In a group $G$ where $\exists!$ nontrivial, proper subgroup, show that $G$ is cyclic and $\lvert G\rvert=p^2$, for $p$ prime. 
Suppose that $G$ is a group that has exactly one nontrivial proper subgroup. Then we have to show that $G$ is cyclic and order of $G$ is $p^2$ where $p$ is prime.

I tried as, if $a$ and $b$ two element of $G$ of order $p$ and $q$ then $\langle a \rangle$ and $\langle b \rangle$ are two proper subgroups – a contradiction. Thus every element of $G$ have order some power of $p$. After this I am stuck. Thanks for help.

Claim:$G$ is cyclic.
Proof: If not, 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, $|G| = p^2$
