Here's a question out of idle curiosity. Let $G$ be a topological group. Is it possible for both $G$ and (a model of) $BG$ to be finite CW complexes? (Apart from the obvious example of $G$ being [up to homotopy] the trivial group.)


A comment is that the fibration sequence $G \to EG \to BG$ shows that $\chi(G)\chi(BG) = 1$ in this case, so $\chi(G) = \pm 1$. This rules out plenty of examples, e.g. Lie groups, finite groups...



Can both G and BG be finite CW complexes?
To expand my comment: No it is not possible. 


First suppose that $G$ is connected and that both $G$ and $BG$ have the homotopy types of finite complexes. If $G$ is not contractible, then let $k>0$ be minimal such that $\pi\_k(G)$ is nontrivial. We have $H\_k(G)=\pi\_k(G)=\pi\_{k+1}(BG)=H\_{k+1}(BG)$. Choose a prime $p$ such that the latter finitely generated abelian group is nontrivial mod $p$. From now on let homology be with mod $p$ coefficients. Let $d$ be maximal such that $H\_d(BG)$ is nontrivial. Let $e$ be maximal such that $H\_e(G)$ is nontrivial. The spectral sequence of the fibration has a nontrivial group at $E^2\_{d,e}$ that will not go away. Contradiction. So $G$ connected implies $G$ contractible.


If $G$ is not connected then $\pi\_0G$ is finite. Let $G\_0$ be the component of the identity in $G$. Then $BG\_0$, the universal cover of $BG$, is also a finite complex, so by the previous paragraph $G\_0$ is contractible. That is, $G$ is (up to homotopy) discrete and $BG$ is the classifying space of a finite group. For any cyclic subgroup $C$ of the finite group $G$, $BC$ is a finite cover of $BG$ and therefore finite. The cohomology of $BC$ is such that it cannot be equivalent to a finite complex unless $C$ is trivial. So $G$ is trivial (contractible). 


