0
$\begingroup$

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?

$\endgroup$
8
  • $\begingroup$ 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

Reset to default
2
$\begingroup$

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 $[\sigma]_{S_n}= [\sigma]_{A_n} \cup [\sigma^{(1 2)}]_{A_n}$ .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.