Normal upper-hook fully normalized implies characteristic

Statement

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

Related facts

Other related facts

• Characteristic upper-hook AEP implies characteristic
• Fully invariant upper-hook EEP implies fully invariant

Applications

• Left transiter of normal is characteristic can be derived from the statement of this page, along with the fact that every group is normal fully normalized in its holomorph.
• Characteristically simple and normal fully normalized implies minimal normal

Proof

Given: $K\leq H\leq G$ 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.