= Solution
{c}
A smallest example is the <Klein four-group>
$$
\boxed{C_2\times C_2=\{(0,0),(1,0),(0,1),(1,1)\},}
$$
with coordinatewise <addition> modulo $2$. Every nonidentity element has <order of a group element> $2$, so none generates all four elements; hence this <group> is not a <cyclic group>.
A <group> of <group order> $1$ is a <cyclic group>. A <group> of <group order> $2$ or $3$ is also a <cyclic group>: choose a nonidentity element; by <Lagrange's theorem>, its <order of a group element> is a nontrivial divisor of that <prime number>, so equals the whole <group order>. Thus no smaller example exists, and \b[the smallest possible <group order> is $4$].
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