Raf. J. of Comp. & Math’s. , Vol. 10, No. 4, 2013 On Maximal Chains In Partially Ordered Sets With Compatible (Left, Right)Group Actions Abdul Aali J. Mohammad Eman M. Tahir Department of Mathematics College of Education University of Mosul, Mosul, Iraq Received on: 09/01/2012 Accepted on: 19/04/2012 ABSTRACT In this paper, we deal with compatible (left, right)-group actions on posets, i.e; (G, H)-posets. Our main purpose in this work is to study the maximal chains in (G, H)posets to observe that this study gives us indications on the type of some (G, H) actions on posets. Therefore, we shall study the behavior of the (G, H) actions on chains. Keywords: maximal chains, partially ordered sets, compatible (left, right) group actions. )‫ أيمن‬،‫جزئيا مع أفعال زمرتية ثنائية (أيسر‬ ً ‫حول السالسل العظمى في المجموعات المرتبة‬ ‫إيمان طاهر‬ ‫عبد العالي محمد‬ ‫ جامعة الموصل‬،‫ كلية التربية‬،‫قسم الرياضيات‬ 2012/04/19 :‫تاريخ قبول البحث‬ ‫الملخص‬ ‫مر ييية ثنائييية يانمييا س انسييرا ملييع الملمومييات المر نيية جائيييا لاميير ا‬ ‫الملمومييات المر نيية‬ 2012/01/09 :‫تاريخ استالم البحث‬ ‫فييه اييلا النتييم تعامييل ميير افعييا‬ ‫ الايية السياييل العجمييع فييه تعي‬. ‫ليية ايينفنا الرئيسييه فييه اييلا النتييم‬.‫ميير ا البييا ل ا للمنا‬ ‫المعنيرات ميا و يية ايلل احفعيا‬ ‫مر ية ثنائيية س حظجنيا اه ايلل الن الاية ع نيا تعي‬ ‫جائيا الته مل ها افعا‬ ‫لااة الوك احفعا الامر ية الثنائية ملع السيال‬. ‫للا انتتاج إلع‬ ‫مر ية ثنائية‬ ‫ أفعا‬،‫ ملمومات مر نة جائيا‬،‫ ايال مجمع‬:‫الكلمات المفتاحية‬ §.1 Introduction: The idea of a group action of groups on sets can be extended on sets with additional mathematical structures, specially on posets. A group action of a group G in a poset P can be considered as a group homomorphism 𝜌 : G isom(P , P) defined by 𝜌(g) = 𝜌g where 𝜌g : P P is an isomorphism defined by 𝜌g(p) = gp for all g ∈ G , p ∈ P . Such a poset P with a left action of G on it, is called a left G-poset, or simply a Gposet. Also, since in general there are many such homomorphism 𝜌, so maybe there is many group actions of the group G on a poset P (at least the trivial action). Hence, by G-poset we mean only left group action of G on P. Equivalently, Let G be a group and P a poset , we say that P is a left G-poset if for every g∈G and p∈P there corresponds a unique element gP∈P such that for all p,q  P and g,g1,g2  G; (i) ep=p (ii )g 2 (g1 p) =(g 2 g1 ) p (iii ) if p  q then g p  gq When condition (iii) is neglected, P is called a left G-set.This definition is slightly different from the definition given in [6].Similarly we define a right H-poset. We can conclude that every G-poset P can be considered as a right H-poset (and conversely) which is defined by: 43 Abdul Aali J. Mohammad & Eman M. Tahir −1 gp =pg for all pP and gG. Also the concept of a group action on sets can be extended to compatible left, right actions on sets. For more details see: [5], [7] and [8]. §.2 Covering (G, H)-posets: In this section, we give the definition of the (left, right) group actions on posets, and the covering poset of a given poset. So, we begin with the formal definition first, before proceeding to explain the intuitive concept behind it. Definition (2-1): [4]. A poset P is called (G,H)-poset if P is a left G-poset, a right H-poset and the two actions are compatible, that is for each gG, hH and pP there corresponds a unique element gph in P such that gph = g(ph) = (gp)h. Equivalently, let G and H be two groups and P a poset , we say that P is a ( G , H )-poset if for every gG , hH and pP there corresponds a unique element gphP such that 1. epe = p 2. P is a left G-poset with the action defined by: gp = gpe ∀ pP, gG. 3. P is a right H-poset with the action defined by: ph = eph ∀ pP, hH. 4. (gp)h = g(ph) ∀ g∈G , h∈H and p∈P . 5. p > q ⟹ gph > gqh ∀ g∈G , h∈H and p,q∈P . When condition (5) is neglected, P is called a (G, H)-set. For more details see: [5], [6] and[8]. Example (2-2) Let the additive group Z acts on the set of the real numbers R by the action : n a= a+n ∀ a∈R , n∈Z , and the additive group Q acts on R from the right by the action an = a-n ∀ a∈R , n∈Q , then R is a (Z , Q)-poset. Also, there exists a G-poset P which is also a right H-poset , but it's not a (G , H)poset , as in the following example; let G = H = C2 = {e,a} and P = {x,y,z,w,t,r} is a poset with x<y , z<w , t<r. Then, P is a G-poset with the action defined by : ax = z , a y = w , at = t , ar = r , ep = p ∀ p∈P, and a right H-poset with the action defined by : xa = x , ya = y , za = r , wa = t , pe = p ∀ p∈P. But, P is not (G, H)-poset, that is since (az)a = za = t and a(za) = ar = r. Remark (2-3): Any G-poset P can be considered as (G, H)-poset with the trivial right action of H on P. Also, from the definition above, we see that a left G-poset P is one left action of G on P. But, for (G, H)-poset P there are two compatible group actions one is from the left and the other from the right. Definition (2-4): [2] Let P be a poset. We say that the element a of P covers the element b of P if a>b and there is no element c∈P such that a>c>b . Proposition (2-5): Let P be a (G, H)-poset and a,bP with a covers b , then gah covers gbh ∀ g∈G and h∈H . 44 On Maximal Chains In Partially Ordered … Proof: Suppose that gah does not cover gbh , then there exists, at least, an element cP g such that gah>c>gbh .Hence, a  covers gbh. ∎ −1 −1 c h  b and this is a contradiction . Therefore, gah Definition (2-6): [1]. Let P be a poset. Then, the set, C (P) = {(a,b) : a covers b} P×P, is called the covering poset of P. Proposition (2-7): [1]. Let (P, ≥) be a poset, then (( P) , ) is a poset such that: for all (a,b) , C (a , b) C(P) , (a, b)  (a, b ) if and only if {( a, b) = (a , b ) or b  a } C Theorem (2-8): Let P be a (G, H)-poset. Then C(P) is also a (G,H)-poset with an action defined by; g(a,b)h = (gah , gbh) ∀ (a,b) ∈ C(P) , g∈G and h∈H . Proof: (i) e(a,b)e = (eae,ebe) = (a,b) ∀ (a,b) C(P). (ii) g1(g2(a,b)h2)h1 = (g1(g2ah2)h1 , g1(g2bh2)h1) = (g1g2ah2h1 , g1g2bh2h1) = (g1g2)(a,b)(h2h1) ∀ (a,b) C(P) (g1 , h1),(g2 , h2)∈G×H . (iii) For all (a,b) , (a',b') ∈ C(P) , (g,h) ∈ G×H , with (a',b')≥(a,b) . Then b'≥a . So gb'h≥gah . Since (a,b),(a',b')∈C(P) . Then (gah , gbh), (ga'h , gb'h) ∈C(P) . That is (ga'h, gb'h) ≥ (gah, gbh). Hence g (a',b')h≥g(a,b)h . Therefore, C(P) is a (G,H)-poset . ∎ §3. (G, H)-Chains: In this section, we study the (left, right) group actions on chains and when the trivial action is the only one. Definition (3-1): [2]. A poset P is called a chain (or totally ordered set) if; for all a,b P : a ≥ b or b≥a. Equivalently, the poset P is called a chain if for every two different elements a,b of P either a > b or b > a . From the definition above, we conclude that every element of a chain covers, at most, one element and covered at most by one element. Also, any chain has, at most, one maximal element 1 and one minimal element 0. Remark (3-2): [2]. Any chain X of n elements is isomorphic to the set of natural numbers n = {1,2,..., n}. that is there exists a bijection function f : X → n such that : f ( x1 )  f ( x 2 ) if and only if x1  x 2 . Theorem (3-3): Let X = {xi }iI be a (G, H)-chain and I be a set of successive integers with … xi−1  xi  xi+1  ... If gxih = xj then, gxhi+r = xj+r ∀ i , j , i+r , j+r∈ I . 45 Abdul Aali J. Mohammad & Eman M. Tahir Proof: (i ) Let i + 1, j + 1 I . Since, X is a chain, then xi +1 covers xi and by proposition (2-3), x i+1 covers gxhi . Since, gxhi = xj then, xj+1 covers gxhi . So gxhi = xj+1. (ii) Now, we shall use the mathematical induction to prove that gxhi+1 = xj+1 for i ℎ =1. Suppose g𝑥𝑖+𝑛 = xj+n for r = n and i+n, j+n∈I. Since, X is a chain, then xi+n+1 covers g h xi+n .So, x i+n+1 covers gxhi+n .Now, from gxhi+n = gxhj+n we have gxhi+n+1 = xj+n+1. Therefore , gxhi+r = xj+r ∀ i , j , i+r , j+r ∈ I . ■ g h Lemma (3-4): Let X be a (G, H)-chain and (g,h)G×H . If gxhi = xt and xi<xt , then xi∈X . g-1 Proof: g g−1 xt g−1 xih = xt ⇒ h−1 = xi . Also, xi < xt ⇒ Definition (3-5): g ( xi h )h g−1 xi h −1 −1 < = g−1 g−1 xt h xt h −1 −1 ⇒ gg−1 . Therefore, g ( xi h )hh g−1 xi h −1 −1 = g−1 < xi ∎ xih-1<xi ∀ xt h −1 ⇒ Let P be a (G, H)-poset. For each p∈P the set: Stab (G, H)(p) = {(g,h)∈G×H : gph = p} is called the stabilizer of p. Proposition (3-6): Let X be a (G, H)-chain and (g,h)G×H with g −1 = g and h-1 = h . Then, (g,h)Stab(G,H)(xi) for all xi  X. g−1 −1 Let gxih = xt. Then, xi = xt h . So, xi = gxth . Suppose that xi ≠ xt. Then, either xi < xt or xt < xi . If xi < xt then gxih < gxth .So, xt < xi . That is a contradiction. Similarly, we have a contradiction if xt < xi . Hence, since X is a chain, then xi = xt. So gxih = xi. Therefore, (g,h)Stab(G,H)(xi) for all xiX. ■ Theorem (3-7): Let (X,) be a (G, H)-chain. Then, the (G, H) action on X is only the trivial action if X has 0 or 1. Proof: Also, (i) Let 0 = x1X and (g,h)G×H . Suppose that gx1h ≠ x1, then x1<gx1h [x1=0] . g−1 xi h g −1 h <x1 = 0 . So, this is a contradiction. = x1. Now, from theorem (3-3) we have gxih = xi for all xiX and So, x1 (g,h)G×H. (ii) Let 1 = x1X and (g,h)G×H . Suppose that gx1h ≠ x1, then gx1h<x1[x1=1]. g−1 −1 Also, x1 < x1 h . So, this is a contradiction. So, gx1h = x1. Now, from theorem (3-3) we have xi ∈ X and (g,h) ∈ G×H . ∎ 46 g xih = xi for all On Maximal Chains In Partially Ordered … Corollary (3-8): Let P = {p1,p2,…,pn} be a (G,H)-chain with p1>p2>…>pn. Then, P is a trivial (G, H)-chain. §.4 Maximal chains: Finally, in this section, we will study the maximal chains in (G, H)-posets and we shall observe that the study of these kinds of chains gives us some indications on the type of some group actions on posets. Definition (4-1): [2]. Let P be a poset and X = {xi , xi+1 ,..., x j }  P be a chain such that xi < xi+1 < … < xj, then X is called a maximal chain in P if and only if: (i) There is no element as c  P such that: xi < xi+1 < … < c < … < xj. (ii) There is no element as k  P such that: k < xi or xj < k. Proposition (4-2): Let P be a (G, H)-poset and Y be a maximal chain in P. Then, gYh is also a maximal chain in P with |gYh| = |Y|. Proof: (i) Since Y is a maximal chain in P , so we can say Y = {xi,xi+1,…,xj}such that xr+1 covers xr for all i< r < j . So ,gYh = {gxih , gxi+1h , . . . , gxjh} for all (g,h)G×H . Hence, gxih < gxi+1h < . . . < gxjh .Suppose that there exists an element as cP such that gxih < gxi+1h < . . . < c < . . . < gxjh . Then, g−1 g ( xi h )h −1 < g−1 g ( xi+1 h )h g−1 −1 −1 < .< g−1 h−1 c < .< g−1 g ( x j h )h −1 That is xi < xi+1 < . . . < c h < . . . < xj and this is a contradiction since Y is a maximal chain. (ii) Suppose that there exists an element bP such that b < gxih then: g−1 h−1 g−1 −1 < xi ⇒ bh = xi ⇒ b = gxih . Similarly, if gxih ≤ a then gxjh = a . Therefore, gYh is a maximal chain. Now, let the map f: Y → gYh is defined by: f(y) = gYh ∀ y∈Y. f is injective map since : f(y1) = f(y2) ⇒ gy1h = gy2h ⇒ y1 = y2 . b ≤ gxih ⇒ b Also f is onto since if x∈gYh then there exits yY such that x = gyh. Hence, f is bijection and |Y| = |gYh| ■ Definition (4-3): [2]. Let P be a poset and x  P. Then, the subset C of P is called a cutset of the element x in P if every element of C is not comparable with x and all the maximal chains in P cut with C∪{x}. We shall denote to this set by cut x. Proposition (4-4): Let P be a (G, H)-poset and C is the cutset of x P. Then, gCh is the cutset of gxh. That is gCh = cut gxh. Proof: 47 Abdul Aali J. Mohammad & Eman M. Tahir Let y  cut gxh then g−1 h−1 y g−1 is not comparable with gxh. So g−1 −1 −1 g−1 h−1 y is not comparable with x. That is y h ∈ C. So, g( y h )h∈ gCh . That is y∈gCh. Hence, cut gxh ⊆ gCh. Now let gsh∈gCh. Then, sC. So, s is not comparable with x. that is gsh is not comparable with gxh. So gsh∈ cut gxh . Therefore, gCh = cut gxh. ∎ Theorem (4-5): Let P be a finite (G, H)-poset with P(M) = {M1,M2,…,Mn} be the set of the maximal chains in P with M i = M j if and only if i = j. Then, the trivial action is the only action of (G, H) on P. Proof: To prove this theorem, we must first prove that g M ih = M i for 1 i  n, after that we must show that x = x for all x  Mi and (g,h)G×H . g h First part: g Mih Our argument proceeds by mathematical induction on the number n to prove that = Mi for all 1 i  n. Let |Mi| = ri, ∀ 1 i  n such that r1<r2<…<rn. (i) Let n=2. That is P(M) = {M1,M2} with M 1 ≠ M 2 . Suppose that gM1h ≠ M1. By proposition(4-2) gM1h is a maximal chain and |gM1h| = |M1|,then gM1h∈P(M) . So gM1h = M2. Hence, |gM1h| = |M2| = |M1|.That is a contradiction.So, gM1h = M1. Similarly, we have gM2h = M2. (ii) Now assume that n=k with gMih = Mi for all 1 i  k. Let n=k+1. Since gMih = Mi for all 1 i  k. Suppose that gMk+1h ≠ Mk+1 then, gMk+1h = Mj for some 1 j  k. So, |gMk+1h| = |Mj| = rj. But |gMk+1h| = |Mk+1| = rk+1. Hence, rj = rk+1, that is j=k+1, and this is a contradiction since k+1>j. So, gMk+1h = Mk+1 . Second part: Since {Mi }n i=1 is the family of the maximal chains in P , then Mi is a finite maximal chain in P. Using corollary (3-8), we get : gxh = x for all xMi , (g , h)G×H with 1 i  n. Therefore, from part one, the action of (G, H) on P is the trivial action only. ■ Definition (4-6): Let (H, *op) be a group. Define Hop to be a group its elements are the element of H and the product h1 *op h2 = h2 * h1 . Proposition (4-7): Let P be a (G,H)-poset , so for all (g,h)∈G×H there exists a permutation g𝜌h : P→ P defined by g𝜌h(p) = gph for all p∈P . Also the map 𝜌 : (G×Hop) →S|P| is defined by: 𝜌(g, h) = g𝜌h for all (g,h)∈G×H is a homomorphism . 48 On Maximal Chains In Partially Ordered … Proof: Similar to the proof in [8] . Hence, from every (G, H)-poset, we can get a (G×Hop)-poset by the action g h p ∀ g∈G, h∈H, p∈P. ∎ (g,h) p= Definition (4-8): A (G, H)-poset is called injective if the corresponding homomorphism 𝜌 is injective. Proposition (4-9): Let P (M) = {M1,M2,…,Mn} be the set of the maximal chains in the (G,H)-poset P. Let gMih = Mi, then, gMjh ≠ Mt for all j  i . Proof: Suppose that gMjh = Mi for some j  i . Then, gMjh = gMih for some j  i . So g−1 g ( Mj h )h −1 g−1 g h −1 = ( Mi )h for some j  i . Hence, Mj = Mi for some j  i . This is a contradiction since j  i implies P(M )  n . Therefore, gMjh ≠ Mt for all j  i . ■ Proposition (4-10): Let P be an injective (G, H)-poset, and P(M) = {M1,M2,…,Mn} be the family of the maximal chains in P. Then: (i) ( M i = M j if and only if i = j), implies that (G, H) = {(e,e)}. (ii) If M 1 = M 2 = ... = M n , then |(G,H)|≤n ! . (iii) If we reordered the maximal chains such that: N1 = N 2 = ... = N r  N r +1 = ... = N t  N t +1 = ... = N n , with NiP(M), 1 i  n , then : |(G,H)|≤r! x(t-r)! x . . . x(n-k)! . Proof: (i) Since ρ((g,h)) = (g ρ h)(p) = p = I(p) for all pP and (g,h)G×H , then (g,h)ker(ρ). But ker(ρ ) = {e,e} because ρ is injective . Then, (g,h) = (e,e) for all (g,h)G×H . So, (G, H) =ker(ρ ) = {(e,e)}. (ii) M 1 = M 2 = ... = M n . So for all MiP(M) and (g,h)G×H there exists some MtP(M) such that gMjh = Mt.From proposition (4-7) ,we have gMjh≠ Mt for all j  i So, the number of permutations on the maximal chains is n!.Now, since P is an injective (G, H)-poset, then | (G, H)|≤n! . (iii) Applying (ii) on every part of equal parts of: N1 = N 2 = ... = N r  N r +1 = ... = N t  N t +1 = ...  N k +1 = ... = N n We get the number of permutations on the equal parts are, r!, (t-r)!,…,(n-k)! respectively. Using the fundamental principle of counting, the number of the permutations on the maximal chains is r! x(t-r)! x … x(n-k)! . Since, P is an injective (G, H)-poset,Then, |(G, H)|≤r! x(t-r)! x . . . x(n-k)! . ■ 49 Abdul Aali J. Mohammad & Eman M. Tahir REFERENCES [1] Behrendt. Gerhard, “Automorphism Groups of Covering Posets and of Dense Posets", Proceeding of Edinburgh Mathematical Society (1992) 35, 115-120. [2] Birkhoff G., “Lattice Theory”, Amer. Math. Soc. Coll. Pub. Vol. XXV, Third Edition, (1967). [3] Hanlon P., “The Incidence Algebra of a Group Reduced Partially Order Set” (1980) Combinatorial Math. 7, Springer. No.829. [4] Mohammad A.J."On colouring maps and semidirect product groupoids on groupposets". J. Edu. & Sci., Vol. 19, No.2, (2007), 99-109. [5] Mohammad A.J. & Khalel L.A. "Kinds of Crossed Modules",J. Edu. & Sci., Vol. 22 No. 2. , (2009) , 354-369. [6] Mohammad A.J. & Mohammad S.A. “β-operations on Finite Posets”, J. Edu. & Sci., Vol. 19, (1994), 104-114. [7] Morris I. & Wensley C.D., “Adams Operations and 2-operations in β-Rings”, Discrete Mathematics, 50, (1984), 253-270. [8] Rose J.S., “ A Course on Group Theory”, Cambridge University Press, Cambridge, (1978). 50