[proofplan] We first prove perfection by using the stable nature of $E(R)$: every elementary generator can be placed in a sufficiently large matrix algebra where an auxiliary index is available. The elementary commutator relation then writes that generator as a commutator of two elementary matrices. The abelianness of $GL(R)/E(R)$ requires a separate argument: a block factorisation shows that $\operatorname{diag}(A,A^{-1})$ is elementary after stabilisation, and this implies that stable block sums commute modulo $E(R)$. [/proofplan]

[step:Write every elementary generator as a commutator inside the stable elementary group]Let $e_{ij}(r)\in E(R)$ be an elementary generator, with $i\neq j$ and $r\in R$. Here $1_R$ denotes the multiplicative identity of $R$. Since $E(R)$ is the stable elementary subgroup, we may regard $e_{ij}(r)$ inside a larger matrix group $GL_m(R)$ with $m\geq 3$ and choose an index $k$ such that $k\neq i$ and $k\neq j$. By the elementary matrix commutator identity from [Elementary Matrix Relations][citetheorem:8650], applied to the distinct indices $i,k,j$ and the elements $r,1_R\in R$, we have \begin{align*} [e_{ik}(r),e_{kj}(1_R)]=e_{ij}(r1_R)=e_{ij}(r). \end{align*} Both $e_{ik}(r)$ and $e_{kj}(1_R)$ lie in $E(R)$, so $e_{ij}(r)\in [E(R),E(R)]$.[/step]

[guided]We want to prove that $E(R)$ is generated by its own commutators. Since $E(R)$ is generated by the elementary matrices, it is enough to take one elementary generator and express it as a commutator of elements already lying in $E(R)$. Fix an elementary generator $e_{ij}(r)$, where $i\neq j$ and $r\in R$. The point of working stably is that we are allowed to enlarge the ambient matrix size. After replacing the matrix by its image under a stabilisation map, we may choose an auxiliary index $k$ distinct from both $i$ and $j$. Now apply the elementary commutator relation from [Elementary Matrix Relations][citetheorem:8650]. Its hypotheses require three distinct indices, and we have chosen $i,k,j$ precisely to satisfy this. With elements $r$ and $1_R$ in the ring $R$, it gives \begin{align*} [e_{ik}(r),e_{kj}(1_R)]=e_{ij}(r1_R). \end{align*} Since $R$ is unital, $r1_R=r$, so \begin{align*} [e_{ik}(r),e_{kj}(1_R)]=e_{ij}(r). \end{align*} The two matrices appearing in the commutator are elementary matrices, hence belong to $E(R)$. Therefore the original generator $e_{ij}(r)$ belongs to the commutator subgroup $[E(R),E(R)]$.[/guided]

[step:Conclude that the elementary subgroup is perfect] The subgroup $[E(R),E(R)]$ is generated by commutators of elements of $E(R)$, hence is contained in $E(R)$. Conversely, the previous step shows that every elementary generator of $E(R)$ belongs to $[E(R),E(R)]$. Since these elementary matrices generate $E(R)$ as a group, we obtain \begin{align*} E(R)\subseteq [E(R),E(R)]. \end{align*} Thus \begin{align*} E(R)=[E(R),E(R)]. \end{align*} [/step]

[step:Use the Whitehead block identity to make inverse block pairs elementary] Let $A\in GL_n(R)$, and let $I_n\in GL_n(R)$ denote the identity matrix. Throughout this step, $R^n$ denotes the free left $R$-module of column vectors, and matrices act on $R^n$ by left multiplication. Define maps $U_A,L_A,P:R^n\oplus R^n\to R^n\oplus R^n$ by \begin{align*} U_A(x,y)=(x+Ay,y) \end{align*} for $x,y\in R^n$, \begin{align*} L_A(x,y)=(x,y-A^{-1}x) \end{align*} for $x,y\in R^n$, and \begin{align*} P(x,y)=(-y,x) \end{align*} for $x,y\in R^n$. Write $A=(a_{pq})_{1\leq p,q\leq n}$. With respect to the ordered decomposition $R^n\oplus R^n$, the map $U_A$ has block matrix \begin{align*} \begin{pmatrix} I_n & A \end{align*} \begin{align*} 0 & I_n \end{pmatrix}. \end{align*} For each pair $(p,q)$ with $1\leq p,q\leq n$, let $T_{pq}$ be the elementary matrix in $GL_{2n}(R)$ that adds $a_{pq}$ times the $q$-th coordinate in the second copy of $R^n$ to the $p$-th coordinate in the first copy. In matrix notation, $T_{pq}=e_{p,n+q}(a_{pq})$. Since all these elementary matrices have their nonzero off-diagonal entries in the upper-right block, their off-diagonal matrix units multiply to zero; hence their product, in any fixed order, is exactly $U_A$. Thus $U_A\in E(R)$. Applying the same argument to the entries of $-A^{-1}$ in the lower-left block gives $L_A\in E(R)$. The map $P$ is elementary because it is the composite of the three elementary block transvections \begin{align*} (x,y)\mapsto (x,y+x), \qquad (x,y)\mapsto (x-y,y), \qquad (x,y)\mapsto (x,y+x). \end{align*} Their composite sends $(x,y)$ first to $(x,y+x)$, then to $(-y,y+x)$, and finally to $(-y,x)$, so it is $P$. For $x,y\in R^n$, direct multiplication of these four maps gives \begin{align*} (U_A L_A U_A P)(x,y)=(Ax,A^{-1}y). \end{align*} Therefore $U_A L_A U_A P=\operatorname{diag}(A,A^{-1})$. Since $E(R)$ is a subgroup and each factor lies in $E(R)$, we obtain \begin{align*} \operatorname{diag}(A,A^{-1})\in E(R). \end{align*} [/step]

[step:Show that every stable commutator lies in the elementary subgroup]By [Elementary Subgroup Is Normal][citetheorem:8655], $E(R)$ is a [normal subgroup](/page/Normal%20Subgroup) of $GL(R)$, so $GL(R)/E(R)$ is a well-defined [quotient group](/theorems/790). Let $A\in GL_n(R)$ and $B\in GL_m(R)$. Choose an integer $N\geq \max\{n,m\}$, and define the stabilisations \begin{align*} \widetilde A:=\operatorname{diag}(A,I_{N-n})\in GL_N(R), \qquad \widetilde B:=\operatorname{diag}(B,I_{N-m})\in GL_N(R). \end{align*} The cosets of $A$ and $B$ in $GL(R)/E(R)$ are represented by $\widetilde A$ and $\widetilde B$, respectively. Apply the previous step to $\widetilde A\in GL_N(R)$. It gives \begin{align*} G:=\operatorname{diag}(\widetilde A,\widetilde A^{-1})\in E(R). \end{align*} Define \begin{align*} H:=\operatorname{diag}(\widetilde B,I_N)\in GL_{2N}(R). \end{align*} Since $E(R)\trianglelefteq GL(R)$ and $G\in E(R)$, the commutator $[G,H]$ lies in $E(R)$: indeed $H G^{-1}H^{-1}\in E(R)$ by normality, and therefore $G(HG^{-1}H^{-1})\in E(R)$. Computing this commutator in the two $N\times N$ diagonal blocks gives \begin{align*} [G,H]=\operatorname{diag}(\widetilde A\widetilde B\widetilde A^{-1}\widetilde B^{-1},I_N). \end{align*} Thus the stable matrix $\widetilde A\widetilde B\widetilde A^{-1}\widetilde B^{-1}$ belongs to $E(R)$. Hence the commutator of the two cosets is trivial: \begin{align*} (\widetilde A E(R))(\widetilde B E(R))(\widetilde A E(R))^{-1}(\widetilde B E(R))^{-1}=E(R). \end{align*} Since $A$ and $B$ represented arbitrary elements of $GL(R)$, all cosets in $GL(R)/E(R)$ commute. Therefore $GL(R)/E(R)$ is abelian.[/step]

[guided]The perfection of $E(R)$ does not by itself imply that $GL(R)/E(R)$ is abelian, so we prove commutativity of the quotient by showing that every stable commutator in $GL(R)$ lies in $E(R)$. First, the quotient must be a group before we discuss whether it is abelian. By [Elementary Subgroup Is Normal][citetheorem:8655], $E(R)\trianglelefteq GL(R)$, so $GL(R)/E(R)$ is defined. Take arbitrary cosets represented by matrices $A\in GL_n(R)$ and $B\in GL_m(R)$. To multiply and commute them inside one finite matrix group, choose an integer $N\geq \max\{n,m\}$ and stabilise both matrices to size $N$: \begin{align*} \widetilde A:=\operatorname{diag}(A,I_{N-n})\in GL_N(R), \qquad \widetilde B:=\operatorname{diag}(B,I_{N-m})\in GL_N(R). \end{align*} These are the standard images of $A$ and $B$ in the direct limit $GL(R)$, so they represent the same cosets as $A$ and $B$. The previous step gives the key elementary inverse-pair identity. Applied to $\widetilde A$, it says \begin{align*} G:=\operatorname{diag}(\widetilde A,\widetilde A^{-1})\in E(R). \end{align*} Now put \begin{align*} H:=\operatorname{diag}(\widetilde B,I_N)\in GL_{2N}(R). \end{align*} Why introduce $H$? Because the commutator $[G,H]$ isolates the commutator $\widetilde A\widetilde B\widetilde A^{-1}\widetilde B^{-1}$ in the first block and leaves the second block equal to the identity. We must first justify that $[G,H]$ is elementary. Since $G\in E(R)$ and $E(R)$ is normal in $GL(R)$, the conjugate $H G^{-1}H^{-1}$ also lies in $E(R)$. Because $E(R)$ is a subgroup, the product \begin{align*} G(HG^{-1}H^{-1})=[G,H] \end{align*} lies in $E(R)$. Now compute the same commutator block by block. In the first $N\times N$ block we get $\widetilde A\widetilde B\widetilde A^{-1}\widetilde B^{-1}$, and in the second $N\times N$ block we get $\widetilde A^{-1}I_N\widetilde A I_N=I_N$. Therefore \begin{align*} [G,H]=\operatorname{diag}(\widetilde A\widetilde B\widetilde A^{-1}\widetilde B^{-1},I_N). \end{align*} Since the left-hand side lies in $E(R)$, the stable commutator $\widetilde A\widetilde B\widetilde A^{-1}\widetilde B^{-1}$ lies in $E(R)$. Passing to the quotient, this says exactly that the commutator of the cosets is the identity coset: \begin{align*} (\widetilde A E(R))(\widetilde B E(R))(\widetilde A E(R))^{-1}(\widetilde B E(R))^{-1}=E(R). \end{align*} Thus $\widetilde A E(R)$ and $\widetilde B E(R)$ commute. Since the original representatives $A$ and $B$ were arbitrary, every two elements of $GL(R)/E(R)$ commute, so $GL(R)/E(R)$ is abelian.[/guided]

[step:Combine the two conclusions] We have proved \begin{align*} E(R)=[E(R),E(R)]. \end{align*} We have also proved that $E(R)\trianglelefteq GL(R)$ and that every pair of cosets in $GL(R)/E(R)$ commutes. Therefore the stable elementary subgroup is perfect, and the Whitehead quotient $GL(R)/E(R)$ is an abelian group. [/step]