# Frattini subgroup is ACIC

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 $G$.