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What does G/Z(G) mean in the below proof? I thought it meant the set of left cosets where $\{gZ(G) : g \in G \}$ If this is true though I don't understand how it can be referred to as being "cyclic" later in the proof. I think I'm missing something fundamental possibly.

http://exwiki.org/mw/index.php?title=If_G/Z(G)_is_cyclic_then_G_is_abelian

From the article

"Let $a,b \in G$ and let $xZ(G)$ be the generator of $G/Z(G)$ so that every element of $G/Z(G)$ is of the form $xkZ(G)$ for some $k ≥ 0$. Since $G/Z(G)$ is cyclic $aZ(G)$ is of the form $xkZ(G)$ for some $k ≥ 0$ and since $a ∈ xkZ(G)$ it follows that $a = xkz$ for some $z ∈ Z(G)$. In the same manner we conclude that $b = xlx′$ for some $l ≥ 0$ and $z′ ∈ Z(G)$. Since $xkxl = xlxk$ and the elements $z, z′$ commute with elements in $G$ we have $ab = xkzxlz′ = xlz′xkz = ba$.

This implies that $G$ is indeed abelian."

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  • $\begingroup$ There's a group structure on $\{ gZ(G):g\in G\}$ because $Z(G)$ is normal in $G$. Did you know that? $\endgroup$
    – user121880
    May 25, 2014 at 2:19
  • $\begingroup$ When $H$ is normal in $G$, $G/H$ has a canonical group structure. Since $Z(G)\lhd G$ always, $G/Z(G)$ is a group under $(hZ)(gZ)=(hg)Z$. $\endgroup$
    – Pedro
    May 25, 2014 at 2:19

2 Answers 2

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$G / Z(G)$ refers to the quotient group formed from studying $Z(G)$ as a normal subgroup of $G$. The elements of this quotient group are indeed the left (or equivalently, right) cosets $\{g Z(G)\}$; the multiplication in this group is given by

$$a Z(G) \ast bZ(G) = (ab) Z(G)$$

In this context, to say the group is cyclic means that there is a coset $x Z(G)$ such that for any other coset $g Z(G)$, there is an integer $n$ with

$$(x Z(G))^n = x^n Z(G) = g Z(G)$$

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$G/Z(G)$ denotes the quotient group, i.e., $G/Z(G)$ is the quotient group of $G$ by $Z(G)$.

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