Exercise 13.1.2

Proof. Let $H$ a subgroup of $S_n$ such that $[S_n:H]=2$. Then $H$ is normal in $S_n$ (by Exercise 12.1.20). Thus $S_n∕H\simeq \{1,-1\}$. So there exists a group homomorphism

Any two transpositions ${\tau }_1=(ab),{\tau }_2=(cd)$ of $S_n$ are conjugate: if $\gamma =(ac)(bd)$, then ${\tau }_2=\gamma {\tau }_1{\gamma }^{-1}$ (even if $b=c$).

Since $\{1,-1\}\simeq \mathbb{Z}∕2\mathbb{Z}$ is abelian,

So ${\tau }_1,{\tau }_2\in H$, or ${\tau }_1,{\tau }_2\in S_n\setminus H$.

If ${\tau }_1,{\tau }_2$ are in $S_n\setminus H$, then $\phi ({\tau }_1{\tau }_2)=\phi ({\tau }_1)\phi ({\tau }_2)=(-1)\times (-1)=1$, so ${\tau }_1{\tau }_2\in H$. In both cases ${\tau }_1{\tau }_2\in H$.

Since every permutation $\sigma$ of $A_n$ is the product of an even number of transpositions, $\sigma \in H$, so $A_n\subset H$. As $|A_n|=|H|=n!∕2$, $H=A_n$.

$A_n$ is the only subgroup of $S_n$ with $n!∕2$ elements. □