TITLE: Does every non-trivial finite group have a subgroup with prime index?
QUESTION [8 upvotes]: Let G be any non-trivial finite group.

Has G always a subgroup, whose index is prime ?

If G is solvable and |G| has a prime divisor $p$, such that $p^2$ does not divide $|G|$, this is the case because of Hall's theorem. 
If $G$ is a $p$-group, the answer is also positive. 
The group $A_5$, for example, is not solvable, but has subgroups with index $5$.
So, I wonder whether we always can find a subgroup with prime index.

REPLY [2 votes]: $A_n$ , has no subgroup of prime index iff $n \geq 6$ and $n$ is not prime .  

Because of this theorm :
Theorem : If $G$ is a simple group with subgroup of index $m>1$, then  $$|G|\quad |\quad m!$$ Proof :  if $G$  has subgroup $H$ , and $G$ act on the cosets of $H$, then we have homomorphism from $G$ to $S_m$ with trivial kernel (because $G$ is simple).
Of course $A_{p-1} \triangleleft A_p$ .$\Box$  
For another examples , because only normal subgroups of $S_n ( n \geq 5)$ are $ 1 , A_n $and $S_n$, by the same way you can show there is no subgroup of index $2<r<n$ in $S_n$ .