Only $1$ Nontrivial Subgroup $\Longrightarrow |G| = p^2$ I am pretty new to this site , so I am not sure how things work, but I am in desperate help with a question that I don't know where to start or finish with. It is for a test I have to study for. 
Here it is 
"If $G$ is a group having only one non-trivial proper subgroup. Prove that $G$ is cyclic and $o(G) = p^2$ where $p$ is a prime number"
Please help me, I have no idea what to do. 
 A: Let $H\subseteq G$ be the unique non-trivial proper subgroup of $G$. Since $H$ is a proper subgroup there is some $g\in G\setminus H$. The subgroup $\langle g\rangle$ generated by $G$ is distinct from $H$ since $g\notin H$ and therefore can not be a proper subgroup. It follows that $G=\langle g\rangle$ for every $g\notin H$, in particular $G$ is cyclic.
Since $G$ has only one non-trivial proper subgroup, it can't be the infinite cyclic group, thus $G$ is finite.
We conclude $G\cong \mathbb Z_n$ for some $n\in\mathbb N$. For every divisor $m\mid n$ we have a subgroup $\langle m\rangle\subseteq\mathbb Z_n$ of order $n/m$ in $\mathbb Z_n$. Since there is only one non-trivial proper subgroup, we can have only one non-trivial proper divisor of $n$, thus $n=p^2$ for some prime $p$.
A: If $G$ were not cyclic, then there must be two distinct elements $a$ and $b$ that generate two distinct cyclic subgroups.  Thus, $G$ must be cyclic.  From here, we know all cyclic groups are isomorphic to $\mathbb{Z}_n$, where $n = |G|$.  At this point, we are done:  For all cyclic groups $H$, there will exist exactly $1$ cyclic subgroup for each divisor of $|H|$.  Thus, $|G| = p^2$.  
A: First show the unique nontrivial proper subgroup is cyclic of prime order $p$. Then argue that any element outside of this subgroup must generate the whole group and have order $p^2$.
