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In $A_7$,

1) Are all subgroups of order 168 are conjugate? ($A_7$ contains a simple group of order 168).

2)Does it contain an abelian group of order 12? What is the largest order of abelian group?

3)What is the Sylow-2 subgroup of $A_7$?

[I am trying to get some other way to prove that there is only one simple group order 168.

In a simple group $G$ of order $168$, the number of Sylow-3 subgroups is $7$, or $28$ . If it is $7$ , then $G$ is contained in $A_7$, and we get an abelian subgroup of order $12$, in $A_7$ . Certainly, $A_7$ does not contain an element of order $12$, hence cyclic subgroup of order $12$. therefore, it may happen that $A_7$ may contain $\mathbb{Z}_2 \times \mathbb{Z}_6$. I want to know whether this is possible? Also, to know about intersection of Sylow-2 subgroups with each other, I want to know their structure.]

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    @Theo: Shifted the comment2011-05-19
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    Thank you, that's much better. As I said in my last comment, maybe you could say which proof you know, since I'm pretty sure there are many ways of doing it. So if it should turn out not to be possible with your strategy, some of the group theory experts here would certainly be able to point you to or outline other proofs of that fact. (and answer the specific question on $A_7$ at the same time)2011-05-19
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    If it's any help, the answer to 1) is no, but there is only one such class in $S_7$.2011-05-19

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