By Azriel Rosenfeld

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25) from follows 3(T3b + + b3b) that + one u = bb2 case, = = 3b3 = b3; 2, (F3, bA 2b + 3T3, b3T + (73, w) = = 5b2 + b2b 26-b + 3bb3 + b2T = = = + 2b 2= 2 101 + 2(b (-b3) U) = 663 b3 + of the + b3 following 3-b3 b + + 5b2 + 2(b + T) + 1 42 b3. Hence T) + + + 3(b3 + 2 = = = = b2b3 2 + 2b 2 (b2 Tb3), = 7 - = = b3(6-b) (b3b)T 73) 3bb3i 101 + 3bb3 Since 3 (-b3, bb2) b2 + 3(b3 + T3) + 2b2 2 5 1 + 4b2 Furthermore, 3b2 + b3 + 73 and b 2 bb3 implies + holds: conditions yields b2T) b3T + b373 + 501 + (i) + 42 + 2 + 4(bb2 20b2 + 3-b + b + N + 101 + 2b + 2-b + 4b2 + 3b + T + bT2=(37b+b+b3)(3b+T+T3-)=96-b+3(b y + b3b + b3T + 2b3b2 + 3-b3b + b3b + 2b3.

It is Even holds property problem counterexamples in are provided an open to are Z. Arad et al. 48 Proof. already If x, y, (a) 2 + If b2 (i) (ii) (iii) (b) [xyl 1). :7 T3, 73, b2T2 = (I First the 1 for = 1 + 1 for = note prime proofs (a) By n = 21 + 1 is = > = of b272 or Icl jb2j 2(n is multiple a 1) - (a) (ii) equivalent: are equivalent: (n2 +I)n 0 (b2 b3) b2 b3) b4), b3 0 b4; + of -==> E a c for n all which (bA, b2b3) < is 1(b3 (c, b2b3) (iii) =, are 2 calculation. B. divides Lemma I there 1n(n = Supp(b2b3)\j1j.

3b + (bc+, bel) that e-) = e = obtain I in the 0 . orthogonal idempotents := 31-1 ; 4 e:= -4 ;j associated with the closed +e jA(DeA. Since jA --- A/M, the structure algebra eA. the algebra eA, we define the following set of A is = if if a is 0; odd; if a is even. ieo+ 2 = 1 following formulae I 2 -' +- e6a 2 The that M: Thus A In holds it = 1 +1 the o-(-a). case a There are two pairwise subset e-, Wedefine 0, then be+ 0 with a o,-,,, + ea e+-,,. 6--a) e,-, = et that Z,,, E a 2o;, bel ff if e,,, et if - It or = 7 show that the a= elements defined completely of elements: a x,,, E Z2n) a E Z2n form.

### An introduction to algebraic structures by Azriel Rosenfeld

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