Topological Groups and the Mapping Class Group

I am currently studying mapping class groups. In particular, I am looking at a relation between the group of topological automorphisms of a topological group (i.e. group automorphisms which are also homeomorphisms of the underlying space) and the mapping class group of the underlying space. Denote the topological automorphism group of a topological group $G$ as $TI(G)$. It is clear that we should look at the mapping class group of the underlying space with a marked point corresponding to the identity. It is the case that if $G$ is $K(\pi,1)$ for some abelian group $\pi$ (since the fundamental group of a topological group is always abelian) then by the Dehn-Nielsen-Baer theorem $Mod(G)\cong Aut(\pi)$. However, I am uncertain as to whether there are many topologcial groups which are in fact Eilenberg-Maclane spaces. So in other words, are there constructions of more complicated topological groups from ones which are Eilenberg-Maclane spaces?

• The product of a $K(G,n)$ and a $K(G',n)$ is a $K(G\times G',n)$ if that helps. – Dan Rust Mar 3 '14 at 15:19
• Maybe I should refine the question to ask whether a classifying space of an abelian group is in fact a topological group? – Joseph Zambrano Mar 3 '14 at 17:59
• Among Lie groups, only virtually solvable ones have contractible universal cover. You also have to decide what do you mean by the mapping class group in dimensions $>2$, as there is no standard definition. Lastly, you are misusing the theorem of Dehn et al since it works only for surfaces. – Moishe Kohan Mar 4 '14 at 0:20
• We will define the mapping class group of a topological space $X$ as $Mod(X)=Homeo^+(X)/Homeo_0(X)$ where $Homeo^+(X)$ denotes the orientation preserving homeomorphisms from $X$ to itself and $Homeo_0(X)$ denotes the connected component of the identity. I was under the impression that $Mod(X)\cong Out(\pi_1(X))$ if $X$ is $K(\pi,1)$ but I realized that this is only true of surfaces. – Joseph Zambrano Mar 4 '14 at 6:45