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Question: Prove that if $G\leq S_n$ of index $2$, then $G=A_n$.

Now, I don't want to appeal, if I can avoid it, to the fact that $A_n$ is simple for $n\geq 5$. Moreover, I want to see if I can use the method below. I know that it can be done by induction on $n$ while assuming $G\neq A_n$ and $|S_n:G|=2$, so there is at least one $3$ cycle not in $G$, then assuming $(123)$ is not in $G$, we get $3$ distinct cosets, which is a contradiction.
However, I want to try and prove it in the following way, because I feel like it is a more "common" technique.....

Suppose $G\neq A_n$, then since $A_n$ is normal in $S_n$ and $G\leq S_n$, we have that $GA_n=S_n$, thus $\frac{|G||A_n|}{|G\cap A_n|}=|S_n|$. Now, I can either divide both sides by $|G|$ or by $|A_n|$, and I get whatever is on the LHS is equal to $2$. However, I don't see how to generate a contradiction whether I consider $|G:G\cap A_n|$ or $|A_n:G\cap A_n|$. I think I am missing something pretty trivial, but I just can't see it. Any help is greatly appreciated! Thank you.

User7238
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  • Not quite. I am aware of that post, and I believe I can get a contradiction to my proof using that $A_n$ is simple for $n\geq 5$, I was just wondering if there was another way to get a contradiction to my proof without appealing to that fact. – User7238 Dec 01 '21 at 03:40
  • Hmm, there are 6 answers on that thread and I do see some of them not using the fact that $A_n$ is simple for $n\ge 5$, eg., this one. For why $A_n$ is generated by the 3-cycles in $S_n$, see this answer. – Prasun Biswas Dec 01 '21 at 03:46
  • I reference that one in my question too. Essentially, I am just curious if I can get a contradiction to $|G:G\cap A_n|$ or $|A_n:G\cap A_n|$ without appealing to the simplicity of $A_n$ for $n\geq 5$. – User7238 Dec 01 '21 at 03:50
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    $G$ cannot contain any transposition (it would contain all of them, being normal); but it contains all products of two transpositions, so it contains $A_n$. – Arturo Magidin Dec 01 '21 at 03:51
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    You get that $G\cap A_n$ has index $2$ in both $G$ and in $A_n$. But unless you use something about the structure of $A_n$, you are not going to get a contradiction: after all, there are plenty of groups that have two distinct proper subgroups of index $2$, like the Klein $4$-group... – Arturo Magidin Dec 01 '21 at 03:52
  • @ArturoMagidin Ah, okay, that makes sense! I really appreciate the help! – User7238 Dec 01 '21 at 03:54
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    Would the fact that $[S_n,S_n]=A_n$ help? I mean, if $G$ has index $2$ in $S_n$ then $G \unlhd S_n$ and $S_n/G$ is abelian, so $[S_n,S_n] \subseteq G$ and you are done ... – Nicky Hekster Dec 01 '21 at 10:54

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Too long for a comment. I don't know if you will accept this as an answer, since it does not follow your method. However, as Arturo says, if you don't use anything about the structure of $A_n$ or $S_n$, you won't be able to onclude.

Here is an argument which does not need anything fancy about $A_n$ ( commutators, or simplicity...)

The only thing you need to know is that there is a unique non trivial group morphism $S_n\to \{\pm 1\}$, namely the signature morphism.

Let me recall the argument: transpositions are all conjugate in $S_n$ ( $\tau (ij)\tau^{-1} =(kl)$, where $\tau\in S_n$ satisfies $\tau(i)=k, \tau(j)=l$), so transpositions under such a morphism all have same image (since $\{\pm \}$ is abelian. Since transpositions generate $S_n$, this ismage is $-1$, and the morphism is the signature.

Once you have that in mind, this is quite easy. Let $H$ be a subgroup of index $2$ of $S_n$; Then $H$ is normal in $S_n$ (classical result, valid for any index $2$ sungroups of a given group $G$), and we have a group isomorphism $S_n/H\simeq \{\pm 1\}$. This isomorphism sends elements of $H$ to $1$, and the other ones to $-1$.

Then the composition $S_n\to S_n/H\simeq\{\pm 1\}$ is a non trivial morphism, with kernel $H$ by definition. But this non trivial morphism is the signature morphism, so its kernel is $A_n$. Consequently, $H=A_n$.

Side remark. More general, there is a $1-1$ correspondence between subgroups of index $2$ of a given group $G$ and the set of non trivial morphisms $ G\to \{\pm 1\}$.

GreginGre
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  • Saying that there is a unique morphism from $S_n$ onto ${1,-1}$ is saying that there is a unique subgroup of index 2. So you are saying that in order to show that there is only one such subgroup, all you need to do is assume that there is only one such subgroup. – Arturo Magidin Dec 01 '21 at 14:29
  • Not quite. The proof that such morphism is unique comes from the fact that the images of any transposition are all equal to -1 (because transposition are conjugate). I have added the argument in my answer for clarification – GreginGre Dec 01 '21 at 17:18
  • So, basically: a proper normal subgroup cannot contain any transposition. A proper normal subgroup of index $2$ must contain the product of any two transpositions, hence contains $A_n$. – Arturo Magidin Dec 01 '21 at 17:41