Any subgroup of a cyclic normal subgroup is also normal. This question has been asked previously here, but two answers use the idea of characteristic which I haven't been introduced to yet and the last one makes no sense to me. Any hints that don't make reference to characteristic.
 A: Hint:


*

*A cyclic group has exactly one subgroup for each divisor of its order.

*Every conjugate of a subgroup has the same order as the subgroup.

*Every conjugate of a subgroup of a normal subgroup is a subgroup of that subgroup.
A: Let $H\lhd G$ be cyclic and $J<H$. We know that any subgroup of a cyclic group must also be cyclic. Furthermore, all cyclic subgroups of the same order are isomorphic. Now, for each $g\in G$ consider the set $gJg^{-1}$. It has the same order as $J$ since $\phi : J \rightarrow gJg^{-1} :x \mapsto gxg^{-1}$ is a bijection between the two sets. Also, $gJg^{-1} < H$ because $gJg^{-1} \subseteq H$ (normality of $H$), and if $gag^{-1}, gbg^{-1} \in gJg^{-1}$, where $a,b\in J$, then $$gag^{-1}(gbg^{-1})^{-1}=gag^{-1}gb^{-1}g^{-1}=gab^{-1}g^{-1}\in gJg^{-1}$$ Therefore, for all $g\in G$, $J=gJg^{-1}$, i.e. $J$ is normal.
A: Another proof:
Let $N \unlhd G$, and let $H \subseteq N$ be a subgroup. For $g \in G$, let  $\varphi_g: N \rightarrow N$ be given by $ n \mapsto gng^{-1}$(that is, conjugation with $g$). This is an automorphism of $N$. If $N$ is infinite, then $N \cong \mathbb{Z}$, so either $\varphi_g = \text{id}|_{N}$ or $\varphi_g (n) = n^{-1}$ for all $n \in N$, since these are the only two automorphisms of $\mathbb{Z}$. In both cases it is easy to see that $\varphi_g(H) = H$, so that $H$ is normal.
Suppose $N$ is finite, let $x$ be its generator, and let $m$ be its order. Now $\varphi_g$ must map $x$ to some other generator of $N$, so $\varphi_g(x) = x^k$ with $\gcd(m, k) = 1$. $H$ is also finite and cyclic, so there exists some $l$ such that $x^l$ generates $H$. Now $\varphi_g(x^l) = \varphi_g(x)^l = x^{kl} = (x^l)^k \in H$. Since $\varphi_g$ is a homomorphism we see $\varphi_g(H) \subset H$, the other inclusion (not necessarily needed) follows from the fact that for any $h \in H$, $h = gg^{-1}hgg^{-1}$.
