# $A_5$ problem with normal subgroup

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.

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Actually, the fact that $A_5$ is simple tells you that the order of $N_G(H)$ cannot be $5!/2$. Because that would tell you that $N_G(H)=A_5$, which implies $H\triangleleft A_5$, which contradicts simplicity. Instead, what you want to observe is that the order of the normalizer is at least $3|H|=12$, must divide $5!/2$, but cannot be $5!/2$. Hence, the order of the normalizer must be $12$. – Arturo Magidin Jan 23 '12 at 17:44
@ArturoMagidin thanks again. Silly mistake. Using simple wrongly. How do you know it's not 20? – simplicity Jan 23 '12 at 17:53
A group of order 20 cannot contain an element of order 3. – Arturo Magidin Jan 23 '12 at 18:05
To be more precise: you know the order is a multiple of $4$ (contains $H$); you know it is a multiple of $3$ (contains an element of order $3$); so you know it is a multiple of $12$ (not just "at least $12$"). – Arturo Magidin Jan 23 '12 at 18:13
Perhaps you can post your final solution as an answer? That way the question won't go on as "unanswered". This also would allow others to help you write it up as nice as possible. – Arturo Magidin Jan 23 '12 at 19:08

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 $\,|N_G(H)|=12\,$
$A_4$ 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 $A_5$