Normal upper-hook fully normalized implies characteristic

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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.

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Applications

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.