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It is a standard exercise to prove that if $G/Z(G)$ is cyclic then $G$ is abelian. I only know one proof of this (the standard one, which is all over this site), which is pretty useless in the sense that it doesn't offer any insight.

Is there a natural way to see that if $G/Z(G)$ is cyclic then $G$ is abelian? Perhaps by looking at the problem from some larger perspective?

I'm pretty much advanced in algebra but I am slowly building up my intuition in algebraic stuff, so it is important for me to understand things of this sort; that's the motivation for the question.

2 Answers2

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I'll give it a shot . . .

If there's only one generator, $x$, say, after modding out by $Z(G)$, then since all powers of $x$ commute with $Z(G)$, and of course, commute with each other, the question arises:

Why isn't $x$ in $Z(G)$?

Of course, it must be.

More generally, if a subset $S$ of a group $G$ is such that the elements of $S$ commute, then the subgroup of $G$ generated by $S$ is abelian.

In the context of the current question, just let $S = Z(G) \cup \{x\}$, and note that by hypothesis, the subgroup of $G$ generated by $S$ is all of $G$.

quasi
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  • I feel a bit dumb, but I don't quite understand this. Can you please try to rephrase your answer? Thanks! –  Jun 14 '17 at 00:42
  • Since $Z(G)$ commutes with $x$, what's left in $G$ to fail to commute? Everything in $G$ is a power of $x$, mod $Z(G)$. – quasi Jun 14 '17 at 00:44
  • @DerekElkins you cant, consider the quaternion group. The center is $1,-1$ and clearly $G/Z(G)$ is abelian as it has order $4$. – Asinomás Jun 14 '17 at 01:00
  • @Derek Elkins: You can generalize only if you can choose commuting elements of $G$ which generate $G$, mod $Z(G)$. That's requiring more than just an abelian quotient. – quasi Jun 14 '17 at 01:07
  • @OpenBall Ball Maybe you can see the point of quasi more clearly with some symbols: Take any $y\in G$. Then $yZ(G)=x^k Z(G)$ for some $k$. So $y=x^kz$ for some $z\in G$. Now $xy=x^{k+1}z=zx^{k+1}=zx^k x= yx$ for all $y\in G$. So $x\in Z(G)$. Therefore $G/Z(G)$ is the trivial group meaning $G=Z(G)$ is abelian. – Hamed Jun 14 '17 at 01:48
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In general you only need a set $S$ of elements such that for each coset of $Z(G)$ there is exactly one $s\in S$ in that coset and such that all elements of $S$ conmute.

If you think about the cayley graph this lets you decompose every element of $G$ into two paths, the first lets you move between cosets and the second inside the coset.

Asinomás
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  • Thank you Jorge! I'll try to understand this later on. Though some elaboration (in both paragraphs) would be great. –  Jun 14 '17 at 00:44