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This may be easier questions; but I did not find a reference.

(If this is trivial or repeated, and if one states the reference, I will delete the question.)

In $A_4$, we know $(1,2,3)$ and $(1,3,2)$ although have same cycle structure, they are not conjugate.

Other situation: suppose for $\sigma\in A_n$, its conjugacy class in $S_n$ splits in $A_n$: $$ \sigma^{S_n} =\sigma^{A_n}\cup \sigma^{\tau A_n} $$ where $\tau$ is an odd permutation.

If $\theta$ is a permutation of same cycle type as $\sigma$, how to detect whether $\theta$ falls in $\sigma^{A_n}$ or $\sigma^{\tau A_n}$?

(Shortly: When two permutations in $A_n$, having same cycle structure are conjugate in $A_n$? Can we find a conjugating element?)

Beginner
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    Yes, there are many duplicates, e.g., this one, or this one, and this one, etc. Did you search already? – Dietrich Burde Sep 02 '22 at 10:49
  • In the first two, I did not found complete answer (before posting question); third link, I will see. (Most of the questions are: when a conjugacy class of $S_n$ splits in $A_n$; but I was looking little different than this!) – Beginner Sep 02 '22 at 10:52
  • You can specify what exactly is missing for you in the answers. This would make your question more to the point. Actually, did you read the GP page? It is also very clear on how to decide whether or not a conjugacy class splits in $A_n$. – Dietrich Burde Sep 02 '22 at 10:54
  • The conjugacy classes of $A_n$ are the same as the conjugacy classes of $S_n$, except when a class of $S_n$ splits in $A_n$, so anything that tells you when a class of $S_n$ splits in $A_n$ tells you everything there is to know. – Gerry Myerson Sep 02 '22 at 10:57
  • @Gerry: So a class of $(1,2,3)$ in $S_4$ splits in $A_4$. So, we ask: whether $(2,3,4)$ is conjugate to $(1,2,3)$ in $A_4$ or not? I could not see any way for this! – Beginner Sep 02 '22 at 10:59
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    OK, I understand now, you want to know, when a class in $S_n$ splits, which members go into each of the two classes in $A_n$. In that particular case, the class of $123$ is $123,134,142,243$. I found those by conjugating $123$ with the four elements of the $(xx)(xx)$ subgroup of $A_4$. I don't know what the general answer is. (But I'd still be surprised if there wasn't already an answer elsewhere on this website.) – Gerry Myerson Sep 02 '22 at 11:30
  • Fix representatives for each conjugacy class of $S_n$. If the class is nonsplit, the class of $g \in A_n$ is determined by the cycle structure. Otherwise, let $x \in A_n$ be the representative for the $S_n$-class of $g$. Then the class of $g$ is determined by the parity of a element conjugating $g$ to $x$. See this question and the answer there which hopefully helps. – spin Sep 02 '22 at 15:22

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