Integral group cohomology of Lie groups pure torsion in odd degree? Let $G$ be a compact finite-dimensional Lie group. Is it true that the integral cohomology $H^k(BG, \mathbb{Z})$ (where $BG$ is the classifying space of $G$) is pure torsion (i.e. finite) whenever $k$ is odd?
 A: The spectral sequence comment outlined in the comments is sufficient. 
When $G$ is connected, use the Hopf theorem that $H^*(G;\Bbb Q) \cong \Lambda(x_{k_i})$, the exterior algebra on some number of generators $x_{k_i}$ with odd degree. Then the cohomological Serre spectral sequence applied to $BG \to EG \to G$ forces $BG$ to be a polynomial algebra on generators of degree $|x_{k_i}| + 1$. (The proof is not trivial but not altogether difficult.) A more complicated approach follows from Chern-Weyl theory. In any case the generators of this polynomial algebra lie in even degree, so $H^*(BG;\Bbb Q)$ is concentrated in even degree.
When $G$ is not connected, the Serre spectral sequence for $BG_0 \to BG \to B(G/G_0)$ takes the form of $$H^*(G/G_0; H^*(BG_0)) \implies H^*(BG),$$ where the $E_2$ page is group cohomology for the action of $G/G_0$ on $H^*(BG_0)$ (induced by conjugation on $G_0$). 
If you work over the rationals, because $G/G_0$ is finite we have that $$H^*(G/G_0; H^*(BG_0)) = H^0(G/G_0; H^*(BG_0)) = (H^*(BG_0))^{G/G_0}.$$ This is again a graded vector space, which is concentrated in even degrees because $H^*(BG_0;\Bbb Q)$ is. So $H^*(BG;\Bbb Q)$ is concentrated in even degrees for all compact Lie groups $G$, which is what you wanted.
