Characteristic Subgroup of Normal Subgroup is Normal
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Theorem
Let $G$ be a group.
Let $N\leq G$ be normal.
Let $H\leq N$ be characteristic.
Then $H$ is normal in $G$.
Proof
Let $g \in G$.
Because $N$ is normal, conjugation by $g$ is an automorphism of $N$.
Because $H$ is characteristic in $N$, $g H g^{-1} = H$.
Thus $H$ is normal in $G$.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Group Homomorphism and Isomorphism: $\S 64 \epsilon$