[proofplan]
The proof is a direct unpacking of the quotient presentation of $K_1(R)$. Stabilization does not change the element of the stable group $GL(R)$ because $GL(R)$ is defined as the colimit under block sum with identity matrices. Elementary row and column operations correspond to multiplication by elements of the stable elementary subgroup $E(R)$, and these elements map to the identity element in the quotient $GL(R)/E(R)$. Finally, the Whitehead group is obtained from $K_1(\mathbb Z[G])$ by imposing the further identifications coming from the units $\pm g$.
[/proofplan]
[step:Identify all stabilizations in the stable general linear group]
For each integer $k \ge 1$, let $1_R \in R$ denote the multiplicative identity of $R$, and let
\begin{align*}
s_k:GL_k(R)\longrightarrow GL_{k+1}(R)
\end{align*}
denote the stabilization homomorphism defined by
\begin{align*}
s_k(B)=\operatorname{diag}(B,1_R).
\end{align*}
By definition,
\begin{align*}
GL(R)=\varinjlim_k GL_k(R)
\end{align*}
with respect to the maps $s_k$. For every integer $m \ge 0$, let $I_m \in GL_m(R)$ denote the $m \times m$ identity matrix over $R$, with $I_0$ interpreted as the empty identity block. Therefore the image of $A \in GL_n(R)$ in $GL(R)$ is equal to the image of
\begin{align*}
\operatorname{diag}(A,I_m)\in GL_{n+m}(R)
\end{align*}
for every integer $m \ge 0$, because $\operatorname{diag}(A,I_m)$ is obtained from $A$ by applying the stabilization maps $m$ times.
Since the theorem statement defines $E(R) \trianglelefteq GL(R)$ as the stable elementary [normal subgroup](/page/Normal%20Subgroup), the [quotient group](/theorems/790) $GL(R)/E(R)$ and its quotient homomorphism are well-defined. Let
\begin{align*}
q_R:GL(R)\longrightarrow K_1(R)=GL(R)/E(R)
\end{align*}
be this quotient homomorphism. Applying $q_R$ to the equality of stable representatives gives
\begin{align*}
[\operatorname{diag}(A,I_m)]_{K_1}=[A]_{K_1}.
\end{align*}
[guided]
The first point is that stabilization is not an operation that changes the stable class and then needs a separate argument to be undone. It is part of the construction of the ambient group. For each integer $k \ge 1$, let $1_R \in R$ denote the multiplicative identity of $R$, and define the stabilization homomorphism
\begin{align*}
s_k:GL_k(R)\longrightarrow GL_{k+1}(R)
\end{align*}
by
\begin{align*}
s_k(B)=\operatorname{diag}(B,1_R).
\end{align*}
The stable general linear group is the colimit
\begin{align*}
GL(R)=\varinjlim_k GL_k(R)
\end{align*}
for this directed system. Thus two matrices represent the same element of $GL(R)$ whenever one is obtained from the other by applying finitely many of the maps $s_k$.
Now fix $m \ge 0$, and let $I_m \in GL_m(R)$ denote the $m \times m$ identity matrix over $R$, with $I_0$ interpreted as the empty identity block. Applying the stabilization maps $m$ times to $A \in GL_n(R)$ gives the matrix
\begin{align*}
\operatorname{diag}(A,I_m)\in GL_{n+m}(R).
\end{align*}
Hence $A$ and $\operatorname{diag}(A,I_m)$ have the same image in $GL(R)$. If
\begin{align*}
q_R:GL(R)\longrightarrow K_1(R)=GL(R)/E(R)
\end{align*}
denotes the quotient homomorphism, then applying $q_R$ to this equality gives
\begin{align*}
[\operatorname{diag}(A,I_m)]_{K_1}=[A]_{K_1}.
\end{align*}
This proves precisely that adding identity blocks does not alter the $K_1$ class.
[/guided]
[/step]
[step:Kill elementary left multiplication in the quotient defining $K_1$]
Let $m \ge 0$, and let $e=e_{ij}(r) \in GL_{n+m}(R)$ be an elementary matrix, where $1 \le i \ne j \le n+m$ and $r \in R$. Define
\begin{align*}
B=\operatorname{diag}(A,I_m)
\end{align*}
as an element of $GL_{n+m}(R)$. Since $e \in E(R)$, its image in the quotient $GL(R)/E(R)$ is the identity element. Therefore
\begin{align*}
q_R(eB)=q_R(e)q_R(B)=q_R(B).
\end{align*}
Equivalently,
\begin{align*}
[e\,\operatorname{diag}(A,I_m)]_{K_1}=[\operatorname{diag}(A,I_m)]_{K_1}.
\end{align*}
Thus elementary row operations do not change the $K_1$ class.
[/step]
[step:Kill elementary right multiplication in the quotient defining $K_1$]
With $B=\operatorname{diag}(A,I_m)$ as above, again $e \in E(R)$ maps to the identity under
\begin{align*}
q_R:GL(R)\longrightarrow GL(R)/E(R).
\end{align*}
Hence
\begin{align*}
q_R(Be)=q_R(B)q_R(e)=q_R(B),
\end{align*}
and so
\begin{align*}
[\operatorname{diag}(A,I_m)e]_{K_1}=[\operatorname{diag}(A,I_m)]_{K_1}.
\end{align*}
Thus elementary column operations do not change the $K_1$ class.
[/step]
[step:Pass the invariance to the Whitehead quotient]
Now take $R=\mathbb Z[G]$. By definition, the Whitehead group is the quotient
\begin{align*}
\operatorname{Wh}(G)=K_1(\mathbb Z[G])/\langle [\pm g] : g \in G\rangle,
\end{align*}
where $\pm g$ denotes the units of $\mathbb Z[G]$ obtained from group elements and signs, and $\langle [\pm g] : g \in G\rangle$ denotes the subgroup of $K_1(\mathbb Z[G])$ generated by their classes.
Let
\begin{align*}
\pi_G:K_1(\mathbb Z[G])\longrightarrow \operatorname{Wh}(G)
\end{align*}
be the quotient homomorphism. Since stable elementary row and column operations leave the class of $A$ unchanged in $K_1(\mathbb Z[G])$, applying $\pi_G$ preserves this equality in $\operatorname{Wh}(G)$. The passage from $K_1(\mathbb Z[G])$ to $\operatorname{Wh}(G)$ imposes exactly the additional identifications generated by the classes $[\pm g]$ with $g \in G$. Therefore the induced Whitehead class is invariant under stable elementary row and column operations, subject only to that prescribed quotient by units.
[/step]
