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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!

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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.

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