K1 of a ring

Remarks

Whitehead's lemma says $[GL(R), GL(R)] = E(R)$, the elementary subgroup, so $K_1(R) = GL(R)/E(R)$. The group $K_1(R)$ is abelian, and the class of a product $AB$ equals the class of $\begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix}$. If $R$ is commutative, the determinant $\det \colon GL(R) \to R^\times$ kills $E(R)$, so it induces a surjection $K_1(R) \to R^\times$, split by $u \mapsto [\mathrm{diag}(u, 1, 1, \dots)]$. Example: for a field $F$, every matrix of determinant $1$ is a product of elementary matrices, so $K_1(F) \cong F^\times$. Non-example: the group $GL(R)$ itself is not $K_1(R)$ for $R = \Q$, since $GL(\Q)$ is not abelian while $K_1(\Q) \cong \Q^\times$ is.