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

Question

How to prove that the only subgroup of the symmetric group $S_n$ of index 2 is $A_n$?

Why isn't there other possibility?

Thanks :)

Answer 1

As mentioned by yoyo: if $H\subset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2=\{1,-1\}$. We thus have a surjective homomorphism $f:S_n\to C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)\in C_2$ is the same element for every transposition $t\in S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $t\in S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.

Answer 2

• Subgroups of index two are normal (exercise).
• $A_n$ is simple, $n\geq 5$ (exercise).

If there were another subgroup $H$ of index two, then $H\cap A_n$ would be normal in $A_n$, contradiction.

[To see that $A_n\cap H\neq\{1\}$ as brought up in a comment, note that if this were the case, then $S_n\backslash(A_n\cup H)=\{x\}$ for some single element $x\in S_n$. Both $A_n$ and $H$ are normal, so $x\in Z(S_n)=\{1\}$, impossible.]

The better answer is probably the accepted one: The sign homomorphism is the only non-trivial homomorphism $\phi:S_n\to \{\pm1\}$. This uniquely identifies $A_n$ as the only index 2 subgroup of $S_n$.

Answer 3

Let $n\geq 2$ and $H \leq S_n$ be such that $\mid S_n:H \mid=2$. Then $H \trianglelefteq S_n$ and $S_n/H$ being isomorphic to $\Bbb Z/2\Bbb Z$, is cyclic. Consider the natural surjection $\pi :S_n\to S_n/H$. Since $S_n/H$ is abelian, the commutator subgroup $[S_n,S_n]\subseteq \ker(\pi)=H$. We know $[S_n,S_n]=A_n$, hence $A_n \subseteq H$. Clealy $\mid A_n \mid=\mid H \mid$, as both of them are subgroups of a finite group of equal index. Therefore we have $H=A_n$.

Answer 4

Let $H\le S_n$ such that $[S_n:H]=2$ and $H\ne A_n$. Denoted $d:=|H\cap A_n|$, we get: $$|HA_n|=\frac{(n!/2)^2}{d} \tag1$$ By Lagrange, $d\mid n!/2$ and, since $HA_n\le S_n$ (as $A_n$ is normal in $S_n$) also $\bigl((n!/2)^2/d\bigr)\mid n!$. Therefore: $$\frac{n!}{2}=kd \tag2$$ and $$1=l\frac{n!}{4d} \tag3$$ for some positive integers $k$ and $l$. From $(2)$ and $(3)$ follows $kl=2$, and hence $k=1$ or $k=2$. But since $H\ne A_n$, it must be $d\ne n!/2$ and hence, by $(2)$, $k\ne 1$. So $k=2$ and hence, again by $(2)$, $d=n!/4$, whence by $(1)$ $|HA_n|=n!$, and finally $HA_n=S_n$: contradiction, by the argument in this post. Therefore, necessarily $H=A_n$.