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The image of F in SL(3,F) consists of matrices of the form / 1 0 r \ 0 10 \0 0 1/ It follows that the center of SL(3, F) which consists of diagonal matrices does not intersect the image of F . By factoring out the center of SL(3, F) we thus obtain a verbal embedding of F into PSL(3,F). Now PSL(3,F) is a simple group which happens to be contained in may non-abelian simple groups. Any such group must be the image of a linkage group L(F,TH) for some subgroup H. We will describe other simple groups that are images of linkage groups in later sections.

4. There does not exist a verbal embedding of a ring R with identity in which the constant group C = (a, b) has order less than 6. Proof. Suppose : R —> G is a verbal embedding with constant group C = (a,b). 1. 2 it follows that R is embedded at each vertex in TcClearly C cannot have order 1. Suppose C has order 2. The encoding word for the embedding is [xa, yb]. Now either b = a or b = e. But if 6 = a the encoding word is [x,y]a and this is impossible since [r^s*]" = e. If b = e then ([a], [e], [b]) = ([a], [e], [e]) is a basic linkage, and [r^a,5^] = e.

Next, if u and v are connected by an edge we have (ru • sv)a = ( r u ) 5 • (sv)a = (ra)„ • (sa)v — = (sa)v ■ (ra)u = (sv)a- (ru)a = (sv ■ ru)a. Finally, if (it, v, w) is a linkage of V then [r u ,5„]a = [(ru)a,(sw)a] = [(ra)u,(sa)w] = = (ra ■ sa)v = ((r ■ s)a)v = (r ■ s)va. 2. This is proven in the same way: (r„ ■ sv)j3 = (r„)/? ■ (sv)P = rvp ■ svp = = (r + s)vp - (r + s)vp. If u and v are connected by an edge we have (r„ ■ sv)~ft = (ru)fi ■ (sv)P = rup ■ svp = = &vp ■ rup = {sv)~\$ ■ (ru)/3 = (sv ■ r„)/3.