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The proof in the other case is similar. If A=C ~) where rJ. = (a 2, ... , an) and I is the identity matrix in Mat n _ 1 (R), then A = e12(a2) .. ·eln(an) and AEEn(R). Return to an arbitrary A and put 1. General Linear Groups, Steinberg Groups, and K-Groups 26 with A' in Mat n- l (R) upper triangular with 1 along the diagonal. Applying induction, assume that A' EE n- 1 (R). Since the product o is in En(R), we are done. Let E: (R) be the subgroup consisting of all upper triangular matrices with 1 along the diagonal.

This is of course the subgroup of GLn(R) that is generated by all the eij(r} and all diagonal elements. 17. Suppose R is commutative. Then GLn(R} = GEn(R} SLn(R} = En(R}. if and only if Proof. If SLn(R} = En(R}, then GLn(R} = GEn(R} follows by applying the determinant. For the converse, let (1 in SLn(R) be arbitrary and put (1 = 't"«5 with 't" in En(R) and «5 in Dn(R}. By repeated application of (E5) write «5 = 't" 1··· 't"m«5', where the 't"i are in En(R} and «5' is a diagonal matrix with 1 everywhere along the diagonal except possibly in the (1,1) entry.

Define GL(M, a) by {~XSIXEM, aEO} flD 34 1. General Linear Groups, Steinberg Groups, and K-Groups GL(M,a)= {aEGL(M)ISs;Ma}. 3, GL(M, a) is a subgroup of GL(M). 4 that GL(M, a) is a normal subgroup of GL(M). The Abelian quotient group MIMa becomes a module over Ria by defining (x + Ma)(r + a) = xr + Ma for all x in M and r in R. We observe in passing that it is a routine matter to check that MIMa is isomorphic to the Ria module M ® R Ria obtained from M by change of rings to Ria via j: R -+ Ria.