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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.2011-06-03
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    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.2011-06-03
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    Hi Theo, thanks for all the informations. It is simple to give an example where $G/Z$ is not compact ?2011-06-03
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    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](http://en.wikipedia.org/wiki/Semi-simple_Lie_group) then $G/Z(G)$ will be non-compact. I hope this answers your follow-up question. Theo2011-06-03
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    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.2011-06-03

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For $G$ discrete, it is a well-known theorem of Schur that those $G$ such that $G/Z(G)$ is finite are finite-by-abelian, i.e., their derived subgroup is finite. See this survey by Dixon, Kurdachenko, and Pypka.

The locally compact version also holds. Namely, if $G$ is locally compact and $G/Z(G)$ is compact, then $\overline{[G,G]}$ is compact. Indeed, the assumption that $G/Z(G)$ is compact implies that all conjugacy classes in $G$ have compact closure. This implies, by a result of T.S. Wu and Y.K. Yu (Michigan Math. J, 1972) that the closure $\overline{[G,G]}$ the derived subgroup of $G$ is a locally elliptic group, in the sense that every compact subset is contained in a compact subgroup. This applies to the set of commutators in $G$, which is compact (because the commutator map $G^2\to G$ factors through $(G/Z)^2$), and therefore $\overline{[G,G]}$ is compact.