## The G/Z Theorem

Theorem: Let $G$ be a group and let $Z(G)$ be the center of $G$. If $G/Z(G)$ is cyclic, then $G$ is Abelian.

Proof: Suppose that $G/Z(G)$ is cyclic. Then there is some $aZ(G)\in G/Z(G)$ such that $\left=G/Z(G)$.

Let $x,y\in G$. We wish to show that $xy=yx$. It follows that there exist integers $j,k$ such that $xZ(G)=(aZ(G))^j$ and $yZ(G)=(aZ(G))^k$. So there exist $z_1,z_2\in Z(G)$ such that $x=a^jz_1$ and $y=a^kz_2$. Now consider their product:

$xy=(a^jz_1)(a^kz_2)$

$=a^j(z_1a^k)z_2$

$=(a^ja^k)(z_1z_2)$

$=(a^ka^j)(z_2z_1)$

$=(a^kz_2)(a^jz_1)=yx$

Note that in the third line above we are able to commute $a^k$ and $z_1$ since $z_1\in Z(G)$. Similarly we are able to commute from line four to five.

Thus, we have shown that $G$ is Abelian.

$\Box$

Reflection: It is necessary that your subgroup is the center of $G$, otherwise we wouldn’t be able to commute and conclude that $G$ is Abelian. The proof itself followed mainly from definitions and helpful manipulation.