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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? –  Mariano Suárez-Alvarez Jan 3 '12 at 2:17
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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? –  Mariano Suárez-Alvarez Jan 3 '12 at 2:19
    
related: math.stackexchange.com/a/1402 –  Grigory M Jan 3 '12 at 8:36

1 Answer 1

The group $G$ acts by conjugation on the set of its own $3$-Sylow subgroups. That means that there is a homomorphism from $G$ into $S_{n_3}$, whose image is a transitive subgroup of $S_{n_3}$. Since $G$ is simple, the kernel of this map must be trivial, so $n_3!$ must be at least $168 = 2^3\times 3\times 7$.

This observation rules out $n_3=4$.

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how can I rule out also the case $n_3=7$? –  Alex M Jan 3 '12 at 2:26
    
Dear Alex, please ask that in the question body. The comments to an answer is not the correct place where to ask a question! –  Mariano Suárez-Alvarez Jan 3 '12 at 2:28
    
@Mariano: To be fair, that is also in the body (third paragraph). –  Arturo Magidin Jan 3 '12 at 2:31
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Eliminating the case of $7$ Sylow 3-groups can be done a variety of ways. If there were $7$, you get a homomorphism into $S_7$, where an element of order $3$ looks like $(123)(456)$. Now what does an element of order $2$ which centralizes this look like? –  user641 Jan 3 '12 at 2:35

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