 Solution

Let $C_n=⟨g⟩$ and take $H=⟨g^p⟩$. 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 $n/m=p$, givingA 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.