1. ## Prove

Suppose that G is a group that has exactly one nontrivial proper subgroup. Prove that G is cyclic and |G|=p^2 where p is prime.

2. Originally Posted by mandy123
Suppose that G is a group that has exactly one nontrivial proper subgroup. Prove that G is cyclic and |G|=p^2 where p is prime.
Let $\displaystyle H$ be the non-trivial proper subgroup.
Then $\displaystyle \{ e \} \subset H \subset G$.

Pick $\displaystyle a \in G - H$. And construct $\displaystyle \left< a \right>$.

3. ## Explain a little further

Note that once you construct $\displaystyle \left< a\right>$ it cannot be $\displaystyle H$ and cannot be $\displaystyle \{ e \}$. Since that is a subgroup and it has no other proper non-trivial subgroups it follows that $\displaystyle \left< a \right> = G$. Since the group is cyclic it is isomorphic to $\displaystyle \mathbb{Z}_n$ for some $\displaystyle n\geq 1$. By the properties of cyclic groups we can show that only $\displaystyle n=p^2$ are the ones with this unique non-trivial proper subgroup property.