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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 $H$?

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    $\begingroup$ Your hypothesis implies that $H/Z(H)$ is cyclic. $\endgroup$
    – Derek Holt
    Jan 7, 2019 at 21:41

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

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