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By Professor Richard Hubert Bruck (auth.)

ISBN-10: 3662428377

ISBN-13: 9783662428375

ISBN-10: 366243119X

ISBN-13: 9783662431191

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Example text

Then ppp =~ p, so p and ef arerelative inverses of p. If p also has a unique relative inverse, then p = ef and p = jpe = (/e) 2 • If, in addition, je has a unique relative inverse q and q has a unique relative inverse, then q =je= q2 and p = (fe) 2 = q2 = q. That is, ef =je. We may draw the following conclusion: I f the regular elements of 5 form a subsemigroup R and if each element of R has a unique relative inverse in 5, then every two idempotent elements of 5 commute. Finally, let us assume that the set R of regular elements of 5 is non-empty and that every two idempotent clements of 5 commute.

Since(e/) (pe) (ef) = efand(pe) (e{) (pe)=pe, then p = pe. Similarly, p = fp. Hence P2 = (pe) (fp) = p. Then ppp =~ p, so p and ef arerelative inverses of p. If p also has a unique relative inverse, then p = ef and p = jpe = (/e) 2 • If, in addition, je has a unique relative inverse q and q has a unique relative inverse, then q =je= q2 and p = (fe) 2 = q2 = q. That is, ef =je. We may draw the following conclusion: I f the regular elements of 5 form a subsemigroup R and if each element of R has a unique relative inverse in 5, then every two idempotent elements of 5 commute.

Moreover, if oc- oc' and ß- ß', then I(Aocßp) 40 II. l); whence ocß- oc' ß'. l); so ß- ß'. Similarly, if ß - ß' and oc ß - oc' ß', then oc - oc'. This is enough to show that () is a congruence relation on 5 and that 5/0 is a cancellation semigroup. In addition, by (C), the identities form an equivalence class which contains the empty sequence. Moreover, for every oc, there is a ß such that oc ß is an identity. Consequently, 5/0 is a group. This proves (i) of the outline. Next suppose that (x) - (y) for some x, y in G.

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A Survey of Binary Systems by Professor Richard Hubert Bruck (auth.)


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