Maximal normal subgroup has prime index

Question

I am trying to solve the following exercise taken from Rotman's An Introduction to the Theory of Groups:

Let $M$ be a maximal subgroup of $G$. Prove that if $M \lhd G$, then $[G:M]$ is finite and equal to a prime.

I am completely lost with this exercise, I would appreciate any suggestions.

Answer 1

You may want to prove that the only groups which have only two subgroups -- the trivial ones: the unit and the the whole group itself -- are finite (and cyclic) groups of order a prime number.

Answer 2

Starter hint: maximality of $M$ translates into some fact about $G/M$.