I have the following question:
For $n\ge 5$ show that the symmetric group $S_n$ cannot have a subgroup $H$ with $3\le [S_n:H]< n$. ($[S_n:H]$ is the index of $H$ in $S_n$).
This is technically not a homework question, since it's from one of the past qual exams, but still I would prefer to get a hint, rather than full solution.
What did I get thus far: suppose $ [S_n:H]=k$, then $|H|=\frac{n!}{k}$. Then by counting argument I proved that $|A_n\cap H|\ge \frac{n!}{2k}$, so either $|A_n\cap H|= \frac{n!}{2k}$ or $H\le A_n$. But here I stuck.
So I have two questions:
1) Am I on the right track? If so, what I should do next?
2) If not, where I should start?
Thanks.