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Let $S_\omega$ be the group of all permutations of $\mathbb{N}$. Given subgroups $H$ and $G$ of $S_\omega$ with $H\cong G$, must $H$ and $G$ be conjugate? That is, must there exist a $p\in S_\omega$ such that $$G=\{p^{-1}\circ h\circ p: h\in H\}$$

Jason Block
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    What have you tried so far? – CyclotomicField Dec 03 '23 at 18:22
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    There are no non-trivial outer-automorphisms of $S_\omega$, so if an isomorphism between $H$ and $G$ can be extended to an automorphism of $S_\omega$, then $H$ and $G$ are conjugate. So I guess the question really comes down to if there exists isomorphic subgroups $H$ and $G$ such that no isomorphism can be extended to an automorphism of $S_\omega$. – Jason Block Dec 03 '23 at 18:34
  • @Jyrki This is an answer to the question, not just a comment. Please post your answer as an answer. This brings extra visibility to the answer. Also notice that comments are not indexed by the full text search and don't offer all the features that answers enjoy. Comments should only be used to clarify the problem. – Martin Brandenburg Dec 03 '23 at 18:35
  • @Martin Ok. I do my usual search first lest I post something that is a duplicate. Valid or not, suspicion of duplicates is a strong reason for me to post comments like that. – Jyrki Lahtonen Dec 03 '23 at 18:39
  • Ah, I understand. – Martin Brandenburg Dec 03 '23 at 18:42
  • The first search pointed at this famous example of non-conjugate copies of $A_5$ inside $S_6$. – Jyrki Lahtonen Dec 03 '23 at 18:44
  • The second search brought up this. A close match. – Jyrki Lahtonen Dec 03 '23 at 18:46

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No. Consider the subgroup of permutations that fix all odd numbers. This is isomorphic to the whole group, but not identical to it (and the whole group always conjugates only to itself).

Martin Brandenburg
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