[proofplan] The proof identifies the kernel of the quotient map $GL(R)\to K_1(R)$ with the commutator subgroup of the stable general linear group. First, every elementary transvection is a commutator after stabilizing far enough to choose a third index, so $E(R)\subseteq [GL(R),GL(R)]$. Conversely, we show directly that $GL(R)/E(R)$ is abelian. The key calculation is that $\operatorname{diag}(A,A^{-1})$ belongs to $E(R)$ for every invertible matrix $A$. This implies that, modulo $E(R)$, the two stabilized placements $\operatorname{diag}(A,I)$ and $\operatorname{diag}(I,A)$ are equal; after moving one factor to the other block, any two stable classes commute. Thus the quotient by $E(R)$ is abelian, so $[GL(R),GL(R)]\subseteq E(R)$. [/proofplan]

[step:Show every elementary transvection is a stable commutator]Let $e_{ij}(r)\in GL_n(R)$ be an elementary transvection, where $1\le i\ne j\le n$ and $r\in R$, and let $1_R$ denote the multiplicative identity of $R$. Choose an integer $m \ge n$ large enough that there exists an index $k \in \{1,\dots,m\}$ distinct from both $i$ and $j$. In $GL_m(R)$ the elementary matrix relation gives \begin{align*} [e_{ik}(r),e_{kj}(1_R)]=e_{ij}(r). \end{align*} Hence every elementary transvection belongs to $[GL(R),GL(R)]$. Since $E(R)$ is generated by the stable elementary transvections, it follows that \begin{align*} E(R)\subseteq [GL(R),GL(R)]. \end{align*}[/step]

[guided]Fix an elementary transvection $e_{ij}(r)$, with $i\ne j$ and $r\in R$. The only small technical point is that the commutator formula for elementary matrices needs a third index. Since $GL(R)$ is stable, we may replace $GL_n(R)$ by a larger $GL_m(R)$ through block stabilization, and we choose $k$ with $k\ne i$ and $k\ne j$. Inside this larger general linear group, the elementary relation is \begin{align*} [e_{ik}(r),e_{kj}(1_R)]=e_{ij}(r). \end{align*} Here the commutator is taken in the group $GL_m(R)$, and both $e_{ik}(r)$ and $e_{kj}(1_R)$ are elements of the stable group $GL(R)$. Therefore $e_{ij}(r)$ is literally a commutator in $GL(R)$. The group $E(R)$ is, by definition, generated by all stable elementary transvections. Since the commutator subgroup $[GL(R),GL(R)]$ is a subgroup and contains each generator of $E(R)$, it contains all of $E(R)$. Thus \begin{align*} E(R)\subseteq [GL(R),GL(R)]. \end{align*}[/guided]

[step:Factor diagonal inverse blocks into elementary block matrices] Let $A\in GL_n(R)$, and write $I_n$ for the identity matrix in $GL_n(R)$. Regard $R^n\oplus R^n$ as the ambient module for the corresponding block automorphisms, and define \begin{align*} U_A: R^n\oplus R^n \to R^n\oplus R^n, \qquad (x,y)\mapsto (x+Ay,y). \end{align*} \begin{align*} L_A: R^n\oplus R^n \to R^n\oplus R^n, \qquad (x,y)\mapsto (x,y-A^{-1}x). \end{align*} \begin{align*} W: R^n\oplus R^n \to R^n\oplus R^n, \qquad (x,y)\mapsto (-y,x). \end{align*} These maps are the usual upper block transvection, lower block transvection, and block rotation. Their composite acts by \begin{align*} U_A L_A U_A W(x,y)=U_A L_A U_A(-y,x)=U_A L_A(-y+Ax,x)=U_A(-y+Ax,A^{-1}y)=(Ax,A^{-1}y). \end{align*} Therefore this composite is exactly $\operatorname{diag}(A,A^{-1})$ on $R^n\oplus R^n$: \begin{align*} U_A L_A U_A W=\operatorname{diag}(A,A^{-1}). \end{align*} Write $A=(a_{pq})_{1\le p,q\le n}$ and $-A^{-1}=(b_{pq})_{1\le p,q\le n}$. The upper map $U_A$ factors as the product, over all entries $a_{pq}$ of $A$, of elementary transvections adding $a_{pq}$ times the $(n+q)$-th coordinate to the $p$-th coordinate. These transvections commute with each other because they all go from the second block to the first block. Similarly, $L_A$ factors as the product, over all entries $b_{pq}$ of $-A^{-1}$, of elementary transvections adding $b_{pq}$ times the $p$-th coordinate to the $(n+q)$-th coordinate. Hence $U_A,L_A\in E(R)$. It remains only to see that $W\in E(R)$. Applying the displayed identity with $A=I_n$ gives $U_{I_n}L_{I_n}U_{I_n}W=I_{2n}$. Since $U_{I_n}$ and $L_{I_n}$ lie in $E(R)$, their product is in $E(R)$, and so \begin{align*} W=(U_{I_n}L_{I_n}U_{I_n})^{-1}\in E(R). \end{align*} Thus $U_A L_A U_A W$ is a product of elements of $E(R)$, and hence \begin{align*} \operatorname{diag}(A,A^{-1})\in E(R). \end{align*} [/step]

[step:Show the quotient by elementary matrices is abelian] Let $Q=GL(R)/E(R)$, and write $\overline{G}$ for the image in $Q$ of a stable matrix $G$. We prove that $Q$ is abelian. Take $A,B\in GL_n(R)$. In the stable group, $A$ and $B$ may be represented by the block matrices $\operatorname{diag}(A,I_n)$ and $\operatorname{diag}(B,I_n)$. The previous step gives $\operatorname{diag}(A,A^{-1})\in E(R)$. Since \begin{align*} \operatorname{diag}(A,I_n)\operatorname{diag}(I_n,A^{-1})=\operatorname{diag}(A,A^{-1}), \end{align*} the two factors $\operatorname{diag}(A,I_n)$ and $\operatorname{diag}(I_n,A)$ have the same image in $Q$. Thus \begin{align*} \overline{\operatorname{diag}(A,I_n)}=\overline{\operatorname{diag}(I_n,A)}. \end{align*} Using this equality, and then using the same equality once more for $A$, we compute in $Q$: \begin{align*} \overline{\operatorname{diag}(A,I_n)}\,\overline{\operatorname{diag}(B,I_n)} =\overline{\operatorname{diag}(I_n,A)}\,\overline{\operatorname{diag}(B,I_n)} =\overline{\operatorname{diag}(B,A)} =\overline{\operatorname{diag}(B,I_n)}\,\overline{\operatorname{diag}(I_n,A)} =\overline{\operatorname{diag}(B,I_n)}\,\overline{\operatorname{diag}(A,I_n)}. \end{align*} Hence the images of any two stable matrices commute in $Q$. Therefore $GL(R)/E(R)$ is abelian. [/step]

[step:Deduce the reverse inclusion] Because $GL(R)/E(R)$ is abelian, the quotient map $GL(R)\to GL(R)/E(R)$ kills every commutator in $GL(R)$. Equivalently, every element of the commutator subgroup lies in the kernel of this quotient map. The kernel is $E(R)$ by definition, so \begin{align*} [GL(R),GL(R)]\subseteq E(R). \end{align*} [/step]

[step:Identify the quotient with the abelianization] Combining the two inclusions gives \begin{align*} E(R)=[GL(R),GL(R)]. \end{align*} By definition, \begin{align*} K_1(R)=GL(R)/E(R). \end{align*} Substituting the equality of subgroups yields \begin{align*} K_1(R)=GL(R)/[GL(R),GL(R)]=GL(R)_{\mathrm{ab}}. \end{align*} Thus the natural quotient map $GL(R)\to K_1(R)$ is precisely the abelianization map followed by this canonical identification. [/step]