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I have this question:

For a group $G=A_5$ and $H=$ $\langle(12)(34),(13)(24)\rangle$ , prove that $(123)\in N_G(H)$ and hence deduce the order of $N_G(H)$.

$A_5$ is defined to be the alternating group and $N_G(H)=\{g\in G :H^g=H \}$, where $H^g=\{g^{-1}hg:h\in H\}$

I have managed to prove that $(123)\in N_G(H)$ simply by calculating $H^{(123)}$ and verifying. I do not know to calculate the order of $N_G(H)$ though, any help would be appreciated.

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1 Answer 1

You have obtained a lower bound on the order of $N_G(H)$ (which is?...). But you also know that $A_5$ is simple, and that will give you upper bounds on the size of $N_G(H)$, and of any proper subgroup, for that matter. Why is $N_G(H)$ a proper subgroup?

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I can't say I understand. $A_5$ is simple which would mean the only normal subgroups of G are $\{1\}$ and $G$ itself. This would mean that the upper bound on the size of $N_G(H)$ is $|A_5|$? –  L.oiler Jan 14 '13 at 22:45
    
What would that tell you about $H$ if $N_G(H)=G$? –  Alex B. Jan 15 '13 at 9:45
    
Could you please write the solution if you're willing? I'm really finding this topic hard and have tried my best to find a solution and can't find any similar examples. –  L.oiler Jan 15 '13 at 15:28
    
Never mind, I figured it out an alternative way. –  L.oiler Jan 18 '13 at 19:15
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