# Centers of quotients of Lie Groups

Exercise 7.11 in Fulton's Representation Theory asks to prove that:

(a) Show that any discrete normal subgroup of a connected Lie group $G$ is in the center $Z(G)$

(b) If $Z(G)$ is discrete, show that $G/Z(G)$ has trivial center.

I was able to solve (a) relatively easily (for $y \in N$, with $N$ a normal discrete subgroup of $G$) by considering the continuous mapping $\phi_y : G \to G$ for $g \mapsto gyg^{-1}y^{-1}$. Since the map is continuous and $G$ is connected, its image must be connected. But since the image is contained in $N$ which is discrete, then the image must be trivial, implying that $gy = yg$ so $y \in Z(G)$ so $N \subset Z(G)$.

However, I am unsure of how to proceed in (b). I know that $G/Z(G)$ is the same as the group of inner automorphisms of $G$, but I don't know if that's how I should proceed or if I should take a different tactic.

Thanks for any help with this problem.

Later Edit:

Since I have to turn this in shortly, I want to say that I eventually ended up using a result from Stillwell's "Naive Lie Theory" that states that $Z(G)$ discrete implies that there are no nondiscrete normal subgroups. Thus, it is pretty easy to show that $G/Z(G)$ is simple, implying that its center must be trivial. However, Stillwell's result uses machinery that won't be introduced in Fulton's text for quite a while, so I'm still unsatisfied with the result. I eventually decided there were at least three possible approaches to the problem:

1. What I ultimately did, except actually using the machinery available up to this point in Fulton to prove Stillwell's result and proceeding from there
2. Possibly something involving $G/Z(G) \cong Inn(G)$, though I still don't know what
3. Proving that $G=[G,G]$ (the commutator of $G$), then using Grun's Lemma to immediately show that the center of $G/Z(G)$ is trivial, though I'm not sure that having $Z(G)$ discrete even implies this fact

I still would like to know a decent solution, so any help/suggestions/proofs would still be more than welcome.

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Are you sure about the result of Stillwell's? Perhas it requires knowing $G$ is simple already? Because if you pick your favorite Lie group $G$ with trivial center, say, $SO(3)$. Then $Z(G\times G) = \{(e,e)\}$ so is discrete but $G\times\{e\}$ is a nondiscrete normal subgroup. –  Jason DeVito Oct 12 '11 at 11:33
Hmm, if you look at the preimage of the center of the quotient, is there some way to conclude that it must be discrete, given that Z(G) is discrete? –  Tobias Kildetoft Oct 12 '11 at 12:35

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 $aZ=Z$.