Exercise 4.6.3 ($A_n$ is the only proper subgroup of index $<n$ in $S_n$ ($n \geq 5$))

Proof. For the sake of contradiction, assume that $H\ne A_n$ is a proper subgroup of $S_n$ of index

If $H\le A_n$, then

Since $H\ne A_n$, this is impossible by Exercise 1.

Therefore $H$ is not a subgroup of $A_n$. Since $A_n⊴S_n$, $H{A}_n$ is a subgroup of $S_n$. Since $A_n\le H{A}_n$, where $A_n\ne H{A}_n$ (otherwise $H\le A_n$), then $S_n=H{A}_n$. By the Second Isomorphism Theorem,

so $|H:H\cap A_n|=2$.

From $|S_n:H\cap A_n|=|S_n:A_n|\cdot |A_n:H\cap A_n|=|S_n:H|\cdot |H:H\cap A_n|$, we obtain

So $H\cap A_n$ is a proper subgroup of $A_n$ of index $k<n$: this is impossible by Exercise 1.

In conclusion, $A_n$ is the only proper subgroup of index less than $n$ in $S_n$ for all $n\ge 5$ □