Let $G$ be a group and $H$ a subgroup of $G$ and $N$ a normal subgroup of $G$. Show that $H \cap N$ is a normal subgroup of $H$ and show by giving a counterexample that $H \cap N$ is not necessarily a normal subgroup of $G$.
Let $x \in H \cap N$, then $x \in H$ and $x \in N$. Since $N$ is normal we have that $g xg^{-1} \in N$ for every $g \in G$. As this holds for every $g$ in particular it holds for every $h \in H$ and so $hxh^{-1} \in N$ for every $h \in H$. But note that since $x \in H \cap N$ we have that $hxh^{-1} \in H$ so $H \cap N$ is a normal subgroup of $H$.
Let $G=\Bbb Z_6$, $H=\langle4\rangle$ and $N=\langle1\rangle = \Bbb Z_6$.
Now I've found online that a subgroup $N=\langle a\rangle$ of $\Bbb Z_n$ is normal if and only if $a \mid n$.
So in this case $1 \mid 6 \implies N$ is normal subgroup of $G$ and since $4 \nmid 6 \implies$ H is not normal subgroup of $G$. However $H \cap N = H \cap \Bbb Z_6 = \{0,2,4\} = H$ which isn't a normal subgroup of $G$ and this concludes the counterexample(?).
I'm trying to prove this result that
A subgroup $N=\langle a\rangle$ of $\Bbb Z_n$ is normal if and only if $a \mid n$.
But I cannot figure out how to approach this. Assuming that $\langle a\rangle$ is a normal subgroup of $\Bbb Z_n$ doesn't seem to be giving me anything to show that $a \mid n$. What can I do here?