# quotient of normalizer and centralizer is cyclic group

It's known that if $$G/Z(G)$$ is a cyclic group, then $$G$$ is Abelian. Since $$G/Z(G)$$ is just the special case $$H=G$$ in the $$N/C$$ theorem $$C_G(H)\triangleleft N_G(H)$$. I wonder if the below statament is true:

If $$H$$ is a subgroup of $$G$$, if $$N_G(H)/C_G(H)$$ is a cyclic group, then $$H$$ is Abelian.

I tried using the proof in the original statement, but it didn't work out. If we set $$N_G(H)/C_G(H)=\langle aC_G(H)\rangle$$, how can i represent an element in $$H$$?

• Your hypothesis implies that $H/Z(H)$ is cyclic. Jan 7, 2019 at 21:41

Hint: $$H \cap C_G(H)=Z(H)$$ and $$H \unlhd N_G(H)$$