Let $G$ be a group, $H$ and $K$ subgroups of $G$ with $H$ a normal subgroup of $K$. The normalizer of $H$ in $G$ is $$N_G(H)=\{g\in G | gHg^{-1}=H\}.$$ Prove that K is a subgroup of $N_G(H)$.

Since $H$ is normal in $K$, then $gHg^{-1}\in K$. Does this fact help me? How can I prove it's a subgroup of the normalizer of H in G? Need a plan for this proof.

  • $\begingroup$ Let $k \in K$. Where is $kHk^{-1}$? $\endgroup$ – Eric Towers Nov 25 '15 at 21:19

You know $K$ and $N_G(H)$ are both subgroups of $G$, so all you need to do is to show that $K \subseteq N_G(H)$. If $k\in K$, then since $H$ is normal in $K$, we have $kHk^{-1}=H$. This is exactly what needs to be true for $k\in N_G(H)$.

  • $\begingroup$ What do I need to do to show $K\subseteq N_G(H)$? $kHk^{-1}\in H$? $\endgroup$ – maidel b Nov 25 '15 at 23:12
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    $\begingroup$ You take a $k$ from the set $K$ and show that it is in $N_G(H)$. This is exactly what I did above is prove $K \subseteq N_G(H)$. $\endgroup$ – CPM Nov 25 '15 at 23:17

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