Q1 Prove that every simple subgroup of $S_4$ is abelian.
Q2 Using the above result, show that if $G$ is a nonabelian simple group then every proper subgroup of $G$ has index at least $5$.
My attempt
For Q1:
$|S_4| = 4\cdot3\cdot2 = 2^3 \cdot 3$.
One can exhibit all possible orders of subgroups of $S_4$ and eliminate subgroups according to the requirement that the subgroup is simple.
For example, any subgroup of order $2\cdot 3$ is not simple so needs not to be considered.
Then it might be possible to show that all remaining subgroups are abelian using results like a group of order $p^2$, where $p$ is a prime, is abelian.
For Q2
I think Cayley's Theorem is involved and I should probably consider the quotient of $G$ on a proper subgroup of $G$. But since $G$ is simple, the quotient is not a group. And I don't really know where to go from here.
My question
For Q1, I'm not sure that the above approach would work and even if it does, I feel like it is too cumbersome and there might be a more principled and more concise way.
Any help would be greatly appreciated.
Best Answer
For the record, here is an elementary way of doing the first part - you might have another way of doing order $6$. It avoids any analysis of what the subgroups of order $8$ and $12$ actually are, but shows that if they exist, they have proper normal subgroups and therefore aren't simple.
The order of a subgroup is a factor of the order of the group.
Possible subgroup orders for $S_4$ are $12, 8, 6, 4,3,2$
The only groups of order $4,3,2$ are abelian.
A subgroup of order $6$ will contain an element of order $3$ and this will generate a normal subgroup (since the subgroup will have index $2$).
A subgroup of order $8$ will contain an odd permutation of order $2$ or $4$, and the even permutations will form a proper normal subgroup. (There aren't enough even permutations of order $2$ or $4$ to make a group of even permutations of order $8$)
A subgroup of order $12$ will either contain an odd permutation (in which case the even permutations form a proper normal subgroup) or will be all even permutations in which case the elements $e, (12)(34), (13)(24), (14)(23)$ will form a proper normal subgroup (union of conjugacy classes closed under multiplication and inverse).
The second part is covered by a link you have already been given.