Prove all products of two disjoint 2-cycles are pairwise conjugate in $A_{n}$

Question

Let $n \geq 4$. Prove that in the group $A_{n}$ of parity-even permutations of $\{1,2,...,n\}$, all products of two disjoint 2-cycles are pairwise conjugate.

This is a past exam question for a course in Group Theory. I think I have an answer for it but it's long and basically just conisders cases. I was wondering if there's a more general way to prove it and also if my method is correct/covers all the cases.

Here's what I did:

I said that w.l.o.g. we want to check whether $(a \ b)(c \ d) = \sigma (1 \ 2)(3 \ 4) \sigma^{-1}$ for some $\sigma \in A_{n}$.

So using that any two disjoint 2-cycles can be swapped and that we can swap elements within a 2-cycle, the cases I considered were the following:

Case 1: $n \geq 8$ and $\{a,b,c,d\}\cap\{1,2,3,4\}=\emptyset$.

Case 2: $n \geq 7$ and w.lo.g. $a = 1$.

Case 3: $n \geq 6$ and w.lo.g. $a = 1, \ b=2$.

Case 4: $n \geq 6$ and w.lo.g. $a = 1, \ c=2$.

Case 5: $n \geq 5$ and w.lo.g. $a = 1, \ b=2, \ c=3$.

Case 6: $n \geq 4$ and w.lo.g. $a = 1, \ b=2, \ c=3, \ d=4$.

Have I considered all the cases here? Or, alternatively is there some better way to approach this problem?

Answer 1

I would do this as follows. Wanting to show that $(12)(34)$ and $(ab)(cd)$ are conjugate whenever $a,b,c,d$ are distinct.

1. We know that they are conjugate in $S_n$, so there exists a permutation $\sigma\in S_n$ that works, i.e. $(ab)(cd)=\sigma(12)(34)\sigma^{-1}$
2. Also observe that $(12)(34)=(1324)^2$, so $(12)(34)$ and $(1324)$ commute.
3. Show that either $\sigma$ or $\sigma'=\sigma(1324)$ is an even permutation, and that conjugating $(12)(34)$ by either one of them gives you $(ab)(cd)$.