A lower bound to the index of a subgroup of a non abelian simple group

Question

Let G be a simple nonabelian group and $p$ is the largest prime number which divides $|G|$, prove that if $H$ is a subgroup of $G$ then $|G:H|\ge p$

I tried to show that if $|G:H|<p$ then $H$ is equal to $G$. I can not understand why this is not true...

Answer 1

Extended hints. Combine the following:

1. The action of $G$ on the left cosets of $H$ gives a homomorphism $f$ from $G$ to $Sym(G/H)\simeq S_n$. Why is this homomorphism non-trivial?
2. If $[G:H]<p$ then $|Sym(G/H)|$ is not divisible by $p$. What does that tell you about the homomorphism $f$?