# Normal upper-hook fully normalized implies characteristic

## Statement

If $K \le H \le G$ are groups, with $K$ normal in $G$, and $H$ fully normalized in $G$, then $K$ is characteristic in $H$.

## Proof

Given: $K \le H \le G$ are groups, with $K$ normal in $G$, and $H$ fully normalized in $G$.

To prove: $K$ is characteristic in $H$.

Proof: Let $\sigma$ be any automorphism of $H$. We need to show that $\sigma(K) = K$.

Since $H$ is fully normalized in $G$, there exists $g \in G$ such that $\sigma$ equals conjugation by $g$. Since $K$ is normal in $G$, conjugation by $g$ leaves $K$ invariant, so $\sigma(K) = K$, completing the proof.