Suppose $X$ is a topological manifold such that $\pi_1(X)\simeq G\times H$ for some nontrivial groups $G$ and $H$. Then can I always find topological manifolds $Y$ and $Z$ such that $X\cong Y\times Z$ and $\pi_1(Y)\simeq G$ and $\pi_1(Z)\simeq H$?
I thought the statement is strange enough to come up with a counterexample easily but I failed. The main reason is I don't know how to show a space cannot be decomposed as a product of two spaces or at least some "usual" argument (but I think one can use some fancy machinery in algebraic topology). I think many people believe this statement is false. But can you provide any counterexample?
Best Answer
Any finitely presented group is the fundamental group of a closed 4-manifold. (See this question.) Consider $𝐺=\mathbb Z^5$. If you could always decompose a group product as a product of manifolds, the corresponding 4-manifold would have to decompose as a product of 5 manifolds with fundamental group $\mathbb Z$, but this is impossible by dimension reasons.