A cyclic subgroup is normal?

Question

if i have a group G, and a normal cyclic subgroup H, any subgroup of H is normal to G? It is to H, but i don't know to G. Can you give me some help with this?

Answer 1

It is indeed normal in $G$. To see this, consider a generator $a$ of $H$. Any subgroup $K$ of $H$ is cyclic, generated by some $b=a^r$.

Lets consider an element $b^i$ of $K$, and $g\in G$. Since $H$ is normal in $G$, $gag^{-1}=a^k$ for some $k$. Then, as conjugation is an automorphism of $G$ $$gb^ig^{-1}=ga^{ri}g^{-1}=\bigl(gag^{-1}\bigr)^{ri}=\bigl(a^{k}\bigr)^{ri}=\bigl(a^{r}\bigr)^{ik}=b^{ik}\in K.$$

Answer 2

Yes. If $\;H=\langle x\rangle\lhd G\;$ , then any $\;K\le H\;$ is of the form $\;K=\langle x^k\rangle\;$ , so for all $\;g\in G\;$ :

$$g^{-1}(x^k)^mg=\left(g^{-1}x^mg\right)^k\stackrel{\text{ normality of}\;H}=(x^r)^k=(x^k)^r\in K\implies K\lhd G$$

Answer 3

If $N$ is any cyclic group, finite or infinite, then every subgroup $H$ of $N$ satisfies $[N : H] < \infty$, and for each $d \in \mathbf{N}$, there is at most one subgroup $H$ of $N$ such that $[N : H] = d$. The only cyclic groups are $\mathbf{Z}$ and $\mathbf{Z}/n\mathbf{Z}$, so you can verify this fact directly for these groups.

It follows that if $N$ is cyclic and normal in $G$, $H$ is a subgroup of $N$, and $g \in G$, then $gHg^{-1}$ is a subgroup of $gNg^{-1} = N$. But $$[N : H] = [gNg^{-1} : gHg^{-1}] = [N : gHg^{-1}]$$

and so $H = gHg^{-1}$.