the group G be the product two nilpotent subgroups A and B. The following example shows that none of the following subgroups of G is necessarily prefactorized: G′, G^{𝔑}, [A, B], O^{π}(G) for all sets π ≠ ∅ of primes. (Observe that since all these subgroups are normal, there cannot exist prefactorized conjugates, and, except possibly for [A, B], no automorphism of G can map these subgroups to a prefactorized subgroup).

Since for every finite group G, the subgroup O^{π}(G) is the unique ℌ-projector of G where ℌ is the Schunck class of π-perfect groups, the example also shows that statement (ii) of Theorem 3.4.3 becomes false when ℌ is a Schunck class that is not a saturated formation.

Observe also that the subgroups A and B are abelian in the following

3.4.4 Example. Let p and q be distinct primes and let N be a q-dimensional vector space over F = GF (p), written additively. Define an automorphism α of N of order q by

α: (x_{1}, …, x_{q}) ↦ (x_{q}, x_{1}, …, x_{q−1}).

Now let G be the semidirect product of N by Q = ⟨α⟩, then G is (isomorphic with) the standard wreath product of a group of order p with a group of order q.

Let D be the diagonal subgroup

D = {(x, …, x) | x ∈ F }

of N. Clearly, α centralizes D (in fact, D = Z(G)) and so A = D × Q is an abelian subgroup of G. Next, let

B = {(x_{1}, …, x_{q−1}, 0) | x_{i} ∈ F },

then B is a q − 1-dimensional F -subspace of N. Since D has F -dimension 1 and clearly B and D intersect trivially, we have N = DB and thus G = QDB = AB, and since B ≤ N, we have A ∩ B = QD ∩ N ∩ B = D(Q ∩ N) ∩ B = D ∩ B = 1.

By [13], A.18.4, the derived subgroup G′ of G equals

M = {(x_{1}, …, x_{q}) | x_{i} ∈ F, ∑_{i=1}^{q} x_{i} = 0}

3.5 ℌ-maximal subgroups 42

and O^{p}(G) = Q^{G} = MQ. Since obviously N = O^{p}(G), we also have that G^{𝔑} = O^{p}(G) ∩ O^{q}(G) = MQ ∩ N = M(Q ∩ N) = M = G′ and [A, B] = [Q, B] = M because [A, B] ≤ G′ = N and [Q, B] ≤ [A, B] contains the elements of the form

(…, −1, 1, …)

which generate M.

Since clearly A∩M ≤ N, we have A∩M = A∩N∩M = QD∩N∩M = D(Q∩N)∩M = D ∩ M; thus

A ∩ M = {(x, …, x) | x ∈ F, q · x = 0}.

Since q ≠ p, we have x = 0 whenever q · x = 0; thus A ∩ M = 1. Furthermore, B ∩ N equals the set

{(x_{1}, …, x_{q−1}, 0) | x_{i} ∈ F, ∑_{i=1}^{q−1} x_{i} = 0}

which has dimension q − 2 while the dimension of N itself is q − 1. This shows that (M ∩ A)(M ∩ B) = M ∩ B < M and so M, the derived subgroup of G as well as its nilpotent residual, is not prefactorized.

Also, if O^{p}(G) = MQ were prefactorized, then also its Sylow p-subgroup M would have to be prefactorized by Lemma 2.1.1. So MQ cannot be prefactorized either.

3.5 ℌ-maximal subgroups

The results about ℌ-projectors obtained in Section 3.4 can also be obtained via a more general result about ℌ-maximal subgroups. The following result will also be used in Section 3.6 to prove certain results for saturated Fitting formations.

3.5.1 Theorem. let ℌ be a Schunck class and suppose that the group G is the product of its nilpotent subgroups A and B. If H is an ℌ-maximal subgroup of G, then:

(i) If char(ℌ) contains σ(A) ∩ σ(B), then H possesses a factorized conjugate in G. (ii) Without any hypothesis on the characteristic of ℌ, if ℌ is a saturated formation,

then H has a prefactorized conjugate in G. Proof. (i) If 1 ≠ N denotes a proper normal subgroup of G, then the ℌ-group HN/N

is contained in an ℌ-maximal subgroup Y/N of G/N. By induction on the order of G, there is a g ∈ G such that Y^{g}/N, and therefore Y^{g}, is factorized. Then H^{g} ≤ Y^{g} is also an ℌ-maximal subgroup of Y^{g}, so if Y^{g} < G, by induction on |G| again, H^{g} possesses a factorized conjugate in Y^{g} which is also a factorized subgroup of G by Lemma 1.1.6. Therefore we may suppose that for all normal subgroups N ≠ 1, the factor group G/N is an ℌ-subgroup. Of course, we may also exclude the case when G ∈ ℌ since then H = G is factorized. But then by Lemma 1.3.1, G must be primitive.

Next, consider the case when A = B (= G): since G is primitive and nilpotent, G is cyclic of prime order p, say, and p is a common prime divisor of A and B, hence ℌ contains G, a case that has already been treated.

So we are left with the case when A ≠ B. Then, however, H possesses a factorized conjugate in G by Lemma 3.2.5 and the proof of (i) is complete.

(ii) Let π = char(ℌ). Since ℌ is a saturated formation, H is a π-group by Proposition 3.4.2, and, replacing H by a suitable conjugate if necessary, we may suppose that H is contained in the Hall subgroup A_{π}B_{π} of G. Now σ(A_{π})∩σ(B_{π}) ≤ π = char(ℌ) whence H, being also an ℌ-maximal subgroup of A_{π}B_{π}, is a factorized subgroup of A_{π}B_{π} by part (i). Thus it is a prefactorized subgroup of G by Lemma 1.1.6.

Observe also that, ℌ-projectors being in particular ℌ-maximal subgroups, Example 3.4.4 shows that Theorem 3.5.1, (ii) cannot be extended to arbitrary Schunck classes.

Remark. The result of Theorem 3.4.3 can also be obtained combining Theorem 3.5.1 (existence) and Proposition 3.3.3 (uniqueness).

Note also that there may well be more than one factorized ℌ-maximal subgroup and that these need not even be isomorphic: consider the symmetric group of degree 3 which is the product of a cyclic group A of order 2 and a cyclic group B of order 3. Let ℌ = 𝔑, then A and B are maximal nilpotent subgroups of G which are factorized but not isomorphic.

As a corollary to Theorem 3.5.1, we obtain the following result which has been proved by Fransman [17] and Amberg and Fransman [5] for Schunck classes containing all finite nilpotent groups. The latter result in turn generalizes a result of Peterson [42] for saturated formations.

3.5.2 Corollary. Suppose that G is a finite group with subgroups A, B and C where A and B are nilpotent and C ∈ ℌ. If G = AB = AC = BC, then G ∈ ℌ, provided that σ(A) ∩ σ(B) ⊆ char(ℌ).

Proof. Let D be an ℌ-maximal subgroup of G containing C. Then also AD = BD = G. Now by Theorem 3.5.1, D^{g} is factorized for some g ∈ G and by Lemma 1.1.8, G = D ∈ ℌ.

This result becomes false as soon as we drop the condition σ(A) ∩ σ(B) ⊆ char(ℌ): let G = A = B be a cyclic group of order p for some prime p and suppose that ℌ is a class of groups (not necessarily a Schunck class) whose characteristic does not contain p.

Then G ∉ ℌ and therefore the unit subgroup is the only ℌ-subgroup of G, and of course G = A · 1 = B · 1 = AB.

3.6 Injectors and radicals

A Fitting set F is a set of subgroups of the group G which satisfies (FS1) If M ∈ F, then M^{g} ∈ F for all g ∈ G, (FS2) If M, N ∈ F and M and N normalize each other, then MN ∈ F. (FS3) If G ∈ F and N ⊴ G, then N ∈ F. Let 𝔉 be a Fitting class, i.e. an ⟨S_{n},N_{0}⟩-closed class of finite groups, then it is easy to see that the set {S ≤ G | S ∈ 𝔉} is a Fitting set of the group G. Moreover, if F is a Fitting set of G and H ≤ G, then the set F_{H} = {S ≤ H | S ∈ F} is a Fitting set of S.

Let F be a set of subgroups of the group G that satisfies (FS1) and (FS2). In analogy to the 𝔉-radical defined in Section 1.3, the F-radical of G is the subgroup of G generated by all subnormal F-subgroups of G. To simplify notation, if H ≤ G, we will denote the F_{H}-radical of H again with H_{F}.

3.6.1 Lemma. Let F be a set of subgroups of the finite group G satisfying (FS1) and (FS2) above. Then:

(i) Every subnormal F-subgroup of G is contained in a normal F-subgroup of G. (ii) G_{F} is the unique maximal normal F-subgroup of G. (iii) If S ◁◁ G, then S_{F} ≤ S ∩ G_{F}, and if F is a Fitting set and S ◁◁ G, then

S_{F} = S ∩ G_{F}. Proof. (i) If S ⊴ G, (i) is trivially true. Therefore by induction on the subnormal

defect of S, the subgroup S is contained in a normal F-subgroup R of S^{G} ⊴ G. Since all G-conjugates of R are normal F-subgroups of S^{G} by (FS1) and their product R^{G} is an F-subgroup by (FS2), R^{G} is a G-invariant F-subgroup containing S as required.

(ii) By (i), G_{F} is generated by all normal F-subgroups of G and so it is an F-subgroup by (FS2).

(iii) S_{F} is a subnormal F-subgroup of G by (ii), therefore S_{F} ≤ S ∩ G_{F}. If F is a Fitting set, by (FS3), S ∩ G_{F} is a subnormal subgroup of G_{F} and so S ∩ G_{F} ∈ F. Therefore S ∩ G_{F} is a normal F-subgroup of S and as such contained in S_{F}.

An F-injector I is a subgroup I of G such that I ∩ S is F-maximal in S for all subnormal subgroups S of G.

By mere definition, it is clear that for every subnormal subgroup S of G, the subgroup I ∩ S is an F_{S}-injector of S.

The following proposition was originally proved for Fitting classes by Fischer, Gasch^{}ütz and Hartley [15]. For a proof, see e.g. Doerk and Hawkes [13], VIII.2.9 and 2.13.

3.6.2 Proposition. Let F be a Fitting set of the finite soluble group G. Then G possesses exactly one conjugacy class of F-injectors. If I is an F-injector of G and I ≤ H ≤ G, then I is also an F_{H}-injector of H. In particular, the F-injectors of G are pronormal subgroups of G.

The next lemma shows that if F-injectors exist and form a single conjugacy class, then their core equals the F-radical.

3.6.3 Lemma. Let G be a finite soluble group. Then the F-radical G_{F} of G equals the intersection of all F-injectors of G.

Proof. Clearly, the intersection of all F-injectors of G is a normal F-subgroup of G, hence it is contained in G_{F}. On the other hand, if I is an F-injector, I ∩ G_{F} is an F-injector of G_{F}, therefore I ∩ G_{F} = G_{F}. This shows that G_{F} is contained in every F-injector of G.

Next, we show that the F-radical of a finite group G which is the product of two nilpotent subgroups is always prefactorized (factorized) if an injector of G is prefactorized (factorized).

3.6.4 Proposition. Let G be the product of its finite nilpotent subgroups A and B and let F be a Fitting set of G such that G possess a prefactorized (factorized) F-injector I. Then the F-radical G_{F} of G is prefactorized (factorized).

Proof. We prove the proposition by induction on the order of G. Observe that the case G = 1 is trivial.

Let X = AG_{F} ∩ BG_{F} be the factorizer of G_{F} in G. X is a factorized subnormal subgroup of G by Corollary 3.1.6, therefore I ∩ X is a prefactorized (factorized) Finjector of X by Lemma 3.6.1. Now G_{F} = X ∩G_{F} = X_{F} which shows that the F-radical of G coincides with that of X. So if X < G, by induction hypothesis, we must have X_{F} = G_{F} which is prefactorized (factorized).

In the other case, we have G = X = AG_{F} = BG_{F} and hence G/G_{F} is nilpotent. This shows that I/G_{F} ◁◁ G/G_{F} is subnormal in G/G_{F}. Thus I is a subnormal F-subgroup of G and therefore I is contained in G_{F}. So G_{F} = I is prefactorized (factorized).

If, with the notation of the preceding Proposition 3.6.4, the F-injector I is contained in a factorized subgroup H of G, then I is also an F_{H}-injector of H by Proposition 3.6.2. This shows that H_{F} is prefactorized (factorized) for all such subgroups H. If, on the other hand, H_{F} is prefactorized (factorized) for all factorized subgroups containing an F-injector of G, the following proposition shows that G has a unique factorized Finjector.

3.6.5 Proposition. Let G be the product of its finite nilpotent subgroups A and B and let F be a Fitting set of G. Let I denote an F-injector of G and define

S = {S ≤ G | S ≤ G is factorized and contains a conjugate of I}.

Then the following statements are equivalent:

(i) Every S ∈ S contains a prefactorized (factorized) F-injector.

(ii) For every S ∈ S, the F-radical S_{F} is prefactorized (factorized).

Proof. By Proposition 3.6.4, we have already seen that whenever S possesses a prefactorized (factorized) F-injector, then the corresponding radical S_{F} must be prefactorized (prefactorized). This proves the necessity of our condition.

Conversely, suppose that the proposition is true for all groups of order smaller than |G| (observe that the statement is trivial if G = 1) and let I be an F-injector of G. Since I is pronormal by Proposition 3.6.2, by Proposition 3.3.11, we may assume w.l.o.g. that N_{G}(I) is factorized and thus that N_{G}(I) ∈ S. Now if N_{G}(I) < G, then by induction hypothesis, I is prefactorized (factorized) in N_{G}(I), hence also in G by Lemma 1.1.6. In the other case, when N_{G}(I) = G, we have I ⊴ G and thus I = G_{F} which is prefactorized (factorized) by hypothesis.

For Fitting classes, we obtain the following result:

3.6.6 Corollary. Let 𝔙 be an S-closed class of finite groups and let 𝔉 be a Fitting class. Then the following statements are equivalent:

(i) For every group G ∈ 𝔙, the 𝔉-radical G_{𝔉} is prefactorized (factorized).

(ii) Every group G ∈ 𝔙 that is the product of two nilpotent subgroups has a unique

prefactorized (factorized) 𝔉-injector Proof. Suppose that the group G ∈ 𝔙 is the product of its nilpotent subgroups A

and B. Let F = {S ≤ G | S ∈ 𝔉}, then F is a Fitting set of G, and since also all subgroups of G belong to 𝔙, the equivalence of (i) and (ii) follows directly from the equivalence of the corresponding statements of Proposition 3.6.5.

Since we know from Theorem 3.1.5 that the Fitting subgroup (and thus F_{n}(G) for all n ≥ 1) is factorized, we obtain the following result about nilpotent injectors:

3.6.7 Corollary. Let the group G be the product of its finite nilpotent subgroups A and B. Then G has a factorized 𝔑^{k}-radical and a unique factorized 𝔑^{k}-injector where 𝔑^{k} is the Fitting class of groups of nilpotent length ≤ k; in particular G has a unique factorized 𝔑-injector.

Proof. By Theorem 3.1.5, F (G), and hence F_{k}(G) for every nonnegative integer k, is factorized for every finite group G that is the product of two nilpotent subgroups. Now F_{k} is the 𝔑^{k}-radical of G, and so by the preceding proposition (with 𝔙 the class of all finite groups, say), G has a unique factorized 𝔑^{k}-injector.

The following example^{1} shows that even when 𝔉 is a Fitting class and the finite soluble group G is the product of two cyclic groups A and B, its 𝔉-radical is not necessarily prefactorized and G need not have prefactorized 𝔉-injectors. Observe also that in the following example, every subgroup of A permutes with every subgroup of B.^{2} The second part of the example shows that for Fitting sets F, a (normally embedded) F-injector is not necessarily prefactorized (factorized), even when the F-radical is prefactorized (factorized).

3.6.8 Example. Let G be a finite soluble group and

1 = G_{0} ≤ G_{1} ≤ ⋯ ≤ G_{n} = G

be a principal series of G. Every p-chief factor G_{i}/G_{i−1} can be regarded as a vector space over GF (p), the field with p elements, on which every g ∈ G acts as a nonsingular linear transformation λ_{i}(g). Let

Δ(g) = ∏ det(λ_{i}(g))

where the product is taken over all i such that G_{i}/G_{i−1} is a p-group. Δ is a homomorphism from G to the multiplicative group of GF (p). (Note that this homomorphism does not depend on the choice of the principal series of G by the Jordan-H^{}ölder theorem). Then 𝔇(p), the class of finite soluble groups G such that Δ(g) = 1 for all g ∈ G, is a normal Fitting class, i.e. a Fitting class such that every finite soluble group has a (unique) normal 𝔇(p)-injector, as has been shown by Blessenhohl and Gasch^{}ütz, [10]; see also Hawkes [27] or Doerk and Hawkes [13], IX.2.14 (b).

Let S and S^{∗} denote the symmetric groups on the sets {1, 2, 3} and {1^{∗}, 2^{∗}, 3^{∗}}. Put G = S × S^{∗}, then the 𝔇(3)-injector of G is the normal subgroup

D = ⟨(123), (1^{∗}2^{∗}3^{∗}), (12)(1^{∗}2^{∗})⟩

1 I wish to thank K. Doerk for pointing out the relevance of the following example in the context of

products of nilpotent groups.

2 Maier [37] calls such subgroups A and B totally permutable; this is, of course, a property much

stronger than just the permutability of A and B.

which has index 2 in G.

Now let A = ⟨(123), (1^{∗}2^{∗})⟩ and B = ⟨(1^{∗}2^{∗}3^{∗}), (12)⟩. It is easy to see that A and B are cyclic groups of order 6 and that A ∩ B = 1.

Therefore |AB| = 36 = |G| and so G = AB. Finally, A ∩ D and B ∩ D are both cyclic of order 3 and so D cannot be factorized (or prefactorized, which is the same in this case).

For the second example, let P = ⟨(12)(1^{∗}2^{∗})⟩ be a Sylow 2-subgroup of D and let F = {1, P^{g} | g ∈ G}. Then it follows from the fact that P is normally embedded in G or simply by direct calculation that F is a Fitting set of G with injector P and radical G_{F} = 1. Thus G_{F} = 1 is a factorized subgroup of G.

On the other hand, since N_{G}(P) = ⟨(12), (1^{∗}2^{∗})⟩ is factorized, by Proposition 3.3.11, if G had a factorized F-injector, then it would have to be contained in N_{G}(P). So P would have to be prefactorized, which is evidently not the case.

Question. Is there an example of a Fitting class 𝔉 and a finite group G which is the product of two nilpotent subgroups such that G does not have a prefactorized (factorized) 𝔉-injectors but nevertheless G_{𝔉} is prefactorized (factorized)?

To obtain a further result in the case when 𝔉 is a saturated Fitting formation, we have to employ the results of Section 3.5. Then we obtain the following proposition, whose first statement has been proved by Amberg and Fransman [5] in the case when ℌ contains all finite nilpotent groups.

3.6.9 Proposition. Let the finite group G be the product of its nilpotent subgroups A and B and let ℌ = N_{0}ℌ be a Schunck class whose characteristic contains σ(A) ∩ σ(B) (an N_{0}-closed saturated formation). Then

(i) G_{ℌ} is factorized (prefactorized).

(ii) If G admits ℌ-injectors, then every ℌ-injector has a factorized (prefactorized)

conjugate. Proof. (i) Suppose first that ℌ is a Schunck class whose characteristic contains

σ(A) ∩ σ(B). Then the factorizer X of R = G_{ℌ} has a triple factorization

X = (A ∩ BR)R = (AR ∩ B)R = (AR ∩ B)(A ∩ BR)

by Lemma 1.2.1, and since (AR ∩ B) and (A ∩ BR) are nilpotent and R ∈ ℌ, we have X ∈ 𝔉 by Corollary 3.5.2. On the other hand, by Corollary 3.1.6, X is subnormal in G, hence X ≤ R and therefore R = X is factorized.

Next, let ℌ be a saturated formation and put π = char(𝔛). Let A_{π}B_{π} be the prefactorized Hall π-subgroup of G. Since ℌ is a saturated formation, every ℌ-group is a π-group, hence R = G_{ℌ} must be contained in the prefactorized subgroup A_{π}B_{π} of G. Hence A_{π}B_{π} contains a prefactorizer X of R, and by Lemma 1.2.4, X = A^{∗}R = B^{∗}R = A^{∗}B^{∗} for suitable subgroups A^{∗} and B^{∗} of A_{π} and B_{π} respectively. Now we have σ(A^{∗}) ∩ σ(B^{∗}) ⊆ π = char(ℌ), hence by Corollary 3.5.2, we have that X ∈ ℌ. Since every prefactorizer of G is subnormal in G by Corollary 3.1.6, it follows that X = R as in the first part.

(ii) The second part follows directly from Section 3.5 since ℌ-injectors are in particular ℌ-maximal subgroups.

Remark. Proposition 3.6.9, (ii) becomes false if N_{0}H = H is only a Schunck class but not a saturated formation by Example 3.4.4, for the classes of π-perfect groups are also N_{0}-closed.

Note also that we do not claim that a (pre)factorized injector be unique in the preceding proposition. However, this is the case when 𝔉 is a saturated Fitting formation, i.e. a Fitting class that is also a saturated formation. Perhaps it is also worth noting that saturated Fitting formations include S-closed Fitting classes, for these are saturated formations by a theorem of Bryce and Cossey [7], [8]. An outline of their proof can also be found in [13], Chapter XI.

3.6.10 Corollary. Let the finite group G be the product of its nilpotent subgroups A and B and suppose that 𝔉 is both a Fitting class and a Schunck class (a saturated Fitting formation whose characteristic contains σ(A) ∩ σ(B)). Then:

(i) G_{𝔉} is prefactorized (factorized).

(ii) G has a unique prefactorized (factorized) 𝔉-injector.

Proof. It follows directly from Proposition 3.6.9 that G_{𝔉} is prefactorized (factorized). Therefore the second statement follows from Corollary 3.6.6, taking 𝔙 to be the class of all finite soluble groups.

Structural properties of a product of two finite nilpotent groups

4.1 Fundamental results about the p-length of a p-soluble group

A group G is said to be π-separable for a set π of primes if the series

1 ⊴ O_{π′}(G) ⊴ O_{π′π}(G) ⊴ O_{π′ππ′}(G) ⊴ …

reaches G after a finite number of steps. Then the number of nontrivial π-factors in that series of G is called the π-length of G. The group G is called π-soluble if the πfactors in the above series are soluble. If π = {p}, the group G is also called p-separable (p-soluble), and we also write l_{p}(G) instead of l_{{p}}(G).

Bounds on the p-length of a finite p-soluble group in terms of certain invariants of its Sylow p-subgroups will play an important role in the sequel. If G is any finite p-soluble group for the prime p and P is a Sylow p-subgroup of G, define the integers b_{p}(G), c_{p}(G), d_{p}(G) and e_{p}(G) as follows: let p^{b_{p}}(G) be the order of P, c_{p}(G) its nilpotency class, d_{p}(G) the derived length of P and p^{e_{p}}(G) its exponent. l_{p}(G) will denote the p-length of G.

The following observations are the basis for the bounds on the p-length of G that we will cite below:

4.1.1 Theorem. Let G be a finite p-soluble group, where p is a prime. Then:

(i) (Hall and Higman [26]) If c_{p}(G) > 0, then c_{p}(G/O_{p′p}(G) < c_{p}(G).

(ii) (Hall and Higman [26]) If p ≠ 2 and d_{p}(G) > 0, then d_{p}(G/O_{p′p}(G) < d_{p}(G). (iii) (Berger and Gross [9]) If d_{2}(G) > 0, then d_{2}(G/O_{2′22′2}(G) < d_{2}(G). If the Sy-

low 2-subgroups or the Sylow 3-subgroups of G are abelian and d_{2}(G) > 0, then already d_{2}(G/O_{2′2}(G) < d_{2}(G).

(iv) (Hall and Higman [26]) e_{p}(G/O_{p′p}(G) < e_{p}(G), provided that e_{p}(G) > 0 and G satisfies one of the following conditions holds:

(a) p ≠ 2 and p is not a Fermat prime;

(b) p is an odd Fermat prime and the Sylow 2-subgroups of G are abelian;

4.1 Fundamental results about the p-lengthof a p-soluble group 51

(c) p = 2 and the Sylow q-subgroups of G are abelian for all Mersenne primes q.

The following bounds on the p-length l_{p}(G) of a finite p-soluble group G in terms of the structure of a Sylow p-subgroup are known. Unless otherwise noted, the bounds are due to Hall and Higman [26].

4.1.2 Theorem. Let the finite group G be a p-soluble group. Then:

(i) b_{p}(G) ≥ \{ \begin{matrix}\frac{p^{l_{p}} − 1}{p − 1} & if p is odd and not a Fermat prime*; \\ \frac{(p − 2)^{l_{p}+1} − l_{p}(p − 3) − p + 2}{(p − 3)^{2}} & if p is a Fermat prime > 3; \\ 2^{l_{p}−1} + l_{p} − 1 & if p = 3; \\ \frac{1}{2}l_{p}(l_{p} + 1) & in any case. \end{matrix}

(ii) c_{p}(G) ≥ \{ \begin{matrix}p^{l_{p}−1} & if p is odd and not a Fermat prime*; \\ \frac{(p − 2)^{l_{p}} − 1}{(p − 3)} & if p is a Fermat prime > 3; \\ min\{l_{p}, 2^{l_{p}−1}\} & if p = 3; \\ l_{p} & in any case. \end{matrix}

(iii) d_{p}(G) ≥ \{ \begin{matrix}l_{p} & if p ≥ 3*; \\ l_{p} & if p = 2 and d_{3}(G) ≤ 1 (Berger and Gross [9]); \\ min\{l_{p}, \frac{1}{2}l_{p} + 1\} & if p = 2 (Berger and Gross [9]). \end{matrix}

(iv) e_{p}(G) ≥ \{ \begin{matrix}l_{p} & if p is neither 2 nor a Fermat prime*; \\ \frac{1}{2}l_{p} & if p is an odd prime*; \\ min\{l_{p}, \frac{1}{2}(l_{p} + 1)\} & if p = 2  (Gross [21]). \end{matrix}

Here [x] denotes the greatest integer ≤ the real number x.

In the same paper, Hall and Higman also show that the inequalities marked * are best possible in the sense that for every integer n, there is a group G of order ≥ n such that the bound is attained. In our context, it is of interest that the examples furnished by Hall and Higman are groups whose order is divisible by only two primes, whence they are the product of their Sylow subgroups and thus products of two nilpotent subgroups.

Observe that all functions of l_{p} in the previous theorem are increasing when l_{p} ≥ 0 so that the inequalities indeed bound l_{p} in terms of b_{p}(G), c_{p}(G), d_{p}(G) and e_{p}(G).

4.2 Bounds on the π-lengthsof products of two finite nilpotent subgroups 52

4.2 Bounds on the π-lengths of products of two finite nilpotent subgroups

We recall that the classes of π-separable groups of π-length ≤ k form saturated S-closed Fitting formations for every nonnegative integer k.^{1} So we will frequently encounter primitive groups as minimal counterexample. In view of Lemma 3.2.5, it is no surprise that the case when G is the product of two nilpotent subgroups of coprime order is of special importance.

4.2.1 Lemma. Let the finite group G be the product of its nilpotent subgroups A and B. If (|A|, |B|) = 1 and O_{π′}(G) = 1 where π is the set of prime divisors of |A|, then:

(i) O_{π′π} = O_{π}(G) = F (G).

(ii) Let F_{0} = 1 and F_{k+1}/F_{k} = F (G/F_{k}) for k > 0. Then

F_{k+1}/F_{k} = \{ \begin{matrix}O_{π}(G/F_{k}) & if k is even \\ O_{π′}(G/F_{k}) & if k is odd; \end{matrix}

moreover every prime divisor of |AF_{k}/F_{k}| divides |F_{k+1}/F_{k}| if k is even and every prime divisor of |BF_{k}/F_{k}| divides |F_{k+1}/F_{k}| if k is odd.

(iii) n(G) = l_{π}(G) + l_{π′}(G); moreover either n(G) = 2l_{p}(G) = 2l_{p′}(G) or n(G) = 2l_{p}(G) − 1 = 2l_{p′}(G) + 1.

Proof. (i) If p divides the order of F (G), then a Sylow p-subgroup of F (G) is a proper normal subgroup of G. Since O_{π′}(G) = 1, we must have p ∈ π and F (G) ≤ O_{π}(G). On the other hand, O_{π}(G) is contained in the nilpotent Hall π-subgroup A of G and is therefore nilpotent. It follows that F (G) = O_{π}(G) = O_{π′π}(G).

(ii) Clearly, O_{π}(G) > 1 since G is soluble by the Kegel-Wielandt theorem (our Theorem 3.1.1). Now G/O_{π}(G) does not contain nontrivial normal π-subgroups and, exchanging π and π′ (observe that the set π′ is the set of primes that do divide the order of B plus the primes that do not divide the order of G, so that we may assume that π′ contains exactly the prime divisors of B), we may suppose that the statement is true for G/F (G). The first part of the statement follows since by part (i), O_{π}(G) = F (G).

For the second statement, observe that F_{1} ≤ A and thus Z(A) ≤ C_{G}(F_{1}) which is contained in F_{1}. Now Z(A) is the product of the (nontrivial) centres of the primary components of A, and so every prime divisor of |A| divides already the order of Z(A) ≤ F_{1}. The general statement follows by considering G/F_{k} instead of G and exchanging A and B if k is odd.

1 The proof is straightforward except possibly for saturation, a proof of which can be obtained easily

from the proof given by Robinson [43], 9.3.4 for π = {p}.

(iii) This follows immediately from (ii), since the Fitting series

1 = F_{0} ◁ F_{1} ◁ ⋯ ◁ F_{n} = G

coincides with the upper π-series which equals the upper π′-series.

1 ⊴ O_{π′}(G) ◁ O_{π′π}(G) ◁ O_{π′ππ′} ◁ ⋯ ◁ G.

We will soon see that (iii) remains true for some prime p also when we remove the condition (|A|, |B|) = 1. This will be proved in Section 4.3. The dual of the previous lemma is likewise true:

4.2.2 Lemma. Let the finite group G be the product of its nilpotent subgroups A and B. If (|A|, |B|) = 1 and O^{π′}(G) = 1 where π is the set of prime divisors of |A|, then:

(i) O^{π′π} = O^{π}(G) = G^{𝔑}.

(ii) Set L_{0} = G and for k > 0, define L_{k+1} = (L_{k})^{𝔑}. Then

L_{k}/L_{k−1} = \{ \begin{matrix}O^{π}(L_{k}) & if k is odd \\ O^{π′}(L_{k}) & if k is even \end{matrix} .

Proof. (i) Obviously, G = AO^{π}(G) so that G/O^{π}(G) ≅ A/A ∩ O^{π}(G) is nilpotent. On the other hand, G/G^{𝔑} must be a π-group because O^{π′}(G) = 1 which proves the other inclusion.

(ii) Note first that every normal subgroup of G is factorized by Corollary 2.2.4 so that in particular O^{π}(G) is factorized (this can be seen more easily in this case observing that B ≤ O^{π}(G)). Now O^{π}(O^{π}(G)) = O^{π}(G) which shows that O^{π}(G) satisfies the hypotheses of this lemma for the set π′ (which we may assume to contain exactly the prime divisors of B; cf. the remark in the proof of Lemma 4.2.1). The full statement now follows by induction on k.

It is easy to see that one can construct from the inequalities stated in Theorem 4.1.2 functions f: 𝔑 → ±⊬ℕ_{0} which satisfy

(BP1) for all finite soluble groups G and all primes p, l_{p}(G) ≤ f(G_{p}), where G_{p} is a

Sylow p-subgroup of G. (BP2) f(P/N) ≤ f(P) for all finite p-groups P and N ⊴ P where p a prime.

4.2.3 Theorem. Let f be a function satisfying (BP1) and (BP2) above, and suppose that the finite group G is the product of its nilpotent subgroups A and B. Then l_{π}(G) ≤ max{f(A_{p}), f(B_{p}) | p ∈ π}.

Proof. Suppose that the theorem is false and let G be a counterexample of minimal order. Let k = max_{p∈π}{f(A_{p}), f(B_{p})}, and denote with 𝔛 the class of nilpotent groups H satisfying f(H_{p}) ≤ k for all p ∈ π. Then we have A, B ∈ 𝔛; moreover, using (BP2), it is easy to see that the class 𝔛 is Q-closed.

If ℌ denotes the Schunck class of finite groups satisfying l_{π}(G) ≤ k, then a finite group of minimal order that is a product of two 𝔛-subgroups but does not lie in ℌ is primitive by Lemma 1.3.3. So G must be primitive with unique minimal normal subgroup N of exponent p and l_{π}(G) > l_{π}(G/N). This shows that p ∈ π. Since in view of (BP1), the theorem is trivially true if G = A = B, we may also assume by Lemma 3.2.5 that A is a Sylow p-subgroup of G and that B is a Hall p′-subgroup.

Since O_{ππ′}/O_{π}(G) is a π′-group and B_{π′} is a Hall π′-subgroup of G, we have

O_{ππ′}/O_{π}(G) ≤ B_{π′}O_{π}(G)/O_{π}(G).

Now B_{π′} is centralized by B_{π} because B is nilpotent, and since moreover

C_{G}(O_{ππ′}(G)/O_{π}(G)) ≤ O_{ππ′}(G)/O_{π}(G)

(see e.g. [26], Lemma 1.2.3) we must have B_{π} ≤ C_{G}(O_{ππ′}(G)/O_{π}(G)) ≤ O_{ππ′}(G)/O_{π}(G); consequently B_{π} ≤ O_{π}(G). This shows that AO_{π}(G)/O_{π}(G) is a p-group with p ∈ π and BO_{π}(G)/O_{π}(G) is a π′-group. In particular, the order of G/O_{π}(G) is divisible only by p and primes in π′, hence every p-series is also a π-series and viceversa, and in particular, l_{p}(G/O_{π}(G)) = l_{π}(G/O_{π}(G)). Since moreover O_{p′p}(G) = N ≤ O_{π}(G), it follows that

l_{π}(G) ≤ 1 + l_{p}(G/O_{π}(G)) ≤ l_{p}(G) ≤ f(A_{p}) ≤ k

by (BP1). This final contradiction proves the theorem.

The following bounds on l_{p} follow directly from Theorem 4.1.2 and Theorem 4.2.3 with π = p:

4.2.4 Corollary. Let the finite group G be the product of its nilpotent subgroups A and B. Write l_{p} for l_{p}(G) and let b_{p} = max{b_{p}(A), b_{p}(B)}, c_{p} = max{c_{p}(A), c_{p}(B)}, d_{p} = max{d_{p}(A), d_{p}(B)} and e_{p} = max{e_{p}(A), e_{p}(B)}, then:

(i) b_{p} ≥ \{ \begin{matrix}\frac{p^{l_{p}} − 1}{p − 1} & if p is odd and not a Fermat prime*; \\ \frac{(p − 2)^{l_{p}+1} − l_{p}(p − 3) − p + 2}{(p − 3)^{2}} & if p is a Fermat prime > 3; \\ 2^{l_{p}−1} + l_{p} − 1 & if p = 3; \\ \frac{1}{2}l_{p}(l_{p} + 1) & in any case. \end{matrix}

(ii) c_{p} ≥ \{ \begin{matrix}p^{l_{p}−1} & if p is odd and not a Fermat prime*; \\ \frac{(p − 2)^{l_{p}} − 1}{(p − 3)} & if p is a Fermat prime > 3; \\ min\{l_{p}, 2^{l_{p}−1}\} & if p = 3; \\ l_{p} & in any case. \end{matrix}

(iii) d_{p} ≥ \{ \begin{matrix}l_{p} & if p ≥ 3*; \\ l_{p} & if p = 2 and d_{3}(G) ≤ 1; \\ min\{l_{p}, \frac{1}{2}l_{p} + 1\} & if p = 2. \end{matrix}

(iv) e_{p} ≥ \{ \begin{matrix}l_{p} & if p is neither 2 nor a Fermat prime*; \\ \frac{1}{2}l_{p} & if p is an odd prime*; \\ min\{l_{p}, \frac{1}{2}(l_{p} + 1)\} & if p = 2. \end{matrix}

Observe that by our remark after Theorem 4.1.2, the inequalities marked * are again best-possible.

It is clear that the bounds for p = 2 above are very bad compared with the results obtained for other primes. However, using the fact that l_{p}(G) ≤ l_{p′}(G)+1, we can obtain better results if we use information about the Hall p′-subgroups of A and B.

In order to obtain handy bounds on π-length, we have to simplify the formulas given in Theorem 4.1.2

4.2.5 Lemma. Let the group G be a finite p-soluble group of p-length l_{p}, where p is an odd prime. Then

(i) b_{p}(G) ≥ 2^{l_{p}−1} + l_{p} − 1;
(ii) c_{p}(G) ≥ min{l_{p}, 2^{l_{p}−1}};
(iii) d_{p}(G) ≥ l_{p};
(iv) e_{p}(G) ≥ \frac{1}{2}l_{p}.

Proof. (i) If p is odd and not a Fermat prime, then by Theorem 4.1.2 we have

b_{p}(G) ≥ \frac{p^{l_{p}} − 1}{p − 1} = ∑_{i=0}^{l_{p}−1} p^{i} ≥ 2^{l_{p}−1} + (l_{p} − 1) · 1,

whereas if p is a Fermat prime > 3, we have

\frac{(p − 2)^{l_{p}+1} − l_{p}(p − 3) − p + 2}{(p − 3)^{2}} = \frac{1}{p − 3}(∑_{i=0}^{l_{p}}(p − 2)^{i} − (l_{p} − 1))

≥ \frac{1}{p − 3}((p − 2)^{l_{p}} + (p − 2) · (l_{p} − 1) + 1 − (l_{p} − 1))

≥ (p − 3)^{l_{p}−1} + l_{p} − 1 ≥ 2^{l_{p}−1} + l_{p} − 1.

The case p = 3 being obvious, it follows that the first formula holds for all odd primes. The other statements should be clear.

Define b_{π}(G) = max{0, b_{p}(G) | p ∈ π} and similarly c_{π}(G), d_{π}(G), e_{π}(G) (note that if G is nilpotent, then c_{π}(G) and d_{π}(G) are the class and derived length of a Hall πsubgroup of G), then we obtain the following bounds on the π-length of a group G which is the product of two finite nilpotent subgroups.

4.2.6 Corollary. Let the finite group G of π-length l_{π} be the product of its nilpotent subgroups A and B, let π be a set of primes, and define b_{π} = max{b_{π}(A), b_{π}(B)}, c_{π} = max{c_{π}(A), c_{π}(B)}, d_{π} = max{d_{π}(A), d_{π}(B)} and e_{π} = max{e_{π}(A), e_{π}(B)}. Then the following inequalities hold:

(i) b_{π}≥ 2^{l_{π}−1} + l_{π} − 1 if 2 ∉ π  and  b_{π′}≥ 2^{l_{π}−2} + l_{π} − 2  if 2 ∈ π;
(ii) c_{π}≥ min{l_{p}, 2^{l_{π}−1}} if 2 ∉ π  and  c_{π′}≥ min{l_{p} − 1, 2^{l_{π}−2}} if 2 ∈ π;
(iii) d_{π}≥ l_{π}  if 2 ∉ π  and  d_{π′}≥ l_{π} − 1  if 2 ∈ π;
(iv) e_{π}≥ \frac{1}{2}l_{π}  if 2 ∉ π  and  e_{π′}≥ \frac{1}{2}(l_{π} − 1)  if 2 ∈ π.

Proof. (i). Suppose first that 2 ∉ π. Then for every finite p-group P, define f(P) by

2^{f(P)−1} + f(P) − 1 = b_{p}(P).

Then f satisfies (BP1) by the preceding lemma, and it satisfies (BP2) because b_{p} satisfies it and the function 2^{x−1}+x+1 is strictly increasing for x ≥ 0. Therefore by Theorem 4.2.3, l_{π}(G) ≤ max_{p∈π}{f(A_{p}), f(B_{p})} and so

\begin{aligned}2^{l_{π}−1} + l_{π} − 1 & ≤ max_{p∈π}\{2^{f(A_{p})−1} + f(A_{p}) − 1, 2^{f(B_{p})−1} + f(B_{p}) − 1\} \\ & = max_{p∈π}\{b_{p}(A), b_{p}(B)\} = b_{π}. \end{aligned}

If 2 ∈ π, we use the fact that l_{π} − 1 ≤ l_{π′}; thus

2^{l_{π}−2} + l_{π} − 2 ≤ 2^{l_{π′}−1} + l_{π′} − 1 ≤ b_{π′}.

The proof of the other statements is similar.

4.3 Connections between Fitting length and π-lengths 57

4.3 Connections between Fitting length and π-lengths

If G is a soluble group of Fitting length n = n(G), then it is always possible to obtain from its Fitting series a series whose factors are π- and π′-groups with \frac{n}{2} π-factors if n is even and \frac{(n+1)}{2} π-factors if n is odd. So 2l_{π}(G) ≤ n+1 for every set of primes π. Since also 2l_{π′} ≤ n + 1, we also have l_{π} + l_{π′} ≤ n + 1. The next theorem shows that products of two finite nilpotent groups have the property that n(G) ≤ 2l_{p}(G) and n(G) ≤ l_{p}(G) + l_{p′}(G) for at least one prime p.

4.3.1 Theorem. Let the group G be the product of its finite nilpotent subgroups A and B. Then

(i)

n(G) ≤ 2 max_{p∈±⊬ℙ}{l_{p}(G)},

(ii)

n(G) ≤ 2 max_{p∈±⊬ℙ}{l_{p′}(G)} + 1 and

(iii)

n(G) ≤ max_{p∈±⊬ℙ}{l_{p}(G) + lp′(G)}.

Proof. Suppose that the group G is a minimal counterexample for one of the above inequalities. Since the classes of groups H such that

2 max_{p∈±⊬ℙ}{l_{p}(H)} ≤ 2 max_{p∈±⊬ℙ}{l_{p}(G)}

2 max_{p∈±⊬ℙ}{l_{p}(H)} ≤ 2 max_{p∈±⊬ℙ}{l_{p′}(G)} + 1

or

max_{p∈±⊬ℙ}{l_{p}(H) + l_{p′}(H)} ≤ max_{p∈±⊬ℙ}{l_{p}(G) + l_{p′}(G)}

are Q-closed and the classes of groups H such that n(G) ≤ k form Schunck classes, the group G can be assumed primitive by Lemma 1.3.1. Now by Lemma 4.2.1, we have n(G) ≤ 2l_{p}(G), n(G) ≤ 2l_{p′}(G) and n(G) ≤ l_{p}(G)+l_{p′}(G) where p is the exponent of the unique minimal normal subgroup of G. So the theorem is true also for primitive groups.

The last theorem generalizes a result of R. Maier [35]:

4.3.2 Corollary. Let G be a finite group G. Then the following statements are equivalent:

(i) G is the product of its nilpotent subgroups A and B and l_{p}(G) ≤ 1 for all primes p.

(ii) G is metanilpotent.

Proof. (i) ⇒ (ii). This follows directly from Theorem 4.3.1 that n(G) ≤ 2, in other words, G is metanilpotent.

(ii) ⇒ (i). Since the class of finite nilpotent groups forms a Schunck class, the soluble group G has a nilpotent projector P (see Section 3.4). Now since G/F (G) is nilpotent, G = P F (G) where P is a nilpotent subgroup, whence G is the product of two nilpotent subgroups. The statement about the p-length is trivial.

Remark. There are finite soluble groups G satisfying l_{p}(G) ≤ 1 for all primes p which are not metanilpotent and therefore do not admit a factorization by two nilpotent subgroups, e.g. if G is the regular wreath product of three groups of orders p, q and r where p, q and r are distinct primes. The group G is also an example of a group satisfying

n(G) ≤ max_{p∈±⊬ℙ}{l_{p}(G) + l_{p′}(G)}

but which is not the product of two nilpotent subgroups.

Question. Does every finite soluble group satisfying

n(G) ≤ 2 max_{p∈±⊬ℙ}{l_{p}(G)} or n(G) ≤ 2 max_{π⊆±⊬ℙ}{l_{π}(G)}

admit a factorization by two nilpotent subgroups?

Recall that a group is called modular if its subgroup lattice is modular. Modular finite nilpotent groups can also be characterized as follows:

4.3.3 Lemma. Let G be a finite nilpotent group. Then G is modular if and only if AB = BA for all subgroups A and B of G.

Proof. Let G be a modular nilpotent group. We show that AB = BA for all subgroups A and B of G. Suppose that this is false and let G be a minimal counterexample. Then G has subgroups A and B such that AB is not a subgroup of G (cf. Lemma 1.1.2), A and B are proper subgroups of G and we have G = ⟨A, B⟩. Now let M be a maximal subgroup of G that contains A, then M = M ∩ ⟨A, B⟩ = ⟨A, M ∩ B⟩ by the modularity of G, and by the minimality of G, we have M = A(B ∩ M). Now A ∩ B ≤ M and since G is nilpotent, the index |G : M| = |B : B ∩ M| = p is a prime. So Lemma 1.1.1 yields

|AB| = \frac{|A| · |B|}{|A ∩ B|} = p · \frac{|A| · |B ∩ M|}{|A ∩ B ∩ M|} = p · |M| = |G|

and therefore we have G = AB. This contradiction shows that in a modular nilpotent group any two subgroups permute.

Conversely, suppose that every two subgroups of the group G permute. We have to show that the subgroup lattice of G is modular, i.e. that for arbitrary subgroups A, B and C of G with A ≤ C, we have ⟨A, B ∩ C⟩ = ⟨A, B⟩ ∩ C. But this follows directly from the ‘usual’ modular law of group theory: since any two subgroups of G permute, we have AB = ⟨A, B⟩ and ⟨A, B ∩ C⟩ = A(B ∩ C), and hence by the modular law, ⟨A, B ∩ C⟩ = A(B ∩ C) = AB ∩ C = ⟨A, B⟩ ∩ C.

This can be used to prove that finite nilpotent modular groups having a complemented abelian maximal normal subgroup are themselves abelian. This follows at once from the following

4.3.4 Lemma (R. Maier [35]). Let P be a modular group of order p^{n} and suppose that N an abelian maximal normal subgroup of P which possesses a complement C Then P is abelian.

Proof. By the maximality of N, we have |C| = p. Let 1 ≠ x ∈ N. then ⟨x⟩ is a subgroup of order p and ⟨x⟩C is a subgroup of order p^{2} by the modularity of P. Thus ⟨x⟩C is abelian so that x commutes with every element of C and P must be abelian.

4.3.5 Theorem (R. Maier [35]). Let the finite group G = AB be the product of the nilpotent modular subgroups A and B. Then G is metanilpotent.

Proof. Let G be a counterexample of minimal order. Since the class of metanilpotent groups forms a saturated formation and the class of modular nilpotent groups is Qclosed, G is a primitive group by Lemma 1.3.3. Denote with N its unique minimal normal subgroup, then N = F (G) is elementary abelian of prime exponent p. Also, we have A ≠ B and by Proposition 3.1.8, A or B is contained in a maximal normal subgroup M of G which must be factorized.

Since G is a minimal counterexample, M is metanilpotent, hence M = F_{2}(G). By Lemma 3.2.5, w.l.o.g. A is a Sylow p-subgroup of G and F_{2}(G)/N is a p′-group and it is easy to see that G/M is a cyclic p-group, whence AM = G. Therefore AM/M ≅ A/A∩M = A/N and since N has a complement C in G, we have A = A∩CN = (A∩C)N which shows that A ∩ C ≅ A/N is cyclic of order p.

So N is an abelian maximal normal subgroup of A which has a complement A ∩ C in A, and by the modularity of A, the subgroup A is abelian by Lemma 4.3.4. But then we have A ≤ C_{G}(N) = N. This final contradiction shows that G must be metabelian.

4.4 Bounds on Fitting length and derived length

Bounds on the Fitting length of a group G which is the product of its finite nilpotent subgroups A and B can be obtained using the bounds on the p-lengths of G obtained in Section 4.2 and the inequalities

n(G) ≤ 2 · max_{p∈±⊬ℙ}{l_{p}(G)}
n(G) ≤ 2 · max_{p∈±⊬ℙ}{l_{p′}(G)} + 1
n(G) ≤ max_{p∈±⊬ℙ}{l_{p}(G) + l_{p′}(G)}

established in Section 4.3. Slightly better bounds on the nilpotent length of G, still based on the bounds in Theorem 4.1.2, are available via a closer analysis of the series described in Lemma 4.2.1,

Because of the great number of possibilities, we will restrict ourselves to bounds on n(G) in terms of d_{p}(A) and d_{p}(G) for all primes p since the methods used there also lead to estimates in terms of b_{p}, c_{p} and e_{p}.

4.4.1 Theorem. Let the finite group G be the product of its nilpotent subgroups A and B. Then

(i)

n(G) ≤ 2d(A) + max{0, 2d_{2}(A) − 4} + 1 and
n(G) ≤ 2d(B) + max{0, 2d_{2}(B) − 4} + 1;

(ii)

n(G) ≤ d(A) + d(B) + max{0, d_{2}(A) − 2, d_{2}(B) − 2};

(iii)

n(G) ≤ 2 max{d_{2′}(A), d_{2′}(B)} + 1;

(iv) (Berger and Gross [9])

n(G) ≤ max\{d(A) + d(B), \frac{4}{3}(d(A) + d(B)) − 1\}.

Proof. Suppose that the theorem is false and let the group G = AB be a minimal counterexample to one of the inequalities. The classes of finite groups of nilpotent length ≤ k clearly form saturated formations, hence Schunck classes, and the class of nilpotent groups H with d_{p}(H) ≤ d_{p}(A) for every prime p is Q-closed, similarly for B. Since in addition the above functions are nondecreasing with d_{p}(A) and d_{p}(B) for all primes p, the minimal counterexample G must be primitive by Lemma 1.3.3. Thus it suffices to prove the above inequalities in case G is primitive.

As all of the above inequalities hold for nilpotent groups G, we can also assume that G ≠ A and G ≠ B and thus we infer from Lemma 3.2.5 that G has a unique minimal normal subgroup N of prime exponent p, that w.l.o.g. A is a Sylow p-subgroup of G and that B is a Hall p′-subgroup.

Since A is a Sylow p-subgroup of G, by Theorem 4.1.2, we have

l_{p}(G) ≤ \{ \begin{matrix}d_{p}(A) & if p is odd \\ max\{d_{2}(A), 2d_{2}(A) − 2\} & if p = 2. \end{matrix}

Therefore we have

(∗)  l_{p}(G) ≤ d(A) + max{0, d_{2}(A) − 2}

in any case.

Similarly, by Theorem 4.2.3, we obtain

(∗∗)

l_{p′}(G) ≤ max_{q∈±⊬ℙ\{p}}{d_{q}(A), d_{q}(B), 2d_{2}(A) − 2, 2d_{2}(B) − 2}

= max_{q∈±⊬ℙ\{p}}{d_{q}(B), 2d_{2}(B) − 2}

= d(B) + max{0, d_{2}(B) − 2},

observing that A is a p-group.

Now by Lemma 4.2.1 and (∗),

\begin{aligned}n(G) & ≤ 2l_{p}(G) ≤ 2d(A) + max\{0, 2d_{2}(A) − 4\} \\ & ≤ 2l_{p}(G) ≤ 2d(A) + max\{0, 2d_{2}(A) − 4\} + 1. \end{aligned}

By (∗∗),

n(G) ≤ 2l_{p′}(G) + 1 ≤ 2d(B) + max{0, 2d_{2}(B) − 4} + 1.

This proves (i).

Next, again by Lemma 4.2.1, n(G) ≤ l_{p}(G) + l_{p′}(G) and so by (∗) and (∗∗),

n(G) ≤ d(A) + max{0, d_{2}(A) − 2} + d(B) + max{0, d_{2}(B) − 2}.

Since either A or B is a 2′-group, we have d_{2}(A) = 0 or d_{2}(B) = 0 and hence

n(G) ≤ d(A) + d(B) + max{0, d_{2}(A) − 2, d_{2}(B) − 2},

proving (ii).

To see that (iii) is true, we have to consider the cases p = 2 and p ≠ 2 separately. Assume first that p = 2 and hence that B is a 2′-group. Then by (∗∗)

n(G) ≤ 2l_{p′}(G) + 1 ≤ 2d(B) + max{0, 2d_{2}(B) − 4} + 1 = 2d_{2′}(B) + 1.

If p ≠ 2, A is a 2′-group and therefore we have

n(G) ≤ 2l_{p}(G) ≤ 2d(A) + max{0, 2d_{2}(A) − 4} + 1 = 2d_{2′}(A) ≤ 2d_{2′}(A) + 1

by (∗). Thus (iii) holds.

To prove (iv), suppose first that d(A) ≤ 2 and d(B) ≤ 2. Then by (ii), n(G) ≤ d(A) + d(B) and (iv) holds. Since we have also excluded the cases G = A and G = B, we may assume that d(A) > 0 and d(B) > 0, hence that d(A) + d(B) ≥ 3. Then d(A) + d(B) ≤ \frac{4}{3}(d(A) + d(B)) − 1 and thus it remains to show that

n(G) ≤ \frac{4}{3}(d(A) + d(B)) − 1.

Consider next the case when p = 2 and 2d(A) − 2 ≤ d(B) + 1. Then

2d(A) − 2 + d(B) ≤ \frac{4}{3}(d(A) + d(B)) − 1

and since we also have d(A) ≤ d(B) + 1, we obtain that

n(G) = l_{p}(G) + l_{p′}(G) ≤ max\{d(A), 2d(A) − 2\} + d(B) ≤ \frac{4}{3}(d(A) + d(B)) − 1.

If p = 2 and 2d(A) − 2 > d(B), then we have

2d(B) + 1 ≤ \frac{4}{3}(d(A) + d(B)) − 1

and hence

n(G) ≤ 2l_{p′}(G) + 1 ≤ 2d(B) + 1 ≤ \frac{4}{3}(d(A) + d(B)) − 1.

Similarly, if p ≠ 2 and 2d(B) − 2 ≤ d(A) + 1, then

n(G) = l_{p}(G) + l_{p′}(G) ≤ d(A) + max\{d(B), 2d(B) − 2\} ≤ \frac{4}{3}(d(A) + d(B)) − 1

and finally if p > 2 and 2d(B) − 2 > d(A) + 1. then

n(G) ≤ 2l_{p}(G) ≤ 2d(A) ≤ \frac{4}{3}(d(A) + d(B)) − 1.

Remark. The difference of the above inequalities is essentially due to the irregular behaviour of the prime 2, for if A and B have odd order, then clearly n(G) ≤ d(A)+d(B). Since in fact l_{2}(G) ≤ d_{2}(G) if d_{3}(G) ≤ 1 by Theorem 4.1.2, it can be checked easily that if d_{3}(A) ≤ 1 when d_{2}(B) ≥ 2 and d_{3}(B) ≤ 1 if d_{2}(A) ≥ 2, then we obtain

n(G) ≤ d(A) + d(B) and n(G) ≤ 2d(A) + 1.

These inequalities, and thus the results of Theorem 4.4.1 for groups of odd order, are best-possible: if p and q are distinct primes, and C_{p} and C_{q} denote cyclic groups of order p and q respectively, then a group of the form G = C_{p} {∩ \atop ∪} C_{q} {∩ \atop ∪} … {∩ \atop ∪} C_{p} {∩ \atop ∪} C_{q} {∩ \atop ∪} C_{p} where C_{q} occurs k times satisfies n(G) = 2k + 1. Moreover, it is the products of its Sylow p- and q-subgroups which have derived lengths k + 1 and k respectively by [26], Theorem 3.5.1.

In this context, it should also be mentioned that Berger and Gross [9] conjecture that l_{2}(G) ≤ d(G) in any case which would, as has been remarked by Berger and Gross themselves, imply that n(G) ≤ d(A) + d(B) and n(G) ≤ 2d(A) + 1 always.

Also, (ii) improves a result of Gross [22] who shows that n(G) ≤ d(A) + d(B) if c_{2}(A) ≤ 3 and c_{2}(B) ≤ 3.

A bound on the Fitting length similar to (i), however based on the inequalities in terms of c_{p} instead of those involving d_{p} has been obtained by Heineken [29].

Next, we will obtain some information about the derived length of a finite group G which is the product of its nilpotent subgroups A and B in terms of the derived lengths or classes of A and B.

The main problem is that hardly anything is known about the derived length of G if G itself is nilpotent. In fact, the only nontrivial result seems to be Itô’s theorem which states that the derived length of a product of two abelian groups has derived length at most 2 (Itô [32]).

In this context, we also mention that the class of the nilpotent group G is not bounded by the classes (or derived lengths) of the subgroups A and B: in fact, for every nonnegative integer n, there exist groups G of order p^{2n} which are the product of two abelian subgroups A and B with A ⊴ G (and thus d(G) ≤ 2) such that c(G) = n, see Dickenschied [12], Beispiel 7.1 for details of the construction.

However, it is possible to obtain bounds on the derived length of certain quotient groups of a group G which is the product of two nilpotent subgroups A and B, such as G/F (G) and G/Φ(G) ∩ O_{π}(G), where π is the set of common prime divisors of |A| and |B|.

To see that groups G that are minimal subject to d(G/Φ(G)) = k are primitive, in view of Lemma 1.3.1, the following lemma is useful:

4.4.2 Lemma. The class of finite soluble groups G that satisfy d(G/Φ(G)) ≤ n equals the class of finite groups G such that G^{(n−1)} is nilpotent, i.e. the class of groups such that d(G/F (G)) ≤ n − 1; this class is S-, Q-, D_{0}- and E_{Φ}-closed; in other words, it is a subgroup-closed saturated Formation.

Proof. We show first that the two classes are equal: if G^{(n−1)} is nilpotent, it is contained in the Fitting subgroup F (G) of G. Since F (G)/Φ(G) is abelian, we have G^{(n)} ≤ Φ(G) and G/Φ(G) has derived length ≤ n. On the other hand, if G^{(n)} = (G^{(n−1)})′ ≤ Φ(G), then G^{(n−1)}Φ(G)/Φ(G) ≤ F (G/Φ(G)) = F (G)/Φ(G) and so G^{(n−1)} ≤ F (G), hence it is nilpotent.

That the class in question is closed with respect to subgroups, homomorphic images and finite direct products can be checked easily using the second definition. Saturation is obvious from the first.

If the group G is the product of its nilpotent subgroups A and B, this can be used to reduce the search for a bound on d(G/Φ(G)) to finding bounds in the special case when A and B have coprime orders.

4.4.3 Proposition. Suppose that f: 𝔑 × 𝔑 → ±⊬ℕ_{0} is a function satisfying

(i) f(A, B) = f(A^{∗}, B^{∗}) if A ≅ A^{∗} and B ≅ B^{∗} for all A, B ∈ 𝔑; (ii) f(A/M, B/N) ≤ f(A, B) for all finite nilpotent groups A and B and for all

M ⊴ A, N ⊴ B and (iii) if the finite group G is the product of its nilpotent subgroups A and B and A

and B have coprime order, then d(G/Φ(G)) ≤ f(A, B). Then we have d(G/Φ(G)) ≤ f(A, B) for every group G that is the product of its

nilpotent subgroups A and B.

Proof. Suppose that the group G is a counterexample of minimal order to the proposition. By the minimality of G, if 1 ≠ N ⊴ G, then we have

d((G/N)/Φ(G/N)) ≤ f(AN/N, BN/N) = f(A/A ∩ N, B/B ∩ N) ≤ f(A, B)

by (i) and (ii). Therefore every proper epimorphic image belongs to the Schunck class described in Lemma 4.4.2 for n = f(A, B) but G does not belong to that class. Thus G must be primitive by Lemma 1.3.1. If A = B (= G), then G is cyclic of prime order, and since G = A · 1, we have

1 = d(G/Φ(G)) = d(G) ≤ f(A, 1) ≤ f(A, B).

Thus we may assume that A ≠ B. But now by Lemma 3.2.5, A and B have coprime orders and so by (iii), we have

d(G/Φ(G)) ≤ f(A, B).

This final contradiction proves the proposition.

For every n ∈ ±⊬ℕ, consider the classes of groups G such that G^{(n)} is a nilpotent π-group. Although these classes are not saturated, groups G that are minimal subject to not belonging to one of these classes and which are the product of two nilpotent subgroups A and B still have a structure very similar to that of a primitive product of two nilpotent subgroups if π is chosen to be the set of common prime divisors of |A| and |B|:

4.4.4 Lemma. Suppose that the finite group G is the product of its nilpotent subgroups A and B and let π = σ(A) ∩ σ(B). If there is an integer n such that (G/K)^{(n)} is a nilpotent π-group for all normal subgroups K ≠ 1 of G but G^{(n)} is not a nilpotent π-group, then G has a unique minimal normal subgroup N of prime exponent p; furthermore G^{(n)} = N, F (G) is a p-group, and (w.l.o.g.) A and B are a Sylow p-group of G and a Hall p′-subgroup of G respectively. In particular, π = ∅.

Proof. Suppose first that G has two distinct minimal normal subgroups N and N^{∗} and let H/N = (G/N)^{(n)} and H^{∗}/N^{∗} = (G/N^{∗})^{(n)}. Then G^{(n)} is contained in H ∩ H^{∗} = (H ∩ H^{∗})/(N ∩ N^{∗}) which is a nilpotent normal π-group by Lemma 1.3.5 and the fact that H/N and H^{∗}/N^{∗} are nilpotent π-groups. Therefore G must have a unique minimal normal subgroup N which is an elementary abelian p-group for some prime p because G is soluble, and also the Fitting subgroup of G must be a p-group.

Next, we show that w.l.o.g. A is a Sylow p-subgroup of G and that B a Hall p′- subgroup: assume first that Φ(G) ≠ 1. Then we must have N ≤ Φ(G), and since (G/N)^{(n)} is nilpotent, we have G^{(n)}Φ(G)/Φ(G) ≤ F (G/Φ(G)) = F (G)/Φ(G). Therefore we have G^{(n)} ≤ F (G) and thus G^{(n)} is a nilpotent p-group. Therefore p ∉ π and we may suppose w.l.o.g. that p divides |A| but not |B|. In particular, B is a p′-group and A contains a Sylow p-subgroup A_{p} of G whence F (G) ≤ A_{p}. Now A_{p′} centralizes A_{p} since A is nilpotent and therefore A_{p′} ≤ C_{G}(F (G)) ≤ F (G). But F (G) is a p-group, from which we deduce that A_{p′} = 1. Hence A is a p-group and we have already observed that B is a p′-group. Then, however, it follows from Lemma 1.1.1 that A and B must be a Sylow p- and Hall p′-subgroups of G, and the lemma is proved in this case.

If Φ(G) = 1, G is primitive by Lemma 3.2.3. If A = B (= G), then G is a nilpotent π-subgroup; therefore we must have A ≠ B and so by Lemma 3.2.5, A and B are a Sylow p- and a Hall p′-subgroup of G.

Therefore we have π = {p} ∩ p′ = ∅ in both cases. Since (G/N)^{(n)} is a π-group, we have (G/N)^{(n)} = 1 and hence G^{(n)} ≤ N. Since G^{(n)} ≠ 1 by hypothesis, we must have N = G^{(n)} by the minimality of G.

Now we are ready to prove the bounds on the derived length of G/O_{π}(G) ∩ Φ(G).

4.4.5 Theorem. Let the group G = AB be the product of its finite nilpotent subgroups A and B. If π is the set of primes that divide the orders of both A and B, then G^{(n)} is a nilpotent π-subgroup contained in Φ(G), where

n = max\{c_{σ′}(A), \frac{1}{2}d_{σ}(A)(d_{σ}(A) + 1)\} + max\{c_{τ′}(B), \frac{1}{2}d_{τ}(B)(d_{τ}(B) + 1)\},

and σ and τ are arbitrary sets of odd primes.

Proof. We show first by way of contradiction that G^{(n)} is a nilpotent π-group, so suppose that G is a counterexample of minimal order. If 1 ≠ N is a normal subgroup of G, then we have

\begin{aligned}n & ≥ max\{c_{σ′}(AN/N), \frac{1}{2}d_{σ}(AN/N)(d_{σ}(AN/N) + 1)\} \\ & + max\{c_{τ′}(BN/N), \frac{1}{2}d_{τ}(BN/N)(d_{τ}(BN/N) + 1)\} \end{aligned}

whence (G/N)^{(n)} is a nilpotent π-group for all normal subgroups N ≠ 1 of G, and so by Lemma 4.4.4, G has a unique minimal normal subgroup N of prime exponent p, w.l.o.g. A is a Sylow p-subgroup and B is a Hall p′-subgroup of G. Moreover, G^{(n)} ≤ N and also F = F (G) is a p-group.

If p ∉ σ, consider the group G/Z where Z = Z(F): Clearly,

1 < Z(A) ≤ Z = Z(F) ≤ C_{G}(F) ≤ F,

therefore c_{p}(A/Z) < c_{p}(A). Since A is a p-group, we have

max\{c_{σ′}(A), \frac{1}{2}d_{σ}(A)(d_{σ}(A) + 1)\} = c_{p}(A) = c(A)

and similarly

c(AZ/Z) = max\{c_{σ′}(AZ/Z), \frac{1}{2}d_{σ}(AZ/Z)(d_{σ}(AZ/Z) + 1)\}.

Therefore

\begin{matrix}n − 1 ≥ & max\{c_{σ′}(AZ/Z), \frac{1}{2}d_{σ}(AZ/Z)(d_{σ}(AZ/Z) + 1)\} \\ + & max\{c_{τ′}(BZ/Z), \frac{1}{2}d_{τ}(BZ/Z)(d_{τ}(BZ/Z) + 1)\} \end{matrix}

which shows that already (G/Z)^{(n−1)} = 1. Since Z is abelian, we must have G^{(n)} = 1. This contradiction shows that we must have p ∈ σ; in particular, p is odd.

Since F = O_{p′p}(G) and p ≠ 2, we have d_{p}(A/F) = d(A/F) ≤ d(A) − 1 = d_{p}(A) − 1 by Theorem 4.1.1 and therefore

max\{c_{σ′}(AF/F), \frac{1}{2}d_{σ}(AF/F)(d_{σ}(AF/F) + 1)\} = \frac{1}{2}d(A/F)(d(A/F) + 1).

Thus

\begin{matrix}n − d(A) ≥ & max\{c_{σ′}(AF/F), \frac{1}{2}d_{σ}(AF/F)(d_{σ}(AF/F) + 1)\} \\ + & max\{c_{τ′}(BF/F), \frac{1}{2}d_{τ}(BF/F)(d_{τ}(BF/F) + 1)\} \end{matrix}

This shows that G/F has derived length ≤ n − d(A), yielding that

d(G) ≤ d(F) + d(G/F) ≤ n

since F ≤ A. This final contradiction shows that G^{(n)} is a nilpotent π-group.

It remains to show that G^{(n)} ≤ Φ(G) for every finite group G that is the product of two nilpotent subgroups A and B. But since our first result implies that G^{(n)} = 1 if A and B have coprime orders, this follows at once from Proposition 4.4.3.

Remark. That we have to treat the prime 2 differently is essentially due to the fact that for a Sylow 2-subgroup P of the finite soluble group G, it is possible to have

d(P O_{2′2}(G)/O_{2′2}(G)) = d(P).

However, this can only happen when P is non-abelian and also the Sylow 3-subgroups of G are non-abelian (cf. Theorem 4.1.1). Transferring these considerations to the proof of the theorem, we obtain that the theorem also holds for arbitrary sets of primes σ and τ, provided that A has an abelian Sylow 3-subgroup if the Sylow 2-subgroup of B is non-abelian and viceversa, i.e. if d_{3}(A) ≤ 1 whenever d_{2}(B) ≥ 2 and d_{3}(B) ≤ 1 if d_{2}(A) ≥ 2.

If we set σ = τ = ∅ in the preceding theorem, we obtain the following

4.4.6 Corollary (Gross [22] and Pennington [40]). If the finite group G is the product of its nilpotent subgroups A and B of classes c and d, then G^{(c+d)} is a nilpotent πgroup contained in the Frattini subgroup of G; in particular G/Φ(G) has derived length ≤ c + d.

4.4.7 Corollary. If the group G is the product of its nilpotent subgroups A and B of coprime order, then d(G) ≤ c(A) + c(B).

The irregular behaviour of the prime 2 can also be compensated by considering a Hall 2′-subgroup of G instead of a Sylow 2-subgroup. The key to this is the observation that most statements of Theorem 4.1.1 can be extended to statements about a nilpotent Hall π-subgroup of a group G if we replace p by π and p′ by π′. We state some of the most important consequences.

4.4.8 Lemma (Berger and Gross [9]). Let G be a finite soluble group.

(i) If O_{π′π}(G)/O_{π′}(G) is nilpotent, then O_{π′π}(G) = ⋂_{p∈π} O_{p′p}(G).

(ii) If H is a nilpotent Hall π-subgroup of G and c(H) > 0, then

c(HO_{π′π}/O_{π′π}(G)) < c(H);

if 2 ∉ π and d(H) > 0, then also

d(HO_{π′π}(G)/O_{π′π}(G)) < d(H).

Proof. (i) Since O_{π′π}(G)/O_{π′}(G) is nilpotent, it is clear that O_{π′π}(G) ≤ O_{p′p}(G) for all p ∈ π. Since O_{π′}(G) ≤ O_{p′p}(G) for all primes p ∈ π, O_{π′}(G) must be contained in O_{p′}(G) for all such primes. On the other hand, ⋂_{p∈π} O_{p′}(G) is a normal π′-subgroup of G, therefore it is contained in O_{π′}(G). This shows that O_{π′}(G) = ⋂_{p∈π} O_{p′}(G). Now a π′-element contained in ⋂_{p∈π} O_{p′p}(G) is contained in O_{p′}(G) for all p ∈ π, whence it is contained in O′_{π}(G). This shows that (⋂_{p∈π} O_{p′p}(G))/O_{π′}(G) is a π-group and therefore ⋂_{p∈π} O_{p′p}(G) ≤ O_{π′π}(G). This proves (i).

(ii) Denote with H_{p} the normal Sylow p-subgroup of H. For all primes p, consider the canonical homomorphisms

α_{p}: H_{p} ⟶ H_{p}O_{p′p}(G)/O_{p′p}(G).

These homomorphisms induce a homomorphism

α: H = /\_{p∈π} H_{p} ⟶ D = /\_{p∈π} H_{p}O_{p′p}(G)/O_{p′p}(G)

whose kernel is H ∩ (⋂_{p∈π} O_{p′p}(G)) = H ∩ O_{π′π}(G) by part (i). This, together with an isomorphism theorem, shows that

\begin{matrix}c(HO_{π′π}(G)/O_{π′π}(G)) = & c(H/H ∩ O_{π′π}(G)) \\ ≤ c(D) = max_{p∈π}\{c(H_{p}O_{p′p}(G)/O_{p′p}(G))\} \end{matrix}

Now by Theorem 4.1.1, for every prime p,

c(H_{p}O_{p′p}(G)/O_{p′p}(G)) < c(H_{p})

because the H_{p} are Sylow p-subgroups of G. This shows that

\begin{matrix}c(HO_{π′π}(G)/O_{π′π}(G)) ≤ & max_{p∈π}\{c(H_{p}O_{p′p}(G)/O_{p′p}(G))\} \\ < & max_{p∈π}\{c(H_{p})\} = c(H) \end{matrix}

as required.

If p ≠ 2 and H_{p} ≠ 1, also

d(H_{p}O_{p′p}(G)/O_{p′p}(G)) < d(H)

so that by the same argument,

d(HO_{π′π}(G)/O_{π′π}(G)) < d(H)

if 2 ∉ π.

The following theorem seems rather similar to the preceding Theorem 4.4.5; however its results are in terms of the derived lengths of A and B only and do not involve c_{2}(A) or c_{2}(B).

4.4.9 Theorem. Let the group G = AB be the product of its finite nilpotent subgroups A and B. If π is the set of primes that divide the orders of both A and B, then G^{(n)} is a nilpotent π-subgroup contained in Φ(G), where

\begin{matrix}n = & \frac{1}{2}d(A)(d(A) + 1) + \frac{1}{2}d(B)(d(B) + 1) \\ + & max\{\frac{1}{2}d_{2}(A)(d_{2}(A) + 1), \frac{1}{2}d_{2}(B)(d_{2}(B) + 1)\}. \end{matrix}

Proof. Like in the proof of Theorem 4.4.5, in view of Proposition 4.4.3, it suffices to show that G^{(n)} is a nilpotent π-group. Suppose that this is false and let G be a counterexample of minimal order. Thus if N ⊴ G, then

\begin{aligned}n & ≥ \frac{1}{2}d(AN/N)(d(AN/N) + 1) + \frac{1}{2}d(BN/N)(d(BN/N) + 1) \\ & + max\{\frac{1}{2}d_{2}(AN/N)(d_{2}(AN/N) + 1), \frac{1}{2}d_{2}(BN/N)(d_{2}(BN/N) + 1)\}. \end{aligned}

Therefore by Lemma 4.4.4, w.l.o.g. A is a Sylow p-subgroup of G containing the unique minimal normal subgroup N of G and also F = F (G) ≤ A; moreover B is a Hall p′- subgroup of G. We may also assume that d(A) = d_{p}(A) > 0 and d(B) > 0 since otherwise G = B or G = A and in these cases the theorem is obviously true.

Consider first the case when p ≠ 2. Then we have d(AF/F) < d(A) by Theorem 4.1.1 and therefore

\begin{matrix}n − d(A) ≥ & \frac{1}{2}d(AF/F)(d(AF/F) + 1) + \frac{1}{2}d(BF/F)(d(BF/F) + 1) \\ + & \frac{1}{2}d_{2}(B)(d_{2}(B) + 1). \end{matrix}

This shows that d(G/F) ≤ n − d(A) and since F ≤ A, we have

d(G) ≤ d(F) + d(G/F) ≤ d(A) + n − d(A) = n.

This contradiction shows that we must have p = 2.

Therefore we have

\frac{1}{2}d(AF/F)(d(AF/F) + 1) + \frac{1}{2}d_{2}(AF/F)(d_{2}(AF/F) + 1) = d_{2}(A)(d_{2}(A) + 1).

Now by Lemma 4.2.1, F_{3} = F_{3}(G) = O_{2′22′2}(G) and by Theorem 4.1.1, we have

d(AF_{3}/F_{3}) < d(A);

moreover, applying Lemma 4.4.8 with π = 2′, we have

d(B) ≥ d(BF/F) > d(BF_{2}/F_{2}) ≥ d(BF_{3}/F_{3})

where F_{2} = F_{2}(G). Hence

\begin{matrix}d(G/F_{3}) ≤ & d_{2}(AF_{3}/F_{3})(d_{2}(AF_{3}/F_{3}) + 1) + \frac{1}{2}(d(BF_{3}/F_{3})(d(BF_{3}/F_{3}) + 1) \\ ≤ & n − 2d(A) − d(B) \end{matrix}

and since by Lemma 4.2.1, F ≤ A, the group F_{2}/F is isomorphic with a section of B and F_{3}/F_{2} is isomorphic with a section of A, we obtain

\begin{matrix}d(G) ≤ & d(F) + d(F_{2}/F) + d(F_{3}/F_{2}) + d(G/F_{3}) \\ ≤ & d(A) + d(B) + d(A) + n − 2d(A) − d(B) \\ = & n. \end{matrix}

This final contradiction proves the theorem.

Remark. As in the remark after Theorem 4.4.5, the bound can be improved significantly if d_{3}(A) ≤ 1 whenever d_{2}(B) ≥ 2 and d_{3}(B) ≤ 1 if d_{2}(A) ≥ 2. In this case, we already obtain that G^{(n)} ≤ Φ(G) ∩ O_{π}(G) where

n ≥ \frac{1}{2}d(AN/N)(d(AN/N) + 1) + \frac{1}{2}d(BN/N)(d(BN/N) + 1).

There is yet another possibility to gain some information about the derived length of a product of two nilpotent subgroups:

4.4.10 Theorem. Let the finite group G be the product of its nilpotent subgroups A and B. Then G/F (G) has derived length at most k where

k = (d(A) + max{0, d_{2}(A) − 1})(d(B) + max{0, d_{2}(B) − 1}).

Proof. Suppose that G is a counterexample of minimal order. Then we have

d((G/N)/F (G/N)) ≤ k

for all normal subgroup N ≠ 1 of G but d(G/F (G)) > k. Since the class of groups H satisfying d(H/F (H)) ≤ k forms a saturated formation, hence is a Schunck class, by Lemma 4.4.2 G must be primitive. Since our statement is trivial if G is nilpotent, we may also assume that d(A) > 0, d(B) > 0 and also that A ≠ B, hence by Lemma 3.2.5, w.l.o.g. A is a Sylow p-subgroup of G and B is a Hall p′-subgroup, where p is the exponent of the unique minimal normal subgroup F = F (G) of G.

Suppose first that p ≠ 2, or that d(AF/F) < d(A) if p = 2. Observe that we also have d(A/F) < d(A) if p ≠ 2 by Theorem 4.1.1 since F = O_{p′p}(G). Therefore if we let F_{2} = F_{2}(G), then

\begin{matrix}d(G/F_{2}) ≤ & d(AF/F)(d(BF/F) + max\{0, d_{2}(BF/F) − 1\}) \\ ≤ & (d(A) − 1)(d(B) + max\{0, d_{2}(B) − 1\}) \\ ≤ & k − d(B). \end{matrix}

Since by Lemma 4.2.1, F_{2}/F ≤ BF/F, we have

d(G/F) ≤ d(F_{2}/F) + d(G/F_{2}) ≤ n.

This contradiction shows that we must have p = 2 and d(AF/F) = d(A). This also shows that d(A) ≥ 2 because if we had d(A) = 1, then A would be abelian, hence A ≤ C_{G}(F) = F which would imply that d(AF/F) = 0 < d(A).

Now by Lemma 4.2.1, F_{2} = O_{22′}(G) and F_{3} = F_{3}(G) = O_{2′2}(G) and so by Theorem 4.1.1, d(AF_{3}/F_{3}) < d(A); moreover by Lemma 4.4.8, d(BF_{3}/F_{3}) < d(B). If F_{4} = F_{4}(G), then we have

d(G/F_{4}) ≤ (2d(AF_{3}/F_{3}) − 1)(d(BF_{3}/F_{3}))
≤ (2d(A) − 1 − 2)(d(B) − 1)
≤ (2d(A) − 1)(d(B) − 1) − (2d(A) − 1) − 2d(B) + 2
≤ (2d(A) − 1)(d(B) − 1) − d(A) − d(B) − (d(B) − 1)

Since F_{2}/F ≤ BF/F, F_{3}/F_{2} ≤ AF_{2}/F_{2} and F_{4}/F_{3} ≤ BF_{3}/F_{3} by Lemma 4.2.1, we have

d(G/F) ≤ d(F_{2}/F) + d(F_{3}/F_{2}) + d(F_{4}/F_{3}) + d(G/F_{4})
≤ d(B) + d(A) + (d(B) − 1)
+(2d(A) − 1)(d(B) − 1) − d(A) − d(B) − (d(B) − 1)
= k.

Thus we have reached a final contradiction which proves the theorem.

Remark. Again, if we have d_{3}(A) ≤ 1 in case d_{2}(B) ≥ 2 and d_{3}(B) ≤ 1 if d_{2}(A) ≥ 2 in the preceding theorem, then we obtain that d(G/F (G)) ≤ d(A)d(B), a result that has already been obtained by Gross [22] under the hypothesis that A and B have coprime orders.

4.4.11 Corollary. If the finite group G is the product of its nilpotent subgroups A and B, then has derived length at most

d(G/Φ(G)) ≤ (d(A) + max{0, d_{2}(A) − 1})(d(B) + max{0, d_{2}(B) − 1}) + 1

and if π = σ(A) ∩ σ(B) and

k = (d(A) + max{0, d_{2}(A) − 1})(d(B) + max{0, d_{2}(B) − 1}) + max{d(A), d(B)}),

then G^{(k)} is a nilpotent π-group contained in the Frattini subgroup of G.

Proof. The first statement follows directly from the fact that F (G)/Φ(G) is abelian. For the second statement, let N be the π-component of F = F (G). Since F =

(A ∩ F)(B ∩ F) by Theorem 3.1.5, we have F/N = (A ∩ F)N/N(B ∩ F)N/N. By the definition of π, (A ∩ F)N/N and (B ∩ F)N/N have coprime orders, and since F/N is nilpotent, F/N = (A ∩ F)N/N × (B ∩ F)N/N whence

d(F/N) ≤ max{d((A ∩ F)N/N), d(B ∩ F)N/N)} ≤ max{d(A), d(B)}.

This shows that G^{(k)} is a nilpotent π-group. Since the corollary is obviously true if G = 1, we may also suppose that max{d(A), d(B)} ≥ 1, and therefore by the first part, also G^{(k)} ≤ Φ(G).

The following theorem describes the special case when one of the factors is abelian.

4.4.12 Theorem. Let the finite group G be the product of an abelian group A and a nilpotent group B. Then

(i) (Franciosi, de Giovanni, Heineken and Newell [16]) AF (G) ⊴ G; (ii) n(G) ≤ 3; (iii) d(G/F (G)) ≤ d(B); (iv) G^{(n)} ≤ Φ(G) ∩ O_{π}(G) where π = σ(A) ∩ σ(B) and n = max{2d(B), 1}. Proof. (i). If G = 1, this is clearly true, so suppose by finite induction that (i) is

true for all groups of smaller order than G. If G/Φ(G) satisfies the theorem, so does G because F (G)/Φ(G) = F (G/Φ(G)). Therefore we must have Φ(G) = 1.

Consider first the case when G possesses two distinct minimal normal subgroups N^{∗} and N^{∗∗}. Denote with F^{∗}/N^{∗} and F^{∗∗}/N^{∗∗} the Fitting subgroups of G/N^{∗} and G/N^{∗∗}, then F^{∗} ∩ F^{∗∗} = F by Lemma 1.3.5 and since by Theorem 3.1.5, F^{∗} and F^{∗∗} are factorized, by Lemma 1.1.9, AF^{∗} ∩ AF^{∗∗} = AF, which is a normal subgroup of G since by induction hypothesis, AF^{∗} and AF^{∗∗} are normal subgroups of G.

Therefore we may assume that G possesses a unique minimal normal subgroup N which is an elementary abelian p-group for a prime p, and, because Φ(G) = 1, the group G is primitive. Since G can be assumed non-nilpotent, A or B is a p-group containing N = F while the other is a p′-group by Lemma 3.2.5. If N = F ≤ A, then A ≤ C_{G}(F) = F and thus A = F, proving that AF = F ⊴ G.

There remains the case when F ≤ B. Then the p′-group F_{2}/F = F (G/F) is contained in the Hall p′-subgroup AF/F of G/F and therefore AF_{2} ≤ AF. The other inclusion AF ≤ AF_{2} is trivial, and by the minimality of G, the subgroup AF_{2}/F = AF/F is normal in G/F, or equivalently, AF ⊴ G. This final contradiction proves (i).

(ii) follows directly from the fact that the series

1 ◁ F (G) ⊴ AF (G) ⊴ G

has nilpotent factors. The same result about n(G) can be obtained as described in the remark after Theorem 4.4.1, observing that we have d_{2}(A) ≤ 1 and d_{3}(A) ≤ 1 since A is abelian.

The remaining statements follow from the remark after Theorem 4.4.10.

4.5 Properties of the Fitting quotient group

4.5.1 Proposition. Let the finite group G be the product of its nilpotent subgroups A and B. Then:

(i) G/[A, B] is nilpotent. The factorizer of [A, B] is A^{G} ∩ B^{G} = (A ∩ B^{G})(B ∩ A^{G}). (ii) Let A_{0} = A, B_{0} = B and G_{0} = G and for all i > 0, define A_{i+1} = A_{i} ∩ B_{i}^{G_{i}}

and B_{i+1} = B_{i} ∩ A_{i}^{G_{i}}. Then G_{i} = A_{i}B_{i} is a factorized subgroup of G for all i and G_{i+1} ⊴ G_{i}; moreover G_{i}/G_{i+1} is nilpotent. If n is the smallest integer such that G_{n} = G_{n+1}, then G_{n} = A ∩ B. Thus the series

G = G_{0} ▷ G_{1} ▷ ⋯ ▷ G_{n} = A ∩ B

consists of factorized subnormal subgroups of G such that G_{i}/G_{i+1} is a nilpotent group.

Proof. (i) First observe that [A, B] ⊴ ⟨A, B⟩ for arbitrary subgroups A and B of G. Now A^{G} = A[A, B] and B^{G} = B[A, B] which shows that G/[A, B] is the product of the normal nilpotent subgroups A[A, B]/[A, B] and B[A, B]/[A, B] and is therefore nilpotent. By Lemma 1.2.1, the factorizer of [A, B] is A[A, B] ∩ B[A, B] = A^{G} ∩ B^{G} = (A ∩ B^{G})(B ∩ A^{G}).

(ii) Suppose by finite induction that G_{i} = A_{i}B_{i} where A_{i} ≤ A and B_{i} ≤ B. (This is clearly true if i = 0). Then G_{i} satisfies the hypothesis of the preceding part; therefore G_{i}/G_{i+1} is a nilpotent group, G_{i+1} = A_{i+1}B_{i+1} and A_{i+1} ∩ B_{i+1} = A_{i} ∩ B_{i} = A ∩ B. as required.

Now suppose that G_{n} = G_{n+1}. Since G_{n+1} = A_{n}^{G_{n}} ∩B_{n}^{G_{n}}, we must have A_{n}^{G_{n}} = G_{n} and B_{n}^{G_{n}} = G_{n} and so neither A_{n} nor B_{n} is contained in a proper normal subgroup of G_{n}. So it follows from Proposition 3.1.8 that G_{n} = A_{n} = B_{n} and so G_{n} = A_{n} ∩ B_{n} ≤ A ∩ B proving that G_{n} = A ∩ B.

Recall that in Example 3.4.4, we have shown that the subgroup [A, B] above is not necessarily prefactorized.

4.5.2 Corollary. Let the finite group G be the product of its nilpotent subgroups A and B. If n is the length of the series defined in Proposition 4.5.1, then G has Fitting length at most n + 1 and derived length at most

n · max{d(A), d(B)} + min{d(A), d(B)}.

Proof. The series

G = G_{0} ▷ G_{1} ▷ ⋯ ▷ G_{n} = A ∩ B ⊵ 1

introduced in Proposition 4.5.1 has nilpotent factors (clearly also A ∩ B is nilpotent) and its length is n + 1. This is clearly an upper bound for the Fitting length of G.

To obtain the bound on the derived length, consider the factor group G_{i}/G_{i+1} for some i: G_{i} = A_{i}B_{i} = A_{i}^{G_{i}}B_{i}; thus

G_{i}/A_{i}^{G_{i}} ≅ B_{i}/B_{i} ∩ A_{i}^{G_{i}} = B_{i}/B_{i+1}

whose derived length is ≤ d(B) and G_{i}/B_{i}^{G_{i}} ≅ A_{i}/A_{i+1} which has derived length ≤ d(A). Since G_{i+1} = A_{i}^{G_{i}} ∩ B_{i}^{G_{i}}, it follows that d(G_{i}/G_{i+1}) ≤ max{d(A), d(B)}. Clearly, d(A ∩ B) ≤ min{d(A), d(B)} and also the second statement follows.

We have already proved in Proposition 3.1.2 that if G is the product of its nilpotent subgroups A and B, then the commutator subgroups [A_{p}, B_{p}] are subnormal p-subgroups and are therefore contained in the Fitting subgroup of G. Therefore the next proposition applies in particular when F (G) ⊴ N:

4.5.3 Proposition. Suppose that the group G is the product of its finite nilpotent subgroups A and B. Suppose that N is a normal subgroup of G such that G/N is nilpotent and that N contains [A_{p}, B_{p}] for every prime p dividing |G/N|. Then the subgroups A_{π}N/N and B_{π}N/N are normal subgroups of G/N for every set of primes π. In particular, AN/N and BN/N are normal subgroups of G. If moreover N is factorized, then G is the direct product of AN/N and BN/N,

\begin{aligned}G/N & = AN/N × BN/N \\ & = /\_{p∈±⊬ℙ} A_{p}N/N × /\_{p∈±⊬ℙ} B_{p}N/N. \end{aligned}

Proof. Consider the Sylow p-subgroup A_{p}N/N of AN/N. Since G/N is nilpotent, A_{p}N/N is centralized by the normal p-complement of G/N; moreover [A_{p}, B_{p}] ≤ N by hypothesis; therefore also B_{p}N/N centralizes A_{p}N/N which shows that A_{p}N/N ⊴ G/N. Therefore also A_{π}N/N = /\_{pππ} A_{p}N/N is a normal subgroup of G.

If N is factorized, then N = AN ∩ BN and therefore AN/N ∩ BN/N = 1, showing that G/N = AN/N × BN/N. The rest of the statement follows.

4.5.4 Corollary. Let G = AB where A and B are finite and nilpotent. If M ⊴ G and N ⊴ M such that M/N is nilpotent, if moreover N contains [A_{p}, B_{p}] for all primes p dividing |M/N|, then AN ∩ M and BN ∩ M are normal subgroups of M. If M contains AN ∩ BN, then (AN ∩ M)(AN ∩ M) = (A ∩ M)(B ∩ M).

Proof. Let X = AM ∩ BM be the factorizer of M. Then X/N is nilpotent by Corollary 3.5.2 and therefore AN ∩ X ⊴ X and BN ∩ X ⊴ X. Hence AN ∩ M = AN ∩ M ∩ X and BN ∩ M are normal subgroups of M.

If M contains AN ∩ BN, also M^{∗} = (AN ∩ M)(AB ∩ M) contains this intersection. Thus, M^{∗}/N is a factorized subgroup of G/N, and M^{∗} is a factorized subgroup of G.

Since by Theorem 3.1.5 the Fitting subgroup and therefore every term of the Fitting series of a product of two nilpotent subgroups is factorized, we obtain

4.5.5 Corollary. Let F_{k} be the k-th term of the Fitting series of G = AB where A and B are finite nilpotent subgroups of G. If k ≥ 1, then F_{k+1}/F_{k} is the direct product of (A ∩ F_{k+1})F_{k}/F_{k} and (B ∩ F_{k+1})F_{k}/F_{k}.

In this context, also the following Proposition is of interest:

4.5.6 Proposition (Heineken [29]). Let the group G be the product of its finite nilpotent subgroups A and B. If F_{k} = F_{k}(G) denotes the k-th term of the Fitting series of G, then for all k ≥ 1, we have that (A ∩ F_{k+1})F_{k}/F_{k} and (B ∩ F_{k+1})F_{k}/F_{k} are normal subgroups of G.

Proof. Since the setting is symmetrical in A and B, it clearly suffices to show that (A ∩ F_{k+1})F_{k} ⊴ G. Since (A ∩ F_{k+1})F_{k} = AF_{k} ∩ F_{k+1} by the modular law and G/F_{k} = (AF_{k})(BF_{k}), we may in addition pass to the quotient group G/F_{k−1}, hence it remains to show that (F_{2} ∩ A)F = AF ∩ F_{2} is normal in G where F = F (G) (= F_{1}).

Suppose by induction on |G| that the proposition is true for all groups of smaller order; in view of F/Φ(G) = F (G/Φ(G)), we may also assume that Φ(G) = 1.

Next, consider the case when G contains two distinct minimal normal subgroups N^{∗} and N^{∗∗}. Now let F^{∗}/N^{∗} = F (G/N^{∗}), F^{∗∗}/N^{∗∗} = F (G/N^{∗∗}) and F_{2}^{∗}/F^{∗} = F (G/F^{∗}), F_{2}^{∗∗}/F^{∗∗} = F (G/F^{∗∗}) then we have F = F^{∗} ∩ F^{∗∗} and F_{2} = F_{2}^{∗} ∩ F_{2}^{∗∗} by Lemma 1.3.5, observing that the classes of nilpotent and metanilpotent groups form Fitting formations. So we have AF = A(F^{∗} ∩ F^{∗∗}) = AF^{∗} ∩ AF^{∗∗} by Lemma 1.1.9. Now by induction hypothesis, AF^{∗} ∩ F_{2}^{∗} and AF^{∗∗} ∩ F_{2}^{∗∗} are normal subgroups of G and thus AF^{∗} ∩ F_{2}^{∗} ∩ AF^{∗∗} ∩ F_{2}^{∗∗} = AF ∩ F_{2} = (A ∩ F_{2})F is a normal subgroup of G as required.

There remains the case when G has a unique minimal normal subgroup; let p denote its exponent. Since also Φ(G) = 1, we conclude that G is primitive. But in this case, by Lemma 3.2.5, A is either a p-group and A ∩ F_{2} = F whence F = (A ∩ F_{2})F ⊴ G or A is a Hall p′-group whence F_{2} is contained in AF and therefore AF ∩ F_{2} = F_{2} ⊴ G.

Next, we report a result of Heineken which will allow to relate products of finite nilpotent subgroups to other classes of groups.

4.5.7 Proposition (Heineken [28]). Assume that the group G is the product of two finite nilpotent subgroups A and B. Then G/F (G) ∈ R_{0}𝔛 where 𝔛 is the class of finite groups whose order is divisible only by two primes one of which divides the order of A while the other divides the order of B.

Indeed, from Heineken’s proof of the above proposition, a stronger result can be obtained.

4.5.8 Proposition. Let the group G be the product of its finite nilpotent subgroups A and B. If H/K is a principal factor of G such that F (G) ≤ K, then |G/C_{G}(H/K)| is divisible at most by one prime divisor of |A| and one of |B|; moreover one of these prime divisors is the exponent of H/K.

Proof. Suppose that the proposition is true for all groups of smaller order than G: this includes, of course, the case when G is nilpotent. Observe also that C_{G/N}(H/K) = C_{G}(H/K)/N for every normal subgroup N of G with N ≤ K.

As a first case, assume that G possesses two distinct minimal normal subgroups N^{∗} and N^{∗∗}. Let F^{∗}/N^{∗} and F^{∗∗}/N^{∗∗} denote the Fitting subgroups of G/N^{∗} and G/N^{∗∗} respectively, then F^{∗} ∩ F^{∗∗} = F = F (G) by Lemma 1.3.5. Refine the series

1 ⊴ F^{∗}/F ⊴ G/F

to a chief series of G. Then by the Jordan-H^{}ölder theorem, H/K is G-isomorphic with a factor of that chief series; therefore we may assume that H/K itself belongs to that chief series. Now if F^{∗} ≤ K, then H/K is already a chief factor of G/F^{∗} and the result follows from the fact that C_{G}(H/K)/N^{∗} = C_{G/N^{∗}}(H/K). If H/K is a chief factor of F^{∗}/F, then since F^{∗}/F ≅_{G} F^{∗∗}F^{∗}/F^{∗}, the chief factor H/K is G-isomorphic with a chief factor of G/F^{∗∗} and the result follows again.

Hence we may assume that G has a unique minimal normal subgroup N. If Φ(G) > 1, we have F/Φ(G) = F (G)/Φ(G) and the result follows by induction.

Therefore there remains the case when G is primitive and non-nilpotent. As before, we may assume that H/K is a principal factor of the chief series of G/F obtained by refining

1 ◁ F_{2} ⊴ G/F,

where F_{2}/F = F (G/F); by finite induction, we may also exclude the case when F_{2} ≤ K < H ≤ G. Now by Lemma 3.2.5, we may w.l.o.g. assume that A is a Sylow p-group containing F and that B is a Hall p′-group of G, we also have F_{2}/F ≤ BF/F. From the last statement, it follows that H/K ≤ B_{q}K/K where q is the exponent of H/K. Now since B is nilpotent, B_{q} is centralized by B_{q′} and therefore B_{q′} ≤ C_{G}(H/K). Since B_{q′} is a Hall {p, q}′-subgroup of G, the order of G/C_{G}(H/K) can be divisible by p and q only.

From this, Proposition 4.5.7 follows with the help of the characterization of the Fitting subgroup of a finite group as the intersection of the centralizers of all principal factors of G.

Following Huppert [30], a finite group G is called a group with many Sylow bases if set of Sylow subgroups of G containing exactly one Sylow subgroup for every prime p is a Sylow basis of G, or equivalently, if every Sylow p-subgroup of G permutes with every Sylow q-subgroup of G when p ≠ q. ^{1}

Of course, the groups of order p^{α}q^{β} (where p and q are primes) are examples of groups with many Sylow bases. In the following, we will denote the class of all such groups by 𝔔; clearly

𝔔 = ⋃_{p,q∈±⊬ℙ} 𝔖_{{p,q}}.

The class 𝔅 of all finite groups with many Sylow bases can be characterized as follows:

1 These groups were first termed ‘Gruppen mit vielen Sylowsystemen’ by Huppert [30] who used the

term ‘Sylowsystem’ for what we call a Sylow basis.

4.5.9 Theorem (Huppert [30]). The following statements about the finite group G are equivalent:

(i) G is a group with many Sylow bases.

(ii) G is soluble; if H/K is a principal factor of G of exponent p, then there are a prime q and nonnegative integers a and B such that the order of G/C_{G}(H/K) is p^{a}q^{b}.

(iii) G is contained in the smallest formation that contains all groups whose order is divisible by at most two primes.

A proof of this can also be found in [31], VI, § 3.

In the following, we will denote the class of groups with many Sylow bases with 𝔅; we also recall that this class is S-closed. Furthermore, by (ii) and the characterization of primitive soluble groups in Lemma 3.2.3 as those groups in which the unique minimal normal subgroup is self-centralized, it is clear that the primitive 𝔅-groups are precisely the primitive groups whose order is divisible by at most two primes.

There follows that primitive 𝔅-groups are products of their nilpotent subgroups since in fact every 𝔔-group is the product of its Sylow subgroups. As a consequence of this, 𝔅 is clearly contained in the smallest Schunck class that comprises all groups that are the product of two finite nilpotent subgroups. To simplify notation, we will for the rest of this section to denote the class of finite groups that are the product of two nilpotent subgroups by 𝔐.

Now proposition can be used to show that 𝔐 ⊆ 𝔑𝔅 in the following

4.5.10 Proposition. Let 𝔑𝔅 be the class of groups G that possess a normal subgroup N such that N ∈ 𝔑 and G/N ∈ 𝔅. Then

(i) 𝔑𝔅 is a saturated formation;

(ii) 𝔑𝔅 = {G | G/F (G) ∈ 𝔅} and

(iii) 𝔐 ⊆ 𝔑𝔅.

Proof. Since 𝔑, the class of all finite nilpotent groups, is S_{n}-closed, 𝔑𝔉 equals the formation product of 𝔑 and 𝔉 and so it is a formation by [13], IV.1.8.

If G ∈ 𝔑𝔅 and N is a nilpotent normal subgroup of G such that G/N ∈ B, we have N ≤ F (G) ∈ 𝔑. On the other hand, G/F (G) is an epimorphic image of G/N and therefore G/F (G) ∈ Q𝔅 = 𝔅. This shows that

𝔑𝔅 = {G | G/F (G) ∈ 𝔅}.

From this and the fact that F (G)/Φ(G) = F (G/Φ(G)) for all finite groups G, it follows immediately that 𝔑𝔅 is saturated. This proves (i) and (ii).

(iii) Let G ∈ 𝔐. Then by Proposition 4.5.8, the chief factors of G/F (G) are of the form described in Theorem 4.5.9, (ii) and so G/F (G) ∈ 𝔅, hence G ∈ 𝔑𝔅.

4.5.11 Example. Let p be an odd prime that is not a Fermat prime. Suppose that N is a cyclic group of order p, and let G be the semidirect product of N with its automorphism group. Then C_{G}(N) = N, and since G is soluble, G is primitive by Lemma 3.2.3. Now the order of G is divisible by more than two primes (namely by p and by all prime divisors of p − 1 which is divisible by 2 and some other prime < p since p is not a Fermat prime). So by the remark after Theorem 4.5.9, G does not belong to the class of groups with many Sylow bases. But obviously, G is the product of N and a subgroup of order p − 1 both of which are abelian.

The next example shows that a group G such that G/F (G) is a group with many Sylow bases is not necessarily the product of two nilpotent subgroups.

4.5.12 Example. Let G and p be as in Example 4.5.11. Now let H = G {∩ \atop ∪} C where C is cyclic of order r where r is a prime ≠ p such that r does not divide p − 1. Then H is primitive by [13], A.18.5 since H is non-abelian and primitive. If H were the product of two nilpotent subgroups, then one of them would have to be a Hall p′-subgroup of H by Lemma 3.2.5, and since the Sylow p-subgroup of H equals its Fitting subgroup F, the quotient group G/F is isomorphic with a Hall p′-subgroup of G. On the other hand, G/F is isomorphic with the regular wreath product of a group of order p − 1 with a cyclic group of order r. But such a group is not nilpotent since its Sylow r-subgroup does not centralize the Hall r′-part. This shows that H is not the product of two nilpotent subgroups. But since the Sylow q-subgroups of G/F are normal for all primes q ≠ r, the factor group G/F is clearly a 𝔅-group.

The last example also shows that PQ𝔐, the smallest Schunck class containing 𝔐, is properly contained in 𝔑𝔅.

Question. Is 𝔑𝔅 the smallest formation (saturated formation) containing all products of two finite nilpotent groups?

If the answer to the first question is negative, is it true that the class 𝔐 of finite groups which are factorized by two nilpotent subgroups closed with respect to subgroups? This would imply that 𝔐 is itself a formation since 𝔐 is D_{0}-closed and Q-closed. Moreover, every group with many Sylow bases would be the product of two nilpotent groups. (Observe that in order to prove that 𝔐 is S-closed, it would suffice to show that a maximal normal subgroup of a product of two finite nilpotent groups is the product of two finite nilpotent groups, for by Corollary 3.3.6, every nonnormal maximal subgroup

\begin{aligned}& 𝔑𝔅 \\ • \end{aligned}

\begin{aligned}& ⟨E_{Φ},Q,S,D_{0}⟩𝔐 \\ • \end{aligned}

• QSD_{0}𝔐 •  \begin{aligned}& ⟨E_{Φ},Q,R_{0}⟩𝔐 \\ • \end{aligned}

•  •
SD_{0}𝔐 = S𝔐 •  \begin{aligned}& QR_{0}𝔐 \\ • \end{aligned}  \begin{aligned}& PQ𝔐 = PQ(𝔐 ∩ 𝔚) \\ • \end{aligned}
•  •
R_{0}𝔐 •

•

𝔚 •  \begin{aligned}& 𝔐 \\ • \end{aligned}
𝔅 •  _{• 𝔚 ∩ 𝔐}

•

\begin{aligned}& QD_{0}𝔔 \\ • \end{aligned}

\begin{aligned}& 𝔔 \\ • \end{aligned}

The ordering of the classes 𝔔, 𝔚, 𝔐 and 𝔑𝔅

of G has a conjugate which is factorized and thus every nonnormal maximal subgroup is the product of two nilpotent subgroups.)

In [30], Huppert also shows that if G ∈ 𝔅 and p ≠ q are prime divisors of |G|, then G has a normal {p, q}-subgroup N such that a Hall {p, q}-subgroup of G/N is nilpotent. It is easy to see that this last property of G is equivalent to the property that G/O_{π}(G) has a nilpotent Hall π-subgroup for every set π of primes. Thus Huppert’s statement is equivalent to the second statement of the following

4.5.13 Proposition. Let 𝔚 be the class of finite groups G such that G/O_{π}(G) has a nilpotent Hall π-subgroup for every set of primes π. Then

(i) 𝔚 is a subgroup-closed formation of finite soluble groups; (ii) 𝔅 ⊆ 𝔚; (iii) 𝔚 ⊆ R_{0}𝔛 and (iv) 𝔚 and 𝔐 contain the same primitive groups, hence the smallest Schunck classes

containing 𝔚 and 𝔐 coincide.

Proof. (i) Let G ∈ 𝔚 and let π be a set of primes. If G_{π}/O_{π}(G) is a Hall subgroup of G/O_{π}(G), then clearly G_{π} is a Hall subgroup of G. Therefore G possesses Hall πsubgroups for all sets of primes π, hence is soluble. Since all Hall subgroups of the soluble group G are isomorphic, all Hall π-subgroups of G/O_{π}(G) are nilpotent.

If N ⊴ G and π is a set of primes, then clearly O_{π}(G)N/N ≤ O_{π}(G/N). Therefore (G_{π}/N)/O_{π}(G/N) is an epimorphic image of G_{π}/N, hence is a nilpotent Hall πsubgroup of G/N, showing that G/N ∈ 𝔚. Therefore 𝔚 is Q-closed. To see that it is S-closed, let S ≤ G. Then O_{π}(G) ∩ S ≤ O_{π}(S), whence S/O_{π}(S) is an epimorphic image of S/S ∩ O_{π}(G) ≅ SO_{π}(G)/O_{π}(G). Now a Hall π-subgroup of SO_{π}(G)/O_{π}(G) is contained in a Hall π-subgroup of G/O_{π}(G), hence is nilpotent. Therefore S ∈ 𝔚 and 𝔚 is S-closed.

Now let M, N ∈ 𝔚 and put G = M × N. If π is any set of primes and M_{π} and N_{π} are Hall π-subgroups of M and N respectively, then G_{π} = M_{π} × N_{π} is a Hall π-subgroup of G. Since O_{π}(G) contains (even equals) O_{π}(M) × O_{π}(N), the group G_{π}/O_{π}(G) is an epimorphic image of M_{π}/O_{π}(M) × N_{π}/O_{π}(N) which is nilpotent. Therefore G ∈ 𝔚 and thus 𝔚 is also D_{0}-closed, hence is a subgroup-closed formation.

(ii) Since clearly 𝔔 ⊆ 𝔚 and 𝔅 is the smallest formation containing 𝔔, it follows that 𝔅 ⊆ 𝔚.

(iii) Let G ∈ 𝔚. Then G/O_{p′}(G) is the product of one of its Sylow p-subgroup and a nilpotent Hall p′-subgroup, hence G/O_{p′}(G) ∈ 𝔐. Since ⋂_{p∈±⊬ℙ} O_{p′}(G) = 1, it follows that G ∈ R_{0}𝔐.

(iv) Let G be a primitive 𝔚-group. Since G is soluble, G has a unique minimal normal subgroup N of prime exponent p, say. Then O_{p′}(G) = 1 and therefore G has a nilpotent Hall p′-subgroup. Since G is the product of a Hall p′-subgroup and a Sylow p-subgroup, we have G ∈ 𝔐. Conversely, let G be a primitive M-group, let p be the exponent of the unique minimal normal subgroup N of G and let π be a set of primes. If p ∈ π, then N ≤ O_{π}(G) and therefore O_{π}(G/N) = O_{π}(G)/N. Now N = F (G) and thus G/N ∈ 𝔅 ⊆ 𝔚 which shows that (G/N)/O_{π}(G/N) has a nilpotent Hall π-subgroup and by an isomorphism theorem, the same is true for G. If p ∉ π, then a Hall π-subgroup of G is contained in a Hall p′-subgroup of G which is nilpotent by Lemma 3.2.5. Hence G/O_{π}(G) has a nilpotent Hall π-subgroup for all sets π of primes.

Remark. Since the class 𝔐 is D_{0}-closed by Lemma 1.3.2, it follows that R_{0}𝔐 ⊆ SD_{0}𝔐 = S𝔐. Therefore every 𝔅-group and also every 𝔚-group is a subgroup of a group which is the product of two nilpotent subgroups. On the other hand, if G ∈ 𝔐, then G/F (G) ∈ 𝔅 ⊆ 𝔚, and since F (G) ≤ O_{π′π}(G) for every set of primes, it follows that G/O_{π′π}(G) has a nilpotent Hall π-subgroup for every set of primes π.

List of symbols

In general, uppercase Latin letters denote groups (A, B, G, H, …) or sets, lowercase (Latin) letters symbolize elements of sets or groups. Uppercase Fraktur stands for classes of groups while script is used for sets of groups. Lowercase Greek letters usually denote homomorphisms of groups (α, β, …) or sets of primes (π, σ, τ …).

In the following, G and H will be groups, A and B are subgroups of G and g, h ∈ G. Ω will be a set acting on G via endomorphisms (the action is written exponentially). Integers are denoted by k, m and n, a prime by p.

±⊬ℕ the set of positive integers ±⊬ℕ_{0} the set of nonnegative integers GF (p^{n}) the finite field of order p^{n} (m, n) the greatest common divisor of the integers m and n ±⊬ℙ the set of primes π′ complement of the set π of primes in ±⊬ℙ p′ = {p}′ = ±⊬ℙ \ {p}

G ≅ H G is isomorphic with H G × H the direct product of G and H G {∩ \atop ∪} H the regular wreath product of G and H ⟨X⟩ the subgroup of G generated by the elements of X ⊆ G ⟨x_{1}, x_{2}, …⟩ the subgroup generated by the set {x_{1}, x_{2}, …} g^{h} the action of h on G: g^{h} = h^{−1}gh X^{ω} the set {x^{ω} | x ∈ X} X^{Ω} the subgroup of G generated by the set {x^{ω} | x ∈ X, ω ∈ Ω} X_{Ω} X_{Ω} = ⋂_{ω∈Ω} X^{ω} [g, ω] the commutator of g and ω; [g, ω] = g^{−1}g^{ω} [A, B] the subgroup of G generated by all [a, b] where a ∈ A and b ∈ B N_{Ω}(X) normalizer of the set X: N_{Ω}(X) = {ω ∈ Ω | [x, ω] ∈ X for all x ∈ X} C_{Ω}(X) centralizer of the set X: C_{Ω}(X) = {ω ∈ Ω | [x, ω] = 1 for all x ∈ X} Z(G) centre of the group G; Z(G) = C_{G}(G). G^{(n)} n-th derived subgroup of G defined recursively by G^{(0)} = G

and G^{(n+1)} = [G^{(n)}, G^{(n)}] for n ≥ 0

G′, G′′ = G^{(1)}, G^{(2)} F (G) Fitting subgroup of G, the subgroup generated by the normal nilpotent

subgroups of G

F_{n}(G) n-th term of the Fitting series of G defined recursively by F_{0}(G) = 1

and F_{n+1}(G)/F_{n}(G) = F (G/F_{n}(G)) for every n ≥ 0

Φ(G) Frattini-subgroup of G, the intersection of all maximal subgroups of G O^{π_{1}π_{2}...π_{k}}(G) defined recursively; O^{π_{1}}(G) is the π-residual of G, i.e. the intersection of

all normal subgroups N of G with G/N a π-group and O^{π_{1}π_{2}...π_{k}}(G) = O^{π_{k}}(O^{π_{1}π_{2}...π_{k−1}}(G)) if k > 1.

O_{π_{1}π_{2}...π_{k}}(G) defined recursively: O_{π_{1}}(G) is the π-radical of G, i.e. the subgroup of G generated by all subnormal π-subgroups of G and O_{π_{1}π_{2}...π_{k}}(G)/O_{π_{1}π_{2}...π_{k−1}}(G) = O_{π_{k}}(G/O_{π_{1}π_{2}...π_{k−1}}(G)) if k > 1.

G_{π} a Hall π-subgroup of the group G X_{G}(H) if G is the product of two subgroups, the factorizer X_{G}(H) of H is the

unique smallest factorized subgroup of G that contains H;

C_{π} G satisfies C_{π} (conjugacy) if it satisfies C_{π} and all Hall π-subgroups of G

are conjugate

D_{π} G satisfies D_{π} (dominance) G satisfies C_{π} and every π-subgroup is con-

tained in some Hall π-subgroup of G

E_{π} G satisfies the property E_{π} (existence) if it possesses a Hall π-subgroup.

|G| the cardinality of the set G σ(G) the set of primes dividing the order of some element of G; if G is finite,

this equals the set of prime divisors of the group order |G|.

c(G) the nilpotency class of the nilpotent group G d(G) the least integer k such that G^{(k)} = 1 b_{p}(G) p^{b_{p}(G)} = |P | where P is a Sylow p-subgroup of G c_{p}(G) the nilpotency class of a Sylow p-subgroup of G d_{p}(G) the derived length of a Sylow p-subgroup of G e_{p}(G) p^{e_{p}(G)} is the exponent of a Sylow p-subgroup of G b_{π}(G) = max_{p∈π} b_{p}(G) c_{π}(G) = max_{p∈π} c_{p}(G) d_{π}(G) = max_{p∈π} d_{p}(G) e_{π}(G) = max_{p∈π} e_{p}(G) n(G) Fitting length of G; the least integer k such that F_{k}(G) = G l_{π}(G) π-length of G; the number of nontrivial π-factors in the series

1 ≤ O_{π′}(G) ≤ O_{π′π}(G) ≤ O_{π′ππ′}(G) ≤ … ≤ G

l_{p}(G) = l_{{p}}(G)

𝔑 the class of all finite nilpotent groups 𝔖 the class of all finite soluble groups 𝔖_{π} the class of all soluble π-groups

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Erkl^{}ärung

Hiermit erkl^{}äre ich, daß ich die vorliegende Arbeit selbst^{}ändig verfaßt und keine anderen als die angegebenen Hilfsmittel verwendet habe.