MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am wondering is there a formula for the whitehead group of product of groups. In other words, if we know the whitehead group of two groups, are we able to calculate the whitehead group of their products. If not, is there a example such that the whitehead group of the product is nontrivial while the whitehead group of both groups vanishes.

share|cite|improve this question
up vote 10 down vote accepted

This is a partial answer. Let $G$ be a group. Then by identifying $\mathbb{Z}G[t,t^{-1}]$ with $\mathbb{Z}[G\times\mathbb{Z}]$ and using Bass-Heller-Swan decomposition you will get $K_1(\mathbb{Z}[G\times\mathbb{Z}])=K_1(\mathbb{Z}G)\oplus\tilde{K}_0(\mathbb{Z}G)\oplus NK_1(\mathbb{Z}G)^2$. Therefore

$Wh(G\times\mathbb{Z})=Wh(G)\oplus\tilde{K}_0(\mathbb{Z}G)\oplus NK_1(\mathbb{Z}G)^2$.

As for the example you are looking for, take $G=\mathbb{Z}/4\mathbb{Z}$ and $H=\mathbb{Z}^{3}$. The Whitehead group of these two groups vanishes but the Whitehead group of its product does not. The reason is that $G\times H$ is the fundamental group of $L\times T\ ^3$ where $L$ is a 3-dimensional lens space and $T\ ^3$ is a 3-dimensional torus. Farrell and Hsiang (F. T. Farrell and W. C. Hsiang, Bull. Amer. Math. Soc. Volume 73, Number 5 (1967), 741-744) proved that there is a non trivial h-cobordism with that space as one of its boundary components.

share|cite|improve this answer

I just realized that the Farrell-Jones conjecture can be treated as a generalization of the Bass-Heller-swan theorem, hence can answer my question. Note that the conjecture has already been verified for a large class of groups.

Given two groups, $H$ and $G$, The Group Ring $\mathbb{Z} (H \times G)$ can be considered as $(\mathbb{Z}H)G$. Denote the Ring $\mathbb{Z}H$ as $R$, then the K-theoretic Farrell-Jones conjecture predicted the following assembly map is an isomorphism $$ H_n^G(E_{\mathcal{VYC}}(G); \mathbf{K}_R) \rightarrow K_n(RG);$$ where $E_{\mathcal{VYC}}(G)$ is the classifying of G respect to the family of virtually cyclic subgroups, $\mathbf{K}$ is the non-connective K-theory spectrum.

Basically, the conjecture says if we know the homology theory of $G$, the K-theory of $R$, then we know the K-theory of $RG$. But in order to calculate $K_1(RG)$, we definitely need a lot K-group information of $R$, not just $K_1(R)$. So only known the whitehead group of $G$ and $H$ is not enough to recover $Wh(G\times H)$. This also explains why there exists two groups with trivial whitehead torsion while the whitehead torsion of their product is non-zero.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.