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By Ben Ayed M., El Mehdi K., Pacella F.

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Proof: By the very definition of Smarandache strong P-groupoid and Smarandache Pgroupoid we see every Smarandache strong P-groupoid is a Smarandache P-groupoid. To prove the converse, consider the example. Consider the groupoid Z12 (5, 10) given by the following table: 0 1 2 3 4 5 6 7 8 9 10 11 ∗ 0 0 10 8 6 4 2 0 10 8 6 4 2 1 5 3 1 11 9 7 5 3 1 11 9 7 2 10 8 6 4 2 0 10 8 6 4 2 0 3 3 1 11 9 7 5 3 1 11 9 7 5 4 8 6 4 2 0 10 8 6 4 2 0 10 5 1 11 9 7 5 3 1 11 9 7 5 3 6 6 4 2 0 10 8 6 4 2 0 10 8 7 11 9 7 5 3 1 11 9 7 5 3 1 8 4 2 0 10 8 6 4 2 0 10 8 6 9 9 7 5 3 1 11 9 7 5 3 1 11 10 2 0 10 8 6 4 2 0 10 8 6 4 11 7 5 3 1 11 9 7 5 3 1 11 9 Now {6} is a semigroup and A = {0, 6} is a Smarandache subgroupoid and the elements in A satisfy the identity (x ∗ y) ∗ x = x ∗ (y ∗ x).

2: Let G = Z12 (3, 9) be a groupoid given by the following table: ∗ 0 1 2 3 4 5 6 7 8 9 10 11 0 0 3 6 9 0 3 6 9 0 3 6 9 1 9 0 3 6 9 0 3 6 9 0 3 6 2 6 9 0 3 6 9 0 3 6 9 0 3 3 3 6 9 0 3 6 9 0 3 6 9 0 4 0 3 6 9 0 3 6 9 0 3 6 9 5 9 0 3 6 9 0 3 6 9 0 3 6 6 6 9 0 3 6 9 0 3 6 9 0 3 7 3 6 9 0 3 6 9 0 3 6 9 0 8 0 3 6 9 0 3 6 9 0 3 6 9 9 9 0 3 6 9 0 3 6 9 0 3 6 10 6 9 0 3 6 9 0 3 6 9 0 3 11 3 6 9 0 3 6 9 0 3 6 9 0 Clearly, A1 = {0, 4, 8} and A2 = {0, 3, 6, 9} are Smarandache subgroupoids of G as P1 = {0, 4} ⊂ A1 is a semigroup and P2 = {0, 6} ⊂ A2 is a semigroup of G.

Hence Zn (t, u) is a semigroup. On the other hand if Zn (t, u) is a semigroup to show t2 ≡ t (mod n) and u2 ≡ u (mod n). Given Zn (t, u) is a semigroup so a ∗ (b ∗ c) ≡ (a ∗ b) ∗ c (mod n) for all a, b, c ∈ Zn. Now a ∗ (b ∗ c) = (a ∗ b) ∗ c implies ta + tub + u2c = t2a + tub + uc. So (t – t2) a + (u2 – u) c ≡ o (mod n) this is possible for all a, c ∈ Zn only if t2 ≡ t (mod n) and u2 ≡ u (mod n). Hence the claim. 2: If in the above theorem when; n is a prime Z (n) never contains a semigroup. 32 Proof: Since t2 ≡ t (mod n) or u2 ≡ u (mod n) can never occur for t, u ∈ Zn \ {0, 1} when n is a prime, Z (n) has no semigroups.

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