Let $G$ be a finite group, $N$ a cyclic normal subgroup of $G$, and $H$ any subgroup of $N$. Prove that $H$ is a normal subgroup of $G$.
This proof has errors probably:
Since $N$ is cyclic, let $N$ be generated by $\langle n \rangle$, for some $n \in N$. Also, $N$ is normal in $G$, so we have $gng^{-1} \in N$. Since $H \le G$, closure holds in $H$. So for some $n \in H$ (since $n \in G$), $gng^{-1} \in H$.
Best Answer
Let $g \in G $.
Observe that $H^{g} \subseteq N^{g} = N$.
N is finite and cyclic, so $H$ is the only subgroup of $N$ with cardinality $|H|$. So $H^{g} = H $ and hence $H$ is normal in $G$.