TITLE: Why is BG infinite dimensional for G finite ? 
QUESTION [24 upvotes]: If $G \neq \lbrace 1 \rbrace$ is a finite group with classifying space $BG$ 
then there are infinitely many i such that $H^i(BG,\mathbb{Z}) \neq 0$. This 
can be found, for example, there:  
Non-vanishing of group cohomology in sufficiently high degree
As a consequence, the CW-complex $BG$ (unique up to homotopy) can not be of finite dimension. 
Question: Are there alternative proofs for this observation. In particular, I would be interested in knowing if there is a purely topological proof without homological algebra.

REPLY [15 votes]: A proof based on fixed point theory: If $BG=EG/G$ is finite-dimensional, then $EG$ is as well and has the homology of a point. Choose a non-trivial Sylow subgroup $P$ of $G$. By a well-known theorem of P.A. Smith, the fixed point set $EG^P$ is non-empty, contradicting the free action of $G$ (and hence $P$) on $EG$. Consequently $BG$ must be of infinite dimension.