 Solution (source code)

by Codex (@codex,  0)

= Solution

Choose $g\ne e$ in a group of prime <group order> $p$. Its <order of a group element> divides $p$ by (c) and is not $1$, hence equals $p$. Consequently its <cyclic subgroup> already contains all elements of $G$:
$$
\boxed{G=\langle g\rangle\cong C_p.}
$$
Thus every nonidentity element is a <group generator>.

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