# Frattini subgroup is normal-monotone

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This fact is an application of the following pivotal fact/result/idea: characteristic of normal implies normal
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## Statement

### Verbal statement

The Frattini subgroup of any normal subgroup is contained in the Frattini subgroup of the whole group, provided the normal subgroup is a group in which every proper subgroup is contained in a maximal subgroup.

(Note that this group property is always satisfied when the normal subgroup is a finite group, so for finite groups, the Frattini subgroup of a normal subgroup is always contained in the Frattini subgroup of the whole group).

### Statement with symbols

Let $N$ be a normal subgroup of a group $G$, where $N$ satisfies the property that every proper subgroup is contained in a maximal subgroup. Then, $\Phi(N)$, the Frattini subgroup of $N$, is contained in $\Phi(G)$, the Frattini subgroup of $G$.

### Property-theoretic statement

The subgroup-defining function that sends a group to its Frattini subgroup is a normal-monotone subgroup-defining function (with some assumptions on the nature of the groups).

## Definitions

### Frattini subgroup

The Frattini subgroup of a group is the intersection of all its maximal subgroups. In other words, an element is in the Frattini subgroup if it is in every maximal subgroup.

## Proof

### Proof outline

The proof uses four facts (for convenience, we denote the group by $G$ and normal subgroup by $N$):

1. Characteristic subgroups of normal subgroups are normal: This fact helps us show that the Frattini subgroup of the subgroup $N$, is in fact normal in the whole group $G$
2. For a group where every proper subgroup is contained in a maximal subgroup, the Frattini subgroup is a Frattini-embedded normal subgroup: a normal subgroup whose product with any proper subgroup is proper. Thus $\Phi(N)$ is a Frattini-embedded normal subgroup inside $N$.
3. Frattini-embedded normal in subgroup and normal implies Frattini-embedded normal: This allows us to show that the Frattini subgroup of the subgroup, is Frattini-embedded normal inside the whole group. In other words, $\Phi(N)$ is a Frattini-embedded normal subgroup inside $G$
4. For any group, any Frattini-embedded normal subgroup is contained inside the Frattini subgroup: This allows us to conclude that the Frattini subgroup of

### Hands-on proof

Given: A group $G$,a normal subgroup $N$ of $G$ with the property that any proper subgroup of $N$ is contained in a maximal subgroup

To prove: $\Phi(N) \le \Phi(G)$

Proof: The Frattini subgroup $\Phi(N)$ is a characteristic subgroup of $N$. Since every characteristic subgroup of a normal subgroup is normal, $\Phi(N)$ is a normal subgroup of $G$.

We need to show that $\Phi(N)$ is contained in $\Phi(G)$. For this, it suffices to show that $\Phi(N)$ is contained in every maximal subgroup of $G$. We prove this by contradiction.

Suppose $M$ is a maximal subgroup of $G$ not containing $\Phi(N)$. Then, the subgroup generated by $M$ and $\Phi(N)$ is the whole of $G$. Since $\Phi(N)$ is normal in $G$, $M\Phi(N) = G$. From that, it follows that $\Phi(N)(M \cap N) = N$ (this is a particular example of the modular property of groups).

Since $M$ does not contain $\Phi(N)$, $M$ does not contain $N$ either and hence $M \cap N$ is a proper subgroup of $N$. Since every proper subgroup is contained in a maximal subgroup of $N$, there is a maximal subgroup of $N$ containing both $M \cap N$ and $\Phi(N)$ . This contradicts the fact that their product is $N$.

Normality of $N$ is thus crucial because it guarantees normality of $\Phi(N)$. This in turn is crucial in converting a subgroup-generated statement to a product of subgroups statement.