= Solution
{c}
Let $N=|G|$ be a <square-free integer>. If $N=1$, then $G$ is the trivial <cyclic group>. Otherwise, finiteness means that a <direct sum of modules> decomposition into <cyclic groups> has only finitely many nontrivial factors, each of finite <group order>. Omit trivial factors and write
$$
G\cong\mathbb Z_{n_1}\oplus\cdots\oplus\mathbb Z_{n_r},\qquad
n_j>1,\qquad N=\prod_{j=1}^r n_j.
$$
The <integers> $n_i,n_j$ are <coprime integers> whenever $i\ne j$. If a <prime number> $p$ divided both, then $p^2\mid N$, contradicting the <square-free integer> hypothesis.
Moreover, $n_1\cdots n_{j-1}$ and $n_j$ are <coprime integers>: any common <prime divisor> would divide one of the earlier factors as well as $n_j$. Thus part (b)(ii) can be applied repeatedly to combine factors, giving
$$
\boxed{G\cong\mathbb Z_{n_1\cdots n_r}=\mathbb Z_N}.
$$
In particular $G$ is a <cyclic group>. Under the displayed identifications, the element whose coordinate in every factor is the class of $1$ is a <generator of a group>.
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