Steinberg group

Remarks

The elementary matrices satisfy the Steinberg relations, so there is a unique homomorphism $\phi_n \colon St_n(R) \to GL_n(R)$ with $\phi_n(x_{ij}(r)) = e_{ij}(r)$. Its image is the elementary subgroup $E_n(R)$. In the limit one gets $\phi \colon St(R) \to E(R)$, which is onto, and its kernel is $K_2(R)$.

Example: for $R = \Z$, the element $w = x_{12}(1) \, x_{21}(-1) \, x_{12}(1)$ of $St(\Z)$ maps to $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, a matrix of order $4$. So $w^4$ lies in the kernel of $\phi$; it is the generator of $K_2(\Z) \cong \Z/2\Z$, so $w^4 \ne 1$ in $St(\Z)$.

Non-example: for $n = 2$ there are no three distinct indices, so only clause 1 of the Steinberg relations has content. The same presentation then gives the free product of two copies of the additive group of $R$, with no relation between $x_{12}(r)$ and $x_{21}(s)$. This is far from $E_2(R)$, which is why the definition takes $n \ge 3$.

Kervaire and Steinberg showed that $\phi \colon St(R) \to E(R)$ is a universal central extension, so $K_2(R)$ is the center of $St(R)$.