Groups with almost Frattini closed subgroups

A subgroup H of a group G is said to be Frattini closed in G if either H=G\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$H=G$$\end{document} or H is the intersection of all maximal subgroups of G containing H. The structure of a soluble group in which every subgroup is Frattini closed is known. In this paper, the behavior of a (generalized) soluble group G in which every subgroup is Frattini closed in a subgroup of finite index of G is studied. Among other results, it is proved that if G is a (generalized) soluble group and there exists a positive integer k such that every subgroup of G is Frattini closed in a subgroup of index at most k in G, then G contains a normal subgroup of finite index in which all subgroups are Frattini closed.


Introduction
Let G be a group, and let H be a subgroup of G. The A group in which all subgroups are Frattini closed is called an I M -group. The structure of soluble I M-groups was described by Menegazzo (1970). It turns out that if G is any soluble I M-group, then any subnormal subgroup of G is normal (see (Schmidt 1994), Lemma 3.3.9); moreover, G is periodic, the derived subgroup G of B Fausto De Mari fausto.demari@unina.it 1 Università degli Studi di Napoli "Federico II", 80126 Naples, Italy G is an abelian Hall subgroup and each non-trivial primary component of the abelian groups G and G/G has prime exponent (see (Schmidt 1994), Theorem 3.3.10).
A subgroup H of a group G is said to be nearly Frattini closed if the index | G (H ) : H | is finite. Groups in which all subgroups are nearly Frattini closed have been studied in de Giovanni and Imperatore (2014). It turns out that if G is an hyperabelian group with nearly Frattini closed subgroups, then G is soluble, G is finite and any subnormal subgroups of G has finite index in its normal closure; further, if there exists a positive integer k such that the index | G (H ) : H | is at most k for every subgroup H of G, then G contains normal subgroups N and K such that N ≤ K , N and |G : K | are finite and the factor K /N is an I M-group.
We shall say that a subgroup H of a group G is almost Frattini closed if there exists a subgroup K of finite index of G which contains H such that K (H ) = H (i.e., if H is Frattini closed in a subgroup of finite index of G), and that the group G is an AM-group (or that it has the AM-property) if every subgroup of G is almost Frattini closed. The aim of this paper is to investigate the behavior of (generalized) soluble AM-groups. In Section 2, among other properties, we prove that if G is an hyperabelian AM-group, then G contains a normal subgroup of finite index whose derived subgroup is abelian and periodic (so that G is soluble and metabelian-byfinite), and that any subnormal subgroup of G has finitely many conjugates. In Section 3, we will consider hyperabelian groups for which there exists a positive integer k such that each subgroup is Frattini closed in a subgroup of index at most k, and we will prove that such a group contains a normal subgroup of finite index which is an I M-group.
Most of our notation is standard and can be found in Robinson (1996). In particular, if G is any group then H G means that H is a maximal subgroup of G; moreover, π(G) denotes the set of all prime numbers p such that G has elements of order p.

General properties of AM-groups
If X is a class of groups, an hyper-X group is a group having an ascending normal series whose factors are X -groups. First step in our investigation is to establish the following.

Lemma 1 Let G be an hyper-(abelian or finite) AM-group and let H be any subgroup of finite index of G. If M is a maximal subgroup of H , then the index |G : M| is finite.
In particular, every maximal subgroup of G has finite index.

Proof
The core H G of H in G has finite index in G, so that it can be assumed that H G is not contained in M. Let {G α } α be an ascending normal series of G whose factors are either abelian or finite. Put L α = G α ∩ H G , then {L α } α is an ascending series of H G , consisting of G-invariant subgroups, whose factors are either finite or abelian. There is a first ordinal α such that L α is not contained in M, and clearly α cannot be a limit ordinal; hence L α−1 ≤ M. Since any factor of G has likewise the AM-property, replacing G by G/L α−1 , it can be assumed that L α−1 is trivial. Then H = L α M and L α is either abelian or finite. Since the index |H : M| is finite if L α is finite, we may suppose that L α is abelian. Let x ∈ L α M, then x G is abelian.
Let X be a subgroup of finite index of G containing x such that X ( x ) = x . Since x is subnormal of defect 2 in G, X ( x ) = x is a normal subgroup of X (see (de Giovanni and Imperatore 2014), Corollary 2.2) and hence |G : N G ( x )| is finite. Therefore x G satisfies the maximal condition. On the other hand, since M is a maximal subgroup of H = x G M, the factor x G / x G ∩ M is an abelian chief factor of H (see (Robinson 1996), 5.4.2) and hence it either is divisible or has prime exponent; thus x G / x G ∩ M is finite. Therefore |H : M| = x G : x G ∩ M is finite and hence also |G : M| is finite.
Clearly any factor of an AM-group is likewise an AM-group, and next result prove that also each normal subgroup of an hyper-(abelian or finite) AM-group has the AM-property.

Lemma 2 Let G be an hyper-(abelian or finite) AM-group and let N be any normal subgroup of G. Then N is an AM-group.
Proof Let H be a subgroup of N and let K be a subgroup of finite index of G containing H such that K (H ) = H . Clearly if H = K , then H has finite index in G and hence H is almost Frattini closed in N . Assume that H = K and note that in this case Let M be any maximal subgroup of K containing H . Then M has finite index in K by Lemma 1 and clearly N ∩ K is a normal subgroup of K , hence M ∩ N = M ∩ (N ∩ K ) is Frattini closed in N ∩ K (see (de Giovanni and Imperatore 2014), Corollary 2.4).

In any case, H is Frattini closed in
Each subnormal subgroup of an I M-group is normal (see (Schmidt 1994), Lemma 3.3.9). In order to prove the corresponding result for AM-groups, we need the following.

Lemma 3 Let G be a group whose maximal subgroups have finite index and let H be a Frattini closed subgroup of G. If H is subnormal in G, then H is normal in G.
Proof Clearly it can be supposed that H = G. Let M be any maximal subgroup of G which contains H , then M has finite index in G. Hence G/M G is finite and so, since

Theorem 4 Let G be an hyper-(abelian or finite) AM-group. Then every subnormal subgroup of G has finitely many conjugates.
Proof Let H be any subnormal subgroup of G. Since G is an AM-group, there exist a subgroup K of finite index of G such that K contains H and H = K (H ). Since all maximal subgroups of K have finite index by Lemma 1, if follows from Lemma 3 that H is normal in K . Hence K ≤ N G (H ) and so |G : Groups in which any subnormal subgroup has finitely many conjugates have been studied by Casolo (1989); it turns out that the structure of soluble groups with this property depends on the behavior of certain automorphism groups. Recall that a power automorphism of a group G is an automorphism which leaves invariant every subgroup of G. It is known that the set of all power automorphisms of G is an abelian residually finite subgroup of the group of all automorphisms of G (see (Cooper 1968)). As a consequence of results of Casolo (1989) we have the following.
Proposition 5 Let G be a group whose subnormal subgroups have finitely many conjugates. If G is hyperabelian, then G is soluble.
Proof The Fitting subgroup F of G is nilpotent (see (Casolo 1989), Lemma 3.1). Hence F/F is the Fitting subgroup of G/F , and replacing G by G/F , it can be supposed that F is abelian. There exist normal subgroups E and N of G with E ≤ F such that both N and C G (E) have finite index in G and N acts as a group of power automorphisms on F/E (see (Casolo 1989), Theorem 2.6). Let is likewise soluble. On the other hand, the factor N /C N (E) is soluble because it is finite and, since the group of all power automorphisms is abelian, N ≤ C N (F/E); hence N /N ∩ C is soluble. Therefore N is soluble. Thus G is soluble-by-finite and so even soluble.
It is known that any soluble I M-group is periodic (see (Schmidt 1994), Theorem 3.3.10). In contrast with this property, notice that the infinite dihedral group is a metabelian AM-group with non-periodic derived subgroup. In particular, it is not true that any soluble AM-group contains a normal I M-subgroup of finite index. However, we have the following result.

Theorem 6 Let G be an hyperabelian-by-finite AM-group. Then G contains a normal subgroup of finite index whose derived subgroup is abelian and periodic; in particular, G is metabelian-by-finite.
Proof The group G contains an hyperabelian normal subgroup of finite index G 1 which is likewise an AM-group by Lemma 2. Then Theorem 4 and Proposition 5 yield that G 1 is a soluble group whose subnormal subgroups have finitely many conjugates, and hence application Theorem 3.3 of Casolo (1989) to the subgroup G 1 proves the statement.

Lemma 7 Let G be an hyper-(abelian or finite) AM-group. Then G has no divisible abelian non-trivial normal subgroups.
Proof Any normal subgroup of an AM-group is likewise an AM-group by Lemma 2, hence the statement follows from the observation that any AM-group contains subgroups of finite index and so also maximal subgroups.
Let A = Dr n≥1 a n be a free abelian group of infinite rank. If p is any prime number and B = Dr n≥1 a p n n , then A/B is a p-group and it contains a proper basic subgroup (see for instance (Fuchs 2015), Lemma 6.1, p.174). Therefore A contains a non-trivial divisible factor and hence it follows from Lemma 7 that A cannot be an AM-group. We use this remark in order to derive the following consequence of Theorem 6.

Corollary 8 Let G be an hyperabelian-by-finite AM-group. If G is torsion-free, then G is an abelian-by-finite group of finite rank.
Proof It follows from Lemma 2 and Theorem 6 that G contains an abelian normal subgroup G 1 of finite index which is an AM-group; in particular, again as a consequence of Lemma 2, any section of the abelian AM-group G 1 has the AM-property. Therefore the previous remark gives that G 1 has finite rank and the proof is completed.

Lemma 9 Let G be an hyper-(abelian or finite) AM-group and let H ba a subgroup of finite index of G. If N is a normal subgroup of H , then (N ) is contained in (H ).
Proof Since any maximal subgroup of H has finite index by Lemma 1, the statement follows from a well-known fact (see for instance (de Giovanni and Imperatore 2014), Lemma 2.5). Abelian I M-groups are elementary abelian (see (Schmidt 1994), Theorem 3.3.10); recall here that it means that they are periodic abelian groups whose elements have square-free order. For abelian periodic AM-groups we have the following.

Lemma 11 Let G be a periodic abelian AM-group. Then G = E × V , where E has finite exponent, V is elementary abelian and π(E) ∩ π(V ) = ∅.
Proof By Lemma 10, there exists a subgroup of finite index N such that (N ) = {1}, so every primary component of N has prime exponent. Since G/N is finite, if π = π(G/N ), it follows that E = Dr p∈π G p has finite exponent. Moreover, if p is any prime outside π , then G p ≤ N and so G p = N p has exponent p. Therefore Using Corollary 8, Lemma 11 and an argument which is essentially the same of Theorem 3.5 of de Giovanni and Imperatore (2014), we can now prove the following.

Theorem 12 Let G be an abelian AM-group. Then G = A×E ×V , where A is torsionfree of finite rank, E has finite exponent, V is periodic with elements of square-free order and π(E) ∩ π(V ) = ∅.
Proof Let T be the subgroup consisting of all elements of finite order of G. It follows from Lemma 2 that all sections of an abelian AM-group are likewise AM-groups, hence Lemma 11 yields that T = E × V where E has finite exponent, V is elementary abelian and π(E) ∩ π(V ) = ∅. By Corollary 8, the torsion-free factor G/T has finite rank, so that there exists a finitely generated torsion-free subgroup U such that G/U is periodic. Again Lemma 11 gives that where L/U has finite exponent, W /U is elementary abelian and π(L/U )∩π(W /U ) = ∅. Let π = π(E) ∪ π(L/U ), then π is finite. If G π is the subgroup consisting of all π -elements of T , then G π is a pure subgroup of finite exponent and hence G = Y ×G π for a suitable subgroup Y (see (Robinson 1996), 4.3.8).
and hence U 0 is torsion-free. Moreover, where H /U 0 has finite exponent which divides the exponent of L/U (and so is a π -number) and K /U 0 is elementary abelian.
where A is torsion-free. Hence application of Corollary 8 yields that A has finite rank and so the theorem is proved.
Finally notice that any locally nilpotent AM-group is nilpotent, and it is close to be abelian.

Proposition 13 Let G be a locally nilpotent AM-group, then G is central-by-finite. In particular, if G is torsion-free, then G is abelian of finite rank.
Proof Any maximal subgroup of a locally nilpotent group is normal (see (Robinson 1996), 12.1.5), so that in a locally nilpotent group any Frattini closed subgroup is normal. Therefore any subgroup of G is normal in a subgroup of finite index, hence G is central-by-finite by a well-known result by B.H. Neumann (see (Neumann 1955)). It follows that if G is torsion-free, then G is abelian (see (Robinson 1996), 5.2.19) and G has finite rank by Lemma 8.

Groups with the bounded AM-property
Let G be a group and let k be a positive integer. We shall say that G is a AM k -group if any subgroup H of G is Frattini closed in a subgroup K such that |G : K | ≤ k. Clearly, if N is a normal subgroup of G and G is an AM k -group, also G/N is an AM k -group.

Lemma 14 Let G be an hyper-(abelian or finite) AM k -group and let N be any normal subgroup of G. Then N is an AM k -group.
Proof Let H be a subgroup of N and let K be a subgroup of G containing H such that K (H ) = H and |G : K | ≤ k. As in the proof of Lemma 2, it can be proved that H is Frattini closed in N ∩ K . Since |N : N ∩ K | ≤ |G : K | ≤ k, the lemma is proved.
A group G will be called B AM-group if it is an AM k -group for some positive integer k. Notice that if C is an infinite cyclic group and k is any positive integer, then C is not an AM k -group. In fact, if p is any prime with p > k, then C p is the unique subgroup in which C p 2 is Frattini closed and |C : C p | = p > k. Therefore the additive group of integers does not have the B AM-property.

Lemma 15 Let G be an hyper-(abelian or finite) B AM-group. Then G is periodic.
Proof Let A and B be normal subgroups of G such that B ≤ A and A/B is abelian, then Lemma 14 yields that A is a B AM-group and so A/B has likewise the B AMproperty. Since A/B is abelian, again Lemma 14 yields that all subgroups of A/B have the B AM-property and so A/B does not have infinite cyclic subgroups, thus A/B is periodic. Since G is hyper-(abelian or finite), it follows that G is periodic.
Recall that a group G is said to be a T -group if all subnormal subgroups of G are normal. Clearly, every simple group is a T -group, but the structure of soluble T -group was investigated by Gaschütz and Robinson (see (Robinson 1964)). If G is any soluble T -group then G is abelian; moreover, if G is periodic and L = [G , G], then there are no odd primes in the set π(L) ∩ π(G/L) and the 2-component L 2 of L is divisible.
The Wielandt subgroup ω(G) of a group G is defined to be the intersection of all the normalizers of subnormal subgroups of G. Clearly, ω(G) is a T -group, and G is a T -group if and only if it coincides with its Wielandt subgroup; thus the size of G/ω(G) can be considered as a measure of the distance of the group G from the property T . Moreover, if G is a group such that G/ω(G) is finite, it is clear that each subnormal subgroup of G has only finitely many conjugates; conversely, Casolo proved that if a soluble group G has boundedly finite conjugacy classes of subnormal subgroups, then the Wielandt subgroup ω(G) has finite index in G (see (Casolo 1989), Theorem 4.8).

Theorem 16 Let G be an hyperabelian-by-finite B AM-group. Then G contains a normal subgroup of finite index which is a soluble I M-group.
Proof The group G contains an hyperabelian normal subgroup G 1 of finite index which is a B AM-group by Lemma 14. Then G 1 is soluble by Theorem 6 and periodic by Lemma 15. If we prove that G 1 contains a normal subgroup G 2 of finite index with the I M-property, then the core of G 2 in G is a normal subgroup of finite index of G which is also an I M-group (see (Schmidt 1994), Lemma 3.3.7). Therefore there is no loss of generality if we replace G by G 1 and so it can be assumed that G itself is soluble and periodic. Let k be a positive integer such that G is an AM k -group. Let H be any subnormal subgroup of G, and let K be a subgroup of G such that H = K (H ) and |G : K | ≤ k. Since all maximal subgroups of K have finite index by Lemma 1, if follows from Lemma 3 that H is normal in K . Hence K ≤ N G (H ) and so |G : N G (H )| ≤ k. Therefore G has boundedly finite conjugacy classes of subnormal subgroups and hence the factor G/ω(G) is finite by the above quoted result by Casolo. Lemma 10 yields that G contains a normal subgroup N of finite index such that Then W is a normal subgroup of the T -group ω(G) and hence W is likewise a T -group; notice also that W has trivial Frattini subgroup since (W ) ≤ (N ) = {1} by Lemma 9. Moreover, W is also a B AM-group by Lemma 14 and hence it cannot contain divisible abelian non-trivial normal subgroups by Lemma 7. Therefore, if L = [W , W ], it follows from the structure of T -groups that L is abelian and π(L) ∩ π(W /L) = ∅ (see (Robinson 1964)). Moreover, if X is any subgroup of L, then X is normal in W and so (X ) ≤ (W ) = {1} by Lemma 9; hence L is elementary abelian. Let I be a subgroup of finite index of W containing L such that L = I (L). Then (I /L) is trivial, and I /L has the B AM-property by Lemma 14. Being a nilpotent T -group, all subgroups of W /L are normal, hence if I /L were not abelian then it would contain a (normal) subgroup Q/L isomorphic to the quaternion group of order 8 (see (Robinson 1996), 5.3.7) and this is not possible being (Q/L) ≤ (I /L) = {1} by Lemma 9. Therefore I /L is abelian. Moreover, if X /L is any subgroup of I /L, then (X /L) ≤ (I /L) = {1} by Lemma 9; hence I /L is elementary abelian. It follows that I is an I M-group (see (Schmidt 1994), Theorem 3.3.10). Since the index |G : I | is finite, the theorem is proved.
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