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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 comment –  user11049 May 19 '11 at 8:17
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) –  t.b. May 19 '11 at 8:20
If it's any help, the answer to 1) is no, but there is only one such class in $S_7$. –  Derek Holt May 19 '11 at 9:31

1 Answer 1

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 $D_8$.

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For (2), the maximal abelian subgroups have type 2×2, 4, 5, 7, 3×3, and 2×3×3, so 12 is the largest order. Derek Holt already answered (1), two conjugacy classes in A7 that fuse to one conjugacy class in S7. Presumably if he is looking at permutation actions, S7-conjugacy is what he wants. –  Jack Schmidt May 19 '11 at 14:45

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