Normal upper-hook fully normalized implies characteristic

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Statement

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

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Proof

Given: KHG are groups, with K normal in G, and H fully normalized in G.

To prove: K is characteristic in H.

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

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