Minimal Clans

International Journal of Engineering Technology, Management and Applied Sciences www.ijetmas.com November 2015, Volume 3, Issue 11, ISSN 2349-4476 Minimal Clans Shinemin Lin, Ph.D. Savannah State University ABSTRACT A minimal clan is a lattice-ordered partial semigroup having the cancellation property and difference property. Wyler (1966 [15]) proved every Abelian clan could be embedded into an Abelian latticeordered group. Conrad (1990) showed that every clan could be embedded into a lattice-ordered group. There are two main results in this paper. First theorem said a clan C is complete if and only if the latticeordered group generated by C is complete. The second theorem said every clan contains a greatest generalized Boolean algebra. That answered Schmit’s question ([16]). The second part of this paper is to that if an l-group G has minimal generating clan such that {0} is the only subsemigroup then G is a Specker group. This theorem makes a connection between Specker groups and Clans. Keywords Clan, Groupoid, l-group, l-ideal, Specker group, subsemigroup, Free generating Boolean Algebra, eequivalent. INTRODUCTION I gather some terms and fundamental results about l-groups. For additional background, refer to [1], [4], and [6]. I follow tradition and use additive notation for the group operation even though most of groups may not be abelian. A partially ordered group (po-group) is a group (G, +) that has a partially order, ≤, defined on it such that a ≤ b implies x + a ≤ x + b and a + x ≤ b + x for all a, b, x of G. If the order is also a lattice order or a total order, then G is called a lattice-ordered group (l-group) or a totally-ordered group (o-group), respectively. The symbols , and  are used to denote the greatest lower bound and the least upper bound. A subgroup A of G is an l-subgroup if A is also a sub-lattice of G. An l-subgroup A is convex, of 0 ≤ g ≤ q ∈ A implies g ∈ A. Suppose H is an l-subgroup of an l-group G, H is large in G or G is an essential extension of H if for each convex l-subgroup L≠ 0 of G, L ∩ H ≠ 0. An l-group G is Archimedean if 0 ≤ ng < h for all n ∈ N implies g = 0. For an Archimedean l-group G, if G admits no proper essential extension then G is essentially closed. An essentially closed essential extension of l-group G is called essential closure of G and is denoted by Ge. An l-group G is complete (laterally complete), if every bounded (disjoint) subset M of G+, ∨ M ∈ G. An l-group H is a completion (lateral completion) of G if H is complete (lateral complete), and G is l-isomorphically dense in H, and no proper l-subgroup of H containing an l-isomorphic copy of G is complete (lateral complete). The completion (laterally completion, divisible hull) of G will be denoted by G^ (GL, Gd). An element h ≤ g is called a component of g if (g – h) ∧ h = 0; an element g > 0 of G is called singular if for any 0 < x ≤ g then x is a component of g. An l-subgroup H of G is saturated if for all h of H, the components of h belong to H. For g ∈ G, g+ = g ∨ 0 (g- = (-g) ∨ 0) is called the positive (negative) part of g, and |g| = g+ + g- is called the absolute value of g. The principal convex l-subgroup of G that is generated by g is denoted by G(g), which is the subset G(g) = {x ∈ G| 0 ≤ |g|≤ n|g| for some 0 < n ∈ N}. A partial groupid is a set E with a relation S  E x E and a map + : S  E such that there exists an element 0 ϵ E satisfying (0,x) ϵ S, (x,0) ϵ S and 0 + x = x+ 0 for all x ϵ E. A partial semigroup is a partial groupoid (E, S, +) which is associative. 33 Shinemin Lin International Journal of Engineering Technology, Management and Applied Sciences www.ijetmas.com November 2015, Volume 3, Issue 11, ISSN 2349-4476 An ordered partial semigroup is a partial semigroup (E, S, +) with ordered relation ≤ such that µ + x + v ≤ v + x + µ holds for all x, y ϵ E satisfying x ≤ y and for all µ, v ϵ E satisfying (µ, x) ϵ S as well as (µ,y) ϵ S and (µ+y, v) ϵ S. An ordered partial semigroup has lattice property if (E, ≤) is a lattice. It has difference property if for all x, y ϵ E having a x  y, and a x  y, there exist µ and v ϵ E+ satisfying µ + x = x  y = x + v and µ + x  y = y = x  y + v. It has cancellation property if x = y holds for all x, y ϵ E satisfying µ + x + v = µ + y + v for some u, v ϵ E. A minimal clan (or a clan) is a lattice-ordered partial semigroup having the cancellation property and difference property. A minimal clan E is called positive if E = E+. An abelian clan C is Archimedean if a ϵ C is such that na ≤ b in [C], for all natural number n and a fixed element b of C, then a ≤ 0. Remarks: 1. (Wyler) Every abelian clan C can be embedded into an abelian I-group G (= [C]). Furthermore, C is Archimedean if and only if [C] is Archimedean. 2. (Conrad) Every clan C can be embedded into an l-group G (=[C]). SECTION 1 Definition 1.1 1. A clan C is called a generating clan of an I-group G, if C is a convex sublattice of G such that for every element g of G, g=c1 + c2 + … + cn, where c 1 ϵ C for all I ϵ {1, 2, 3, …, n}. We write G = [C]. 2. A subset D of C is called a subclan of C if D is a sublattice of C and for a, b ϵ D such that a ≤ b there exists c ϵ D such that a + c = b. If D is also convex then D is called a convex subclan of C. Lemma 1.2. Let C be a positive clan, and D be a subset of C then D’ = {x ϵ C| x  d = 0 for all d ϵ D} is a convex subclan. Proof. Clearly, D’ is a sublattice of C. If a ≤ b, a, b ϵ D’  C, then there is c ϵ C satisfying a + c = b. For all d ϵ D, 0 ≤ x  d ≤ (a + c)  d = b  d = 0; therefore, c  d = 0 which implies c ϵ D. Definition 1.3. D’ is called a polar subclan (or polar) of C. Lemma 1.4. Let R(C) be the set of all polars of C. Then R(C) is a complete Boolean algebra. Proof. The proof is similar to the proof of the theorem that the set of all polars of an I-group is a complete Boolean algebra ([4]). Proposition 1.5. Let C be a positive abelian clan. Then G = [C] is a free abelian I-group over C. Proof. C g [C]=G F F(C) f f* [f(C)] H, any abelian I-group Let f be an additive lattice homomorphism from C into any abelian I-group H, and g be the natural embedding of C into [C]. Then f(C) is an abelian clan that is in H. [f(C)] is an l-ideal of H. Now define f*: g1+ g2+… +gn  f(g1) + f(g2) + … + f(gn), then f* is an l-homomorphism from G into H. Lemma 1.6. A clan C freely generates an abelian l-group G. Then C is a generating clan of G. 34 Shinemin Lin International Journal of Engineering Technology, Management and Applied Sciences www.ijetmas.com November 2015, Volume 3, Issue 11, ISSN 2349-4476 Proof. By Wyler’s construction, G is a free abelian l-group on C, it suffices to show that C is a convex sublattice of G. Clearly C is a sublattice of G. Now suppose 0 ≤ g ≤ c, where c ϵ C, and g ϵ G. Then g = g1+ g2+… +gn , where g 1 ϵ C for all i’s and there is 0 ≤ h = h1+ h2+… + hm ∈ G, hi∈ C such that g + h = c. This implies the addition of g1+ g2+… + gn is defined in C hence g ϵ C. Definition 1.7. A clan C is complete if T= { 0 ≤ ci ϵ C | i ∈ I} is a bounded subset of C then  T ϵ C. Proposition 1.8. Let C be a complete clan that generates the l-group G, and let T = { 0 ≤ ci ϵ C | i ∈ I} be a bounded subset of C. Then ˅GT=˅CT. Proof. Let v= v1ci be the join of T in C. Then v ≥ ci in G for all i ϵ I. Suppose u ∈ G satisfying u ≥ ci in G for all i ϵ I, then u ≥ ˅GT. Now v ≥ v  u = (˅Ici)  u = t ≥ 0. Therefore t ϵ C, and v ≥ ˅I(u  ci) = t  t ≥ ci  u = ci for all i ∈ I. That implies t ≥ ˅Ici= v, and hence t = v. Since u  (˅Ici) = t, u ≥ t = v. Therefore v = ˅GT. Lemma 1.9 ([4]) Let G be an l-group and let {gi | i ∈ I} be a subset of G and g ∈ G. Then ˅I(gi + g) exists if and only if ˅Igi exists. If this is the case then ˅I(gi + g) = ˅Igi + g. Theorem 1.10. A Clan C is complete if and only if G = [C] is complete.  T = {c i ϵ C | i ϵ I} ϵ  Proof. ( ) Let be a bounded subset of C, say bounded by g C G. Then, in G, ˅GT = v ≤ g exists. By convexity of C, v ϵ C. Let s be an element of C  G such that s ≥ ci for all i ϵ I. ˅ T ˅ T Then s ≥ G = v. Hence v = C . ( ) Let T = {a i ϵ C | i ϵ I} be a bounded subset of G, say bounded by g = c1 +c2 +…. + cn> 0, where a ϵ C for all i ϵ I. We apply induction on n. If n =1, then g = c1 ϵ C implying ˅I ai exists. Now suppose that T is bounded by g = c1 + c2 + …. +cn-1, then ˅ai exists in G. If T is bounded by g = (c1 +c2+….+cn-1) + cn, cn ≥ 0, then ai - cn ≤c1 +c2 + …. + cn-1. By induction hypothesis, ˅I(ai - cn) exists in G and is equal to (˅ai) – an. Therefore ˅GT exists. i Remark. Let G = [C], C is literally complete but [C] is not necessarily laterally complete. Example. Let C be a complete Boolean algebra of all subset of natural numbers. Then C is laterally complete. But [C] = set of all bounded sequences of integers, which is not laterally complete. Corollary 1.11. Let C be a positive clan, then C is an Archimedean Clan. Proof. C is complete  [C] is complete  [C] is Archimedean  C is Archimedean. Corollary 1.12. If C is a positive Archimedean clan then there exists a complete clan C^ such that C is dense in C^. Proof. C is a convex sublattice of G = [C] which is Archimedean, so G has a unique completion G^. Let C^ = (G^)+ then C  C^ and C is dense in C^. Theorem 1.13. ([16]) Let C be a clan. If x ≤ 𝑛𝑖=1 yi, then there exists xi ϵ C satisfying x = 𝑛𝑖=1 xi and xi ≤ yi for all i ϵ {1, 2,…, n}. Moreover, if x ≥ 0 and y1,….,yn ≥ 0 then xi can be chosen such that 0 ≤ xi ≤ x  yi hold for all i ϵ {1, 2,….,n}. 35 Shinemin Lin International Journal of Engineering Technology, Management and Applied Sciences www.ijetmas.com November 2015, Volume 3, Issue 11, ISSN 2349-4476 Let C be a positive clan and let &(C) be the set of all convex subclans of C. For A, B ϵ &(C), let A + B denote the set {a +b | if a + b is defined, a ϵ A, b ϵ B}. Proposition 1.14. If D ϵ &(C) then [D] is a convex 1-subgroup of [C]. Proof. It is sufficient to show that [D] is convex in [C]. Let g be an element of [C] such that 0 < g < d ϵ [D]. We have 0 < g < d = d1 + d2 + ….+ dn, where di ϵ D for all i = 1, 2, ..., n. By the theorem 1.13, g = g1 +g2 + …. + gn with 0 ≤ gi ≤ di for all i = 1, 2, …, n. Since D is convex, gi ϵ D for all i = 1, 2, … , n. This shows g ϵ [D]. Proposition 1.15. If H is a convex 1-subgroup of [C] and D is a generating clan of H, then D  C ϵ &(C) and [D  C] = H. Proof. It is sufficient to show that D  C is convex in C. Let c be an element of C such that 0 < c < d ϵ D  C  [D] = H. Since [D] is convex, c ϵ [D], we can write c = c1 + c2 + ...+ cn, where ci ϵ D for all i = 1, 2, … , n. Hence c ϵ C  D. Clearly, H  [D  C]. Now if 0 < h ϵ H  [C], then h= h1+h2+…+ hn, where hi ϵ D for all i ϵ {1, 2,….,n}. But hi ϵ [C] for all i ϵ {1, 2, … , n}, which implies hi = hi1 +….+ Him, where hij ϵ C for all j ϵ {1, 2,…,m}. We have hi ϵ D  C. This shows h ϵ [C  D]. Therefore, H= [C  D]. Schmit asked in [16] whether each minimal clan contains a greatest generalized Boolean algebra. Here we answer this question. Proposition 1.16. Suppose C is a clan. Then C contains a greatest generalized Boolean algebra. Proof. Let G be the 1-group generated by C. i.e. G = [C]. Let H be the Specker kernel of G and X be the set of singular elements of H. Then D = X  C is a convex subclan of C that generates H. Hence D is a generalized Boolean algebra that is contained in C. Since H is the Specker kernel of G, D is the greatest generalized Boolean algebra that is contained in C. Proposition 1.17. Let A, B be the convex subclans of C, then A + B = B + A if and only if A + B ϵ &(C) and B + A ϵ &(C). Proof. ( ) a) If a ϵ A and b ϵ B such that a+b is defined in C, then a = a1+b1 and b= a2+b2 where ai ϵ A and bi ϵB for i= 1, 2. a+b = (a1+b1) + (a2+b2) = a1 + (b1+a) + b2 = a1 + (a2’+b1’) +b2 = (a1 +a2’) + (b1’ + b2) ϵ A+B. b) If a, b ϵ A+B  C then there exist c1 and c2 ϵ C such that a +c1 = a˅b= c2 + a and c1 + a  b = b = a  b + c2. If a  b=0 then a˅b = a + b, so a˅b ϵ A +B. If a  b > 0 then 0 < c1< b, so c1 ϵ B. Therefore, a ˅ b = a + c1 ϵ A+B. Since 0 < a  b < a ϵ A+B, a  b ϵ A  A+B. This proves that A+B is a lattice. c) If g ϵ A+B and 0 < x <g then 0 < x < a+b = g. By theorem 1.13, x= a1+b1, where 0< a1 < a and 0 < b1< b. This shows that x ϵ A + B. Therefore, A + B is convex in C. Likewise, we can prove B+A is a convex subclan of C. (  ) Let A˅B denote the convex subclan of C that is generated by the set {A, B}. Clearly A˅B  A+B but A  A˅B and B  A˅B  A+B  A˅B, so A+B = A˅B = B˅A = B+A. Definition 1.18. A positive clan is normal valued if for all A, B ϵ (C), we have A+B = B+A. 36 Shinemin Lin International Journal of Engineering Technology, Management and Applied Sciences www.ijetmas.com November 2015, Volume 3, Issue 11, ISSN 2349-4476 Proposition 1.19. Let C be a generating clan of an 1-group G. Then G is normal valued if and only if C is normal valued. Proof. (  ) If A, B ϵ (C) then [A] + [B] = [B] + [A] If 0 < x ϵ A +B, then x=a +b ϵ [A] + [B] = [B] + [A], and hence 0 < x = b1 +a1 = (b1 +a1) +< b1+ + a1+ that implies that there exists 0 < a2 ϵ [A], 0 < b2 ϵ [B], such that x = b2 + a2 , this implies a2, b2 ϵ C. Therefore, a2 ∈ A and b2 ∈ B. This proves A+B  B+A. Likewise, we can show that B+A  A+B. (  ) If A, B ϵ (G) then [A  C] = A and [B  C] = B. For any a ϵA, and b ϵB, a + b = (a1+….an) + (b1+…..bm), where ai’s ϵ A  C and bj’s ϵ B  C. We know that (A  C) + (B  C) = (B  C) + (A  C). Therefore (a1+...an) + (b1+….bm) ϵ B+A, which implies A+B  B+A. Likewise, we can show B+A  A+B and hence A + B = B+ A. Proposition 1.20. Suppose C is a generating clan of an l-group G. If for each G(g) there exists a c ϵ C such that G(c) = G(g), then C =  [0,c] and G =  G(c). Proof. 0< g ϵ G implies g ∈ [C]+, the semigroup generated by C.  g = c1 + c2 + …. + cn, ci’s ∈ C+  G(g) = G (c1+c2+….+cn) = G(c1 + c2…..+ cn) = G(c1  c2  c3  …  cn) = G(c), where c = c1  c2  …..  cn∈ C. SECTION 2 From now on, let C be a positive abelian clan. Thus C is a convex sublattice of the positive cone G of an abelian l-group G. Moreover, the lattice order and the partial addition on C is that induced by G. Let S be a subsemigroup of C and let SC = {x ∈C | x ≥ s for some s ∈S} denotes the convex hull of S in C. + Lemma 2.1. SC is a subsemigroup of C. Thus each maximal subsemigroup of C contains 0 and is convex. Proof. For x, y ∈ SC  x < s and y < t for some s, t ∈ S  x+y is defined in C and x+ y  s + t so x + y ϵ SC. Lemma 2.2. If G(a)+  G(b)+  C then G(a˅b)+  C. Proof. For all n ϵ N, na˅nb = n( a˅b). Since a + b = b + a and since na, nb ϵ C, we have n(a˅b)= na˅nb ϵ C. Let K= {a ϵ C | G(a)+  C}. Then by above lemma {G(a)+ | a ϵK} is a directed set of subsets of C. Lemma 2.3. S =  {G(a)+ | a ϵ K} is the largest subsemigroup of C. Proof. Clearly S is a subsemigroup of C and, of course S is convex and contains 0. Now suppose B is a subsemigroup of C. Then, without loss of generality, we may assume 0 ϵ B. B is convex in C. Thus if 0 < b ϵ B then G(b)+  B  C, so b ϵ B and hence B  S. In particular [S] is an l-ideal of G = [C]. If C is not an abelian clan then C may not have a largest subsemigroup. The next example is provided by Dr. David Nelson. Example. Let G = A(R) be the group of all o-permutations of the real numbers and let α(x) = x+1 for all x ϵ R. Then [e, α] is a clan with maximal convex subsemigroups. For let β ϵ G with orbit < 1 then [e, α]  G(β)+, but the product of two such semigroups need not belong to [e, α]. Thus for a nonabelian clan need not have a largest subsemigroup. 37 Shinemin Lin International Journal of Engineering Technology, Management and Applied Sciences www.ijetmas.com November 2015, Volume 3, Issue 11, ISSN 2349-4476 Definition 2.4. A clan C is called a principal clan if C = [0,g] for some g ϵ [C]. Let G be an l-group that is generated by a principal clan C and let M be the intersection of all maximal l-ideas of G(g) (i.e. all the values of g in G(g)). Lemma 2.5. M+ is the greatest subsemigroup of C = [0, g]. Proof. For a, b ∈ M+ we have a + b ∈ M+ which implies a + b << g implies a + b ∈ C. Therefore, M+ is a subsemigroup of C. If B is a subsemigroup of C then, without loss of generality, we may assume 0 ∈ B and B is convex in C. Thus [B] is an l-ideal of G with positive cone B. Suppose, by way of contradiction, that x ∈ B\M+ then x does not belong to Gγ, a value of g, so nx is not comparable with g, which is a contradiction. Thus B  M+. Note. G/[M] is Archimedean. In fact it is a subdirect product of real numbers, so C is an extension of the largest subsemigroup M+ be an Archimedean principal clan. Once again let C be an arbitrary positive abelian clan. Then for each a ∈ C, [0, a] is a principal subclan. Let M(a) be the largest subsemigroup of [0, a], then a < b implies [0, a]  [0, b] implies G(a)  G(b). Also M(a) = {b ∈ G+| g << a} so M(a)  M(b). Lemma 2.6. M =  {M(a)| a ∈ C} is a convex subsemigroup of C that contains 0. Proof. a, b ∈ C implies M(a)  M(B)  M(a  b), so we have an upper directed set. Note. M need not to be the largest subsemigroup of C.   Example. Let G = R  R , and C = G+ then M = R  0 Lemma 2.7. If C is a positive abelian clan with {0} the only subsemigroup, then [C] is Archimedean. Proof. If [C] is not Archimedean, then we can pick 0 < a << b in C. Then 0  a ∈ M(a), which is a semigroup. Let S =  {G(a)+ | a ∈ C} be the largest subsemigroup of the clan C. Lemma 2.8. [C]/[S] =  {([S] + G©)/[S] | c ∈ C}. Proof. x∈ ([C]/[S])+ implies x = [S] + g, g >0, but g ∈ G(c ) for some 0 < c ∈ C so x ∈ ([S] + G(c))/[S]. Lemma 2.9. [C]/[S] is the join of convex l-subgroup ([S]+G(c ))/[S]  G(c )/([S]  G(c )) Proposition 2.10. Let G =[X] be a Specker group with the set of all singular elements X. Then X is the smallest generating clan of G. Proof. Suppose Y is a subclan of X and x ∈ X. If Y generates G then it must have an element y that is a-equivalent to x. But then y and x are singular that implies x = y. Now we need to show that any generating clan Y must contains X. We may assume, without loss of generality, G    Z  and X is the set of all characteristic elements of G. Since Y is convex, for each y ∈ Y the characteristic function on the support of y belongs to Y. Therefore, Y  X generates G as a group and it is a convex sublattice of G. Therefore, Y  X is a generating clan so Y  X = X and hence Y  X. 38 Shinemin Lin International Journal of Engineering Technology, Management and Applied Sciences www.ijetmas.com November 2015, Volume 3, Issue 11, ISSN 2349-4476 The converse is not true. For it we let G be a totally ordered abelian group with no maximal convex subgroup, then G+ is the only generating clan and hence the smallest generating clan. But G is not a Specker group. Proposition 2.11. Let C1 and C2 be generating clans for the representable 1-group G. Then C1  C2 is a generating clan. Proof. We know that [C1] = [C2] = G and want to show that [C1  C2] = G. Let ~ denote a-equivalence. 0 < g ∈ G g ~ a ∈ C1 and g~ b ∈ C2. There is an integer n such that ng > a and b and na, nb > g  ng > a  b and nan b = n(a  b) > b. Therefore, b  a ~ g. Remark. Suppose G is an o-group. Then, since the subclans are the convex subset of G+, they form a chain. So if G does not have a largest proper convex subgroup then G+ is the only generating clan. Suppose M is the largest convex subgroup of G then the generating clans are of the form [0,g], where g G+ \M. Corollary 2.12. If G = [X] is Specker with X the generalized Boolean algebra of singular elements, then G is the free I-group over X. Proof. X is a generating clan of G. Corollary 2.13. If C = [X] is Specker with generating clan X of singular elements, then any two addition on G so that it is an l-group are connected by an isomorphism t that is an identity on X. Proof. The only addition on X is the join of disjoin elements. So each addition of G must agree on X. Definition 2.14. A  &(C) is regular if A is maximal with respect to not containing c ∈ C. Proposition 2.15. A  &(C) is regular if and only if [A] is regular in G = [C]. Proof: (  ) Let Gα be a value of g  G = [C]. Then there is a element c  C such that c ~ g since G(g) = G[c] for some c  C. Therefore c  Gα hence c  Gα  C = A  &[C]. If B  A such that B  &[C] then [A]  [B]. Therefore, c ~ b for some b  [B]. Therefore, c  [B] hence c  [B]  C = B, Therefore A is maximal with respect to not containing c. (  ) Suppose A  &[C] is maximal with respect to not containing c  C. Then pick g  G such that g ~ c then g  [A] for if g  [A] implies c  [A] which implies c  A which contradicts our assumption. Now for any X  &(G) such that [A]  X then A  X  C  &(C)  c  X  C  g  [X  C] = X. Therefore, [A] is a value of g. Lemma 2.16. Let N be a convex sublattice of C, then [N] is an l-ideal of [C]. Consider the subset T of the l-group [C]/[N], where T={c + [N] | c ∈ C} then (1) T is a convex sublattice of [C]/[N] and hence a clan. (2) [T] = [C]/[N]. Proof. (1) T is a lattice, since (c + [N])  (d + [N]) = c  d + [N]  T, and (c + [N]  (d + [N]) = c  d + [N]  T, Now we want to show that T is convex. If 0 < g + [N]  d + [N]  T, then there is m  [N]+ such that g  d + m. Since 0 < g + [N] so 0 < (g+ - g-) + [N] = (g+ + [N]) – (g- + [N]) so 0 < g- + [N] < g++ [N] implies g- + [N] = 0 hence g-  [N]. Without loss of generality, we may assume 0 < g  G = [C].  g = d1 +m1 where 0< d1< d and 0< m1< m. g + [N] = d1 +m1 + [N] = d1 + [N]  T. To show (2) Suppose 0 < s  [C]/[N]. Then S = c + [N]/[C], where c  [C] 39 Shinemin Lin International Journal of Engineering Technology, Management and Applied Sciences www.ijetmas.com November 2015, Volume 3, Issue 11, ISSN 2349-4476 = c1 + c2 + … + cn + [N], ci  C for i=1,….,n = (c1+ [N]) +….+ (cn + [N])  [T] On the other hand, suppose 0 < t  [T]. Then t = (t1 + [N]) +……+ (tn + [N]), ti  C for i = 0,…..,n. = (t1 + t2 + … + tn) + [N]  [C]/[N]. Proposition 2.17. Let N be a convex subsemigroup of C, hence a convex sublattice. Then there is a one to one correspondence between the set of all convex subsemigroups C(&[C]) of C that contain N and the set of all convex subsemigroups of T, where T = [c +[N]| c  C]. Proof. (1) Suppose M is a convex subsemigroup of T. Let α : [C]  [C]/[N] be the natural l-homomorphism. Then α maps C onto T. Let M* = [c  C| cα  M]. Then (a) For all n  N, nα = 0 hence N  M*. (b) If h1, h2  M* then h1α, h2α  M, h1α+h2α  M T. h1α + h2α = (h1 + h2)α . Hence h1 + h2 is defined in C. Hence h1 + h2  M*. (c ) If 0 < h  h1  M* then 0 <hα  h1α  M*  hα  M  h  M*. Therefore M* is a convex subsemigroup of C that contains N and (M*)α = M. For if m*  M*, then m*α  M and if m  M  T, there is c  C such that cα = m, so c  M*. (2)If J is a convex subsemigroup of C that contains N then J’ = Jα = [j + [N] | j  J] is a convex subsemigroup of T and [J’] * = J. For if c  (J’)* then cα  J’. There is j  J such that cα = jα  c – j  N  c = j + m  J. If j  J then jα  J’  j  (J’)*. Now define f : C(&[C])  T by f(J) = J’ and g : T  C(&[C]) by g(M) = M*, then by (2) g(f(J)) = J and by (1) f(g(M)) = m. Corollary 2.18. If S is a maximal subsemigroup of C, then T = [c + [S] | c  C] has no proper nonzero convex subsemigroup. Lemma 2.19. Let S be the largest subsemigroup of C. Then [S] = G = [C] if and only if S = G + (if and only if C= G+). Proof. (  ) Clear. ( ) Suppose, by way of contradiction, that S is a proper subset of G+. Hen pick g ∈ G+\S then there is a value Gα of g such that S  Gα hence [S]  G which is a contradiction. Proposition 2.20. Let C be a proper generating clan of G. S be a maximal subsemigroup of C, and P is a value of G such that P  [S]. Then 1) G/P is an Archimedean o-group 2) G/[S] is Archimedean. Proof. 1) G/P is generated by {c + P | c ∈ C}, which is a convex sublattice of G/P and hence is a clan. If {c + P| c ∈ C} is not Archimedean then there is 0 < x + P << y + P. Let K = {c ∈ C | c + P << y + P}. Then K  C and K = [0]. If k1 + P << y + P and k2 +P << y + P then (k1+k2) + P << y +P hence k1 + k2 + P << y + P and hence k1 + k2∈ K. So K is a subsemigroup of C and for all s  S, s + P = 0 + P << y + P hence S is properly contained in K which contradicts that S is maximal subsemigroup. Therefore [c + P | c ∈ C] is Archimedean so G/P is an Archimedean o-group. {G / P | [ S ]  P and P is a value of G} by 2) Now define  : G   40 Shinemin Lin International Journal of Engineering Technology, Management and Applied Sciences www.ijetmas.com November 2015, Volume 3, Issue 11, ISSN 2349-4476  (g)      (G)  {G / P | [S ]  P and P is a value of G} . Clearly [S]   {P  | [S] c P  and P  is a value of G}. If g  [S] then there is a value P of g such that [S]  P g   {P  | [S] c P  and P  is a value of G}  [S]   {P  | [S] c P  and P  is a value of G}. Therefore ker  =  {P  | [S] c P  and P  is a value of G} = S.  G/[S]   (G)  {G / P | [ S ]  P and P is a value of G} . (….g +P,….) then is an l-homomorphism and G/ker Therefore, G/[S] is a subdirect product of Archimedean o-group hence Archimedean. Corollary 2.21. M = {M  |  ∈  , M  is a maximal 1-ideal of G} then  {M  | M  ∈ M }  C. Proposition 2.22. For an abelian I-group G. The following are equivalent. (1) G has proper generating clan. (2) G has maximal I-ideal. Proof. (2)  (1) Let M be a maximal 1-ideal of G. Then G/M is an archimedian o-group. Pick 0 < g  M then [M, r+M] generates G/M, so C =  [0, g  m] | m ∈ M] is a proper generating clan of G. (1)  (2) Let C be a proper generating clan of G. Since G is abelian, G is a subdirect product of O-groups.  ,11l momomorphism  Gi, Gi is a 0-group for all i ∈ I. C  G    pi, a projection map Gi Identify C as a subset {<c> | c ∈ C} of G. Let Ci = pi(  (C )), then for 0 < gi∈ Gi there is g ∈ G such that pi(  (g )) = gi, and since C generates G there is c ∈ C such that c is e-equivlent to g. Hence g  nc for some n ∈ N. Therefore 0 < gi = pi(  (g ))  pi(  (nc )) = npi(  (c )), which implies gi is a-equivalent to pi(  (c )), so Ci generates Gi. Similarly  (C ) generates  (G) properly since  is one to one. Gi   (C )   (G ). Let [S] =  (G) Now, it Ci = Gi for all i ∈ I, then Gi = Ci   (C )   then  (G ) = S =  (C ), which is a contradiction. Therefore [S] is a proper subset of  (G) = [  (C )]. Then pick 0 < g ∈  (G)\[S]. Let P be a value of g such that S  [S]  P, then  (G)/P is an 0-groupp generated by {  (c ) + P| c ∈ C} = T. If there is  (c1) + P <<  (c2) + P ∈ T, then let K = {  (c)|  (c) <<  (c2)}  0. Claim K is a subsemigroup of  (C ). For if  (a),  (b) ∈ K, then  (a) + P<<  (c2) + P,  (b) + P <<  (c2) + P, so  (a)+  (b) + P <<  (c2) + P, so S  K which contract to that C/P is archimedean. It follows that T is an Archimedean clan, hence  (G)/P is an Archimedean o-group. Therefore P is a maximal l-idea of  (G). Let M = {x ∈ G |  (x) ∈ P}, then M is an l-idea of G. Suppose M  J, an l-ideal of G, then P =  (M)   (J)   (G) so  (M) =  (J), hence M = J. This proves that M is a maximal l-ideal of G. If there is i ∈ I such that Ci  Gi, since Gi is an o-group, Gi has maximal l-ideal Mi. Let M = {g ∈ G| pi(  (g)) ∈ Mi} then Mi is an l-ideal of G. Define  : G/M  Gi/Mi by g + M  pi(  (g))+ Mi. Then  is onto and ker  = {g + M | pi(  (g)) ∈ Mi} = {g + M | g ∈ M} = 0 so  is one to one. So G/M is an Archimedean o-group. That means M is a + 41 Shinemin Lin International Journal of Engineering Technology, Management and Applied Sciences www.ijetmas.com November 2015, Volume 3, Issue 11, ISSN 2349-4476 maximal l-ideal of G. In particular, if S = [0] then all values of G are maximal l-ideal, so G is an Archimedean l-group. Theorem 2.23. G admits min generating clan C in which {0} is the only subsemigroup then G is Specker. Proof. (1). G is Archimedean and hence abelian. (2). G+ is not the only generating clan so G has maximal l-ideals. (3). Let M = {M  |  ∈  } be the set of all maximal l-ideal. For each M  , T  = {c + M  |c ∈ C} generates G/M  and T  is a minimal generating clan of G/M  . For if there is H* = {h + M  | h ∈ H}  T  such that H* is also a generating clan of G/M  , then N = M  C, then N  H. For g ∈ G if and only if g ∈ M   g ∈ [N]  [H]. If g  M  , g + M  = n(h + M  )  g = nh + m ∈ [H], m ∈ M   H is properly contained in C and H generates G. This is a contradiction. So G/M  has least element d + M  and hence G/M   Z. we may assume, without loss of generality, T  [0, 1] as clans. (4).  {M  |  ∈  }  C, hence  {M  |  ∈  } = 0. If g  C then there is a value Gα of g such that g  Gα. Let Cα =Gα  C, then Cα is a value of C, hence a convex sublattice of C. Let Tα = {c + [Cα]| c ∈ C} = {c + Gα | c ∈ C} then Tα is a convex sublattice of G/Gα and generates G/Gα. If [C]/Gα is not Archimedean then [c +G| c ∈ C] is not an Archimedean clan. Then there exists 0 < c + Gα << d + Gα] in T. Let K ={c ∈ C | c + Gα << d + Gα} then K is a nonzero subsemigroup of C which is a contradiction. Therefore T is an Archimedean clan. Therefore, Gα is a maximal l-ideal. Hence g   all maximal l-ideals   all maximal l-ideals  C. (G/M  )  Z  by α(g) = (…g  +M  ….). Then ker α =  {M  |  ∈ (5). Define α : G     }= 0, so α is a one to one l-isomorphism and α|C : c   I  where I  = [0,1]. Therefore C is the set of all singular elements of G and hence G is a Specker group. REFERENCES [1] M. Anderson, and T. Feil, Lattice-Ordered Groups: An Introduction, D. Reidel, 1987 [2] Bigard, A., Keimel, K., and Wolfenstein, S., Groupes et Anneaux Reticules. Lecture Notes in Mathematics, 608. Ed. A. Dold an B. Eckmann. Berlin: Springer-Verlag, 1977. [3] Birkhoff, G., Lattice Theory, Third Edition, Amer.Math. Soc. Coll. Publ. 25, New York, 1968 [4] Darnel, M., Lattice-Ordered Groups, Preprint. [5] Fuchs, L., Infinite Abelian Groups, 2 vols. 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