Consider a finite group $G$ and two conjugacy classes $H$ and $I$ of isomorphic subgroups of $G$.
Question. Is there some finite overgroup of $G$ which fuses $H$ and $I$ into a single conjugacy class?
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3$\begingroup$ You are asking about finite overgroups. But just to note that with HNN extension you can find an infinite overgroup of $G$ where $H$ and $I$ are conjugate. $\endgroup$– spinCommented Apr 29, 2021 at 1:19
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3$\begingroup$ If $H$ and $I$ are cyclic, embedding $G$ into $S_{|G|}$ with the regular permutation representation works. Would that work for any $H$ and $I$? $\endgroup$– spinCommented Apr 29, 2021 at 1:26
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4$\begingroup$ @spin Yes - $G$ is a union of $|G|/|H| = |G|/|I|$ orbits of $H$ with trivial stabilizer, and also the union of the same number of orbits of $I$ with trivial stabilizer. These are isomorphic, giving an element in $S_{|G|}$ conjugating one to the other. $\endgroup$– Will SawinCommented Apr 29, 2021 at 1:36
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$\begingroup$ $H,I$ are conjugacy classes of subgroups (hence are sets of subgroups), or are subgroups? $\endgroup$– YCorCommented Apr 29, 2021 at 5:38
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2$\begingroup$ Perhaps it's worth mentioning a less direct/constructive point of view that fits into a broader conceptual framework. As mentioned by @spin, the "universal" method of making $H$ and $I$ isomorphic produces the HNN extension $\Gamma=G*_{H\sim I}$. This is always infinite, but for $G$ finite it is not too hard to prove (using ideas similar to Will Sawin's comment) that $\Gamma$ is virtually free, and hence residually finite. Once you know $\Gamma$ is residually finite, it has a finite quotient into which $G$ embeds, and this is the group you want. $\endgroup$– HJRWCommented Apr 29, 2021 at 7:55
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1 Answer
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The answer is yes, as noted in the comments above by Will Sawin.
Let $H$ and $I$ be isomorphic subgroups of the finite group $G$.
Consider the regular permutation representation $G \leq \operatorname{Sym}(\Omega)$, where $\Omega = G$. Then for $H$ and $I$ the action on $\Omega$ splits into $[G:H] = [G:I]$ orbits. Furthermore, for both groups the action on every orbit is regular. So since $H$ and $I$ are isomorphic, their actions on $\Omega$ are equivalent, which means that they are conjugate in $\operatorname{Sym}(\Omega)$.
In general, for any group $G$, with HNN extension you can find an infinite overgroup $G^*$ of $G$ such that $H$ and $I$ are conjugate in $G^*$.
