[proofplan] We prove normality by showing that every conjugate of an elementary generator by an arbitrary stable invertible matrix again lies in the stable elementary subgroup. The key point is the standard block factorization expressing $\operatorname{diag}(g,g^{-1})$ as a product of elementary block transvections. After stabilizing $g$ and the elementary generator into a common finite size, this factorization converts conjugation by $g$ into conjugation by an element of $E(R)$. [/proofplan] [step:Embed the conjugation problem into one finite matrix size] Let $g \in GL(R)$ and let $x=e_{ij}(r)$ be an elementary generator of $E(R)$, where $i \ne j$ and $r \in R$. By the stable direct-limit definition of $GL(R)$, there exists an integer $N \ge 1$ such that $g$ is represented by a matrix $g \in GL_N(R)$ and $x$ is represented in the same group $GL_N(R)$. It is enough to prove that $gxg^{-1}$ represents an element of $E(R)$, because $E(R)$ is generated by elementary matrices. Indeed, if every conjugate of every elementary generator lies in $E(R)$, then for any product $y=x_1\cdots x_k$ of elementary generators and any $g \in GL(R)$, \begin{align*} gyg^{-1}=(gx_1g^{-1})\cdots(gx_kg^{-1}) \end{align*} lies in $E(R)$. [/step] [step:Write block transvections as products of ordinary elementary matrices] For a matrix $A=(a_{pq}) \in M_N(R)$, define $U_{12}(A) \in GL_{2N}(R)$ to be the $2 \times 2$ block matrix with diagonal blocks $I_N$ and $I_N$, upper-right block $A$, and lower-left block $0$. Define $U_{21}(A) \in GL_{2N}(R)$ to be the $2 \times 2$ block matrix with diagonal blocks $I_N$ and $I_N$, upper-right block $0$, and lower-left block $A$. For $1 \le p,q \le N$, the ordinary elementary matrix $e_{p,N+q}(a_{pq})$ has its nonzero off-diagonal entry in the upper-right block. Multiplying these matrices over all pairs $(p,q)$ gives \begin{align*} U_{12}(A)=\prod_{p=1}^{N}\prod_{q=1}^{N} e_{p,N+q}(a_{pq}). \end{align*} There are no extra cross terms in this product, since the column index $N+q$ of each off-diagonal entry is never equal to the row index $p'$ of another upper-right entry. Hence $U_{12}(A)\in E_{2N}(R)\subset E(R)$. Similarly, \begin{align*} U_{21}(A)=\prod_{p=1}^{N}\prod_{q=1}^{N} e_{N+p,q}(a_{pq}), \end{align*} so $U_{21}(A)\in E_{2N}(R)\subset E(R)$. [guided] We need a finite-dimensional way to see that block matrices with one off-diagonal block are elementary. Let $A=(a_{pq}) \in M_N(R)$ be an arbitrary $N \times N$ matrix over $R$. By definition, $U_{12}(A) \in GL_{2N}(R)$ is the $2 \times 2$ block matrix with diagonal blocks $I_N$ and $I_N$, upper-right block $A$, and lower-left block $0$. Hence $U_{12}(A)$ has the entry $a_{pq}$ in row $p$ and column $N+q$ of the full $2N \times 2N$ matrix. Therefore the ordinary elementary matrix $e_{p,N+q}(a_{pq})$ contributes exactly that entry and leaves all diagonal entries equal to $1$. When we multiply all these elementary matrices, no unwanted products appear. The only possible extra term in multiplying two elementary matrices occurs when the column index of the first off-diagonal entry equals the row index of the second. Here every column index in the upper-right block has the form $N+q$, while every row index in the same block has the form $p'$ with $1 \le p' \le N$. Thus $N+q \ne p'$ for all such indices. Consequently the product is exactly \begin{align*} U_{12}(A)=\prod_{p=1}^{N}\prod_{q=1}^{N} e_{p,N+q}(a_{pq}). \end{align*} This proves $U_{12}(A)\in E_{2N}(R)$. The lower-left case is identical in structure but uses the entries in rows $N+p$ and columns $q$. Thus \begin{align*} U_{21}(A)=\prod_{p=1}^{N}\prod_{q=1}^{N} e_{N+p,q}(a_{pq}), \end{align*} and so $U_{21}(A)\in E_{2N}(R)$ as well. [/guided] [/step] [step:Factor $\operatorname{diag}(g,g^{-1})$ as a product of elementary block matrices] Define $W \in GL_{2N}(R)$ to be the $2 \times 2$ block matrix with upper-left block $0$, upper-right block $-I_N$, lower-left block $I_N$, and lower-right block $0$. By the previous step, \begin{align*} W=U_{12}(-I_N)U_{21}(I_N)U_{12}(-I_N) \end{align*} lies in $E_{2N}(R)$. Now multiply the block matrices in the displayed order. Since block multiplication preserves the order of multiplication in the possibly noncommutative ring $R$, the product $U_{12}(g)U_{21}(-g^{-1})U_{12}(g)$ is the $2 \times 2$ block matrix with upper-left block $0$, upper-right block $g$, lower-left block $-g^{-1}$, and lower-right block $0$. Multiplying on the right by $W$ gives \begin{align*} U_{12}(g)U_{21}(-g^{-1})U_{12}(g)W=\operatorname{diag}(g,g^{-1}). \end{align*} Each factor on the left belongs to $E_{2N}(R)$, so \begin{align*} h:=\operatorname{diag}(g,g^{-1}) \end{align*} belongs to $E_{2N}(R)\subset E(R)$. [/step] [step:Convert the stable block factorization into normality] View $x\in GL_N(R)$ in the first block of $GL_{2N}(R)$ and define \begin{align*} \widetilde{x}:=\operatorname{diag}(x,I_N)\in GL_{2N}(R). \end{align*} Since $x=e_{ij}(r)$ is elementary in the first block, $\widetilde{x}$ is the ordinary elementary matrix $e_{ij}(r)$ in $GL_{2N}(R)$, hence $\widetilde{x}\in E_{2N}(R)$. Using the matrix $h=\operatorname{diag}(g,g^{-1})$ from the previous step, block multiplication gives \begin{align*} h\widetilde{x}h^{-1}=\operatorname{diag}(gxg^{-1},I_N). \end{align*} Because $h\in E(R)$ and $\widetilde{x}\in E(R)$, the element $h\widetilde{x}h^{-1}$ lies in $E(R)$. Therefore the stabilized matrix $\operatorname{diag}(gxg^{-1},I_N)$ represents an element of $E(R)$, which means that $gxg^{-1}\in E(R)$ in the stable group. Thus every conjugate of every elementary generator of $E(R)$ by every element of $GL(R)$ lies in $E(R)$. Since $E(R)$ is generated by these elementary matrices, $gE(R)g^{-1}\subseteq E(R)$ for every $g\in GL(R)$. Applying this inclusion to $g^{-1}$ gives the reverse inclusion, so $gE(R)g^{-1}=E(R)$. Hence $E(R)\trianglelefteq GL(R)$. [/step]