Internal Structure of A_7 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.]
 A: Part of answer:
(2) $A_7$ has an abelian subgroup of order $12$, namely $\langle (123)\rangle \times \langle (45)(67), (46)(57)\rangle \cong \mathbb{Z}_3 \times K_4$, $K_4$ is the Klein-$4$ group.
(3) The Sylow-$2$ subgroup of $A_7$ has order $8$ and it should be $D_8$: as $A_7$ has a subgroup $H=\langle (12)(34), (13)(24) \rangle \cong K_4$, it should be contained in some Sylow-$2$ subgroup $K$ of $A_7$. So the Sylow-$2$ subgroup should can not be $\mathbb{Z}_8$ or $\mathbb{Q}_8$, as they do not contain Klein-$4$ group.
If $\sigma \in K$ commutes with every element of this subgroup $H$, then we can see that $\sigma$ should permute the symbols $\{5,6,7\}$, within themselves, and it should permute the symbols $\{1,2,3,4\}$, within themselves; i.e. $\sigma=\sigma_{1} \sigma_{2}$, where $\sigma_1$ permutes $\{1,2,3,4\}$, and $\sigma_2$ permutes $\{5,6,7\}$.
We can see that elements of type $(12)(56)$ (i.e. a transposition on $\{1,2,3,4\}$ followed by a transposition on ${5,6,7}$) can not be in the centralizer of the subgroup $H$; such type of elements commutes with only two elements of $H$- the identity and $(12)(34)$. 
So $\sigma=\sigma_1 \sigma_2$, where $\sigma_1 \in H$, and $\sigma_2 = Id$. 
Hence the centralizer of the subgroup $H$ in Sylow-$2$ subgroup $K$ should be itself; i.e. Sylow-$2$ subgroup of $A_7$ is non-abelian; hence it must be $D_8$.
