By B. Levitan, et. al.,

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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. Finally, if (u, v, w) is a linkage of T then K , sjfi = [(r„)/3, [sw)P) = [rup, swf3] = = (r ■ s)vp = (r ■ s)vp. Linkage Graphs and Linkage Groups 28 3.

We will need the linkage graphs in the next definition. 2-4 Constant Groups that Cannot Arise 39 Definition 1. IfTi and T2 are linkage graphs then Ti is a linkage subset 0/T2 if the set of vertices, the set of edges, and the set of linkages 0/T1 are subsets of the set of vertices, set of edges, and set of linkages (respectively) ofT2. We say that Tx is a linkage subgraph of T2 if Ti is a linkage subset 0/T2 and 1. Ifu,v € V'(r'i) and {u,v} is an edge in T2 it is also an edge in IV 2. If u,w £ V(Ti) and (u,v,w) (u,v,w) is a linkage ofYi is a linkage of T2 then v £ V(T^) and Definition 2.

By the previous lemma we cannot have a = b2 or b = a2. As in the first parts of this proof we cannot have a = e or b = e or a = 6. Hence, 6 = a - 1 . Thus Tc contains the linkage ([a], [e], [b])a = ([a2], [a], [e]) and also the linkage ([a2], [a], [e])a. = ([&], [a2], [a]). 2. Suppose that C is the Klein 4-group. Then Tc contains the linkage ([a],[e],[6])6=([afc],[6],[e]). Since (ab)2 = e, the result follows as in the previous case. Finally, suppose that C is the cyclic group of order 5. We cannot have b = e, or b = a, or b — a2, or b2 = a.

### Almost Periodic Functions and Differential Eqns. by B. Levitan, et. al.,

by Charles

4.5