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By J. F. Berglund, K. H. Hofmann
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6 (x, [y,x] -I, y); its maximal groups, Straightforward. 5 every Idempotent is primitive; is completely simple. 3)), Proof: simple Let I & Se be a left is a group, so x -I exists Then I = Ie. Let x ~eI. in eSe. 3, Now & x - l e I @ SeI c I. So Se = (Se)e _~ Sel c_ I ~_ Se. Let J be a right a right (a left group). ideal. eI = ele ~- eSe ~ I . 8 (hence, then Se is left simple Consequently, eSe 4T- ideal Thus in S and I = Se. I a right ideal in J. , J is simple; If I is minimal ideal Proof: in S IJ ~_ I; (ii) If J is minimal, (iii) ideal in J, then I is a right in S.
Are equivalent: simple (c) S is a right semigroup, group if for all G is a group, is non-empty, Specifically, no zero semigroup. statements (b) E(S) if S contains ideals. 3 zero semigroup. 'or an idempotent H(e) 45 = (se, See Clifford (se)-Is), where (se) -I is the G = Se. & Preston [I], p. 38. 4 46 - If I ~ S is a minimal right ideal, and if E(I) is not empty, Proof: then I is a right group. Let x ~ I. Since (xI)S = x(IS) G xl, xI is a right ideal in S. minimal, so xl = I. 3(b), I is a right group.
10. (a) ===> (e) : Trivial. (b) < = = = 7 (c): Straightforward computation in a paragroup. (e) ==='~ (d): If f , g , h ~ E ( S ) , and efghe then ~'(efghe) = e 5 = e= H(e); so efghe = e by the inJectlvlty of ~'IH(e). It follows that So f e f g E ( S ) ~ H ( f ) , (Eef) 2 = ~efgef = g e f c E ( S ) whence fef = f. Then (fg)2 = fgfg = (fef)g(fef)g = f(efgfe)f~ = fefg = fg. Thus f g 6 E ( S ) . 9. 3 is the case if and only if ~: S --, eSe is a morphism. Remark: By considerinE any paragroup which is not a direct product, one sees that (f) does not imply (c).