K2 of the integers

Example

Milnor showed that $K_2(\Z) \cong \Z/2\Z$. The nonzero element is the Steinberg symbol $\set{-1, -1}$. In the Steinberg group $St(\Z)$ it is the element $w^4$, where $w = x_{12}(1) \, x_{21}(-1) \, x_{12}(1)$.

Check that $w^4$ lies in $K_2(\Z)$. The image of $w$ in $GL(\Z)$ is $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},$$ and this matrix has order $4$, so $w^4$ maps to the identity.

Check that $\set{-1, -1}$ is not trivial: the inclusion $\Z \to \R$ induces $K_2(\Z) \to K_2(\R)$, which sends $\set{-1, -1}$ to $\set{-1, -1}$. The real Hilbert symbol is a Steinberg symbol with value $-1$ on $(-1, -1)$, so this image is not trivial.

The upper bound, that $K_2(\Z)$ has at most two elements, uses the Euclidean algorithm to reduce words in the Steinberg group.