What is the Non-Simplest Group of Order mp?

[playa007]

Homework Statement

Let G be a group with |G|= mp where p is prime and 1<m<p. Prove that G is not simple

Homework Equations

The Attempt at a Solution

I have proven the existence of a subgroup H that has order p(via Cauchy's Theorem), but I don't know how to use the representation of G on cosets of H or another method to somehow deduce that H is a normal subgroup of G thus forcing G to be not simple.

 

[morphism]

Let G act on the cosets of H by left translation. This induces a homomorphism f:G->S_[G]=S_m (why?). Since kerf sits in H (why?), we can consider [H:kerf]. Since |H|=p, it follows that either [H:kerf]=1 or [H:kerf]=p (why?). If it's the former, we're done (why?). So suppose that [H:kerf]=p and deduce a contradiction.