Elements of $A_8$ commuting with $A_5$

Question

How would we go about showing that in $A_8$, an element of order $3$ commuting with a subgroup isomorphic to $A_5$ is necessarily a $3$-cycle?

I know that an element of order $3$ in $A_8$ must be a product of disjoint $3$-cycles, and therefore can be either a $3$ cycle or a product of two disjoint $3$-cycles.

Answer 1

$A_5$ contains $20$ elements of order $3$. Given a disjoint product of two 3-cycles, you simply cannot find a set of $20$ elements of order $3$ in $A_8$ that commute with it. So finding a whole subgroup isomorphic to $A_5$ that commutes with said element is also necessarily impossible.