= Solution
For the intended finite nontrivial case, let $G=\langle g\rangle$ have composite order $N$. Choose a prime divisor $p$ of $N$, with $1<p<N$. Then
$$
\boxed{H=\langle g^{N/p}\rangle}
$$
is a nontrivial proper <cyclic subgroup> of order $p$. Indeed $(g^{N/p})^p=e$, and no smaller positive power is the identity because $g$ has order $N$. Thus $1<|H|=p<N$.
If the source also allows an infinite cyclic group, $\langle g^2\rangle$ is a proper cyclic subgroup: its powers have even exponents and do not include $g$. The trivial group, of order one, is an exception to the literal formulation “not of prime order”: it has no proper subgroup at all. The nontriviality assumption is therefore necessary for that general wording. These facts are instances of <subgroups and quotients of a cyclic group>.
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