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By Charles F. Miller III (auth.), Gilbert Baumslag, Charles F. Miller III (eds.)

The papers during this quantity are the results of a workshop held in January 1989 on the Mathematical Sciences study Institute. issues lined contain determination difficulties, finitely provided easy teams, combinatorial geometry and homology, and automated teams and similar issues.

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Example text

But if w =a 1 then G w = G and of course u =F 1 in G w . Clearly G w is again recursively presented. To decide whether an arbitrary word w is equal to 1 in G begin recursively enumerating two lists of words. F. Miller equal to 1 in G. The second list consists of all words equal to 1 in G w . If W =e 1 then W will appear in the first list. But w i-e 1 if and only if U appears in the second list. By examining the lists until one of these events occurs we can determine whether or not w is equal to 1 in G.

P. g. p. p. g. p. resid. P. Miller As we have mentioned before, the structure of a finitely generated abelian group can be completely and effectively determined from a finite presentation for such a group. In particular this enables one to solve the word problem in each such group and to solve the ismorphism problem for the class of such groups. Now for abelian groups conjugacy is the same as equality so the conjugacy problem is also solvable. The generalized word problem for a subgroup H of an abelian group G is equivalent to the word problem for G / H so GW P( G) is also solvable.

Be new generating symbols and form the presentation u(G) =< G,x,tl,t2,··. I t;l[Ui,X]ti = ViX-1UiX,i = 1,2, ... >. Then 0-( G) is an HNN-extension of the free product of G with the infinite cyclic group generated by x. The associated subgroups are just the infinite cyclic groups genrated by the [Ui,X] and the ViX-1UiX. A routine argument shows that the word problem for u( G) in the indicated presentation can be solved using the given solution to the word problem for G. Also observe that in 0-( G) each Vi lies in the normal closure of the corresponding Ui.

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