I learned that the conjugacy class $a^{S_n}$ of $a$ in $S_n$ splits into 2 in $A_n$ if and only if the centralizer $C_{A_n}(a) = C_{S_n}(a)$.
Taking $S_5$ as an example, we see that the conjugacy class in $S_5$ of $(1\;2\ 3)$ has $20$ elements, that of $(1\ 2\ 3\ 4\ 5)$ has $24$ elements, and that of $(1\ 2)(3\ 4)$ has $15$ elements (these can be calculated using counting arguments, using the fact that conjugates have same cycle type). I want to show their conjugacy classes in $A_5$ have size $20$, $12$, and $15$ respectively. In other words, I want to show that the centralizer of $(1\ 2\ 3\ 4\ 5)$ is contained in $A_5$, and that this is not the case for the other two.
I know that $(1\ 2\ 3)$ commutes with $(4\ 5)$ which is odd, so this shows that the centralizer of $(1\ 2\ 3)$ is not contained in $A_5$. I also know that $(1\ 2)(3\ 4)$ commutes with $(1\ 2)$ and $(3\ 4)$, so the centralizer of $(1\ 2)(3\ 4)$ is not contained in $A_5$. However, I don't know how to show that the centralizer of $(1\ 2\ 3\ 4\ 5)$ doesn't contain any odd permutations.
Let's say $\pi$ commutes with $\tau=(12345)$. Then by definition, $\tau(\pi(j))=\pi(\tau(j))$ for all $j\in\{1,2,3,4,5\}$. Using this, one can show that if $\pi(1)=k$, then $\pi=(k,\tau(k),\tau^2(k),\tau^3(k),\tau^4(k))$. This is a five cycle since $\tau$ has order $5$, and hence is even.