# nontrivial subgroup

If $G=\{e\}$ we're done. So, assume not. Then, let $g\in G-\{e\}$ we must clearly have that $\langle g\rangle\leqslant G$ but it isn't trivial and so it must be improper. Thus, $G=\langle g\rangle$. Now, if $m\mid |G|$ then by $G$'s cyclicness we must have that there exists some $H\leqslant G$ such that $|H|=m$. Since $H$ cannot be nontrivial or proper it follows that $m=1,|G|$. The conclusion follows.