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It can be proven (e.g., here) that permutations in $S_n$ are conjugate iff they have the same cycle structure. However, the same theorem does not, in general, hold for subgroups of $S_n$ (see this).

Can we make any statements about the cycle structure of elements of conjugacy classes for general permutation groups? I expect that having the same cycle structure is necessary but not sufficient for two permutations to be conjugate. Stated another way: if $G$ is a subgroup of $S_n$, then any conjugacy class of $G$ must be a subset of a conjugacy class of $S_n$. This seems sensible since removing elements from $S_n$ to form $G$ should not allow permutations of different cycle structure to be conjugate but it can make elements that were previously conjugate no longer so since the element that allowed two elements to be conjugate may not be present in $G$.

Is this correct? Can we say anything more? Is my explanation above enough of a proof, or is more required?

MattHusz
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More generally, if $G$ is a subgroup of some group $S$, the conjugacy class in $G$ of any element $x\in G$ is obviously a subset of its conjugacy class in $S$: $$G\subset S\implies cl_G(x)=\{y^{-1}xy\mid y\in G\}\subseteq\{y^{-1}xy\mid y\in S\}=cl_S(x).$$ This inclusion may be strict and for some elements $x$, and may be an equality for other elements $x$: see for instance Find the conjugacy classes of $A_5$.

Anne Bauval
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  • See also Cayley's theorem for relating this statement (about general groups) to the statement about subgroups of the symmetric group in the question. Thanks for pointing this out, I had not made the connection to general groups. – MattHusz Jan 02 '25 at 16:03
  • My generalization of your post (replacing your $S_n$ by some group $S$) is not related to Cayley's theorem: though $G\subset S\subset S_m$ for some $m$, we are only interested in $cl_G(x)$ and $cl_S(x)$. – Anne Bauval Jan 02 '25 at 16:14
  • You're right, my mistake. – MattHusz Jan 02 '25 at 16:49