Vrftsjtryk

Suppose that $N$ is a normal and cyclic subgroup of $G$, and $H$ is a subgroup of $N$. Show that $H$ is normal in $G$.

What should I do when $N$ is finite?
I have tried to show that it is right for infinite $N$.

Let $N=langle nrangle$ and $H=langle n^arangle$. To show that $H$ is normal in $G$, it is enough to show that for all $gin G$ we have $gn^ag^{-1}in H$.

First since $nin N$ and $N$ is normal in $G$, then $gng^{-1}in N$. Write $gng^{-1}=n^b$ for some $b$. Now
$$gn^ag^{-1}=(gng^{-1})^a=(n^b)^a=(n^a)^bin H$$
since $n^ain H$ and we are done.

An abelian subgroup is not necessarily normal.

Hence, the assumption that $N$ is cyclic matters.

Indeed, let $g in G$, $h in H$. Then $ghg^{-1} in N$ because $N$ is normal, and $(ghg^{-1})^{|H|} = gh^{|H|}g^{-1}=e$, thus (this is a standard fact for cyclic groups) $ghg^{-1} in H$.

Edit: if I understood the comment correctly, I need to prove that if $N$ is a cyclic group and $H$ is a cyclic subgroup of $N$, then $H={h in N,,h^{|H|}=e}:=H_0$.
First, $H subset H_0$ (Clearly $H_0$ is a subgroup of a $N$, containing a generator of $H$).
Conversely, let $n$ be a generator of $N$. Let $|N| > k geq 1$ be such that $n^k$ is a generator of $|H|$. Then the set of all $n^{kl}$ for integers $l$ is exactly $H$ and has cardinality $|H|$. Besides, the value of $n^{kl}$ is uniquely deterlined by that of $kl$ mod $|N|$. So $|H|=|N|/gcd(|N|,k)$, and, for every integer $l$, $n^l in H_0$ iff $n^{|H|l}=e$ iff $|N| | |H|l$ iff $gcd(|N|,k)|l$, iff $l=a|N|+bk$ for some $a,b$, iff $n^l=(n^k)^b$ for some $b$ iff $n^l in H$.

Let's start you off,

we know

G > N > H

N is normal

wts

H is normal in G. 

that is, wts hG=Gh for all h in H

We know,

for all n in N, nG=Gn... 

what next?