= Groups with no proper nontrivial subgroups
{title2=$G\cong C_p$}
A nontrivial <group> has no proper nontrivial <subgroups> exactly when it is a <cyclic group> of prime <group order>. To prove necessity without presupposing finiteness, take $g\ne e$. Its cyclic subgroup is nontrivial and hence equals $G$. An infinite-order generator would give the proper nontrivial subgroup $\langle g^2\rangle$, so its order $N$ is finite. If $N$ were composite, a proper divisor $d$ with $1<d<N$ would make $\langle g^d\rangle$ another proper nontrivial subgroup. Thus $N$ is prime. Conversely <Lagrange's theorem> excludes intermediate subgroup orders in a prime-order group. The qualification “nontrivial” is necessary because every group with more than one element has the proper <trivial subgroup> $\{e\}$.
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