By Andrew H Wallace
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But this is an easy c o n s e q u e n c e of the two relations ~2i+1 = w2i+1 + decomposable; sqlw2i = w2i+1. Finally, we will show that T h e o r e m 2 implies T h e o r e m 1. Recall from  that the maps induce rational by Borel w~iff(*) isomorphisms on ~. 1 ~ A(*) for i > I. and A(*) ~ K(Z) Furthermore, , wi ~KfZ%", ® Q ~ ~i G/O" ® Q ~ [~Q O if so that it is enough to show that the map is injective after t e n s o r i n g with Q. i = 4j, j ~ I else ~i(G/O) ~ wi(~K(Z)) Since ~i(G/O) is finite- ly generated, this is e q u i v a l e n t to the statement that ~i(G/O~[1/2]) 211/2]) ~ ~i(~K(Z) ^ is injective .
Finally, we will show that T h e o r e m 2 implies T h e o r e m 1. Recall from  that the maps induce rational by Borel w~iff(*) isomorphisms on ~. 1 ~ A(*) for i > I. and A(*) ~ K(Z) Furthermore, , wi ~KfZ%", ® Q ~ ~i G/O" ® Q ~ [~Q O if so that it is enough to show that the map is injective after t e n s o r i n g with Q. i = 4j, j ~ I else ~i(G/O) ~ wi(~K(Z)) Since ~i(G/O) is finite- ly generated, this is e q u i v a l e n t to the statement that ~i(G/O~[1/2]) 211/2]) ~ ~i(~K(Z) ^ is injective .
But this is cearly impossible. the proof of Theorem B. last example of Q(8a,b,c) shows that the finiteness presents new problems when compared to the finiteness o b s t r u c t i o n of Q(8a,b). In [BI] we saw that the plus part of o (e) could always be "made rational" for 4 groups of type Q(8a,b). The above example shows that this is not possible for groups of type Q(8a,b,c). 24 R E F E R E N C E S [BM] S. Bentzen: groups. Bentzen: A. Fr6hlich: Aarhus Locally A. Fr6hlich: angew. (1976) [Ha] H.