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Let $G$ be a simple group of order 168, I have to show that it has at least 14 elements of order 3.

Using Sylow's theorems I proved that if $n_3$ is the number of 3-Sylows then $n_3\in\{4,7,28\}$, but now I don't know how to continue, I have to exclude the possibility of $n_3=4$, could you help me?

Reading this question Prove there is no element of order 6 in a simple group of order 168 I saw that actually $n_3=28$ could you explain me why?

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    Why do you *have* to do this? Is anyone pointing at you with a gun? Should we call the police?2012-01-03
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    If there were $4$ $3$-Sylow groups, your simple group $G$ would act on them by conjutation, giving a map $G\to S_4$. Can you get a contradiction from this?2012-01-03
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    related: http://math.stackexchange.com/a/1402/2012-01-03

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