Date modified: Thu 2023-05-11 . 11:51 AM
Date created: Sat 2023-10-07 . 09:02 PM
# $G/Z(G)$ cyclic implies $G$ is abelian. This is a classic exercise, but has a cute application in the [[0 inbox/0.625 theorem|5/8 theorem]]. And since when $G$ abelian, the center $Z(G)=G$, this implies either $G/Z(G)$ is trivial (when $G$ abelian), or $G/Z(G)$ non-cyclic (when $G$ non-abelian). Recall the center $Z(G)=\{g\in G: xg=gx \text{ for all }x\in G\}$, namely the set of all elements in $G$ that would commute with any element in $G$. This is a normal subgroup, so the quotient $G/Z(G)$ is well-defined. Ok, now let us prove our main claim! --- Let us write $Z=Z(G)$, and suppose $G/Z$ is generated by some coset $gZ$. So $G/Z = \langle gZ \rangle$. Take any $x,y\in G$. This means $x\in g^a Z$ and $y \in g^b Z$ for some integers $a,b$. In particular we can write $x = g^a z_1$ and $y = g^b z_2$ for some $z_1,z_2 \in Z$. So $xy = g^az_1 g^bz_2 = g^bz_2 g^az_1=yx$, as any power of $g$ commutes with any power of $g$, and any element in $Z$ commutes with everything. Done! #group-theory