Frattini subgroup is ACIC

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 $$H$$.

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
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
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.

To Frattini-embedded normal subgroups
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
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 $$G$$.