# Frattini-embedded normal in subgroup and normal implies Frattini-embedded normal

## Statement

Suppose $H \le K \le G$ are groups, such that:

• $H$ is a normal subgroup of $G$
• $H$ is a Frattini-embedded normal subgroup of $K$. In other words, $H$ is normal in $K$ and for any proper subgroup $M$ of $K$, $HM$ is a proper subgroup

Then $H$ is a Frattini-embedded normal subgroup of $G$.

## Definitions used

### Frattini-embeddded normal subgroup

Further information: Frattini-embedded normal subgroup

A subgroup $H$ of a group $G$ is termed Frattini-embedded normal in $G$ if $H \triangleleft G$ and, for any proper subgroup $N$ of $G$, $HN$ is proper.

## Proof

Given: $H \le K \le G$ are groups, such that:

• $H$ is a normal subgroup of $G$
• $H$ is a Frattini-embedded normal subgroup of $K$. In other words, $H$ is normal in $K$ and for any proper subgroup $M$ of $K$, $HM$ is a proper subgroup

To prove: For any proper subgroup $N$ of $G$, $HN$ is a proper subgroup of $G$

Proof: Pick a proper subgroup $N$ of $G$. We need to show that $HN$ is proper in $G$. Suppose $HN = G$. We'll derive a contradiction.

We have $HN \cap K = K$, which, by the modular property of groups, gives: $H(N \cap K) = K$

Now, clearly, $H$ is not contained in $N$ (otherwise $HN = N \ne G$), and since $H \le K$, $K$ is not contained in $N$. So $N \cap K$ is a proper subgroup of $K$, and thus we have found a proper subgroup of $K$ whose product with $H$ equals $K$. This contradicts the assumption that $H$ is a Frattini-embedded normal subgroup of $K$.