[proofplan] We first record the subgroup-theoretic fact that conjugation by an elementary transvection preserves $E(R)$, since the conjugating element, the generator being conjugated, and the inverse conjugating element all lie in the subgroup generated by elementary transvections. The main point is then to reduce conjugation by an arbitrary matrix $g\in GL_n(R)$ to conjugation by an element already in $E(R)$ after one stabilization. This is done by explicitly factoring the block diagonal matrix $\operatorname{diag}(g,g^{-1})$ as a product of elementary block transvections in $GL_{2n}(R)$. Conjugating a stabilized elementary generator by this block diagonal matrix gives the desired conjugate by $g$, and the colimit definition of $GL(R)$ identifies the stabilized result with the original conjugate. [/proofplan]

[step:Show that elementary conjugation preserves the stable elementary subgroup]Let $x=e_{ij}(a)$ and $y=e_{kl}(b)$ be elementary transvections over $R$, viewed as elements of $GL(R)$. Since $E(R)$ is the subgroup of $GL(R)$ generated by all elementary transvections, the elements $x$, $x^{-1}=e_{ij}(-a)$, and $y$ all lie in $E(R)$. Hence the product \begin{align*} xyx^{-1}\in E(R) \end{align*} by closure of a subgroup under multiplication. Thus conjugation by the elementary transvection $x$ sends every elementary generator $y$ of $E(R)$ into $E(R)$. Since conjugation by $x$ is a [group homomorphism](/page/Group%20Homomorphism) from $E(R)$ to $GL(R)$ on products, it sends every finite product of elementary generators and their inverses into $E(R)$. Applying the same argument to $x^{-1}$ gives the reverse inclusion, so every elementary transvection normalizes $E(R)$.[/step]

[guided]The point of this step is purely subgroup-theoretic. Let $x=e_{ij}(a)$ be an elementary transvection. Its inverse is $x^{-1}=e_{ij}(-a)$, because \begin{align*} (I_m+aE_{ij})(I_m-aE_{ij})=I_m \end{align*} inside any common stabilization $GL_m(R)$ with $m\ge 2$ containing the indices $i$ and $j$. Therefore both $x$ and $x^{-1}$ are among the elements used to generate $E(R)$. Now let $y=e_{kl}(b)$ be another elementary transvection. Since $E(R)$ is a subgroup and $x,y,x^{-1}\in E(R)$, closure under multiplication gives \begin{align*} xyx^{-1}\in E(R). \end{align*} No classification of the commutator $[x,y]$ is needed here; this avoids the opposite-root case, where the commutator of two elementary transvections need not itself be a single elementary transvection over an arbitrary ring. Because the elementary transvections generate $E(R)$ as a group, every element of $E(R)$ is a finite product of elementary transvections and their inverses. Conjugation by $x$ carries each elementary generator into $E(R)$, and hence carries each such finite product into $E(R)$. Repeating the same argument with the elementary transvection $x^{-1}$ gives the reverse inclusion. Thus conjugation by $x$ preserves $E(R)$ exactly.[/guided]

[step:Factor a stabilized block diagonal matrix into elementary matrices] Fix $n\ge 1$ and let $g\in GL_n(R)$. Set $V=R^n\oplus R^n$. Define endomorphisms $U(g)$, $L(g^{-1})$, and $S$ of $V$ by \begin{align*} U(g)(u,v)=(u-gv,v), \end{align*} \begin{align*} L(g^{-1})(u,v)=(u,v+g^{-1}u), \end{align*} \begin{align*} S(u,v)=(v,-u). \end{align*} A direct substitution shows that \begin{align*} U(g)\circ L(g^{-1})\circ U(g)\circ S=\operatorname{diag}(g,g^{-1}). \end{align*} Indeed, for $(u,v)\in V$ one has \begin{align*} S(u,v)=(v,-u), \end{align*} \begin{align*} U(g)(S(u,v))=(v+gu,-u), \end{align*} \begin{align*} L(g^{-1})(U(g)(S(u,v)))=(v+gu,g^{-1}v), \end{align*} \begin{align*} U(g)(L(g^{-1})(U(g)(S(u,v))))=(gu,g^{-1}v). \end{align*} The map $S$ is a product of elementary block transvections. Define maps $U_0:V\to V$ and $L_0:V\to V$ by \begin{align*} U_0(u,v)=(u+v,v) \end{align*} and \begin{align*} L_0(u,v)=(u,v-u). \end{align*} Then \begin{align*} U_0\circ L_0\circ U_0=S. \end{align*} To expand block transvections into ordinary elementary transvections, write $A=(a_{pq})\in M_n(R)$. The upper block transvection $T_A:V\to V$ defined by $T_A(u,v)=(u+Av,v)$ is represented in $GL_{2n}(R)$ by the product of the commuting elementary transvections $e_{p,n+q}(a_{pq})$ over all $1\le p,q\le n$. The lower block transvection $T'_A:V\to V$ defined by $T'_A(u,v)=(u,v+Au)$ is represented by the product of the commuting elementary transvections $e_{n+p,q}(a_{pq})$ over all $1\le p,q\le n$. Applying these expansions to $A=-g$, $A=g^{-1}$, $A=I_n$, and $A=-I_n$ shows that $U(g)$, $L(g^{-1})$, $U_0$, and $L_0$ all lie in $E(R)$. Therefore $S\in E(R)$, and the same is true for $U(g)$ and $L(g^{-1})$. Consequently \begin{align*} \operatorname{diag}(g,g^{-1})\in E(R)\subset GL(R). \end{align*} [/step]

[step:Conjugate a stabilized elementary generator by the block diagonal factorization] Let $e_{ij}(r)\in GL_n(R)$ be an elementary generator, after increasing $n$ if necessary so that $1\le i\ne j\le n$. Define the stabilized elementary matrix \begin{align*} x=\operatorname{diag}(e_{ij}(r),I_n)\in GL_{2n}(R). \end{align*} Let \begin{align*} h=\operatorname{diag}(g,g^{-1})\in GL_{2n}(R). \end{align*} By the previous step, $h\in E(R)$. Since every elementary transvection normalizes $E(R)$ and $h$ is a finite product of elementary transvections, conjugation by $h$ preserves $E(R)$. Because $x\in E(R)$, it follows that \begin{align*} hxh^{-1}\in E(R). \end{align*} Computing the block conjugate gives \begin{align*} hxh^{-1}=\operatorname{diag}(g e_{ij}(r)g^{-1},I_n). \end{align*} In the stable group $GL(R)$, the element $\operatorname{diag}(g e_{ij}(r)g^{-1},I_n)$ represents the same stable element as $g e_{ij}(r)g^{-1}\in GL_n(R)$, because the transition maps are precisely stabilization by identity blocks. Hence \begin{align*} g e_{ij}(r)g^{-1}\in E(R) \end{align*} as an element of $GL(R)$. [/step]

[step:Conclude that every stable invertible matrix normalizes $E(R)$] Let $\gamma\in GL(R)$. By definition of the filtered colimit, choose $n\ge 1$ and $g\in GL_n(R)$ representing $\gamma$. Let $y\in E(R)$. Since $E(R)$ is generated by elementary transvections, it is enough to prove that $\gamma x\gamma^{-1}\in E(R)$ for every elementary generator $x=e_{ij}(r)$. After stabilizing $g$ and $x$ to a common group $GL_m(R)$, the preceding step applies and gives \begin{align*} \gamma x\gamma^{-1}\in E(R). \end{align*} Thus conjugation by $\gamma$ sends every generator of $E(R)$ into $E(R)$, and therefore sends all of $E(R)$ into $E(R)$. Applying the same argument to $\gamma^{-1}\in GL(R)$ gives the reverse inclusion \begin{align*} \gamma^{-1}E(R)\gamma\subset E(R), \end{align*} which is equivalent to $E(R)\subset \gamma E(R)\gamma^{-1}$. Hence \begin{align*} \gamma E(R)\gamma^{-1}=E(R). \end{align*} Since $\gamma\in GL(R)$ was arbitrary, $E(R)$ is normal in $GL(R)$. [/step]