By Professor Richard Hubert Bruck (auth.)

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**Example text**

For any left ideal A, ßn CA n 1\ AnS for every positive integer n. If A is nilpotent, An= 0 and hence ßn= 0 for some positive integer n. Similarly for right ideals. This completes the proof. 8. Ideal structure of semigroups 45 The nil radical N = N (5) is the set of all properly nilpotent elements of 5. An element a of 5 is properly nilpotent if a5 is a nil subset. Since a2 E a5, a itself is nilpotent. 1 that N contains every nil or nilpotent left or right or two-sided ideal. Moreover N is itself a nil ideal and hence is the (unique) maximalnilideal of 5.

Hence oc- oc'. 1. It is easy to prove that G* is the free covering group. That is: if K is a covering group, there exists a homomorphism of G* upon K which induces the identity mapping on G. Although the Coset Theorem by no means exhausts the subject of polyadic groups, it seems a proper stopping point in a survey of binary systems. For many interesting topics, see the references cited above. BAER [328] has investigated a connection linking Brandt groupoids and mixed groups with the flocks of PRÜFER [332].

Let G be a semigroup such that aG = Ga = G for every a in G. Then ( and only then) Gis a group. (Cf. ) Proof. x, y in G suchthat ae = a, fb = b, xa=f, by=e. Then fe=(xa)e=x(ae)=xa=f and fe=f(by) = (fb) y = by = e. Hence e = f. Therefore G has an identity e = f, which we denote by 1. 1. A subsemigroup G of a groupoid (5, ·) is called a subgroup of S if (G, ·) is a group. 3. A caSnSa. Thenthereexists a unique subgroup Gof S, maximal in the property that A C G. Specifically, if a E A, the set a S contains exactly one element e such that a e = a; 29 3.