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Jul 2, 2019 at 9:12 comment added IJL Without putting any constraint on the finite group, problems like this can always be solved. If $G$ is any finite group, and $G\rightarrow S$ is the embedding of $G$ into the group of all permutations of the set $G$, then any isomorphism between subgroups of $G$ is a conjugation inside $S$. I'm not sure whether $S_{16}$ is simpler than Derek's $PSL(2,15)$ as a non 2-group solution to the original question. This doesn't help with Geoff's question.
Jun 29, 2019 at 20:13 comment added Geoff Robinson Thanks @RichardLyons . I had forgotten that result of John's (if I ever knew it). So it follows that if $S$ is a $2$-group which embeds in a finite solvable group with a single class of involutions, then all involutions of $S$ commute with each other.
Jun 28, 2019 at 12:40 comment added Richard Lyons @Geoff: If I recall correctly, Thompson has proved that a solvable group with one class of involutions has $2$-length one.
Jun 28, 2019 at 11:22 comment added Geoff Robinson The question, and Derek's elegant answer and example me wonder whether it is possible to give a reasonable answer to "Which 2-groups S can be embedded in a finite group G with one conjugacy class of involutions?" More ambitiously, one might try to insist that $G$ be solvable. I think these are different questions ( or give different answers anyway): I think a semidihedral $2$-group of order 16 embeds in $M_{11}$ which has one class of involutions, but not in any finite solvable group with one class of involutions.
Jun 28, 2019 at 9:41 history edited Derek Holt CC BY-SA 4.0
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Jun 28, 2019 at 0:38 comment added Nicholas Kuhn Also thanks for the composite order group example!
Jun 27, 2019 at 23:44 vote accept Nicholas Kuhn
Jun 27, 2019 at 23:44 comment added Nicholas Kuhn Thanks for the friendly proof! I really have a more general question of the form "Can this happen?" and this question here was my attempt to find an example. If I can't easily resolve my general question with arguments in the style of what you did, I'll be back with more!
Jun 27, 2019 at 22:04 history edited verret CC BY-SA 4.0
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Jun 27, 2019 at 21:58 history answered Derek Holt CC BY-SA 4.0