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Proof that if group $G/Z(G)$ is cyclic, then $G$ is commutative
If $G/Z(G)$ is cyclic, then $G$ is abelian

If $G$ is a group and $Z(G)$ the center of $G$, show that if $G/Z(G)$ is cyclic, then $G$ is abelian.

This is what I have so far:

We know that all cyclic groups are abelian. This means $G/Z(G)$ is abelian. $Z(G)= \{z \in G \mid zx=xz \text{ for all } x \in G \}$. So $Z(G)$ is abelian.

Is it sufficient to say that since $G/Z(G)$ and $Z(G)$ are both abelian, $G$ must be abelian?

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marked as duplicate by Chandrasekhar, Jack Schmidt, JavaMan, Chris Eagle, Asaf Karagila Nov 3 '11 at 23:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

It's certainly not sufficient just to say it. You would have to prove it. Don't waste too much time trying to prove this, though, since it's false. For example, if $G$ is the quaternion group, then $G/Z(G)$ and $Z(G)$ are both abelian, but $G$ is not. You really do need the assumption that $G/Z(G)$ is cyclic, not just abelian. – Chris Eagle Nov 3 '11 at 19:35
Oh, okay. How do I use the fact that G/Z(G) is cyclic to show that G is abelian? – Lily Nov 3 '11 at 19:42

Here's part of the proof that $G$ is abelian. Hopefully this will get you started...

Let $Z(G)=Z$. If $G/Z$ is cyclic, then it has a generator, say $G/Z = \langle gZ \rangle$. This means that for each coset $xZ$ there exists some $i \in \mathbb{Z}$ such that $xZ=(gZ)^i=g^iZ$.

Suppose that $x,y \in G$. Consider $x \in xZ=g^iZ$ so that $x=g^iz$ for some $z\in Z$.

Represent $y$ in a similar manner and consider $xy$ and $yx$. Why are they equal?

Edit: !!!Spoiler alert!!! :) Here's the rest of the story.

$yZ \in G/Z = \langle gZ \rangle$ so that $yZ=(gZ)^j=g^jZ$ for some $j \in \mathbb{Z}$. Therefore, $y \in yZ=g^jZ$ so that $y=g^jz_0$ for some $z_0 \in Z$.

Finally, $xy=g^izg^jz_0=g^ig^jzz_0=g^{i+j}zz_0=g^{j+i}zz_0=g^jg^izz_0=g^jz_0g^iz=yx$

The second equality follows because $z$ is in the center and thus commutes with everything. Then we're just messing with powers of $g$ (which commute with themselves). The next to last equality follows because $z_0$ is in the center and thus commutes with everything.

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They're equal because G is abelian? – Lily Nov 3 '11 at 20:22
You're trying to prove $G$ is abelian (so that can't be assumed). They will commute since $x$ and $y$ are made from powers of $g$ and elements of the center. Powers of $g$ and elements of the center commute with each other. – Bill Cook Nov 3 '11 at 20:29

I don't much like that phrasing of the problem (though it is quite standard), since in fact we end up concluding that $G/Z(G)$ is trivial; which, granted, is cyclic, but still...

Generally, I prefer the phrasing:

If $N\leq Z(G)$ and $G/N$ is cyclic, then $G$ is abelian.

Here's a stronger conclusion, due to Baer:

Theorem. (R. Baer, 1938) Let $G$ be a finitely generated abelian group, $$G \cong C_{a_1}\oplus\cdots C_{a_k},$$ where $C_r$ is the cyclic group of order $r$, infinite cyclic with $r=0$, and $1\lt a_1|a_2|\cdots|a_k$. Then $G$ is isomorphic to $H/Z(H)$ for some $H$ if and only if $k=0$, or $k\geq 2$ and $a_{k-1}=a_k$.

(In fact, Baer characterized all abelian groups that can be written as direct sums of cyclic groups and are central quotients, no just the finitely generated ones, and went even further, describing exactly when, given abelian groups $G$ and $K$ that are direct sums of cyclic groups, you can find a group $H$ with $H/Z(H)\cong G$ and $Z(H)\cong K$. See this question for the citation and other related results.)

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