Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 8 / v / Solution

For the intended finite nontrivial case, let $G=⟨g⟩$ have composite order $N$. Choose a prime divisor $p$ of $N$, with $1<p<N$. Thenis 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, $⟨g^2⟩$ 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.