The commutativity of $(12)(34)$ and $(12)$ sufficient to say that : the $S_5$-conjugacy class of $(12)(34)$ is also an $A_5$-conjugacy class?

Question

I read in a document about the symmetric group. I came across a paragraph that I didn't understand why, it's the following:

One checks that $(123)$ commutes with the odd permutation $(45)$. Therefore, the $S_5$-conjugacy class of $(123)$ is also an $A_5$-conjugacy class. Similarly,the permutation $(12)(34)$ commutes with the odd permutation $(12)$. Therefore, the $S_5$-conjugacy class of $(12)(34)$ is also an $A_5$-conjugacy class.

Why is the commutativity of $(12)(34)$ and $(12)$ sufficient to say that : the $S_5$-conjugacy class of $(12)(34)$ is also an $A_5$-conjugacy class?

Answer 1

Let $p$ be an odd permutation that commutes with permutation $q$. Then $S$-conjugacy class of $q$ is equal to $A$-conjugacy class of $q$.

As usual, to show equivalence we need to show two inclusions. One is trivial: if $s$ is $A$-conjugated to $q$ then of course it's also $S$-conjugated.

To prove the other direction, assume $s = h q h^{-1}$. We want to show that also $s = h' q h'^{-1}$ where $h'$ is even.

If $h$ is even, we can take $h' = h$. Otherwise, we can take $h' = hp$: $(hp) q (hp)^{-1} = hpqp^{-1}h^{-1} = hpp^{-1}qh^{-1} = hqh^{-1} = s$.