[proofplan] Identify the group ring of $G\times \mathbb Z$ with the Laurent [polynomial ring](/page/Polynomial%20Ring) $R[t,t^{-1}]$. Then apply the [Bass-Heller-Swan decomposition](/theorems/8686) to split $K_1(R[t,t^{-1}])$ into the four summands $K_1(R)$, $K_0(R)$, and two copies of $NK_1(R)$. The only remaining task is to track the subgroup generated by the standard Whitehead units $\pm g t^m$: the factors $\pm g$ lie in the $K_1(R)$ summand, while the Laurent unit $t$ contributes the class $[R]$ in the $K_0(R)$ summand and contributes nothing to the two nil summands. Taking the quotient therefore changes $K_1(R)$ to $\operatorname{Wh}(G)$, changes $K_0(R)$ to $\widetilde K_0(R)$, and leaves both nil summands unchanged. [/proofplan]

[step:Identify the group ring of $G\times \mathbb Z$ with a Laurent polynomial ring] Let $z$ denote the standard generator of the infinite [cyclic group](/page/Cyclic%20Group) $\mathbb Z$. Define the ring homomorphism \begin{align*} \Phi:\mathbb Z[G\times \mathbb Z]\to R[t,t^{-1}] \end{align*} on group elements by \begin{align*} \Phi((g,m))=g t^m \end{align*} for $g\in G$ and $m\in\mathbb Z$, and extend $\mathbb Z$-linearly. Its inverse sends $g t^m$ to $(g,m)$ and extends $\mathbb Z$-linearly, so $\Phi$ is a ring isomorphism. Under this identification, the Whitehead group of $G\times\mathbb Z$ is \begin{align*} \operatorname{Wh}(G\times\mathbb Z)=K_1(R[t,t^{-1}])/\langle [\varepsilon g t^m] : \varepsilon\in\{\pm1\},\ g\in G,\ m\in\mathbb Z\rangle. \end{align*} [/step]

[step:Apply Bass-Heller-Swan with the standard normalization] The Bass-Heller-Swan decomposition from [citetheorem:8686] applies to every unital ring. Since $R=\mathbb Z[G]$ is a unital ring with identity element $1_R$, its hypothesis is satisfied. Let $e_0:R[t]\to R$ be the unital ring homomorphism given by evaluation at $t=0$, and define \begin{align*} NK_1(R)=\ker\bigl(K_1(e_0):K_1(R[t])\to K_1(R)\bigr). \end{align*} Let \begin{align*} \beta:K_1(R[t,t^{-1}])\to K_1(R)\oplus K_0(R)\oplus NK_1(R)\oplus NK_1(R) \end{align*} denote the normalized Bass-Heller-Swan isomorphism in which the inclusion $R\hookrightarrow R[t,t^{-1}]$ gives the $K_1(R)$ summand, the class $[t]$ of the Laurent unit $t\in R[t,t^{-1}]^\times$ maps to $[R]\in K_0(R)$, and the two nil summands are the positive and negative Bass nil summands. Thus \begin{align*} K_1(R[t,t^{-1}])\cong K_1(R)\oplus K_0(R)\oplus NK_1(R)\oplus NK_1(R). \end{align*} [/step]

[step:Track the standard units under the Bass-Heller-Swan decomposition]Let $U$ be the subgroup of $K_1(R[t,t^{-1}])$ generated by all classes $[\varepsilon g t^m]$, where $\varepsilon\in\{\pm1\}$, $g\in G$, and $m\in\mathbb Z$. Let $U_G$ be the subgroup of $K_1(R)$ generated by all classes $[\varepsilon g]$ with $\varepsilon\in\{\pm1\}$ and $g\in G$. For each $\varepsilon\in\{\pm1\}$, $g\in G$, and $m\in\mathbb Z$, multiplicativity in $K_1$ gives \begin{align*} [\varepsilon g t^m]=[\varepsilon g]+m[t]. \end{align*} Under the normalized Bass-Heller-Swan isomorphism $\beta$, the class $[\varepsilon g]$ lies in the $K_1(R)$ summand, the class $[t]$ maps to $[R]$ in the $K_0(R)$ summand, and neither class has a component in either nil summand. Therefore \begin{align*} \beta(U)=U_G\oplus \langle [R]\rangle \oplus 0\oplus 0. \end{align*}[/step]

[guided]The quotient defining the Whitehead group kills only the standard group-ring units, so we must compute exactly where those units land in the Bass-Heller-Swan splitting. Let $U\le K_1(R[t,t^{-1}])$ be generated by the classes $[\varepsilon g t^m]$, where $\varepsilon\in\{\pm1\}$, $g\in G$, and $m\in\mathbb Z$. Also let $U_G\le K_1(R)$ be generated by the classes $[\varepsilon g]$. The equality \begin{align*} [\varepsilon g t^m]=[\varepsilon g]+m[t] \end{align*} holds because $K_1$ is written additively and the class of a product of units is the sum of their $K_1$-classes. The factor $\varepsilon g$ is a unit of $R$, so its image in $K_1(R[t,t^{-1}])$ comes from the inclusion $R\hookrightarrow R[t,t^{-1}]$. Under the Bass-Heller-Swan decomposition, this is precisely the $K_1(R)$ summand. The remaining factor is the Laurent unit $t$. The chosen normalization of the Bass-Heller-Swan isomorphism identifies the class of $t$ with the free rank-one class $[R]\in K_0(R)$. Hence the class of $t^m$ contributes $m[R]$ to the $K_0(R)$ summand. This is the point at which the infinite cyclic factor is detected: the generator of $\mathbb Z$ contributes the rank-one free module class. Finally, neither $\varepsilon g$ nor $t$ lies in the positive or negative nil part of the Bass-Heller-Swan decomposition. The nil summands are complementary summands measuring the kernel of evaluation in the polynomial directions, while the displayed units are accounted for entirely by the $K_1(R)$ and $K_0(R)$ summands under the normalized decomposition. Therefore \begin{align*} \beta(U)=U_G\oplus \langle [R]\rangle \oplus 0\oplus 0. \end{align*}[/guided]

[step:Take the quotient and identify the resulting summands] Using the description of $\beta(U)$, quotienting the Bass-Heller-Swan decomposition by $U$ gives \begin{align*} K_1(R[t,t^{-1}])/U\cong (K_1(R)/U_G)\oplus (K_0(R)/\langle [R]\rangle)\oplus NK_1(R)\oplus NK_1(R). \end{align*} By definition, \begin{align*} K_1(R)/U_G=\operatorname{Wh}(G) \end{align*} and \begin{align*} K_0(R)/\langle [R]\rangle=\widetilde K_0(R). \end{align*} Since $K_1(R[t,t^{-1}])/U=\operatorname{Wh}(G\times\mathbb Z)$ under the group-ring identification from the first step, we obtain \begin{align*} \operatorname{Wh}(G\times \mathbb Z)\cong \operatorname{Wh}(G)\oplus \widetilde K_0(R)\oplus NK_1(R)\oplus NK_1(R). \end{align*} The construction uses only the canonical group-ring identification and the normalized natural Bass-Heller-Swan decomposition. More explicitly, a [group homomorphism](/page/Group%20Homomorphism) $G\to H$ induces compatible ring homomorphisms $\mathbb Z[G]\to\mathbb Z[H]$ and $\mathbb Z[G][t,t^{-1}]\to\mathbb Z[H][t,t^{-1}]$, carries the unit subgroup generated by $[\varepsilon g t^m]$ into the corresponding unit subgroup for $H$, and preserves the $K_1$, $K_0$, positive nil, and negative nil summands under naturality of Bass-Heller-Swan. Hence the resulting isomorphism is natural in $G$. [/step]