Let $\sigma \in A_n$ be a permutation whose cycle decomposition contains a 2-cycle.
1) Prove the conjugacy class of $\sigma$ in $S_n$ is equal to its conjugacy class in $A_n$.
2) Find a permutation $\sigma \in A_7$ such that its conjugacy class in $S_7$ is larger than its conjugacy class in $A_7$.
My attempt: for 1, we should look at the centralizer of $\sigma$ in $S_n$ and notice that since $\sigma$ commutes with at least one odd permutation, we have an odd permutation $\tau$ in its centralizer. Then its centralizer isn't contained in $A_n$. I'm not sure if this is a good diretion or how to continue from here. For 2, we should consider two permutations conjugate by a single transposition, but then how would you show there aren't any other elements which they are conjugate by?
Any help is much appreciated!
Best Answer
First of all let's call the $2$-cycle $(a,b)$ and write $\sigma=(a,b)\sigma_0$ when $\sigma_0$ is a permutation in which $a,b$ are fixed points. Now we are going to prove the first part.
Assume $\tau$ is conjugate to $\sigma$ in $S_n$. We want to show they are conjugate in $A_n$ as well. We know there is a permutation $\lambda\in S_n$ such that $\tau=\lambda\sigma\lambda^{-1}$. If $\lambda\in A_n$ then $\sigma$ and $\tau$ are conjugate in $A_n$, nothing left to prove. Now suppose $\lambda\notin A_n$, which means $\lambda$ is an odd permutation. Now let $\mu=\lambda(a,b)$. It is an even permutation as a product of odd permutation, which means $\mu\in A_n$. And now note that:
$\mu\sigma\mu^{-1}=\lambda(a,b)(a,b)\sigma_0(a,b)\lambda^{-1}=\lambda\sigma_0(a,b)\lambda^{-1}=\lambda(a,b)\sigma_0\lambda^{-1}=\lambda\sigma\lambda^{-1}=\tau$
I simply used the fact that disjoint cycles commute, and of course that $(a,b)^{-1}=(a,b)$. So $\sigma$ and $\tau$ are conjugate in $A_n$ as well.
As for the second part, I'll show two possible ways to solve it. The first way: take the permutation $(1234567)$. The permutations which are conjugate to it have the form $\lambda(1234567)\lambda^{-1}=(\lambda(1)\lambda(2)...\lambda(7))$. Now try to find what $\lambda$ needs to be in order to get $\lambda(1234567)\lambda^{-1}=(2134567)$. Such a permutation must satisfy $(\lambda(1)...\lambda(7))=(2134567)$. There are $7$ such permutations. Find them and you will get they are all odd. Hence $(1234567)$ and $(2134567)$ are conjugate in $S_7$ but not in $A_7$.
Now the second way to solve the second part: again, let $\sigma=(1234567)$. Suppose its conjugacy class in $S_7$ equals to its conjugacy class in $A_7$. That means $(12)\sigma(12)=\lambda\sigma\lambda^{-1}$ for an even permutation $\lambda$. From here we get that $\sigma$ commutes with $(12)\lambda$, which means $\sigma$ commutes with an odd permutation. On the other hand, we know that $\sigma=(1234567)$ is conjugate to $6!$ different permutations in $S_7$, which are all the permutations with same cycle structure as $\sigma$. By the Orbit-Stabilizer theorem we get that $\sigma$ commutes with $\frac{7!}{6!}=7$ permutations. But we also know it commutes with $id,\sigma,\sigma^2,...,\sigma^6$ which are $7$ different permutations. But as $\sigma$ commutes only with $7$ permutations we conclude that it can't commute with anything else. Hence $\sigma$ commutes only with even permutations which is a contradiction.