Prove that $A_5$ cannot have a normal subgroup of order $2$
Suppose that $H$ is a normal subgroup of $A_5$ and $|H|=2$ .
It can be shown that $H$ is contained in the center of A 5 . So elements of $H$ commutes with every element of $A_5$ .
Say $(ab)(cd) \in H$ . Since every element of $H$ is in the center , $(ab)(cd)$ commutes with every element of $A_5$ .
But we can check that $(abc)(ab)(cd) \neq (ab)(cd)(abc)$ .
Therefore $A_5$ doesn't have a normal subgroup of order $2$.
Is this proof correct?
Can someone please provide some alternative proof for this question .