Let G=$A_5$ and $H=\bigl\langle (12)(34),(13)(24)\bigr\rangle$. Prove $(123) \in N_{G}(H)$ and hence deduce the order of $N_{G}(H)$.
I know you claim that $A_5$ is simple, then $N_{G}(H)$ has order $\frac{5!}{2}$. But, the problem is this.
Since $(12345) \in A_{5}$ because $(12345)=(12)(13)(14)(15)$
But, $(12345)^{-1}(12)(34)(12345)$ is not an element in $H$. So, then I can't see how $N_{G}(H)$ can be the whole group.