K1, K2 of orders and grouprings

Hence [GL(R), GL(R)] ⊂ E(R) by 3.1.2. Also, from (i) above, E(R) ⊂ [E(R), E(R)] ⊂ [GL(R), GL(R)]. Hence E(R) = [GL(R), GL(R)].

Definition 3.1.1

K1(R) := GL(R)/E(R) = GL(R)/ [GL(R), GL(R)] = H1 (GL(R),Z) .