Proving that $A_n$ is the only subgroup of $S_n$ of index 2.

Question

The problem was to prove that $A_n$ is the only subgroup of $S_n$ of index 2. I am aware of the fact that there are multiple posts on this question, but I was wondering why this particular method I thought of fails.

Suppose that $G \not= A_n$ is a subgroup of $S_n$ of index 2. Then there is one other coset of $G$, which is $S_n - G$ ($-$ is the set difference operator). Then we must have that there is some $x$ in $S_n - G$ that is even. Then the coset $x^{-1}(S_n - G) = G$, and since the sign map is a homomorphism, we have that $S_n - G$ and $G$ have the same number of even and odd elements. So $G$ contains half of $A_n$.

However, in order to achieve the contradiction, we must show that half of $A_n$ will generate $G$, that is, there is no subgroup of $A_n$ of index 2. However, from searching on google, apparently this is only true of $n$ is prime. My question is exactly where did I make the flaw and is there any way to resolve it?

It is known (though presumably not by you) that $A_n$ is simple for $n\ge 5$, hence there is no subgroup of index $2$ at least for $n\ge 5$; I don't know where you found that primeness might be involved. — Hagen von Eitzen, Aug 18 '15 at 06:15
@HagenvonEitzen Hm.. I knew that $A_n$ is simple for $n \ge 5$ but how does that imply that there is no such subgroup? By the way, I saw the primeness in another M.SE post. — Tae Hyung Kim, Aug 18 '15 at 06:16
The fact that $A_n$ is simple for $n \geq 5$ means that $A_n$ cannot have a subgroup of index $2$, because any subgroup of index $2$ is a proper normal subgroup. — coldnumber, Aug 18 '15 at 06:20
See also here and here and here. (All of them were listed among related questions in the sidebar.) — Martin Sleziak, Aug 18 '15 at 08:36
In particular this one is, to some extent, related to your approach. — Martin Sleziak, Aug 18 '15 at 08:39

score 1 · Answer 1 · answered Aug 18 '15 at 07:51

Let us consider n>=5. A subgroup H of S_n of index 2 either contained in A_n and then we are done or half of its elements is even. In the latter case we get a normal subgroup in the simple A_n which cannot hold (A_n is simple). For the case n=4 it is known that there are no subgroups of order 6 in A_4( the book of Rotman). The remaing cases n=3 and n=2 are easy to prove.

Proving that $A_n$ is the only subgroup of $S_n$ of index 2.

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