# $G/Z(G)$ is cyclic useful for proving groups abelian?

It's a common exercise to prove in an abstract algebra book that if $G/Z(G)$ is cyclic then $G$ must be abelian. But I've always found the exercise strange because if $G$ is abelian then $Z(G)=G$ and the quotient is trivial.

Is there a specific example of this being a useful technique to proving a group is abelian? As it seems you must know enough about a specific group $G/Z(G)$ to proves it's cyclic, but not enough to notice that it the trivial group, which would prove the commutativity of $G$ immediately.

• Usefulness is irrelevant. It's a test of students' ability to use tools they've been given. – anon Dec 7 '14 at 4:57
• @anon: I don't agree. – Martin Brandenburg Dec 7 '14 at 9:10
• I can't understand how to miss that tags when posting a group theory problem asking for the abelian property. – user26857 Dec 7 '14 at 9:21
• I find that the usefulness becomes slightly more apparent when the center is replaced by any central subgroup. – Tobias Kildetoft Dec 7 '14 at 10:02
• Possible duplicate of math.stackexchange.com/q/999247/589. – lhf Dec 9 '14 at 16:45

## 3 Answers

A very good example to serve its usefulness is

Is it possible to have a group $G$ such that 0$(G/Z(G))=91$?solve this and get the beauty of the result

If $G$ is a group of order $p^2$, then $G$ is abelian. It follows immediately from the exercise. And then one knows $G \cong \mathbb{Z}/p^2$ or $G \cong \mathbb{Z}/p \oplus\mathbb{Z}/p$ by the structure theorem.

If $G$ is a group of order $p^3$, then $G$ doesn't have to be abelian. But it is abelian if the center of $G$ has order $\neq p$. In other words, in order to classify (non-abelian) groups of order $p^3$, we may assume that the center has order $p$. This is the first step in their classification.

• Well, from the exercise and the class equation. But it is a good example. – Tobias Kildetoft Dec 7 '14 at 10:00

If $\overline{G}=G/Z(G)$ is cyclic, then, exist $\overline{x}$ such that $\overline{G} = <\overline{x}>$. (Note that, $G/Z(G) = <x>Z(G)$).

Thus for all $g \in G$, exist $i$ such that $\overline{g}=\overline{x}^i$. Consequentely, $g = x^iz$, where $z \in Z(G)$.

Then $g_1,g_2 \in G$, we have that $g_j = x^{i_j}z_j, j=1,2$, where $z_j \in Z(G)$. Now is very simple see that $g_1g_2 = g_2g_1$, ie, $G$ is abelian.