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.
Best Answer
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, $|G| = p^2$