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Let $G$ be a simple group and $H_1$ and $H_2$ be abelian subgroups of $G$ such that $H_1\cong H_2$ and $H_1,H_2\subseteq Cl_G(e_G)\cup Cl_G(x)$ for some $x\in G$, where $Cl_G(\cdot)$ denotes the conjugacy class on $G$.

Is it true that $H_1$ and $H_2$ are conjugate subgroups in $G?$ That is, is there is an element $g\in G$ so that $g^{-1}H_1g=H_2?$

Is it true that if $H_1$ and $H_2$ are finitely generated abelian subgroups?

pharazphazel
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  • There are finite counterexamples – Derek Holt Feb 13 '24 at 15:47
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    @DerekHolt such as? – pharazphazel Feb 13 '24 at 15:54
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    The question is very good. Why close it? I think there is a counterexample for infinitely generated groups. Consider the group $G=\sum_{k\in\mathbb{N}}\mathbb{Z}$. It has a continuum $\Delta$ pairwise isomorphic subgroups. We embed $G$ in a countable group $G^*$ in which any two nontrivial elements are conjugate (see here). It is clear that there are two non-conjugate subgroups of $\Delta$. – kabenyuk Feb 13 '24 at 16:57
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    @kabenyuk As I said, three are finite counterexamples but I am trying to maintain the conventions of the site by not answering questions completely lacking in context. – Derek Holt Feb 13 '24 at 17:06
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    Derek Holt, I agree with you that there is no context. But there are many such questions here. For example, the question I referred to. It's two lines. @pharazphazel please provide some context, such as how this problem arose and how you tried to solve it. In a nutshell. – kabenyuk Feb 13 '24 at 17:26
  • Is $e_G$ the identity element of $G$? If yes, the question is obviously true. – Brauer Suzuki Feb 13 '24 at 19:00
  • @BrauerSuzuki As I keep saying, there are finite counterexamples. – Derek Holt Feb 13 '24 at 19:45
  • Ah right, now I see it. Thanks. – Brauer Suzuki Feb 13 '24 at 20:19
  • Also the final sentence of the post doesn't make sense – Derek Holt Feb 13 '24 at 21:19

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As Derek indicated, there are counterexamples. Take $G=SL(2,16)$ and $x$ an involution. Then your union contains a full Sylow $2$-subgroup $P$, which is elementary abelian. Subgroups of $P$ are conjugate in $G$ if and only if they are conjugate in $N_G(P)$ by a theorem of Burnside. But there are too many subgroups of order $4$ to form a single conjugacy class (use the Gaussian binomial coefficient to compute their number).

Brauer Suzuki
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  • Yes that's exactly the example I had in mind. You get similar examples in ${\rm PGL}(2,p^n)$ for prime $p$ and large enough $n$. – Derek Holt Feb 14 '24 at 07:10