It's widespread that for any group $G$, whenever $G/Z(G)$ is cyclic, $G$ is abelian. But on this site, I have also seen on here the corollary that if $N \leq Z(G)$ has $N \trianglelefteq G$ such that $G/N$ is cyclic, then $G$ is abelian. But I cannot find this result anywhere online. Maybe because it's obvious(?). It's not clear to me though how this is immediate. If there was some direct understanding that if $A \leq B$ then $G/B \leq G/A$, then I would agree that $G/Z(G)$ would inherit its cyclicity from $G/N$, but that's a nonsensical approach. Any help here?
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I am confused. You say that you have found the statement on this site; and it is, see the linked duplicate. But in the very next sentence you say that you cannot find this result online. – Martin Brandenburg Dec 30 '24 at 23:05
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Btw, note that "$G/Z(G)$ cyclic $\Longrightarrow$ $G$ abelian" is a corollary of "$N\unlhd G$, $N\le Z(G)$, $G/N$ cyclic $\Longrightarrow$ $G$ abelian", not vice versa. – Kan't Dec 31 '24 at 08:02