Why is BG infinite dimensional for G finite ?  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.
 A: I believe there's an argument using Euler characteristic. Let $G$ be a finite group, $BG=K(G,1)$ the classifying space, and $EG=\widetilde{BG}$ the universal cover, which is contractible. Then $\chi(EG)=1$. Now, if $H_{\ast}(G)=H_{\ast}(BG)$ were finite, then $\chi(BG)$ would be an integer (use whatever field coefficients you prefer). But by the multiplicativity of Euler characteristic, then $\chi(BG)|G|=\chi(EG)=1$, so $\chi(BG)=1/|G|$, a contradiction. I forget who this argument is attributed to. Also, one may see geometrically that any finite group has a classifying space with finitely many cells in each dimension, so if $H_{\ast}(BG)$ is infinite, it must be non-vanishing in infinitely many dimensions (i.e. not infinite rank in a single dimension). 
A: For every subgroup $H\subset G$, $BH$ occurs as a covering space of $BG$. If $BG$ were finite-dimensional then every covering space would be finite-dimensional. But for $C_p$ cyclic of prime order $p$ the space $BC_p$ has nontrivial mod $p$ cohomology in infinitely many (in fact all) dimensions. This can be made pretty geometric: there is a nice cell structure with one cell in every dimension and manifolds (lens spaces) as the odd-dimensional skeleta ...
EDIT  By the way, this also yields the more general statement that $BG$ cannot be finite-dimensional unless $G$ is torsion-free.
EDIT  In response to a comment here are some details: Make $C_p$ act on $S^{2n-1}$, the unit sphere in $\mathbb C^n$, freely by $p$th roots of $1$. This sphere is $(2n-2)$-connected and the union as $n\to\infty$ is contractible, so the orbit space is a model for $BC_p$. One can describe a cell structure in $S^{2n-1}$ with $p$ cells in every dimension up to $2n-1$ yielding a cell structure on the orbit space with one cell in every dimension up to $2n-1$, so that $BC_p$ gets one cell in every dimension. You can work out the boundary maps and see that the mod $p$ cohomology is nontrivial in all dimensions. Or you can save some trouble by using Poincare duality, since these odd-numbered skeleta are manifolds.
A: 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.
