I'm interested in a proof (or counter-example) of the following:
Let $G$ be a topological group. If $G$ contains torsion then $H^n(BG)\neq 0$ for infinitely many $n$.
I know this is true for discrete groups, but I'm wondering if an analogous argument can be made for groups with a topology.
In the simplest case $G=C_k$ is cyclic of order $k<\infty$, and one computes $H^*(BC_k)\cong \mathbb{Z}[u]/(ku)$ for a class $u$ of degree $2$. The approach I know is to compute the Serre spectral sequence for the universal bundle $S^1 \to S^\infty \to \mathbb{C}P^\infty$, and notice that if we include $C_k \to S^1$ as the $k$-th roots of unity then $BC_k \simeq S^\infty/C_k$ and we get a fibre bundle $S^1/C_k \to BC_k \to \mathbb{C}P^\infty$ along with a map of fibrations from the universal bundle which is a map of degree $k$ in the fibres; now study the induced map on spectral sequences.
Now if $G$ is an arbitrary group with $k$-torsion, it has a subgroup isomorphic to $C_k$. If $G$ is discrete we can complete the argument using group cohomology $H^*_{Grp}(G)\cong H^*(BG)$, since in this context it's easy to see that $cd(G) \geq cd(C_k)=\infty$. However if $G$ is not discrete then group cohomology is more subtle (see this question), because the usual definition computes $H^*(K(G,1))$ instead of $H^*(BG)$. Is there an analogous argument using an appropriate definition of $H^*_{TopGrp}(G)$?
Alternatively, is there an argument that avoids group cohomology altogether, and instead uses geometry (for Lie groups) or algebraic topology? I tried thinking about how the inclusion $C_k \to G$ induces a map on spectral sequences (like in the case $G=S^1$), but if $G$ is not connected then the coefficients are potentially twisted, and in either case there are too many unknowns for me to see what to do.
There's an argument involving the Euler characteristic that may or may not apply in your situation. Namely, there's a fibration $$G \to EG \to BG$$ with $EG$ contractible. So if $H^* BG$ is zero in all but finitely many degrees and thus $\chi(BG)$ is a finite number, we would have $$\chi(BG) = \frac{1}{\chi(G)}.$$ This would only make sense if $\chi(G) = \pm 1$. However, for example if $G$ is a connected compact Lie group, then $\chi(G) = 0$ and $\chi(BG)$ wouldn't make sense.