I know there are questions that may look similiar at first glance, but they ask about different aspects.

In a book I read there is a certain result proven($[\sigma]$ denotes the conjugacy class of $\sigma$):

Let $n \geq 2$, and let $\sigma \in A_n$. Then $[\sigma]_{A_n} = [\sigma]_{S_n}$ or the size of $[\sigma]_{A_n}$ is half the size of $[\sigma]_{S_n}$, according to whether the centralizer $Z_{S_n}(\sigma)$ is not or is contained in $A_n$.

Then it says that in case if $Z_{S_n}(\sigma)$ lies in $A_n$(that is, $|[\sigma]_{S_n}| = 2|[\sigma]_{A_n}|$ ) then $[\sigma]_{S_n}$ actually splits into two conjugacy classes in $A_n$(one of which is, obviously, $[\sigma]_{A_n}$).

But how come $[\sigma]_{S_n} \setminus [\sigma]_{A_n}$ is even a conjugacy class in $A_n$?

What I know is that $g \in [\sigma]_{S_n} \setminus [\sigma]_{A_n}$ iff $g = \tau \sigma \tau^{-1}$ for some odd permutation such that for all $\rho \in A_n \ \ \ \rho^{-1} \tau \notin Z_{S_n}( \sigma ) \Leftrightarrow \rho^{-1} \tau \sigma \tau^{-1} \rho \neq \sigma \Leftrightarrow \tau \sigma \tau^{-1} \neq \rho \sigma \rho^{-1}$.

It's obvious that $g \in A_n$. But why all such $g$'s are conjugates to some even permutation?

  • $\begingroup$ Let $\sigma$ be an element of $A_n$. Then for any $\tau\in S_n$ we have that $\tau\sigma\tau^{-1}$ is in $A_n$ because, conjugation preserves cycle type. Essentially all the elements elements in $[\sigma]_{S_n}\setminus[\sigma]_{A_n}$ are of the form $\rho=\tau \sigma \tau^{-1}$ for some odd $\tau\in S_n$. So now conjugate these by any odd element and you should find that they are in $[\sigma]_{S_n}$. (In other words, for an odd $\tau'$ we have that $\tau'\tau \sigma \tau^{-1} \tau'^{-1}$ is in $[\sigma]_{A_n}$ since $\tau' \tau$ is even. ) $\endgroup$
    – daruma
    Jul 30, 2016 at 1:23
  • $\begingroup$ I hope it's now clear that this is then a conjugacy class because we have just concluded, $[\sigma]_{S_n}\setminus[\sigma]_{A_n}=\tau [\sigma]_{A_n} \tau^{-1}=[\tau \sigma \tau^{-1}]_{A_n}$ $\endgroup$
    – daruma
    Jul 30, 2016 at 1:23
  • $\begingroup$ @daruma I see that if $\tau \sigma \tau^{-1} \in [\sigma]_{S_n} \setminus [\sigma]_{A_n}$, then $\tau$ is odd. That is true. Then $\tau' \tau \sigma \tau^{-1} \tau'^{-1}$ is in $[\sigma]_{A_n}$ if $\tau'$ is odd. But it doesn't tells us that $\tau' \tau \sigma \tau^{-1} \tau'^{-1}$ is in $[\sigma]_{S_n} \setminus [\sigma]_{A_n}$ if $\tau'$ is even. What is more, even if we prove that $[ \tau \sigma \tau^{-1}]_{A_n} \subseteq [\sigma]_{S_n} \setminus [\sigma]_{A_n}$, we don't know the converse is true. $\endgroup$
    – Jxt921
    Jul 30, 2016 at 9:08
  • $\begingroup$ If $\tau' \sigma \tau^{-1}\in [\sigma]_{S_n}\setminus [\sigma]_{A_n}:=B$ and conjugating by an even element doesn't send it to $B$ then it belongs in $[\sigma]_{A_n}:=C$, we get a contradiction because then conjugating by its inverse(which is even) means it belongs in $C$. $\endgroup$
    – daruma
    Jul 30, 2016 at 9:44
  • $\begingroup$ If $\tau \sigma \tau^{-1}\in B$ then the containment $[\tau \sigma \tau^{-1}]_{A_n}\subset B$ is by definition true. If conjugating by an even element $\rho$ sends it to $[\sigma]_{A_n}$ then, $\rho (\tau \sigma \tau^{-1}) \rho^{-1}=\rho' \sigma \rho'^{-1}$ for some $ \rho'$. Then, conjugate by $\rho^{-1}$ on both sides of the equation and we find that $\tau \sigma \tau^{-1}$ is a conjugate of $\sigma$ by an even element. This contradicts our initial assumption that it belonged to $B$. $\endgroup$
    – daruma
    Jul 30, 2016 at 9:50

1 Answer 1


Note that we can write any odd permutation in the form of $(1 2) \omega$ where $\omega$ is an even permutation, if $\sigma$ does not commute with any odd permutation in $S_n$ then, $[\sigma^{(1 2)}]_{A_n} = \{ \sigma^\tau \;|\;\tau \in S_n - A_n \}$ and $[\sigma]_{S_n}= [\sigma]_{A_n} \cup [\sigma^{(1 2)}]_{A_n}$ .


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