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This is an exercise in Dummit and Foote (4.6.3) I'm doing for revision: prove that for $n \geq 5$, $A_n$ is the only proper subgroup of $S_n$ such that $|S_n/G| < n$. ($A_n$ is the alternating group on $n$ elements and $S_n$ is the symmetric group on $n$ elements).

Not sure how to get started - given the condition that $n \geq 5$ it seems that the fact that $A_n$ is simple should be useful.

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I assume that you did the exercise above it that asks you to find all the normal subgroups of $S_n$ ($n\geq 5)$. You should get that the only normal subgroups are the trivial, the alternating, and the whole group.

Now to tackle your problem, let $K$ be a proper subgroup of $S_n$ such that $|S_n/K|=k. Then, note that your group $S_n$ acts on the set of left cosets of $K$ via left multiplication. Thus you get the permutation representation given by: $\pi:S_n\longrightarrow S_k$Note that $\ker(\pi)\leq K$, and recall that kernel of homomorphisms are normal, so you have that $\ker(\pi)$ is a normal subgroup of $S_n$, and thus you have that $\ker(\pi)$ is either trivial, $A_n$ or $S_n$. Note that it cannot be $S_n$ because $K$ has to be proper, and $\ker(\pi)$ cannot be trivial since then you would have an injection from $S_n$ into $S_k$ where $k, a contradiction. The only remaining option is that $\ker(\pi)=A_n$.