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Let $G$ be an infinite group. Suppose that $H, C, D \leq G$ where $C$ and $D$ are $H$-invariant.

If $d\in D$ such that $dHd^{-1}\leq HC$ and $dCd^{-1}=C$. I need to show that $d\in N_G(HC)$

Firstly, let $x\in d(HC)d^{-1}$ then $x =dhcd^{-1}$ where $h\in H$ and $c\in C$. Now $x= dhd^{-1}dcd^{-1} \in HC$ since $dhd^{-1} \in dHd^{-1}\leq HC$ and $dcd^{-1} \in d(HC)d^{-1} \leq HC$. Thus $d(HC)d^{-1} \leq HC$

I need to show the other inclusion $HC \leq d(HC)d^{-1}$ but I am running into problems.

R Maharaj
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  • Sorry I erroneously added that in there. I have edited my post to reflect what i was supposed to answer – R Maharaj Mar 05 '17 at 06:37
  • As was pointed out in comments to your previous question http://math.stackexchange.com/questions/2161979 this is not true in general in the case $C=1$ and $D=G$, so you will need some further assumptions. – Derek Holt Mar 05 '17 at 07:02
  • I think if $d \in D$ is arbitrary then $HC \leq d(HC)d^{-1}$ will be true? – R Maharaj Mar 05 '17 at 07:14
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    I don't understand what you mean. Try and state the problem clearly and unambiguously. At the moment you have have a sentence beginning "If $d \in D$" which is not a complete sentence. Are you saying that you want make these assumptions for all $d \in D$. If so then you need to use the words "for all". – Derek Holt Mar 05 '17 at 09:06

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