Show that a group G has no nontrivial subgroups if and only if it is a cyclic group of prime order
If $\displaystyle G=\{e\}$ we're done. So, assume not. Then, let $\displaystyle g\in G-\{e\}$ we must clearly have that $\displaystyle \langle g\rangle\leqslant G$ but it isn't trivial and so it must be improper. Thus, $\displaystyle G=\langle g\rangle$. Now, if $\displaystyle m\mid |G|$ then by $\displaystyle G$'s cyclicness we must have that there exists some $\displaystyle H\leqslant G$ such that $\displaystyle |H|=m$. Since $\displaystyle H$ cannot be nontrivial or proper it follows that $\displaystyle m=1,|G|$. The conclusion follows.