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 $H$?

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