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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 $12$.2012-01-23
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    @ArturoMagidin thanks again. Silly mistake. Using simple wrongly. How do you know it's not 20?2012-01-23
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    A group of order 20 cannot contain an element of order 3.2012-01-23
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    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 $12$").2012-01-23
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    @ArturoMagidin Thanks again that was really helpful. I understand the question now. I can see where I went wrongs. Thanks2012-01-23
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    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.2012-01-23
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    $(12345)\neq(12)(13)(14)(15)$ because $(12345)$ sends $1$ to $2$, whereas the permutation on the right sends $1$ to $5$. The permutation on the right is actually $(154)(23)$.2012-08-26

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