Combinatorics

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By Professor Richard Hubert Bruck (auth.)

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In connection with (I), if M is a multigroupoid, let G be the set of all non-empty subsets of M, let (~) be the relation of inclusion (as subsets of M) among the elements of G and, for each ordered pair (x, y) of elements of G, let xy be the element of G consisting of all elements c in M for which there exist elements a, b in M with a in x, b in y and c in ab. We note that multiplication in G is single-valued, so that G is a groupoid. Moreover, for x, x', y, y' in G, if x ~ x', y ~ y' then xy ~ x' y'.

Let S be a semigroup given in termsofafinite set of generators and a finite set of relations between the generators. Then the elements of S are classes of equivalent words in the generators, the rules for equivalence being explicitly set up in terms of the relations between the generators. The word problem for S consists in describing an algorithm which will show in a finite nurober of steps whether two (arbitrarily chosen) words are equivalent. PosT [240] constructed a semigroup for which the word problern is recursively unsolvable; another example with two generators was later given by HALL [199].

Hence, by {d) with y = 1, k- 1 k' = 1, so that k = k'. Again, if k E K and x, y E M and kx = ky, then x = k-1 (kx) = k- 1 (ky) = y. At this stage we see that there mustexist at least one non-empty subset E of M (which we may and do assume to contain the identity 1 of K) such that each x in M has a unique representation x = ke in M. where k E K, e E E. It is now clear that M has a representation of type (I) and therefore a representation of type (II). ) For thii--see BAER [325], [326]. 6. Polyadic groups.

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