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.
 A: 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 $\langle g\rangle$?
A: Since G has only three subgroups, H has only two - $\lbrace e\rbrace$ and itself (otherwise we would have a fourth subgroup in G). So H is cyclic, i.e., $H=\langle h\rangle$ for some $h \in H$. More importantly, though, consider $g \in G$ not in H, and consider $\langle g \rangle$. Since $g \not\in H$, we have $\langle g \rangle \neq H$. What else could $\langle g \rangle$ be, then?
As an aside, to clarify what a generator is, remember that we define a generator to be an element $g$ of the group such that every element of the group is a multiple of it, i.e. $\forall x \in G, x = g^k$ for some $k \in \mathbb Z$. So $e$ cannot be a generator, since the only multiple of $e$ is $e$ itself. Similarly, $G$ cannot be a generator, since being a generator is something we define for elements, not sets.
A: Let $H$ be the third subgroup, so by assumption $\{e\}\ne H\ne G$. Then there exists $x\in G\setminus H$. Then $\langle x\rangle$ must be one of $\{e\}, H, G$ and it certainly is neither of the first two. (The argument can be generalized as follows: A group with a proper subgroup that contains all proper subgroups is cyclic)
