= Solution
A <cyclic group> is generated by one element: $G=\langle g\rangle=\{g^n:n\in\mathbb Z\}$. The hypothesis needs the qualification \b[no proper nontrivial <subgroup>]: every nontrivial <group> has the proper <trivial subgroup> $\{e\}$, so the literal wording admits no examples.
Under that intended hypothesis, choose $g\ne e$. The <subgroup> $\langle g\rangle$ is nontrivial and must therefore be all of $G$, proving cyclicity. If $g$ had infinite order, then $\langle g^2\rangle$ would be nontrivial and proper, since $g$ is not an even power of an infinite-order generator. Thus $g$ has finite order $N>1$ and $|G|=N$.
If $N$ were composite, choose a divisor $d$ with $1<d<N$. Then $\langle g^d\rangle$ would have order $N/d$, strictly between one and $N$, again contradicting the hypothesis. Consequently
$$
\boxed{G\cong C_p\text{ for a prime }p.}
$$
Conversely, <Lagrange's theorem> shows that a <group> of prime order has only itself and its trivial <subgroup>. This proves the characterization of <groups with no proper nontrivial subgroups>, even without assuming finiteness at the outset.
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