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Suppose the $G$ is a group and that $H$, $D$ are subgroups of $G$. If $d\in D$ such that $dHd^{-1}\leq H$, Can we conclude that $d \in N_D(H)$?

R Maharaj
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    No, you can't. For that you need $dHd^{-1} = H$. Of course that would follow if $H$ was finite since $|dHd^{-1}| = |H|$, but not in general. – Derek Holt Feb 26 '17 at 10:10
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    This is essentially a duplicate of http://math.stackexchange.com/questions/107862 – Derek Holt Feb 26 '17 at 11:38
  • If $dHd^{-1}<H$ then there exists $h_1,h_2 \in H$ and a $d \in G$ such that $dh_1=dh_2$. But then this implies $h_1=h_2$. Therefore only equality can exist. – marshal craft Feb 26 '17 at 23:15

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