Proof that $A_n$ the only subgroup of $ S_n$ index $2$.

Question

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.

Answer 1

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 $H = S_n$.

Answer 2

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).

Answer 3

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??

Answer 4

Assume that $A_n$ is a simple group for $n\neq 4$. If $H$ is any subgroup of index $2$ in $S_n$, then $H$ must be normal in $S_n$ (because the only non-trivial coset in $S_n/H$ must be the set-theoretic complement of $H$, and must therefore be both the right and left coset obtained by multiplying on the right, or the left by any particular $g\notin H$), and so $(H\cap A_n)$ must be a normal subgroup of $A_n$. Since $A_n$ is simple for $n\neq 4$, it follows that $(H\cap A_n)$ is either equal to $\{1\}$ (the subgroup of $S_n$ consisting only of the identity element) or $A_n$ itself, these two possibilities being mutually exclusive. If $(H\cap A_n)=\{1\}$, then ${\it every}$ element of $H$ (except for the identity permutation) must be an odd permutation. Furthermore, since $H$ has index $2$ in $S_n$, $|H|={\frac{n!}{2}}> 2$ for $n\geq 3$. Let $\sigma$ and $\tau$ be any two distinct odd permutations in $H$. Then, since ${\sigma}^2$, and ${\sigma}{\tau}$ are both even permutations in $H$, we would be forced to conlcude that both are equal to the identity permutation which is the only even permutation in $H$ by hypothesis. This in turn forces us to conclude that $\sigma = \tau$, which is a contradiction. Thus, $(H\cap A_n) = A_n$. In particular, $A_n\subset (H\cap A_n)$ and as both $A_n$ and $H$ have the same cardinality, they must be equal.