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Let $G$ be a group and suppose $G/Z(G)$ is cyclic. Prove that G is abelian. where $Z(G)$ is the center of the group. That is,

$$ G(Z) = \{a \in G \ | \ \forall x \in G: ax = xa\} $$

Any ideas?

So far what I have. Consider two arbitrary elements $a,b \in G$, and consider their left cosets $aG(Z)$ and $bG(Z)$. Suppose that $G/Z(G)$ is generated by $yZ(G)$, then $aG(Z) = (yZ(G))^n = y^nZ(G)$ for some $n$ and $bG(Z) = (yZ(G))^m = y^mZ(G)$ for some $m$. Hence $abZ(G) = aZ(G) \cdot bZ(G) = y^nZ(G) \cdot y^m Z(G) = y^ny^mZ(G) = y^{n+m}Z(G) = y^{m+n}Z(G) = y^my^nZ(G) = y^mZ(G) \cdot y^n Z(G) = bZ(G) \cdot aZ(G) = baZ(G)$.

So $abG(Z) = baG(Z)$.

Now consider $e = b^{-1}b$. Clearly $e \in Z(G)$. So we have

$ab = ab(b^{-1}b) = bax$ for some $x \in Z(G)$.

But how to proceed from here? And this the idea so far correct?

DHMO
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Doler017
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    In your first paragraph you are proving that $G/Z(G)$ is abelian however you already know that since every cyclic group is abelian. And then you are trying to use only the fact that $G/Z(G)$ is abelian, however there are groups where $G/Z(G)$ is abelian and $G$ is not abelian. – clark Apr 17 '17 at 05:59
  • Your beginning is promising, but things become easier if you leave the "modulo $Z(G)$" universe earlier. You got that $a=y^nz_1$ and $b=y^mz_2$ for some elements $z_1,z_2\in Z(G)$. After all, you want to show that $a$ and $b$ commute as elements of $G$. Looking at only the cosets of $Z(G)$ is not going to give you that, because then you are identifying elements modulo the center. – Jyrki Lahtonen Apr 17 '17 at 06:06
  • Thanks guys! Those tips are very helpful to improve my proof. – Doler017 Apr 17 '17 at 06:10
  • See also https://math.stackexchange.com/questions/999247/if-g-zg-is-cyclic-then-g-is-abelian-what-is-the-point. – lhf Apr 17 '17 at 10:23

1 Answers1

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$G/Z(G)$ is cyclic so each $g/Z(G) = z ^ {i}/Z(G)$ for some $z \in G$.

Let $x,y \in G$.

We have $x/Z(G) = z \ ^ {i} /Z(G) $ and $y/Z(G) = z \ ^{j}/Z(G)$.

This means $x = z \ ^{i} z_1$ for $z_1 \in Z(G)$ and $y = z \ ^{j} z_2$ for $z_2 \in Z(G)$.

Remember that $z_1g = gz_1$ for all $g \in G$ , because $z_1 $ in the center , the same for $z_2$.

Now we have $xy = z \ ^{i} z_1 z \ ^{j} z_2 = z \ ^{i} z \ ^{j} z_1z_2 = z \ ^{j} z \ ^{i} z_2z_1 = z \ ^ {j} z_2 z \ ^ {i} z_1 = yx$ , so $G$ is abelian.