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By Lucchini A.

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PROBLEM 3: Find subgroupoids of Q+ (11/3, 7/3). PROBLEM 4: Can R+ (√2, √3) have subgroupoids, which are conjugate with each other? PROBLEM 5: Is Q (7, 11) a P-groupoid? PROBLEM 6: Can R (2, 4) be a Moufang groupoid? PROBLEM 7: Can Q (7, 49) have conjugate subgroupoids? PROBLEM 8: Find (t, u) so that Q (t, u) has conjugate subgroupoids which partition Q (t, u). PROBLEM 9: If Z (m, n) has conjugate subgroupoids that partition Z (m, n) find m and n. Supplementary Reading 1. Bruck, A Survey of Binary Systems, Springer Verlag, (1958).

Hence we have proposed in Chapter 7 an open problem about the possible construction of a groupoid which is not a semigroup using Zn. The only way to obtain groupoids with identity is by adjoining an element with Zn. ∗∗ DEFINITION: Let Zn = {0, 1, 2, ... , n – 1} n ≥ 3, n < ∝. Let G = Zn ∪ {e} where e ∉ Zn. Define operation ∗ on G by ai ∗ ai = e for all ai ∈ Zn and ai ∗ e = e ∗ ai = ai for all ai ∈ Zn. Finally ai ∗ aj = tai + uaj (mod n) t, u ∈ Zn. It can be checked (G, ∗) is a groupoid with identity e and the order of G is n + 1.

Further A2 = 2A1 = 2 {0, 3, 6, 9} = {2, 5, 8, 11}. Hence, A1 and A2 are Smarandache conjugate subgroupoids of G. 5: Let G be a SG. If P and K be any two Smarandache subgroupoids of G which are Smarandache conjugate then they are Smarandache semiconjugate. But Smarandache semiconjugate subgroupoids need not in general be Smarandache conjugate subgroupoids of G. Proof: By the very definition of Smarandache semiconjugate subgroupoids and Smarandache conjugate subgroupoids we see every Smarandache conjugate subgroupoid is Smarandache semiconjugate subgroupoid.

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(2, 3, k)-generated groups of large rank by Lucchini A.

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Categories: Symmetry And Group