G be a finite group and H be a non trivial proper subgroup of index 3, then G is not simple. If the index of a subgroup is the smallest prime dividing the order of group then that subgroup is normal, with this, I am done with the case order of group is odd. How to proceed when order is even.

In general: if $G$ is a finite non-abelian simple group and $H$ is a subgroup, then $|G:H| \geq 5$.
Proof (Sketch) Assume $|G:H| \leq 4$. Let $G$ act by left multiplication on the left cosets of $H$. The kernel of this action is $core_G(H)=1$, since $G$ is simple. Hence $G$ can homomorphically be embedded in $S_{|G:H|}$. Since $S_4$ is solvable and $G$ is simple, it follows that $G=1$, a contradiction. 
Observe that $A_5$ is the first example of a non-abelian simple group and that $|A_5:A_4|=5$.
