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# 半群理论作业代做 MT5863代写 半群理论代写 数学作业代写

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## MT5863 Semigroup theory: Problem sheet 3

Binary relations, equivalences, homomorphisms, and isomorphisms

### Binary relations and equivalences半群理论作业代做

3-1. Let X = {1, 2, 3, 4, 5, 6}, let ρ be the equivalence relation on X with equivalence classes {1, 4}, {5} and {2, 3, 6}, and let σ be the order relation on X given by the following Hasse diagram:

Write both ρ and σ as sets of ordered pairs. Find ρ σ, ρ σ, σ1 , ρ σ and σ ρ.

3-2. Prove the following statements about a binary relation ρ on a set X.

(a) ρ is reflexive if and only if ∆X ρ where ∆X = { (x, x) : x X };

(b) ρ is symmetric if and only if ρ1 ρ;

(c) ρ is transitive if and only if ρ ρ ρ.

3-3. Prove that the intersection ρ∩σ of two equivalence relations on a set X is again an equivalence relation. Describe the equivalence classes of this relation. 半群理论作业代做

3-4. Find examples that show that neither the union nor composition of two equivalence relations needs to be an equivalence relation.

3-7. Let S(n, r) (1 n r) be the number of equivalence relations on X with precisely r equivalence classes. (The numbers S(n, r) are called Stirling numbers of the second kind.) Prove that

S(n, 1) = S(n, n) = 1

S(n, r) = S(n 1, r 1) + rS(n 1, r) (2 r n 1).

Use this to calculate S(n, r) for 1 ≤ r ≤ n ≤ 6.

### Homomorphisms and isomorphisms半群理论作业代做

3-8. Let f : S T be a homomorphism, and let x S. Prove that if x is an idempotent, then so is xf. Is it true that if x is the identity of S, then xf is the identity of T? Prove that if x is the identity and f is onto, then xf is the identity of T. If P S, then prove that P f = {pf : p P} is a subsemigroup of T.

3-9. Let S be a semigroup such that x2 = x and xyz = xz for all x, y, z ∈ S. Fix an arbitrary element a ∈ S. Let I = Sa = {sa : s ∈ S} and Λ = aS = {as : s ∈ S}. Define a mapping f from S into the rectangular band I × Λ by xf = (xa, ax). Prove that f is an isomorphism.

### Further problems

3-10. Prove that a semigroup S is a rectangular band if and only if

(∀a, b ∈ S)(ab = ba ⇒ a = b).

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