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If $K$ is a normal subgroup of $H$, and $H$ is a normal subgroup of $G$, is it true that $g^{-1} K g$ is a normal subgroup of $H$ for all $g \in G$?

I think I know the answer, but I just want to double check! Thanks.

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up vote 6 down vote accepted

Yes. First note that since $H$ is normal in $G$, $g^{-1}Kg$ is certainly a subgroup of $H$. Now suppose $h \in H$, and consider $h^{-1}g^{-1}Kgh$. Since $H$ is normal in $G$, we have $gh=lg$ for some $l \in H$. But then $h^{-1}g^{-1}Kgh$=$g^{-1}l^{-1}Klg$. Since $K$ is normal in $H$ and $l$ is in $H$, we have $l^{-1}Kl=K$, and so $h^{-1}g^{-1}Kgh$=$g^{-1}l^{-1}Klg=g^{-1}Kg$, as desired.

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Great. Thanks Chris. – Mark Apr 25 '11 at 21:14

Yes. $h^{-1}(g^{-1}Kg)h = (h^{-1}g^{-1})K(gh) = g^{-1}u^{-1}Kug = g^{-1}Kg$, for some $u\in H$ since $H$ is normal in $G$.

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