Cyclic $G/Z(G)$ implies that $G$ is abelian

Question

Suppose that $G$ is a group where $|G| = pq$ for prime $p$ and $q$. Suppose further that $|Z(G)| = q$. Thus, $Z(G) \cong Z_q$ (it is a cyclic group). Moreover, we see that

$$ |G/Z(G)| = \frac{|G|}{|Z(G)|} = \frac{pq}{q} = p. $$

Thus $G/Z(G)$ is also a cyclic group.

Why does this imply that $G$ is an abelian group? Our professor stated this today in lecture, and I can't seem to find a concrete reason why.

If $|G|$ is prime, then one can use Lagrange's theorem to show that $G$ has no non-trivial subgroups. Thus every element other than $e$ generates $G$, in other words $G$ is cyclic. This might be easier than your current method. — hrkrshnn, Oct 29 '14 at 13:40

score 2 · Answer 1 · answered Oct 28 '14 at 07:36

2

Let $G/Z=\langle u Z\rangle$. Let $a,b\in G$, $\exists i,j$ such that $aZ=u^iZ$ and $bZ=u^jZ$. There exists $z_1,z_2 \in Z(G)$ such that $a=u^iz_1$ and $b=u^jz_2$. Clearly, $ab=ba=u^{i+j}z_1z_2$.

answered Oct 28 '14 at 07:36

hrkrshnn

6,287

Can we generalize this to say that if a group $G$ contains a cyclic subgroup, then it is abelian? – Ryan Oct 28 '14 at 07:37
@Ryan. I do not think so. Take any element $a\in G$, then $\langle a \rangle$ is a cyclic subgroup. – hrkrshnn Oct 28 '14 at 07:39
Oh, that makes sense. Thanks for the help. – Ryan Oct 28 '14 at 07:52

Cyclic $G/Z(G)$ implies that $G$ is abelian

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