How many subgroups of $S_8$ with index $2$? I was wondering how to prove that the only subgroup of $S_8$ (symmetric group on $8$ letters) with index $2$ is $A_8$.
I thought we can show that if we have a subgroup of index $2$ in $S_8$, it has to contain $A_8$. (And then because they have the same order they must be the same).
So if I try to take a subgroup $H$ in $S_8$ with index $2$, then there exist two left cosets, say $H$ and $aH$ for $a$ not in $H$. (Then $S_8 = H + aH$)
And because $A_8$ in $S_2$ has index $2$ there also exist $2$ left cosets $A_8$ and $sA_8$, where $s$ is some odd permutation. (Then $S_8 = A_8 + sA_8$).
And now I need to show that $H$ and $A_8$ are conjugate as well as $aH$ and $sA_8$.
How can I do that? Can I prove it like this? Any hints would be greatly appreciated :)
Thank you in advance for any help. 
 A: Let me suggest the following:
If $H \subset S_8$ is a group of index $2$, then $H \cap A_8$ is a normal subgroup of $A_8$. Now we know that $A_8$ is simple, such that the intersection is trivial or $A_8$. What can we therefore say about $H$?
In case that you do not know that $A_8$ is simple (or don't want to use that fact) let us consider this which is basically a similar idea as yours (just not writting it down as cosets):
If the index is $2$, we know that we have a surjective map $f \colon S_8 \rightarrow C_2$, such that $H = \text{ker}(f)$. That means that all transpositions get mapped to the same element as they are conjugate. If they were sent to the trivial element in $C_2$, the morphism would not be surjective as the transpositions generate $S_8$. Therefore they are sent to $-1$ and thus we get the sign map and $H = A_8$. 
The latter proof works for all $n$ by the way whereas the first one only for $n \geq 5$.
Another nice approach is to show that subgroups of index $2$ contain all elements off odd order (which also yields counterexamples for the converse of Lagrange's theorem).
A: In general, any subgroup $H$ of index $2$ of a group $G$ is normal, and in that case $G/H \cong C_2$ is abelian, hence the commutator subgroup $[G,G] \subseteq H$. Since $[S_8,S_8]=A_8$, and $|S_8:A_8|=2$, it follows that $A_8=H$.
A: It is well known that for $n \geq 5$, the normal subgroups of $S_n$ are precisely $\{1, A_n, S_n\}$ (see for example Isaac's book: Algebra: A graduate course; essentially this follows because $A_n$ is simple for $n \geq 5$). Since any subgroup of index $2$ is normal, there is precisely one subgroup of index $2$: $A_n$.
