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Let $G$ be a group with subgroups $H$ and $K$. Is it true that the normal closure of $H$ in $K$, $H^K = \langle aHa^{-1} | a \in K \rangle$, is a subgroup of $K$?

I can see this is true if $H \leq K$ but does it hold in general?

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    $\begingroup$ I can very well imagine it not being true if $G$ is the internal direct product of $H$ and $K$ $\endgroup$ – tehjh Mar 10 '17 at 10:44
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Counterexample

Let $G$ be any abelian group. Then $H^{K}=\langle H\rangle =H$ which is not necessarily a subgroup of $K$.

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  • $\begingroup$ Great! it makes sense. Suppose that for your case $G$ is non-abelian and $H^K = H$, would it be correct to say that $K \leq N_G(H)$? $\endgroup$ – R Maharaj Mar 10 '17 at 10:58
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    $\begingroup$ @RMaharaj Yes you are right. $\endgroup$ – Alan Wang Mar 10 '17 at 11:01

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