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I have what seems to me a very simple proof that $A_{n}$ is the only subgroup of $S_{n}$ of index 2. Since I’ve seen other people prove it with what feel like really complicated methods (Like here.), I’m wondering if I’ve overlooked something.

**Proof**: Let $H$ be any subgroup of index 2 in $S_{n}$. Then $H\cap A_{n}$ is a normal subgroup of $A_{n}$ and since $A_{n}$ is simple, $H$ is either the trivial or the improper subgroup of $A_{n}$. If it’s the trivial subgroup it doesn’t have index 2 in $S_{n}$, and otherwise the theorem is proved.

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Your proof looks **simple** because you assumed not so **simple** result that $A_n $ is **simple** for $n\geq 5$…

Actually something more is true…

Suppose that $H\leq S_n$ of index $m $ with $m< n$ then we have homomorphism $\eta: S_n \rightarrow S_m$.

As $Ker(\eta)$ is a normal subgroup of $S_n$ we should have $Ker(\eta)=1$ or $Ker(\eta)=A_n$.

Suppose $Ker(\eta)=(1)$ then we should have $S_n$ embedded in $S_m$, which is not possible as $m<n$.

So,$Ker(\eta)=A_n$ and we know that $Ker(\eta)\subset H$ i.e., $A_n\leq H< S_n$.

As $A_n$ is maximal subgroup of $S_n$ we have $H=A_n$.

So.. Do you see what i am concluding??

The following idea is super-simple. However, it only works if $n$ is odd (so it works precisely half the time!). But as it is so simple I thought it would be nice to record it here.

Suppose $n$ is odd. Note that $S_n$ can be generated by the elements $\alpha:=(1, 2)$ and $\beta:=(1, 2, \ldots, n)$. Now, every subgroup $H$ of index 2 is the kernel of some homomorphism $S_n\rightarrow \mathbb{Z}_2$. The key point is the following:

Homomorphisms are defined by the images of the generators. As $\beta:=(1, 2, \ldots, n)$ has odd order, it is killed by every homomorphism $\phi:S_n\rightarrow \mathbb{Z}_2$, so $\beta\in\ker\phi$ for all such $\phi$.

Therefore, if $\phi$ has non-trivial image it must be *precisely* the map defined by $\alpha\mapsto 1$, $\beta\mapsto0$. Hence, there is a unique homomorphism $\phi:S_n\rightarrow \mathbb{Z}_2$, and hence $S_n$ contains a unique subgroup of index 2 (for $n$ odd).

As suggested, I am writing the complete answer.

First we use the not so simple result:

**Theorem:** Let $n = 3$ or $n\geq 5$. Then $A_n$ is simple.

**Statement:** Now for $n=3$ or $n \geq 5$ let’s show that $\{id\}, A_n$ and $S_n$ are the only normal subgroups of $S_n$.In particular, $A_n$ is the **only** sugbroup of $S_n$ of index $2$.

Proof: It is clear that $\{id\}, A_n$ and $S_n$ are normal subgroups of $S_n$. Now let $H$ be a normal subgroup of $S_n$ and consider the group homomorphism $$\psi: H \to \{-1,+1\}$$

defined by $$\psi(\alpha)= \begin{cases}1&, \text {if}\ \ \alpha\ \ \text{is even}\\-1&, \text {if}\ \ \alpha\ \ \text{is odd}\end{cases} $$

Naturally, $\ker \psi = H \cap A_n$ and $(H: \ker \psi) = |\psi(H)| = 1$ or $2$. Thus, $(H: H \cap A_n) = 1$ or $2$.

**$1^{st}$ case:**

$(H: H\cap A_n) = 1$, that is, $H \subset A_n$. As $H \lhd S_n$, then a fortiori $H \lhd A_n$, then it follows from the theorem that $H=\{id\}$ or $H=A_n$.

**$2^{nd}$ case:**

$(H:H\cap A_n) = 2$. As $H \lhd S_n$, then $H \cap A_n \lhd A_n$, then from the theorem, $H \cap A_n = \{id\}$ or $H \cap A_n = A_n$. Therefore $|H| =2$ or $|H| = S_n$.

Let’s suppose $|H|=2$. As $H \cap A_n = \{id\}$, then $H$ contains an odd permutation $\tau$ of order $2$. Such permutation $\tau$ is necessarily a product of disjoint transpositions, say $\tau = (12)\rho_{2}\ldots\rho_{s}$. Then

$$\tau ‘ = (13)\tau(13)^{-1} \in H , \text{because}\ \ \tau \in H \lhd S_n$$

As $$\tau ‘ (2) = [(13)\tau(13)](2) = [(13)\tau](2) = [(13)](1) =3$$

and $\tau(2) =1$, we obtain $\tau \neq \tau ‘$ and it follows that $|H| \geq 3$. So assuming that $|H| = 2$ leads us to a contradiction and then $H = S_n$.

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