A group with 3 subgroups. prove it is cyclic Let G be a group with exactly three subgroups. Prove that G is cyclic.
So I know 2 of the subgroups: e (the identity) and G.
And {e} and G are distinct. 
My first thought is to show that G has a generator and that's both e and G right?
Not sure what/how to find the third subgroup and what to do after that. 
This is what I have so far:
PROOF: Let G be a group with exactly three subgroups.
Then by definition, {e} is a subgroup of G and a generator of G.
Also, G is a subgroup of G and is also a generator.

You're using the term generator a little too loosely here. The term generator means that if $g$ generates a subgroup $H$, then $H = \{g^n:n\in\Bbb Z\}$. $H$ is then denoted $\langle g\rangle$.
As for the problem at hand: suppose $g\neq e\in G$. What can you say about $\langle g\rangle$? Can there be any elements not in $\langle g\rangle$?
