Rational group homology of an infinite product of finite groups

Question

Let $G_{1}, G_{2}, \cdots$ be a countably infinite sequence of finite groups. It is well-known that the group homology $H_{n}(BG_{i};\mathbb{Q})=0$ for any $n\geq 1$.

Let $X=\prod^{\infty}_{i=1}BG_{i}$ be the product space with product topology. Then $X$ is just the classifying space of $\prod^{\infty}_{i=1}G_{i}$.

My question: Do we know $H_{n}(X;\mathbb{Q})=0$?

The difficulty is that the Kunneth formula doesn't work for an infinite product. I am specifically interested in the case $n=3$ and $G_{i}$ are the even permutation groups $A_{*}$.

Answer 1

The higher homology is not necessarily zero. For example $H_1(X,\mathbb Q)$ is $\mathbb Q \otimes \pi_1(X)^{\rm ab} $, so if one takes an infinite product of cyclic groups $\prod_{n \in \mathbb N} \mathbb Z/n \mathbb Z$ then the element $1 \otimes (1,1 \dots) \in H_1(X)$ is nonzero.