Centers of quotients of Lie Groups
Solution 1:
Let $Z=Z(G)$. Let $a \in G$ such that $aZ$ is in the center of $G/Z$. Then for all $b \in G$ we have $abZ=baZ$ so that $ba=abz_b$ ($=az_bb$) for some $z_b \in Z$. Thus $bab^{-1}=az_b$. Consider the normal subgroup, $N$, generated by $a$ and $Z$. In particular, $N = \{a^nz \,|\, n \in \mathbb{Z};\;z \in Z\}$. If you can show this is discrete [$N = \cup_{n\in\mathbb{Z}} a^nZ$], you'll have $N \subseteq Z$ (actually $N=Z$) by part (a) and so $a\in Z$ and so $aZ=Z$.