ON COLOURING MAPS & SEMIDIRECT PRODUCT GROUPOIDS ON GROUP-POSETS

In this paper we will study the (G,H)posets , to give a condition of freeness on colouring maps . Also we will look at the group – posets from the view point of the category theory . That is to give some results on semi-direct product poset groupoid.

Much of group theory , particularly that part which deals with finite groups , originated from the study of groups of permutations groups between 1844 and 1900 .Some of the first workers in this area were Lagrange , Galois , and Cauchy [8].
Hence it is natural that the concept of the group actions on sets began as : A group action of finite group G on finite set X to be a group homomorphism from G to S /X/ , the group of permutations on X .The set X is called a G-set [9].
Equivalently , a group G acts on X on the left (X called left G-set) if , to each g∈G and each x ∈ X there corresponds a unique element denoted by g x in X such that , for all x ∈ X and g 1 ,g 2 ∈ G : Similarly , we define a right H-set .In this paper we shall study a left and a right compatible actions on posets and we shall look at the group actions on posets from the viewpoint of category theory .
Finally , a semi-direct product poset groupiod P H related to a right H-poset P is considered .§1 (G,H) -posets : For any group G and any poset P , we say that G acts on P from the left if to each g∈G and p∈P there corresponds a unique element in P denoted by g p such that for all p,q ∈ P and g 1 ,g 2 ∈ G : Such poset P with a left action of G on it , is called a left G-poset , or simply a G-poset .If we agnor condition (iii), a G-poset P becomes G-set .
Similarly we can define the right group-posets.The condition (iii) is different from that which is given in [7] , which is : We can conclude that every G-poset P can be considered as a right G-poset (and conversely) which is defined by :

Definition (1.1) :
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 each g∈G , h∈H and p ∈ P there corresponds a unique element g p h in P such that g p h = g (p h ) = ( g p) h .
In [1] there is another condition on the definition of (G,H)-set, that is ; the right action of H must be regular .
Equivalently, following [10], a poset P is call (G,H)-poset if for all p∈P , g∈G and h∈H there exist a unique element g p h ∈P such that : (i) e p e = p (ii) P is a left G-poset with the action defined by : g p = g p e for all p∈P and g∈G .
(iii) P is a right H-poset with the action defined by : p h = e p h for all p∈P and h∈H.(iv) ( g p h ) = ( g p) h = g (p h ) for all p∈P , g∈G and h∈H.
From any group H we will define another group denoted by H op such that H op = H and h 1 x op h 2 in H op is h 2 h 1 in H for all h 1 ,h 2 ∈ H op = H.

Lemma (1.2) :
If P is a (G,H)-poset , then it is (G × H op )-poset and conversely as the following action : (g,h) p = g p h for all p∈P , g∈G and h∈H.

Proof :
It is routine chack , following [10 ; proposition (1.15)].■ The following theorem gives us an equivalent definition to (G,H)-poset which considers the action as a group homomorphism.

Theorem (1.3) :
(1) let P be a (G,H)-poset , then for every g∈G and h∈H there exist an isomorphism .g ρ h : P P defined by : g ρ h (p) = g p h for all p ∈P. Also the map ρ : (G×H op ) Isom (P,P) defined by : ρ (g,h) = g ρ h for all g∈G and h∈H is a homomorphism ,called the (G,H)-action on P.
(2) let P be a poset and a homomorphism σ: (G×H op ) Isom (P,P) , then P is a (G,H)-poset with an action defined by : g p h = (σ(g,h))(p) for all p∈P , g∈G and h∈H.

Lemma (1.4):
Let X and Y be any two posets , then Y X the set of all the maps from X to Y is a poset with a binary relation defined by : α ≤ β if and anly if α (x) ≤ β(x) for all x∈X.

Proof :
The proof is straight forward .Let X be a G-poset and Y be a right H-poset , then the poset Y X is a (G,H)-poset with an action defined by : for any f ∈ Y X , g ∈G and h ∈ H : (ii) Now for any g∈G , h∈H and f∈ Y X ; Therefore Y X is a (G,H)-poset.■ §.2 : Freeness on colouring maps: Let X be a G-poset and a poset C be a set of colours which is a right H-poset .
In this section we shall give the conditions of freeness on C X , the set of colouring maps from X to C .

Definition (2.1):
A G-poset X is called free G-poet if : Stab G (x) = {g∈G : g x = x} = {e} for all x ∈ X .[5] Similarly , the definition of free right H-poset and free (G,H) -poset .
In genral , if a (G,H)-set X is free , then X is a free G-set and a free right H-set.But the converse is not true.[10] In the following propositions we shall prove that a (G,H)-poset C X is free if and only if it is free G-poset with G = {e} and the right H-poset C is free.

Proposition (2.2):
The right H-poset C is free if and only if C X is a free right H-poset .

Proof :
Let C be a free right H-poset.Let α ∈ C X such that α h = α for some h ∈ H. Hence (α h )(x) = α(x) for all x ∈ X.So, (α(x)) h = α(x) for all x ∈ X.Since C is a free right H-poset , then h = e .Therefore C X is a free right H-poset .
Conversely , suppose that C X is a free right H-poset.Let c∈C and c h = c for some h∈H.Let α∈C X defined by α(x)=c for all x∈X.
Therefore C is a free right H-poset.■

Proposition (2.3):
The G-poset C X is free if and only if G = {e}.

Proof :
It is obvious that C X is free G-poset when G= {e}.Now suppose that C X is a free G-poset.Let c∈C and consider the map α ∈ C X defined by : α(x)=c for all x∈X .So, ( g α)(x) = α ( g-1 x) = c = α (x) for all x∈X.Hence g α = α for all g∈G.Now since C X is a free G-poset .Therefore G= {e}.■

Proposition (2.4):
The (G,H)-poset C X is free if and only if G = {e} and C is a free right H-poset.

Proof :
Suppose that C X is a free (G,H)-poset.Let α∈C X with g α = α for some g ∈G.Then g α e = g α = α .So, g=e , that is since C X is a free (G,H)-poset.Hence C X is a free G-poset and by [prop (2.3)] G = {e}.Similarly we prove that C X is a free right H-poset.
Conversely , suppose that G = {e} and C is a free right H-poset.So, C X is a free G-poset since G = {e}.Now let α∈C X and g α h = α for some g∈G and h∈H.Since G = {e} , α h = α .Hence h = e.Therefore C X is a free (G,H)-poset.■ §.

G-posets as G-objects of the category Posets
In this section we look at group -Posets from the view point of category theory.

Proof :
Let P be a poset .Then we consider P to be the category P with :

Proposition (3.3):
The set of all the posets is a category denoted by Posets.Proof : (i) ob (Posets) is the set of all posets .(ii) If P,Q ∈ ob (Posets) , then f ∈ (Posets) (P,Q) if and only if f ∈ Hom (P,Q) .(iii) If f ∈ Posets (P,Q) and g ∈ Posets (Q,W) , then fg = gof.(iv) for any P∈ Posets , 1p is the identity map on P. ■ Let (G , *) be a finite group , then G is a category with a single object G and the arrows are the elements of G .So we can consider a functor F from G to Posets .Hence F(G) is an object in Posets , a poset P say .For each arrow g of G the image F(g) : P P is a poset isomorphism with (F(g)) -1 = F(g -1 ).So {F(g): g ∈ Hom (G,G)} is a subgroup of the group of isom (P,P).Therefore , F determines a G-action on the poset P.
For more generalization , we can define a group action on poset a s group action on object of Posets .For more details see [4] , [6], [7].

Definition (3.4) :
A left G-object in the category Posets is a pair (α , P) with P is an object of Posets and α a left action on P defined by ; to each g ∈ G , there corresponds a unique morphism α g ∈ Isom (P,P) such that : (ii) α e = 1 P where 1 P : P P is the identity map.

Proposition (3.5):
The definition of G-object (α , P) in the category Posets is equivalent to the definition of a G-poset P.

Proof :
Let (α , P) be a G-object in the category Posets .For each p ∈ P and g ∈ G let g p = α g (p).Hence ; (i) e p = α e (p) = 1 P (p) = p.
(iii) For p,q ∈ P , if p > q then α g (p) > α g (q) , that is , g p > g q.
Conversely ; suppose that P is a G-poset .Then there exists a homomorphism α : G Isom (P,P) such that : α(g) = α g ∈ Isom (P,P) with ; (α(g))(p) = g p for all p∈P .So ; (ii) α e (p) = e p = p = 1 P (p) for all p ∈ P Therefore ; (α , P) is a G-object in the category Posets.■ §.4 Semi-direct product poset groupoid : A groupoid should be thought of as a group with many objects , or with many identities .
The definition of groupoid were introduced by Brands in his 1929 paper [2] , is given with extra condition on the definition bellow , such a groupoid we nowadays called connected or transitive .[3]

Definition (4.1):
A groupoid G is a small category in which every arrow is invertible (every morphism is an isomorphism) .
Hence G has a set of morphims , may be called elements of G , a set ob(G) of objects , together with maps s,t : G ob(G), i:ob(G) G such that si = ti = 1 the identity map .The maps s,t called the source and target maps respectively .
If x , y ∈ G and t(x) = s(y) , then a product xy is exists such that s(xy) = s(x) , t(xy) = t(y) , and this product is associative .For every a ∈ ob(G) , the element i(a) is the identity morphism of a .
Hence we can consider any group G to be a grouoid G with one object G and the arrows are the elements of G.So the definition of groupoids is an extension of that of groups.
The basic reference for the theory of groupoids is Higgins' book [4] , a survy of the wide range of applications of the theory is given by Brown in [3].

Example [4.2] :
An equivalence relation R on a set X becomes a groupoid with X is the set of the objects , R is the set of arrows and product : (x,y) (y,z) = (x,z) whenever (x,y) , (y,z) ∈ R.
This example is due to croisot [3] .A special case of this groupoid is the coarse groupoid X x X , which obtained by taking R = XxX .

Lemma (4.3):
Let P be a right H-poset .Then PxH is a right H-poset.Proof : 1. PxH is a poset defined by (p,a) ≥ (q,b) if and only if p ≥ q. 2. PxH is a right H-poset with action defined by : (p,a) h = (p h ,a) for all p∈P , a , h ∈ H ; that is since : (i) (p , a) e = (p e , a) = (p,a) (iii) (p,a) > (q,b) ⇒ p > q ⇒ p h > q h ⇒ (p h ,a) > (q h ,a) ⇒ (p,a) h > (q,b) h ■ Proposition (4.4) : Let P be a right H-poset .Then we have a groupoid denoted by P H with P is the object set and arrows (p,h) : p p   ■ The groupoid P H may be called semi-direct product poset groupoid because this groupoid is a special case of the semi-direct product groupoid obtained from an action of a group on a set .[8] Note that for any right H-poset P we have two groupoids , P x P considering P as a set and P H , we shall prove that ≅ P x P P H if and only if P is a regular right H -poset .

Proposition (4.5):
Let F : P H P x P defined by ; F(p) = p and F(p,h) = (p,p 1 h − ) for all p ∈ P and h ∈H.Then F is a functor .Proof : ■

Definition (4.6) :[5]
A right H-poset P is called right transitive H-poset if P ≠ φ and for any p , q ∈ P there exists h ∈ H such that p h = q. (p Equivalently , if there is some element p ∈ P such that for any q ∈ P there exists h ∈ H with p h = q.

Definition (4.7) :[5]
A right H-poset P is called right regular H-poset if it's free and transitive right H-poset .

Proposition (4.8) :
P H P x P ≅ by the functor F defined above if and only if the right H-poset P is regular .

Proof :
Suppose that P H P x P ≅ by the functor F. Then F is an isomorphism functor .That is F is onto and 1-1.Now let p,q ∈ P. Since F is onto then there exists h∈H such that F(p,h) = (p,q) .So , (p,p Conversely , suppose that P is regular right H-poset .Then P is transitive and free right H-poset .Let (p,q) ∈ PxP .Since P is transitive right H-poset, then there exists h ∈ H such that q = p h .So, (p,h -1 ) ∈ (P H) and F(p,h -1 ) = (p,p h ) = (p,q) Also , let p ∈P then F(p) = p .Hence F is onto .Now let F(p,h) = F(q,k) .Then (p,p 1 h − ) = (q,q 1 h − ) So, p =q and p 1 h − = q 1 k − .Hence p 1 h − k = q.Since the right H-poset P is free, h -1 k = e.So , h=k .Also , let F(p) = F(q) then p =q .Hence F is injective .Therefore F is an isomorphism and P x P H P ≅ . ■ (i) ob (P) = P .(ii) for any a,b ∈ ob (P) ; there exist a unique arrow denoted (a,b) if and only if a ≤ b .(iii) For any arrows (a , b) , (b , c) , (a , b) (b , c) = (a , c) , since (a ≤ b , b ≤ c) implies a ≤ c. (iv) For every object a of P , the arrow (a , a) denoted by 1 a is an identity , since 1 a (a , b) = (a , b) and (x , a)1 a = (x , a).
with p ∈ P and h ∈H , and the product : (p,h) (p 1 h − ,t) = (p, th) with p ∈ P , h,t ∈ H .

1 h
arrow (p,h) is invertible and it's inverse is the arrow (p − , h -1 ).