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I am reading an introductory material about topological groups and the question in the tittle comes up. Due this Proposition

Proposition. A locally compact Hausdorff topological group $G$ is compact, if and only if, $\mu(G)<+\infty\qquad $ ($\mu$ is the Haar measure of $G$).

it is enough to know what are the groups $G$ for which $G/Z$, where $Z$ is the center of $G$, is a compact group. Are those groups well-known ?

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You are asking about the classification up to isomorphism of (topological) central extensions $0\to A\to G\to K\to1$ with $A$ locally compact abelian and $K$ compact. These extensions with $K$ and $A$ fixed are classified by the second continuous cohomology group $H_{c}^2(K,A)$. Phrasing it this way it is rather clear that full answer is rather impossible. For specific (nice) groups $K$ (e.g. semi-simple Lie) and $A$ not too bad (e.g. $\mathbb{R,Z,R/Z}$ and finite products thereof) the answer should be known but I don't have the relevant texts (Guichardet, Borel-Wallach) handy at the moment. –  t.b. Jun 3 '11 at 6:28
    
In my first sentence I was a bit ambiguous. I meant: you are asking in particular about the classification up to isomorphism of all $G$ arising as such topological central extensions. –  t.b. Jun 3 '11 at 6:37
    
Hi Theo, thanks for all the informations. It is simple to give an example where $G/Z$ is not compact ? –  Leandro Jun 3 '11 at 7:05
    
Hi Leandro This should be a comment rather than an answer, but I can't currently login from this computer. A very simple example of a non-compact quotient group is $PSL_2 (\mathbb{R}) = SL_2 (\mathbb{R})/\{\pm 1\}$ where $\pm 1$ is the center of $ SL_2 (\mathbb{R})$ consisting of $\pm$ the identity matrix. More generally, if $G$ is a non-compact semi-simple linear Lie group then $G/Z(G)$ will be non-compact. I hope this answers your follow-up question. Theo –  t.b. Jun 3 '11 at 8:57
    
Hi Theo, thanks a lot. I first thought about some matriz groups but I wrong convinced myself that they were compacts. The collection of example you provided let more clear that is in pratical very hard to classify all the groups with that property. –  Leandro Jun 3 '11 at 16:16

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