Let $N$ be a cyclic normal subgroup of a group $G$ and $H$ is any subgroup of $N$.
Prove $H$ is a normal subgroup of $G$
Guessinng that exists $a\in G$ where $\langle a\rangle=N$ of some order.
$H \subset N$, $ H$ is a subgroup of $N$.
$gH =Hg \equiv g h_1 =h_2 g \equiv g a^{k_1} =a^{k_2}g $.
*Not sure if on the right track just spit balling *
Hints:
In a cyclic group of order $n$, for each $k$ dividing $n$, there is a unique subgroup of that order.
Let $g\in G$. $gHg^{-1}$ is a cyclic group of the same order as $H$ (and $H$ is a cyclic group as it is a subgroup of a cyclic group).
$gHg^{-1}$ is a subgroup of $N$ by the normality of $K$ ($gHg^{-1}$ is a subset of $N$ by normality of $N$ and $gHg^{-1}$ is a group, so its a subgroup of $K$).
Can you use these three facts to prove that $gHg^{-1}=H$?
