# $G$ is a finite group, prove that $G$ is abelian when $G/C(G)$ is cyclic. [duplicate]

Is it true that when $G/C(G)$ is cyclic, then $G$ is abelian?

The original question is

Prove that if $G$ is a finite non-abelian group, then $4|C(G)|\le G$.
Proof from the answer book: suppose that $1<G/C(G)<4$, then $G/C(G)$ is of order 2 or 3 and therefore cyclic. Hence $G$ would be abelian, a contradiction.

As is said in the title, I have no idea how the implication in the last sentence goes...

## marked as duplicate by Najib Idrissi, Jeremy Rickard, user1551, TravisJ, SiminoreSep 21 '15 at 14:07

• (Note that $Z(G)$ is another notation for the center $C(G)$ of $G$. You don't even need $G$ to be finite.) – Najib Idrissi Sep 21 '15 at 8:34
Let $aC(G)$ be a coset which generates $G/C(G)$, so that $$G=\bigsqcup_{n\geq 0}a^nC(G).$$ Consider any two elements $g,h\in G$. We can write them as $g=a^nx$ and $h=a^my$ where $x,y\in C(G)$. Then $gh=a^nxa^my=a^{n+m}xy=hg$, where we have commuted $x$ and $y$ past everything (because they belong to the center). Hence $G$ is abelian.
A slight variation on pre-kidney's answer: Suppose $a$ is a cyclic generator of $G/C(G)$. Then for any $g\in G$ we have $g=a^nx$ where $x\in C(G)$, so $ag=aa^nx=a^nax=a^nxa=ga$, so $a\in C(G)$ and $G=C(G)$. Hence $G$ is abelian.