Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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

1 Answer 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 $aZ=Z$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.