On Equivalent Sets of Elements in a Free Group

TLDR: In this article, the authors give a mechanical process for deciding whether or no two sets are equivalent and also a process for reducing (a) to one of a finite number of normal forms.

Abstract: together with the 'simple automorphism' which replaces a, by its inverse. Relative to this kind of equivalence we have very little to add to a paper by J. Nielsen,2 in which he gives a mechanical process for deciding whether or no two sets are equivalent and also a process for reducing (a) to one of a finite number of normal forms. When reduced in this way we shall describe a set of elements as reduced (N), and we recall that (a) is reduced (N) if it contains no two words of the form (AB)" and (AC)" respectively,3 where 1(A) > 1(B) or > I(C), and if the last half of every word with an even number of letters is an 'isolated ending.' That is to say, if a AB and 1(A) = 1(B) no other word in (a) ends with B or begins with B'. In ?2 it is assumed that any empty words which appear' during the process of reduction are discarded, as in J. N., while in ?4 they are retained. Theorem 1 below is essentially a restatement of various arguments used by Nielsen, while Theorem 2 adds a detail to J. N. The second kind of equivalence refers to the effect on (a) of automorphisms5 of G, two ordered sets of elements (a) and (I), both of which contain the same number of words, being equivalent if ax, corresponds to #X(X = 1, 2, *. * ) in some automorphism of G. That is to say they are equivalent if there is an automorphism