Finite groups with multiple factorizations

by Bernhard Amberg, Burkhard Höfling and Lev S. Kazarin

1. Introduction

Since the fundamental work of P. Hall on Sylow systems of finite soluble groups and of Kegel and Wielandt on groups which can be written as products of pairwise permutable nilpotent subgroups, groups which are the product of permutable subgroups have been studied by many authors (see e. g. [1], [4] and [6]). A group G is the product of its pairwise permutable subgroups A_{1}, …, A_{n} if G = A_{1}…A_{n} and A_{i}A_{j} = A_{j}A_{i} for all integers i and j with 1 ≤ i, j ≤ n. This implies that for every choice of indices 1 ≤ i_{1} ≤ i_{2} ≤ … ≤ i_{k} ≤ n, the product A_{i_{1}}…A_{i_{k}} is a subgroup of G. In the sequel, we consider the following question: Suppose that every product of only k of the subgroups A_{i} belongs to a group class 𝔛. When does the whole group G belong to 𝔛?

Let π be a set of primes. It follows from a result of Kegel [7] and Lemma 2.2 below that the finite group G = A_{1}…A_{n} has a normal Hall π-subgroup if every product A_{i}A_{j} has a normal Hall π-subgroup. In particular if all A_{i}A_{j} are nilpotent, then the group G is nilpotent. In [4, Satz VI.10.2], Huppert shows that the finite group G has a Sylow tower of type σ if every product A_{i}A_{j} has a Sylow tower of type σ, where σ is an ordering of the set of all primes, and G is supersoluble if every product A_{i}A_{j}A_{k} is supersoluble. Recently, Kazarin proved in [5] that if every product A_{i}A_{j} is finite and soluble, then so is the group G = A_{1}…A_{n}.

This motivates the following definition: Let 𝔛 and 𝔙 be classes of groups and k a positive integer. We shall say that the class 𝔛 has property P_{k} for 𝔙-groups if a 𝔙-group G belongs to 𝔛 if G can be written as product of n subgroups A_{1}, …, A_{n} such that for every choice of indices 1 ≤ i_{1} ≤ i_{2} ≤ … ≤ i_{k} ≤ n, the group A_{i_{1}}…A_{i_{k}} belongs to 𝔛.

Clearly, the class of all finite groups satisfies P_{1} (for the class of all groups), and also the classes of all finite π-groups satisfy P_{1} for the class of all groups, where π is a set of primes. Moreover, the classes of finite soluble π-groups are the only subgroup-closed classes satisfying P_{1} for finite soluble groups (see Proposition 5.1). Some further examples of subgroup-closed classes of nonsoluble finite groups satisfying P_{1} for finite groups are given in Section 5. In the following, 𝔙 will always be a class of finite soluble groups which is closed with respect to subgroups and epimorphic images.

If 𝔛 and 𝔜 are group classes, 𝔛𝔜 denotes the class of all groups which have a normal subgroup N such that N is an 𝔛-group and G/N is a 𝔜-group. Furthermore, let 𝔛^{0} be the class of all trivial gropus and for every nonnegative integer k, let 𝔛^{k+1} = 𝔛^{k}𝔛. The class of all finite nilpotent groups will be denoted by 𝔑.

Theorem A. Let k be a positive integer and 𝔉 a subgroup-closed formation of finite soluble groups which satisfies P_{k} for 𝔙-groups. Then the subgroup-closed saturated formation 𝔑^{m}𝔉 satisfies P_{k+m} for 𝔙-groups for every nonnegative integer m.

If 𝔉 is the class of all trivial groups, we obtain the following

Corollary A1. Let k be a nonnegative integer, then the class 𝔑^{k} of all groups of Fitting length ≤ k satisfies P_{k+1} for finite soluble groups but does not satisfy P_{k} for finite soluble groups.

A formation function f is a function which assigns to every prime p a (possibly empty) formation of finite groups. The support π of f is the set of all primes such that f(p) is nonempty. We say that f has property P if f(p) satisfies P for every p ∈ π. Recall that f is full if f(p) = 𝔖_{p}f(p) for every p ∈ π, where 𝔖_{p} is the class of all finite p-groups.

Our second theorem allows the construction of saturated formations satisfying P_{k+2} from subgroup-closed formations satisfying P_{k}.

Theorem B. Let k be a positive integer and 𝔉 a saturated formation of finite soluble groups with subgroup-closed formation function f and suppose that f has property P_{k} for 𝔙-groups. Then 𝔉 satisfies P_{k+2} for 𝔙-groups. If f is full, then 𝔉 even satisfies P_{k+1} for 𝔙-groups.

If the formation function f in Theorem B satisfies f(p) = f(q) for all primes p and q, then Theorem A shows that the local formation 𝔉 = 𝔑f(p) even has property P_{k+1} for 𝔙-groups. We will show in Proposition 3.2 that a saturated formation 𝔉 satisfying P_{k} for 𝔙-groups has a formation function (namely the canonical one) which likewise satisfies P_{k} for 𝔙-groups.

If f is not full, the formation 𝔉 in Theorem B need not satisfy P_{k+1} for 𝔙-groups, as the following corollary shows. Here 𝔖 denotes the class of all finite soluble groups. If π is a set of integers, 𝔖_{π} denotes the class of all finite soluble π-groups.

Corollary B1. Let π be a set of primes and k a positive integer. If π is neither empty nor the set of all primes, then the class (𝔖_{π′}𝔖_{π})^{k} satisfies P_{2k} but not P_{2k+1} for finite soluble groups, while the class (𝔖_{π′}𝔖_{π})^{k}𝔖_{π′} satisfies P_{2k+1} but not P_{2k} for finite soluble groups.

For saturated formations of metanilpotent groups (which are always subgroup-closed), Theorem B can be improved when the formation function is not full. The following theorem extends the above-mentioned result of Huppert on finite supersoluble groups.

Theorem C. Let k be a positive integer and 𝔉 a saturated formation of finite metanilpotent groups and suppose that 𝔉 is locally defined by a formation function f having property P_{k} for finite nilpotent groups. Then 𝔉 satisfies P_{k+1} for finite metanilpotent groups and P_{m} for finite soluble groups, where m = max(3, k + 1).

Corollary C1. Let 𝔉 be a saturated formation of finite metanilpotent groups. If 𝔉 is closed with respect to products of normal subgroups, then 𝔉 satisfies P_{2} for finite metanilpotent groups. Moreover 𝔉 satisfies P_{3} for finite soluble groups.

Our notation is standard and can for instance be found in [2]. All groups considered are finite.

2. Preliminary results

The following observation is elementary.

2.1 Lemma. If the class of groups 𝔛 satisfies P_{k} for some positive integer k, then 𝔛 also satisfies P_{k+1}.

Proof. Let the group G be the product of its pairwise permutable subgroups A_{1}, …, A_{n} and suppose that A_{i_{1}} … A_{i_{k+1}} is an 𝔛-group for every choice of indices i_{1}, …, i_{k+1}. Let B_{i} = A_{i}A_{n+1} for i = 1, …, n − 1, then B_{i_{1}} … B_{i_{k}} = A_{i_{1}} … A_{i_{k}}A_{n} is an 𝔛-group for all indices 1 ≤ i_{1} ≤ … ≤ i_{k} ≤ n − 1. Since G = B_{1}B_{2} … B_{n−1} and the subgroups B_{i} permute, it follows from property P_{k} that G is an 𝔛-group.

The next lemma gives a sufficient criterion for a class of groups to have property P_{2}. Similar criteria can be obtained for the other properties P_{k}.

2.2 Lemma. Suppose that the class of groups 𝔛 satisfies the following property: whenever a 𝔙-group G has 𝔛-subgroups A, B and C such that G = AB = AC = BC, then G is an 𝔛-group. Then 𝔛 satisfies P_{2} for 𝔙-groups.

Proof. Suppose that the group G is the product of n ≥ 2 pairwise permutable subgroups A_{1}, …, A_{n} and that every product A_{i}A_{j} is an 𝔛-group. If n = 2, there is nothing to prove, so suppose that n > 2 and that the theorem holds for n − 1. Then A = A_{1} … A_{n−1} is an 𝔛-group, B = A_{2} … A_{n} is an 𝔛-group and also C = A_{1}A_{n} is an 𝔛-group. Moreover, G = AB = AC = BC and hence G belongs to the class 𝔛.

The following proposition describes the structure of a minimal counterexample to our theorems.

2.3 Proposition. Let 𝔙 be a class of finite soluble groups which is closed with respect to subgroups and epimorphic images and suppose that 𝔉 is a subgroup-closed formation. If k is a positive integer such that 𝔉 does not satisfy P_{k} for 𝔙-groups, then there exists a 𝔙-group G such that

(1) G is the product of its subgroups A_{1}, A_{2}, …, A_{k+1};

(2) G has a unique minimal normal subgroup N and G/N is an 𝔉-group;

(3) N is an elementary abelian p-group for some prime p and O_{p′}(G) = 1; moreover the Fitting subgroup F (G) of G is a p-group and F_{2}(G)/F (G) = F (G/F (G)) is a p′-group;

(4) for each integer i with 1 ≤ i ≤ k + 1, let Â_{i} = A_{1}…A_{i−1}A_{i+1}…A_{k+1}. If H is a proper subgroup of G containing Â_{i} for some index i, then H is an 𝔉-group; in particular Â_{i} is an 𝔉-group for every i;

(5) the subgroups Â_{1}N, …, Â_{k}N are proper 𝔉-subgroups of G;

(6) if 𝔉 is saturated, then N = C_{G}(N) = F (G). Moreover O_{p′}(Â_{i}N) = 1 and F (Â_{i}N) is a p-group for every i with 1 ≤ i ≤ k + 1;

Proof. (1) Since 𝔉 does not satisfy P_{k}, there exists a finite soluble group G which is the product of its subgroups A_{1}, A_{2}, …, A_{n}, such that A_{i_{1}}…A_{i_{k}} is an 𝔉-group for every choice of indices 1 ≤ i_{1} ≤ … ≤ i_{k} but G does not belong to 𝔉. If B_{i} = A_{i} for i ≤ k and B_{k+1} = A_{k+1}…A_{n}, then by induction on |G| + n, the product B_{i_{1}}…B_{i_{k}} is an 𝔉-group for every choice of indices 1 ≤ i_{1} ≤ … ≤ i_{k} ≤ k + 1. Therefore it suffices to consider the case when n = k + 1.

(2) By the minimality of G, every proper epimorphic image of G belongs to 𝔉. Since 𝔉 is a formation, it follows that G has a unique minimal normal subgroup N and G/N is an 𝔉-group.

(3) Since G is soluble, N is an elementary abelian p-group for some prime p. Clearly, F (G) = O_{p}(G) and F_{2}(G)/F (G) = F (G/F (G)) is a p′-group.

(4) Suppose that H is a proper subgroup of G which contains one of the subgroups Â_{i}. By the modular law, H = H ∩ Â_{i}A_{i} = Â_{i}(A_{i} ∩ H). Now A_{j}(A_{i} ∩ H) = A_{i}A_{j} ∩ H which is a subgroup of G for every j ≠ i. Therefore A_{i} ∩ H permutes with all A_{j} and so H is the product of A_{1}, …, A_{i−1}, A_{i} ∩ H, A_{i+1}, …, A_{k+1}. Since 𝔉 is subgroup-closed, H belongs to 𝔉 by induction on |G| + n.

(5) Suppose that G = Â_{i}N = Â_{j}N for two indices i ≠ j. Then Â_{i} and Â_{j} are maximal subgroups which complement N. Since Â_{i} and Â_{j} have trivial cores, it follows from a result of Ore (see e.g. [2, Theorems A, 16.1 and A, 16.2]) that Â_{i} and Â_{j} are conjugate, contradicting G = Â_{i}Â_{j}. So we may assume that Â_{i}N is a proper 𝔉-subgroup of G if 1 ≤ i ≤ k.

(6) If 𝔉 is saturated, we must have Φ(G) = 1. Therefore N = F (G) which contains C_{G}(F (G)). Since O_{p′}(Â_{i}N) centralizes N, it follows that F (Â_{i}N) = O_{p}(Â_{i}N).

3. Proof of Theorem A and Theorem B

Proof of Theorem A. If m = 0, the statement is trivial. Therefore suppose that m ≥ 1. Then 𝔑^{m}𝔉 = 𝔑(𝔑^{m−1}𝔉). By induction, the theorem holds for 𝔑^{m−1}𝔉. Thus, replacing 𝔑^{m−1}𝔉 by 𝔉, it suffices to consider the case m = 1. Since 𝔑𝔉 is the local formation defined by the formation function f, where f(p) = 𝔉 for all primes p, it follows that the class 𝔑𝔉 is saturated.

Assume that the theorem is false and let G be a counterexample as in Proposition 2.3. If Y/N = F (G/N), then Y/N is a nilpotent p′-group and G/Y is an 𝔉-group. Therefore Â_{i}N/(Â_{i}N ∩ Y) ≅ Â_{i}Y/Y is an 𝔉-group. Moreover, Â_{i}N/F (Â_{i}N) is also an 𝔉-group. It follows that Â_{i}N/(Y ∩ F (Â_{i}N)) is an 𝔉-group. Now Y ∩ F (Â_{i}N) is a p-subgroup of Y containing N for every i ≤ k + 1. Thus we have N = F (Â_{i}N) ∩ Y and Â_{i}N/N is an 𝔉-group for every i ≤ k + 1.

Now choose indices i_{1}, …, i_{k} ≤ k + 2, then there is an i ≤ k + 1 such that A_{i_{1}} … A_{i_{k}} is contained in Â_{i} and so A_{i_{1}} … A_{i_{k}}N/N is an 𝔉-group. Since 𝔉 satisfies P_{k}, it follows that G/N belongs to 𝔉 and so G is an 𝔑𝔉-group. This contradiction proves Theorem A.

Proof of Corollary A1. By Theorem A, the class 𝔑^{k} satisfies P_{k+1} for finite soluble groups. We have to show that 𝔑^{k} does not satisfy P_{k}:

Let p and q be distinct primes and put G_{0} = 1. For each nonnegative integer n, define G_{n+1} = C_{p} {∩ \atop ∪} G_{n} if n is odd and G_{n+1} = C_{q} {∩ \atop ∪} G_{n} if n is even. Moreover, let A_{n+1} be the base group of G_{n+1}. Identifying G_{n} with a subgroup of G_{n+1} for each n in the usual way, we obtain that for each n, G_{n} = A_{n} ⋊ A_{n−1} ⋊ ⋯ ⋊ A_{1}. Clearly, A_{i}A_{j} = A_{j}A_{i} for all i, j. Now for every choice of indices i_{1} … i_{n−1}, the subgroup A_{i_{1}}A_{i_{2}} … A_{i_{n−1}} belongs to the class 𝔑^{n−1}. But G_{n} does not belong to 𝔑^{n−1}.

Observe that Corollary A1 implies that the class 𝔑^{m}𝔉 in Theorem A need not satisfy P_{k+m−1}.

Theorem B is an immediate consequence of the following result.

3.1 Theorem. Let k be a positive integer and suppose that 𝔉 is a subgroup-closed formation satisfying P_{k} for 𝔙-groups. If π is a set of primes, then 𝔖_{π}𝔉 satisfies P_{k+1} for 𝔙-groups.

Proof. Assume that the theorem is false, and let the 𝔙-group G be a minimal counterexample as in Proposition 2.3. Note that the exponent p of N does not belong to π.

We show that Â_{i}N is an 𝔉-group for i ≤ k + 1. If Â_{i}O_{p}(G) < G, then O_{π}(Â_{i}O_{p}(G)) ≤ C_{G}(O_{p}(G)) ≤ O_{p}(G) since p does not belong to π. It follows that O_{π}(Â_{i}O_{p}(G)) = 1 and Â_{i}O_{p}(G) is an 𝔉-group. If Â_{i}O_{p}(G) = G but Â_{i}N < G, then Φ(G) ≠ 1 since otherwise N = F (G) = O_{p}(G). Thus N ≤ Φ(G) and O_{π}(G/N) = 1. It follows that G/N is an 𝔉-group. Therefore also Â_{i}N/N and Â_{i}N/O_{π}(Â_{i}N) belong to 𝔉. Since 𝔉 is residually closed, Â_{i}N is an 𝔉-group.

Let 1 ≤ i_{1} ≤ i_{2} ≤ … ≤ i_{k} ≤ k + 2. Then there is an index i ≤ k + 1 such that H = A_{i_{1}} … A_{i_{k}} is a subgroup of the 𝔉-group Â_{i}N and so also H is an 𝔉-group. Since 𝔉 satisfies P_{k} for 𝔙-groups, it follows that the 𝔙-group G belongs to the class 𝔉 which is contained in 𝔖_{π}𝔉. This contradiction proves Theorem 3.1.

Proof of Theorem B. Let π be the support of f, then

𝔉 = 𝔖_{π} ∩ ⋂_{p∈π} 𝔖_{p′}𝔖_{p}f(p)

(see e.g. [2, Theorem IV, 3.2(b)]). It follows from Theorem 3.1 that the classes 𝔖_{p′}𝔖_{p}f(p) satisfy P_{k+2} for 𝔙-groups for every p ∈ π. Also, 𝔖_{π} satisfies P_{1} for 𝔙-groups. Since intersections of classes satisfying P_{k+2} for 𝔙-groups likewise satisfy P_{k+2} for 𝔙-groups, this shows that 𝔉 satisfies P_{k+2} for 𝔙-groups.

If f is full, then 𝔖_{p′}𝔖_{p}f(p) = 𝔖_{p′}f(p) for every prime p. Therefore 𝔖_{p′}𝔖_{p}f(p) satisfies P_{k+1} for 𝔙-groups. By the above argument, 𝔉 has P_{k+1} for 𝔙-groups.

The following corollary shows in particular that the class 𝔉 in Theorem B need not have property P_{k+1}.

Proof of Corollary B1. Let f(p) = (𝔖_{π′}𝔖_{π})^{k−1}𝔖_{π′} for every p ∈ π and f(p) = 𝔖 otherwise. Then f is a local definition for (𝔖_{π′}𝔖_{π})^{k}𝔖_{π′}. By induction on k, it follows from Theorem B that (𝔖_{π′}𝔖_{π})^{k}𝔖_{π′} has property P_{2k+1}.

Now let p ∈ π and q ∈ π′ and set

G = C_{p} {∩ \atop ∪} C_{q} {∩ \atop ∪} … {∩ \atop ∪} C_{p} {∩ \atop ∪} C_{q} {∩ \atop ∪} C_{p}_{⟵2k+1 factors ⟶}.

Then G = A_{1} ⋊ A_{2} ⋊ … ⋊ A_{2k+1} where A_{i} is a p-group if i is odd and a q-group otherwise. Clearly Â_{i} belongs to (𝔖_{π′}𝔖_{π})^{k}𝔖_{π′} for every i but G is not contained in the class (𝔖_{π′}𝔖_{π})^{k}𝔖_{π′}.

Similarly, if f(p) = (𝔖_{π′}𝔖_{π})^{k−1} for every p ∈ π and f(p) = 𝔈 otherwise, then f is a local definition for (𝔖_{π′}𝔖_{π})^{k}. Therefore (𝔖_{π′}𝔖_{π})^{k} has P_{2k}. Let

G = C_{p} {∩ \atop ∪} C_{q} {∩ \atop ∪} … {∩ \atop ∪} C_{p} {∩ \atop ∪} C_{q}_{⟵2k factors ⟶}.

Then G = A_{1} ⋊ A_{2} ⋊ … ⋊ A_{2k} where A_{i} is a p-group if i is odd and a q-group otherwise. Moreover, the groups Â_{i} belong to the class (𝔖_{π′}𝔖_{π})^{k} for every i but G does not belong to (𝔖_{π′}𝔖_{π})^{k}.

On the other hand, one might ask whether the canonical local definition F of a saturated formation 𝔉 satisfies some P_{m} when 𝔉 has P_{k}. An argument similar to those in [2, Proposition IV, 3.16] yields:

3.2 Proposition. Suppose that 𝔉 is a saturated formation of finite soluble groups and let F be the canonical local definition of 𝔉. If k is a positive integer such that 𝔉 satisfies P_{k} for finite soluble groups, then F also satisfies P_{k}.

Proof. Let p a prime such that 𝔉 contains a group of order p and suppose that the group G is the product of its pairwise permutable subgroups A_{1}, …, A_{n}. Further, suppose that A_{i_{1}} … A_{i_{k}} belongs to F (p) for every choice of indices 1 ≤ i_{1} ≤ i_{2} ≤ … ≤ i_{k} ≤ n. Let W = C_{p} {∩ \atop ∪} G and N denote the base group of W. Put B_{i} = A_{i}N for every i, then B_{i_{1}} … B_{i_{k}} = A_{i_{1}} … A_{i_{k}}N is in the class 𝔖_{p}F (p) = F (p), which is contained in 𝔉 for every choice of indices 1 ≤ i_{1} ≤ i_{2} ≤ … ≤ i_{k} ≤ n. Since 𝔉 satisfies P_{k}, it follows that W is contained in 𝔉 and hence in the class 𝔖_{p′}𝔖_{p}F (p). Now O_{p′}(W) ≤ C_{W} (N) = N which shows that O_{p′}(W) = 1 and W even belongs to 𝔖_{p}F (p) = F (p). Therefore G = W/N belongs to the class F (p).

The next example shows that the formation function F in the above proposition need not satisfy P_{k−1}.

3.3 Example. Let p be a prime and k ≥ 1 an integer and set F (p) = 𝔖_{p}(𝔖_{p′}𝔖_{p})^{k−1} and F (q) = (𝔖_{p′}𝔖_{p})^{k} if q ≠ p. Then F is the canonical local definition of 𝔉 = (𝔖_{p′}𝔖_{p})^{k}. So by Corollary B1, both 𝔉 and F satisfy P_{2k} but not P_{2k−1}.

4. Proof of Theorem C and Corollary C1

Proof of Theorem C. We show first that 𝔉 satisfies P_{k+1} for metanilpotent groups. Assume that the theorem is false. Since every saturated formation of finite metanilpotent groups is subgroup-closed (see e. g. [2, Theorem IV.3.18]), there exists a minimal counterexample G as in Proposition 2.3.

The property P_{1}  7

If i ≤ k + 1, the 𝔉-group Â_{i}N belongs to the class 𝔖_{p′}𝔖_{p}f(p). Since O_{p′}(Â_{i}N) = 1 and G/N = F_{2}(G)/N is a p′-group, we have N = O_{p}(Â_{i}N). It follows that Â_{i}N/N is an f(p)-group for every i ≤ k + 1.

Let 1 ≤ i_{1} ≤ i_{2} ≤ … ≤ i_{k} ≤ k + 2. Then there exists an index i ≤ k + 1 which is not among the i_{j} and so A_{i_{1}} … A_{i_{k}}N/N ≤ Â_{i}N/N which belongs to f(p). Since f(p) satisfies P_{k} for finite nilpotent groups, it follows that the nilpotent group G/N is in the class f(p) and so G belongs to the class 𝔖_{p′}𝔖_{p}f(p). Since G/N ∈ 𝔉, the group G also belongs to 𝔖_{q′}𝔖_{q}f(q) for every q ≠ p. It follows that G is an 𝔉-group. This contradiction shows that 𝔉 satisfies P_{k+1} for metanilpotent groups.

The second statement of the theorem follows from the first statement because m ≥ 3 and so the group G is metanilpotent by Theorem A.

Proof of Corollary C1. If F is the canonical local definition of 𝔉, then F is closed with respect to subgroups and products of normal subgroups by [2, Proposition IV, 3.16]. Therefore this is also true for the local function f defined by f(p) = F (p)∩𝔑 for every prime p. Since f(p) is a class of nilpotent groups for every prime p, it is closed with respect to products of (subnormal) subgroups. This shows that f satisfies P_{1}. As in the proof of [2, Theorem IV, 3.18], f is a local definition of 𝔉, and so by Theorem C, 𝔉 satisfies P_{2} for metanilpotent groups and P_{3} for finite soluble groups.

Huppert’s result on finite supersoluble groups can now be deduced from Theorem C.

4.1 Corollary (Huppert [4, Satz VI.10.2]). The class 𝔘 of all finite supersoluble groups satisfies P_{3} for finite soluble groups.

Proof. The class 𝔘 of all supersoluble groups can be defined by the formation function u defined by u(p) = 𝔄(p − 1) for every prime p, where 𝔄(p − 1) is the class of all abelian groups of exponent dividing p − 1. It is easy to see that u satisfies P_{2} for all finite groups and so 𝔘 satisfies P_{3} for the class of all finite groups by Theorem C.

The following example shows that for every positive integer k, there is a subgroup-closed saturated formation of metanilpotent groups which does not satisfy P_{k} for metanilpotent groups.

4.2 Example. Let k > 0 be an integer and p and q distinct primes. Define G_{0} = C_{q} and G_{n} = G_{n−1} {∩ \atop ∪} C_{p} for every positive integer n. If G = G_{k+1}, then G = Q ⋊ P_{1} ⋊ P_{2} ⋊ … ⋊ P_{n} where Q = O_{q}(G) and the P_{i} are elementary abelian p-groups. For every i ≤ n, let A_{i} = QP_{i}, then A_{i}A_{j} = A_{j}A_{i} for all i, j and G_{n} = A_{1}A_{2}…A_{n}. Then A_{i_{1}}…A_{i_{k}} belongs to 𝔑𝔄^{k} for every choice of indices i_{1}, …, i_{k} but G does not belong to 𝔑𝔄^{k}.

5. The property P_{1}

There are not many subgroup-closed classes satisfying P_{1} for finite soluble groups.

5.1 Proposition. Suppose that 𝔉 is a class of finite soluble groups which satisfies P_{1} for finite soluble groups. If 𝔉 is subgroup-closed or saturated, then 𝔉 = 𝔖_{π} for some set of primes π.

Proof. Let π be the set of all primes p such that 𝔉 contains a cyclic group of order p. If 𝔉 is subgroup-closed, it is clear that every 𝔉-group is a π-group. If 𝔉 is saturated, this follows from [2, Theorem IV, 4.3].

Let G be a finite soluble π-group. By induction on the group order, we may suppose that every proper subgroup of G belongs to 𝔉. If G has two maximal subgroups L and M which are not conjugate in G, then G = LM by a theorem of Ore (see e.g. [2, Theorem A, 16.2]). Since L and M are 𝔉-groups, also G is an 𝔉-group by the property P_{1}. Therefore suppose that G has a unique maximal subgroup M. Then G is cyclic of prime-power order p^{n} and G/Φ(G) is cyclic of prime order p, hence is an 𝔉-group. Thus G belongs to 𝔉 if the class 𝔉 is saturated.

Suppose now that 𝔉 is subgroup-closed. By [2, Theorem A, 18.10], G can be embedded into a wreath product W of n cyclic groups of order p. Since the property P_{1} implies that 𝔉 is closed with respect to split extensions, W belongs to 𝔉. Thus G is also an 𝔉-group whenever 𝔉 is subgroup-closed.

To investigate classes of groups satisfying P_{1} for all finite groups, we need the following

5.2 Lemma. Let 𝔎, 𝔏 and 𝔐 be non-empty classes of groups which are closed with respect to subgroups and epimorphic images. In addition, assume that 𝔎 is extension-closed and contains all finite soluble groups. If not every product of an 𝔏-group and an 𝔐-group is a 𝔎-group, then there exists a group G = AB where A is an 𝔏-group and B is an 𝔐-group such that the following holds:

(1) G does not belong to 𝔎;

(2) G has a uniquely determined minimal normal subgroup N such that G is isomorphic with a subgroup of Aut N and G = AN = BN = AB.

(3) N = N_{1} × … × N_{k} where the N_{i} are isomorphic simple groups not belonging to the class 𝔎 and A and B act transitively on the set { N_{1}, N_{2}, …, N_{k} }.

(4) If H is a maximal subgroup of G containing A (or B), then H belongs to the class 𝔎 and

N ∩ H = (N_{1} ∩ H)(N_{2} ∩ H) … (N_{k} ∩ H);

for each i, the subgroup N_{i} ∩ H is conjugate to N_{1} ∩ H by an element of A (or B).

(5) If Ã and B̃ are maximal subgroups of G containing A and B, respectively, then |N_{1}| divides |OutN_{1}||N_{1} ∩ Ã||N_{1} ∩ B̃|.

Proof. (1) By hypothesis there exists a group G = AB which is the product of an 𝔏-group A and an 𝔐-group B but is not a 𝔎-group. Choose such a group G of minimal order.

(2) If N is a nontrivial normal subgroup of G, the factor group G/N = (AN/N)(BN/N) is a 𝔎-group. Thus if M and N are nontrivial normal subgroups of G with M ∩ N = 1, then G can be embedded into the direct product G/M × G/N which is a 𝔎-group. This contradiction shows that G has a unique minimal normal subgroup N which is not a 𝔎-group. Since every finite soluble group is in 𝔎, the soluble radicals of G and N are trivial. Therefore C_{G}(N) ∩ N = Z(N) = 1 and so C_{G}(N) = 1. It follows that G ≤ Aut(N).

If X = AN ∩ BN, then X = (A ∩ X)(B ∩ X) = (A ∩ X)N = (B ∩ X)N (see e.g. [1, Lemma 1.1.4]). Assume that X is a proper subgroup of G. Then X is an 𝔎-group and so N belongs to 𝔎, a contradiction. Hence G = AN = BN.

(3) Since N is a nonsoluble minimal normal subgroup of G, we have N = N_{1} × … × N_{k}, where the N_{i} are isomorphic minimal normal subgroups of N.

(4) If H is a maximal subgroup of G containing A, then H = A(H ∩ B) and so H is a 𝔎-group by the minimality of G. Let L = H ∩N. Since N = N_{1}×N_{2}×…×N_{k}, each element y of L can be written in a unique way as y = x_{1}x_{2} … x_{k}, where x_{i} ∈ N_{i} for 1 ≤ i ≤ k. Define a subgroup K_{i} of N by

K_{i} = { x_{i} | y = x_{1}x_{2} … x_{k} ∈ H ∩ N }.

If K_{i} = N_{i} for some i, then there is an epimorphism from H ∩ N to N_{i}. As H belongs to 𝔎, it follows that N_{i} is a 𝔎-group. Thus N is contained in 𝔎, a contradiction. Therefore K_{i} is a proper subgroup of N_{i} for every i. Since the K_{i} are permuted transitively by A, the direct product K = K_{1} × K_{2} × … K_{k} is an A-invariant subgroup of N. Hence also H = A(H ∩ N) normalizes K. As K_{i} is properly contained in N_{i}, the product KH is a proper subgroup of G. So the maximal subgroup H of G contains K and H ∩ N = K_{1} × K_{2} × … × K_{k}, where the K_{i} are conjugate to K_{1} and K_{i} = H ∩ N_{i}. This proves (4).

(5) We have N ∩ Ã = (N_{1} ∩ Ã) … (N_{k} ∩ Ã) and N ∩ B̃ = (N_{1} ∩ B̃) … (N_{k} ∩ B̃). The set G_{0} of all g ∈ G which normalize N_{i} for every i is a normal subgroup of G such that the factor group G/G_{0} is isomorphic with a subgroup of the symmetric group S_{k} of degree k. Furthermore, G_{0} can be embedded into Aut N_{1} × … × Aut N_{k}. Hence |G_{0}| divides |OutN_{1}|^{k}|N_{1}|^{k}. Since G = ÃB̃ = ÃN = B̃N, it follows that |G : G_{0}||G_{0} : N||N ∩ Ã||N ∩ B̃| = |N||Ã ∩ B̃|. Hence |N_{1}|^{k} divides

|N_{1} ∩ Ã|^{k}|N_{1} ∩ B̃|^{k}|OutN_{1}|^{k}|G : G_{0}|.

It is enough to prove that for each prime p dividing |N_{1}|, the k-th power p^{k} does not divide |G : G_{0}|. Since G/G_{0} ≤ S_{k}, the highest power of p dividing |S_{k}| is p^{m}, where m = [\frac{k}{p}] + [\frac{k}{p^{2}} ] + … < k. This proves (5).

It is obvious that for every set π of primes, the class of all finite π-groups satisfies P_{1} for the class of all finite groups. The next theorem exhibits another class of nonsoluble finite groups which is closed under products of subgroups.

5.3 Theorem. For every positive integer q_{0}, let S(q_{0}) be the class of all 3′-groups in which every nonabelian composition factor belongs to the class of Suzuki groups Sz(q), where q ≤ q_{0} is a power of 2. Then S(q_{0}) has property P_{1}.

Proof. By a result of Thompson the Suzuki groups are the only nonabelian simple 3′-groups; see [3, Corollary 7.3]. Moreover, by Suzuki [8, Theorem 9], the only nonsoluble subgroups of a Suzuki group Sz(2^{r}) are of type Sz(2^{s}), where s divides r. Therefore the class S(q_{0}) is closed with respect to subgroups, epimorphic images and extensions.

Let G = AB be a counterexample of minimal order as in Lemma 5.2 and choose the subgroups A and B as large as possible. Then A and B are maximal subgroups of G. Since the class S(q_{0}) is closed under extensions, it follows that N is not in S(q_{0}). This implies that the N_{i} are Suzuki groups Sz(q) with q > q_{0}. The order of N_{1} is

|N_{1}| = q^{2}(q − 1)(q^{2} + 1) = q^{2}(q − 1)(q − \sqrt{2q} + 1)(q + \sqrt{2q} + 1),

where q = 2^{r} and r is an odd number. Furthermore |Out(N_{1})| = r by [8, Theorem 11] and by [8, Theorem 9], the orders of the maximal subgroups of N_{1} are 2(q − 1), q^{2}(q − 1), 4(q−\sqrt{2q}+1), 4(q+\sqrt{2q}+1), and q_{1}^{2}(q_{1}−1)(q_{1}^{2}+1) where q_{1} = 2^{s} and s divides r. Comparing this list with Lemma 5.2 (5), we obtain a contradiction.

Remark. (1) In the same way, one can prove the following. Let S_{1}(r) be the subclass of S(q_{0}) consisting of all groups in S(q_{0}) whose nonabelian composition factors are isomorphic with a Suzuki group of type Sz(q) where q = 2^{s} for some integer s dividing r. Then the subgroup-closed formation S_{1}(r) has property P_{1} for the class of all finite groups.

(2) Let R(q) denote the class of all finite groups whose nonabelian conposition factors are simple Ree groups of type ^{2}G_{2}(q) and define Re(q_{0}) to be the class of all finite groups whose nonabelian composition factors are isomorphic with ^{2}G_{2}(q) or L_{2}(q) for some q ≤ q_{0}. Then the subgroup-closed formation Re(q_{0}) has property P_{1} with respect to R(q).

References

[1] Amberg, B., Franciosi, S. and de Giovanni, F.: Products of groups. Oxford University Press, Oxford (1992).

[2] Doerk, K. and Hawkes, T.: Finite soluble groups. de Gruyter, Berlin (1992).

[3] Glauberman, G.: Factorizations in local subgroups of finite groups. Reg. Conf. Ser. Math. 33 (1977).

[4] Huppert, B.: Endliche Gruppen I . Springer, Berlin (1967).

[5] Kazarin, L. S.: Factorizations of finite groups by soluble groups. Ukrainian Mat. J. 34 (1991), 947–950.

[6] Kazarin, L. S. and Kurdačenko, L. A.: Finiteness conditions and factorizations in infinite groups. Uspehi Mat. Nauk. 47 (1992), 75–114.

[7] Kegel, O.: On the solvability of some factorized linear groups. Illinois J. Math 9 (1965), 535–547.

[8] Suzuki, M.: On a class of doubly transitive groups. Ann. of Math. 75 (1962), 105–145.

Contact Address:

Prof. Dr. Bernhard Amberg Fachbereich Mathematik Universität Mainz 55099 Mainz Germany

                      e-mail: Amberg@mat.mathematik.uni-mainz.de