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.
Best Answer
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)