Is there a finitely presented simple group with exactly 8 conjugacy classes of finite subgroups, which have the following respective isomorphism types?
- $C_1$
- $C_2$
- $C_3$
- $C_2^2$
- $C_6$
- $S_3$
- $A_4$
- $D_6$
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3$\begingroup$ @Satan'sMinion: While I agree that the question would benefit from some motivation, your comment is way out of line. $\endgroup$– Andy PutmanCommented Jan 21 at 12:41
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2$\begingroup$ Do you have any theoretical reason to believe there might? Or do you have a reason why you would like to see such a group to highlight some particular property? $\endgroup$– Geoff RobinsonCommented Jan 21 at 13:08
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$\begingroup$ @Andy Putman: Ok, I deleted my comment. $\endgroup$– Satan's MinionCommented Jan 21 at 14:47
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4$\begingroup$ The only f.p. simple groups I know about are either torsion-free (Burger-Mozes, Hyde-Lodha), contain unbounded torsion (Higman-Thompson, Roever-Nekrashevych, Brin-Thompson, etc etc), or have infinitely many torsion conjugacy classes (Caprace-Remy following Kac-Moody). So we don't even have an example with exactly $n$ conjugacy classes of finite subgroups for any $1<n<\infty$, much less with any prescribed isomorphism types. $\endgroup$– Matt ZaremskyCommented Jan 21 at 15:00
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2$\begingroup$ Are you aware of Osin's construction of groups with prescribed conjugacy classes? It is on the arXiv here. This can be applied to give finitely generated simple groups which contain each of the subgroups you wish, and such that all elements of the same order are conjugate. (So misses on a few counts, but suggests hope...) I suspect promoting the finite generation to finite presentability in his construction would be extremely difficult though. $\endgroup$– ADLCommented Jan 21 at 20:16
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