Solution
= Solution
Let $C_n=\langle g\rangle$ and take $H=\langle g^p\rangle$. Since $n=mp$, the order of $g^p$ is $m$, so $H\cong C_m$. All subgroups of a <cyclic group> are normal. The quotient is generated by $gH$ and has order $n/m=p$, giving
$$
\boxed{C_n/H\cong C_p.}
$$
A group of prime order has no nontrivial proper <subgroup>. By the <subgroup> correspondence theorem, no <normal subgroup> lies strictly between $H$ and $C_n$. Thus $H$ is a maximal proper <normal subgroup>.