# A permutation in $S_n$ whose cycle decomposition contains a transposition

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!

## 1 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.