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 learn Institute. issues coated comprise choice difficulties, finitely provided uncomplicated teams, combinatorial geometry and homology, and automated teams and similar subject matters.

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For then one can systematically try the bounded number of R-sequences with conjugating elements of length at most the above estimate. Indeed it is easy to see that being able to compute such a bound is equivalent to solving the word problem in the case that R is finite. 2. Suppose that G has finite presentation P =< X I R > where R is a symmetrized set of words on X. Let F be the free group with basis X and N the normal closure of R. Then the word problem for G = F / N is recursively solvable if and only if there is a recursive function f such that Ap(w) :::; f(lwl) for all wEN.

We call such a device a synchronous two-tape automaton. If instead such a two tape automaton is allowed to read its input tapes at different rates, we call such a device an asynchronous two-tape automaton. An asynchronous two tape automaton can recognize far more complicated sets of pairs than a synchronous one. Let X be a set of generators for a group G. Let cP be the free monoid with basis X U X-I. For any word w E cP define f-L(w) E G to be the element represented by w. An (synchronously) automatic structure for G with respect to the generating set X is a regular language L in cP such that f-L(L) = G together with a synchronous two-tape automaton M which accepts the collection of pairs of elements of L which represent elements of G lying at most a unit apart in the Cayley graph r(G, X).

It can be shown that if the finitely presented group G is defined by two finite presentations P and P1 , then there are constants Cl, C2 and C3 such that OPl (n) :::; C10p(C2n) + C3n. In a different direction, suppose that the finite presentation P2 =< X I R2 > of G is obtained from P by adding some elements of N to R so that R ~ R2. Then A p2 (w) :::; Ap(w) for all wEN and Op2(n) :::; Op(n). The above corollary can be restated now as follows: the word problem for G is solvable if and only if there is a recursive function f such that Decision Problems for Groups: Survey and Reflections 43 Dp(n) ::::; f(n) for all n > O.