This is a sideways answer.

Let $E_{ij}(a)=I+a E_{ij}$, for $i\neq j$ and $a\in A$.These matrices generate the conmutator subgroup $$E(n, A)=[\mathrm{GL}(n, A),\mathrm{GL}(n, A)]\subseteq\mathrm{GL}(n, A).$$ One can easily check that the obvious relations satisfied by these elements are $$E_{ij}(a)E_{ij}(b)=E_{ij}(a+b),$$ $$[E_{ij}(a),E_{jk}(a)]=E_{ik}(a) \mbox{ if $i\neq k$,}$$ $$[E_{ij}(a),E_{kl}(b)]=1 \mbox{ if $i\neq l$ and $j\neq k$.}$$ Yet the group presented by generators and this relations is not $E(n,A)$, but what we call the $*n$-th unstable Steinberg group* $\mathrm{St}(n, A)$ of $A$. In general, this is larger than (precisely, an extension of) $E(n,A)$.

This is seen, for example, because the map $\mathrm{St}(n, A)\to E(n,A)$ has a non-trivial kernel. Indeed, after passing to the direct limit as $n$ goes to infinity, the kernel of that map is precisely the third algebraic $K$-theory group of $A$, $K_3(A)$. Quillen famously computed that $K_3(\overline{\mathbb{F}}\_p)$ is not zero. (Another cute example is that $K_3(\mathbb Z)$ is cyclic of order 48, by a theorem of Lee and Szczarba; this group can already be seen before passing to the limit for large $n$; of course, this example does not involve a field)

A nice reference for this is Jonathan Rosenberger's Algebraic $K$-theory and its applications. A short intuitive description for $K_3(A)$ is: it measures how much more information is there in the elementary matrices of a ring which does not follow formally from the Steinberg relations.

This is a sideways answer.

Let $E_{ij}(a)=I+a E_{ij}$, for $i\neq j$ and $a\in A$.These matrices generate the conmutator subgroup $$E(n, A)=[\mathrm{GL}(n, A),\mathrm{GL}(n, A)]\subseteq\mathrm{GL}(n, A).$$ One can easily check that the obvious relations satisfied by these elements are $$E_{ij}(a)E_{ij}(b)=E_{ij}(a+b),$$ $$[E_{ij}(a),E_{jk}(a)]=E_{ik}(a) \mbox{ if $i\neq k$,}$$ $$[E_{ij}(a),E_{kl}(b)]=1 \mbox{ if $i\neq l$ and $j\neq k$.}$$ Yet the group presented by generators and this relations is not $E(n,A)$, but what we call the $*n$-th unstable Steinberg group* $\mathrm{St}(n, A)$ of $A$. In general, this is larger than (precisely, an extension of) $E(n,A)$.

(NB: The following paragraph has been edited to make it match reality. Thanks to Allen for pointing the mistake in the comment bellow)

This is seen, for example, because the map $\mathrm{St}(n, A)\to E(n,A)$ has a non-trivial kernel. Indeed, after passing to the direct limit as $n$ goes to infinity, the kernel of that map is precisely the second algebraic $K$-theory group of $A$, $K_2(A)$. Milnor shows in his book that $K_2(\mathbb{R})$ is uncountable, and describes $K_2(\mathbb Q)$ (he also shows that $K_2(\mathbb Z)$ is cyclic of order two, so this can be done for rings that are not fields too...)

A nice reference for all this is Jonathan Rosenberger's Algebraic $K$-theory and its applications, and there is John Milnor's Introduction to algebraic $K$-theory, which is also extremely nice. 

A short intuitive description for $K_2(A)$ is: it measures how much more information is there in the elementary matrices of a ring which does not follow formally from the Steinberg relations.

This is a sideways answer.

Let $E_{ij}(a)=I+a E_{ij}$, for $i\neq j$ and $a\in A$.These matrices generate the conmutator subgroup $$E(n, A)=[\mathrm{GL}(n, A),\mathrm{GL}(n, A)]\subseteq\mathrm{GL}(n, A).$$ One can easily check that the obvious relations satisfied by these elements are $$E_{ij}(a)E_{ij}(b)=E_{ij}(a+b),$$ $$[E_{ij}(a),E_{jk}(a)]=E_{ik}(a) \mbox{ if $i\neq k$,}$$ $$[E_{ij}(a),E_{kl}(b)]=1 \mbox{ if $i\neq l$ and $j\neq k$.}$$ Yet the group presented by generators and this relations is not $E(n,A)$, but what we call the $*n$-th unstable Steinberg group* $\mathrm{St}(n, A)$ of $A$. In general, this is larger than (precisely, an extension of) $E(n,A)$.

This is seen, for example, because the map $\mathrm{St}(n, A)\to E(n,A)$ has a non-trivial kernel. Indeed, after passing to the direct limit as $n$ goes to infinity, the kernel of that map is precisely the third algebraic $K$-theory group of $A$, $K_3(A)$. Quillen famously computed that $K_3(\overline{\mathbb{F}}\_p)$ is not zero.

A nice reference for this is Jonathan Rosenberger's Algebraic $K$-theory and its applications.

This is a sideways answer.

Let $E_{ij}(a)=I+a E_{ij}$, for $i\neq j$ and $a\in A$.These matrices generate the conmutator subgroup $$E(n, A)=[\mathrm{GL}(n, A),\mathrm{GL}(n, A)]\subseteq\mathrm{GL}(n, A).$$ One can easily check that the obvious relations satisfied by these elements are $$E_{ij}(a)E_{ij}(b)=E_{ij}(a+b),$$ $$[E_{ij}(a),E_{jk}(a)]=E_{ik}(a) \mbox{ if $i\neq k$,}$$ $$[E_{ij}(a),E_{kl}(b)]=1 \mbox{ if $i\neq l$ and $j\neq k$.}$$ Yet the group presented by generators and this relations is not $E(n,A)$, but what we call the $*n$-th unstable Steinberg group* $\mathrm{St}(n, A)$ of $A$. In general, this is larger than (precisely, an extension of) $E(n,A)$.

This is seen, for example, because the map $\mathrm{St}(n, A)\to E(n,A)$ has a non-trivial kernel. Indeed, after passing to the direct limit as $n$ goes to infinity, the kernel of that map is precisely the third algebraic $K$-theory group of $A$, $K_3(A)$. Quillen famously computed that $K_3(\overline{\mathbb{F}}\_p)$ is not zero. (Another cute example is that $K_3(\mathbb Z)$ is cyclic of order 48, by a theorem of Lee and Szczarba; this group can already be seen before passing to the limit for large $n$; of course, this example does not involve a field)

A nice reference for this is Jonathan Rosenberger's Algebraic $K$-theory and its applications. A short intuitive description for $K_3(A)$ is: it measures how much more information is there in the elementary matrices of a ring which does not follow formally from the Steinberg relations.