Frattini subgroup is ACIC

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This fact is related to the problem of realization related to the following subgroup-defining function: Frattini subgroup
Realization problems are usually about which groups can be realized as subgroups/quotients related to a subgroup-defining function.
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Statement

Verbal statement

The Frattini subgroup of a finite group (or more generally, of a group where every proper subgroup is contained in a maximal subgroup)is an ACIC-group.

Symbolic statement

Let G be a finite group (or more generally, a group where every subgroup is contained in a maximal subgroup), and H = \Phi(G) be its Frattini subgroup. Then, H is an ACIC-group: whenever K is an automorph-conjugate subgroup of H, K is characteristic in H.

Definitions used

Frattini subgroup

Further information: Frattini subgroup

The Frattini subgroup of a group is defined as the intersection of all its maximal subgroups.

Group where every subgroup is contained in a maximal subgroup

For infinite groups, it could happen that there are no maximal subgroups, and it could also happen that not every proper subgroup is a maximal subgroup. The proof we give here does not work for such groups. We require our group to have the property that every proper subgroup is contained in a maximal subgroup.

This property is satisfied by finite groups, and more generally, by slender groups (the equivalent in group theory of Noetherian ring: a group where every subgroup is finitely generated).

Automorph-conjugate subgroup

Further information: automorph-conjugate subgroup

A subgroup K in a group H is automorph-conjugate if for any automorphism \sigma of H, K and \sigma(K) are conjugate.

Any characteristic subgroup is automorph-conjugate.

ACIC-group

Further information: ACIC-group

An ACIC-group is a group in which every automorph-conjugate subgroup is characteristic. Equivalently, every automorph-conjugate subgroup is normal (The two ar eequivalent because normal automorph-conjugate subgroups are characteristic).

Facts used

The proof is a generalized version of Frattini's argument.

Generalizations

To Frattini-embedded normal subgroups

Further information: Frattini-embedded normal-realizable implies ACIC The result generalizes to the following result for arbitrary groups (with no finiteness assumptions): any Frattini-embedded normal subgroup of a group is ACIC. A Frattini-embedded normal subgroup is a normal subgroup whose product with any proper subgroup is proper.

Proof

Given: G a group with the property that every proper subgroup is contained in a maximal subgroup, and H = \Phi(G). K is an automorph-conjugate subgroup of H.

To prove: K is a normal subgroup of H

Proof: We will in fact show that K is normal in G. Normality in H will then follow.

Frattini's argument step

Further information: Frattini's argument

We first show that HN_G(K) = G. This part is called the Frattini's argument. It only uses the hypothesis that K is automorph-conjugate in H and H is normal in G.

Suppose g \in G is any element. Then, since H is normal in G, conjugation by g restricts to an automorphism of H. Thus, K and gKg^{-1} are related via an automorphism in H.

Using the hypothesis that K is automorph-conjugate inside H, we see that there exists h \in H such that gKg^{-1} = hKh^{-1}. Hence, the element h^{-1}g lies in the normalizer N_G(K). Rearranging, we see that any element of G can be written as a product of an element of H and an element of N_G(K). So G = HN_G(K).

Clinching step

This part uses the assumption we made about G (that every proper subgroup is contained in a maximal subgroup) and the fact that H is the Frattini subgroup of G.

We assume that N_G(K) \ne G and obtain a contradiction. If N_G(K) \ne G, i.e. it is a proper subgroup, then by hypothesis it is contained in some maximal subgroup M. By definition of Frattini subgrou, H = \Phi(G) \le M. So HN_G(K) \le M, yielding G \le M, a contradiction.

Thus, N_G(K) = G, hence K is normal in G.