= Solution
Write $C_n=\langle g\rangle$ and set $N=\langle g^p\rangle$. The <order of a group element>[order] of $g^p$ is $n/\gcd(n,p)=m$, so $N\cong C_m$. Every <subgroup> of an <abelian group> is a <normal subgroup>. The <quotient group> $C_n/N$ is generated by $gN$ and has order $n/m=p$, hence is $C_p$ and is a <simple group>. By the <subgroup correspondence theorem>, there is no proper <normal subgroup> strictly between $N$ and $C_n$. Therefore \b[$N$ is a maximal <normal subgroup>] with the required <quotient group>.
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