$A_5$ problem with normal subgroup

Question

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.

Answer 1

Just rephrasing the solution given in the comments by Arturo, hoping the question won't remain unanswered. With $\,G:=A_5\,$: $$\begin{align*}(i)&\,\,\,|H|=4\Longrightarrow 4\;\;|\;\; |N_G(H)|\\ (ii)&\,\,\,(123)\in N_G(H)\Longrightarrow 3\;\;|\;\; |N_G(H)|\end{align*}$$

$$(i)-(ii)\Longrightarrow 12\;\;|\;\;|N_G(H)|$$

But since also $\,|N_G(H)|\;\;|\;\;|A_5|=60\,$ , then the only possible orders for the normalizer are $\,12\,,\,60\,$ , and since $\,|N_G(H)|=60\Longleftrightarrow N_G(H)=G=A_5\Longleftrightarrow H\triangleleft A_5$ , which is impossible as $\,A_5\,$ is simple and $\,H\neq \{1\}\,$ , so we finally get $\,|N_G(H)|=12\,$

Answer 2

$A_4$ is a subgroup of $G=A_5$ and contains the normal subgroup $H=\{1,(12)(34),(14)(23),(13)24\}$; thus $A_4\subseteq N_G(H)$. Any element which moves $5$ or its inverse will not normalize $H$. Thus $A_4$ is the normalizer since it is the maximal subgroup of $S_4$ contained in $A_5$