[proofplan] We use the standard convention that $\operatorname{Wh}(G)$ is the quotient of $K_1(\mathbb Z[G])$ by the subgroup generated by the classes of the units $\pm g$ with $g\in G$. For the group with one element, the group ring $\mathbb Z[1]$ is canonically $\mathbb Z$, and the units coming from the identity element are exactly $\pm 1$. The determinant identifies $K_1(\mathbb Z)$ with $\mathbb Z^\times=\{\pm 1\}$, so quotienting by the subgroup generated by $\pm 1$ leaves the zero group. [/proofplan] [step:Identify the group ring and the units coming from the identity element] Let $e$ denote the unique element of the group $1$. The group ring $\mathbb Z[1]$ consists of finite sums $n e$ with $n\in \mathbb Z$, and multiplication is determined by $e e=e$. Define the ring homomorphism \begin{align*} \Phi:\mathbb Z[1]\to \mathbb Z \end{align*} by \begin{align*} \Phi(n e)=n. \end{align*} Its inverse is the ring homomorphism \begin{align*} \Psi:\mathbb Z\to \mathbb Z[1] \end{align*} defined by \begin{align*} \Psi(n)=n e. \end{align*} Thus $\mathbb Z[1]\cong \mathbb Z$. Under this identification, the units $\pm g$ with $g\in 1$ are exactly $e$ and $-e$, hence correspond to $1$ and $-1$ in $\mathbb Z^\times$. [guided] Let $e$ be the unique element of the group $1$. An element of the group ring $\mathbb Z[1]$ is a finite integral linear combination of elements of $1$, so it has the form $n e$ for a unique integer $n\in\mathbb Z$. The multiplication rule in the group ring uses the group law, and because $e e=e$, we have \begin{align*} (n e)(m e)=(nm)e \end{align*} for all $m,n\in\mathbb Z$. Define \begin{align*} \Phi:\mathbb Z[1]\to \mathbb Z \end{align*} by \begin{align*} \Phi(n e)=n. \end{align*} This map preserves addition because $\Phi(n e+m e)=\Phi((n+m)e)=n+m$, and it preserves multiplication because \begin{align*} \Phi((n e)(m e))=\Phi((nm)e)=nm=\Phi(n e)\Phi(m e). \end{align*} It sends the multiplicative identity $e$ of $\mathbb Z[1]$ to $1\in\mathbb Z$, so it is a unital ring homomorphism. Its inverse is the unital ring homomorphism \begin{align*} \Psi:\mathbb Z\to \mathbb Z[1] \end{align*} given by \begin{align*} \Psi(n)=n e. \end{align*} Therefore $\mathbb Z[1]\cong\mathbb Z$ as unital rings. The definition of the Whitehead group quotients by the classes of the units $\pm g$ for $g\in G$. When $G=1$, the only possible $g$ is $e$, so the units are precisely $e$ and $-e$. Under the isomorphism $\Phi$, these become $1$ and $-1$ in $\mathbb Z^\times$. [/guided] [/step] [step:Compute $K_1(\mathbb Z)$ by the determinant] Since $\mathbb Z$ is the ring of integers in the number field $\mathbb Q$, [citetheorem:8689] applies with $A=\mathbb Z$. Hence the determinant induces an isomorphism \begin{align*} \det:K_1(\mathbb Z)\xrightarrow{\cong}\mathbb Z^\times. \end{align*} The only units of $\mathbb Z$ are $1$ and $-1$, so \begin{align*} K_1(\mathbb Z)\cong \mathbb Z^\times=\{1,-1\}. \end{align*} Under this isomorphism, the classes of the units $1$ and $-1$ generate all of $K_1(\mathbb Z)$. [/step] [step:Apply the definition of the Whitehead group] By definition, \begin{align*} \operatorname{Wh}(G)=K_1(\mathbb Z[G])/\langle [\pm g]:g\in G\rangle, \end{align*} where $\langle [\pm g]:g\in G\rangle$ denotes the subgroup generated by the classes in $K_1(\mathbb Z[G])$ of the units $\pm g$. For $G=1$, the preceding steps identify $\mathbb Z[1]$ with $\mathbb Z$ and identify the subgroup generated by these units with all of $K_1(\mathbb Z)$. Therefore \begin{align*} \operatorname{Wh}(1)=K_1(\mathbb Z)/K_1(\mathbb Z). \end{align*} This quotient is the zero group, so \begin{align*} \operatorname{Wh}(1)=0. \end{align*} [/step]