If $G$ is a topological group that contains a torsion element, then the classifying space $BG$ is infinite-dimensional? We know that if $G$ is a topological group that contains a torsion element and $G$ satisfies additional conditions such as $G$ discrete or $G$ finite-dimensional, then the classifying space $BG$ is infinite-dimensional.
My question: can we remove the additional condition such that the statement still holds? Namely, is there any counterexample for $G$ infinite-dimensional?
Edit: Sorry, G contains a torsion element should be G is not torsion-free. Thank you for your thoughtful inputs.
 A: Let $A$ be a discrete group with a torsion element. Let $EA$ be the geometric realization of the action groupoid of $A$ acting on itself by left multiplication. $EA$ is the geometric realization of a simplicial group, so it is a topological group. It is well-known that $EA$ is a contractible space, and therefore the classifying space of $EA$ is contractible too. Because $A$ has torsion elements, $EA$ has torsion elements too.
If you want an example where the group itself is not contractible, let $G$ be any group such that $BG$ is homotopy equivalent to a finite space. Then $G\times EA$ is a topological group with torsion elements, and its classifying space is still homotopy finite.
A: Here is sort of a canonical example.
Consider $GL(\mathbb H)$ the group of invertible operators on a Hilbert space. By Kuipers theorem it is contractible. But $GL(\mathbb H)$ acts freely and properly on itself. Hence the classifying space is $BGL(\mathbb H)=GL(\mathbb H)/GL(\mathbb H)$ a point. Consequences are that Hilbert bundles are trivial over paracompact spaces, K-theory is represented by Fredholm operators etc.
Of course $GL(\mathbb H)$ contains a lot of torsion. Any finite group is a subgroup of $GL(\mathbb H)$ for example.
