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Given a group $G$ and $K<G$. Let $H_1,H_2$ be subgroups in $G$ satisfying:

\begin{align} &H_1\triangleleft K;\\ &H_2\triangleleft K;\\ &H_1\sim_G H_2.\quad\mbox{(They are conjugate to each other in $G$)} \end{align}

Is it true that

\begin{align} H_1\sim_{N_G(K)}H_2, \end{align}

where $N_G(K)$ denotes the normalizer of $K$ in $G$? Or is there a counterexample?
Any idea would be appreciated.

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$G=S_6$, $H_1=\langle (1,2,3) \rangle$, $H_2=\langle (4,5,6) \rangle$, $K = \langle (1,2,3),(4,5,6),(4,5) \rangle \cong C_3 \times S_3$.

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  • $\begingroup$ BTW, in case you have not seen, there is now a chatroom (chat.stackexchange.com/rooms/19567/finite-group-theory) for finite group theory in case you are interested (sorry for the somewhat off-topic comment). $\endgroup$ Jan 19, 2015 at 9:25
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    $\begingroup$ It would be interesting to explain how this was conceived. =) $\endgroup$
    – Pedro
    Jan 19, 2015 at 9:32
  • $\begingroup$ @PedroTamaroff Is your comment referring to my answer or to Tobias Kildetoft's comment? $\endgroup$
    – Derek Holt
    Jan 19, 2015 at 10:26
  • $\begingroup$ @DerekHolt Your answer. $\endgroup$
    – Pedro
    Jan 19, 2015 at 10:36
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    $\begingroup$ Well I knew the answer was going to be no (since there is no reason I could think of why it should be true), so it was just a matter of thinking of an example. I decided I would look for a subgroup $H_1 \times H_2$ of a group $G$ with $H_1$ and $H_2$ conjugate in $G$, and then try and embed $H_2$ normally in a bigger group $K_2$ that still commuted with $H_1$. Then $H_1$ and $H_2$ would be normal in $K=H_1 \times K_2$ but might not be conjugate in the normaliser of $K$, because we have distorted one of the two similar looking subgroups. $\endgroup$
    – Derek Holt
    Jan 19, 2015 at 11:35

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