Algebraic properties of configuration sets for groups
aa r X i v : . [ m a t h . C O ] J a n ALGEBRAIC PROPERTIES OF CONFIGURATION SETS FORGROUPS
CESAR A. IPANAQUE ZAPATA
Abstract.
Let G be any group and k ≥ be an integer number. The ordered configu-ration set of k points in G is given by the subset F ( G, k ) = { ( g , . . . , g k ) ∈ G × · · · × G : g i = g j for i = j } ⊂ G k . In this work, we will study the configuration set F ( G, k ) inalgebraic terms as a subset of the product G k = G × · · · × G . As we will see, we developpractical tools for dealing with the configuration set of k points in G , which, to ourknowledge, can not be found in literature. Contents
1. Introduction 12. The configuration set F ( G, k )
23. Configuration groups 64. Cayley graphs 75. The set F ( G − { } , k ) Introduction
Let G be a group and k ≥ be an integer number. The (ordered) configuration set of G (see [1] for the case G a topological space) is the subset of the product G k = G × . . . × G ( k times) given by F ( G, k ) = { ( g , . . . , g k ) ∈ G k | g i = g j for all i = j } . Notice first that F ( G, k ) = ∅ if and only if the order of the group | G |≥ k . In thiswork we will assume that | G |≥ k . By definition F ( G,
1) = G .For k ≥ , notice that F ( G, k ) is not a subgroup of G k (all do not even contain theidentity (1 , . . . , , where denotes the identity element of the group G ).The intersection of a family of subgroups of G is a subgroup. There is therefore asmallest subgroup of G containing a given subset X of G ; it is called the subgroup generated by X , denoted by h X i . Recall that h X i = { x · · · x n : x i ∈ X ∪ X − for ≤ i ≤ n, n ≥ } , Mathematics Subject Classification.
Primary 55R80, 20F05; Secondary 20F99.
Key words and phrases.
Configuration set, groups, generated set.The author would like to thank grant where X − = { x − : x ∈ X } . If G = h X i then X is called a generated system (orgenerated set) of G . Remark 1.1.
Recall that if f : G → H is an epimorphism of groups and X is a generatingset of G then f ( X ) is a generating set of H . Hence, if f : G → H is an isomorphism, then f k : G k → H k , ( g , . . . , g k ) ( f ( g ) , . . . , f ( g k )) is an isomorphism and f k ( F ( G, k )) = F ( H, k ) . In particular, F ( G, k ) is a generating set of G k if and only if F ( H, k ) is agenerating set of H k .The object of this paper is to describe the set F ( G, k ) (see Section 2, Theorem 2.8,Proposition 2.11 and Proposition 2.17). In Section 3, Proposition 3.3, we determine thesubgroup generated by F ( Z p , p ) for any prime number p ≥ . We use our results to knowthe Cayley graph Cay ( G k , F ( G, k )) (see Section 4, Theorem 4.1). Moreover, in Section 5,Theorem 5.2, we study the set F ( G − { } , k ) in terms of the configuration set F ( G, k + 1) .2.
The configuration set F ( G, k ) Let G be a group and k ≥ be an integer number. The (ordered) configuration set of G (see [1] for the case G a topological space) is the subset of the product G k = G × . . . × G ( k times) given by F ( G, k ) = { ( g , . . . , g k ) ∈ G k | g i = g j for all i = j } . Because F ( G, k ) is not a subgroup of G k ( k ≥ ), it is interesting to explore this. It iseasy to see that set F ( G, k ) is symmetric , i.e., F ( G, k ) = F ( G, k ) − . Example 2.1.
For G = Z = { , } , the integer numbers module 2, and k = 2 , we have F ( G, k ) = { (0 , , (1 , } ⊂ Z = Z × Z . Furthermore, we have (0 , / ∈ F ( G, k ) and (0 ,
1) + (1 ,
0) = (1 , / ∈ F ( G, k ) . Note that a subset X of a group G is a generating set of G if and only X satisfies thefollowing statement: if H is a subgroup of G such that X ⊂ H , then H = G . Proposition 2.2.
Let G be a group. (i) For any subgroup H of G with F ( G, ⊂ H , we have H = G . In particular, F ( G, is a generating set of G . (ii) If X is a generated system of G , then F ( G, ∩ X = ∅ .Proof. ( i ) Let ( g , g ) ∈ G . If g = g then ( g , g ) ∈ F ( G, ⊂ H . On the other hand,if g = g = 1 we have ( g , g ) ∈ H . Now, if g = g = 1 we have (1 , g ) ∈ F ( G, ⊂ H and ( g , ∈ F ( G, ⊂ H and thus ( g , g ) = (1 , g ) · ( g , ∈ H . ( ii ) Suppose that F ( G, ∩ X = ∅ , and so X ⊂ G − F ( G,
2) = D ( G ) where D ( G ) = { ( g, g ) ∈ G × G : g ∈ G } is the diagonal. Note that D ( G ) is a subgroup of G ,then G = h X i ⊂ D ( G ) , and so G = D ( G ) , which is a contradiction. Therefore, F ( G, ∩ X = ∅ . (cid:3) ONFIGURATION SETS FOR GROUPS 3
For a point g = ( g , . . . , g k ) ∈ G k , we define the norm of g by | g | = g · · · g k ∈ G . Notethat G is an abelian group if and only if | gh | = | g || h | for any g, h ∈ G k ( k ≥ . Remark 2.3.
For G = Z = { , , } , the integer numbers module 3, and k = 3 . Onehas that the configuration set F ( Z ,
3) = { (0 , , , (1 , , , (0 , , , (2 , , , (2 , , , (1 , , } is not a generating set of Z , because each element g in F ( Z , satisfies | g | = 0 . Hence,any element in the subgroup h F ( Z , i has norm , in particular (0 , , / ∈ h F ( Z , i (note that | (0 , , | = 1 ) and so h F ( Z , i ( Z .Furthermore, for G = Z n = { g , g , . . . , n − g n } , the integers numbermodulo n , and k = n ≥ , we have the configuration set F ( Z n , n ) = { ( g σ (1) , . . . , g σ ( n ) ) : σ ∈ S n } , where S n is the symmetric group on n letters. Any g ∈ F ( Z n , n ) satisfies | g | = ( n − n .If n is odd, | g | = 0 for any g ∈ F ( Z n , n ) . Hence, any element in the subgroup h F ( Z n , n ) i has norm . If n = 2 l , | g | = l for any g ∈ F ( Z n , n ) , in this case any element in thesubgroup h F ( Z n , n ) i has norm or l . Note that l has order . Therefore, h F ( Z n , n ) i ( Z nn .Furthermore, by Remark 1.1, for any finite cyclic group G , say | G | = k ≥ , theconfiguration set F ( G, k ) is not a generating set of G k .Generally we have the following statement. Proposition 2.4.
Let G be an abelian finite group with | G | = k ≥ . Then F ( G, k ) isnot a generating set of G k .Proof. Suppose G = { g , g , . . . , g k } . Note that the configuration set F ( G, k ) = { ( g σ (1) , . . . , g σ ( k ) ) : σ ∈ S k } and so any g ∈ F ( G, k ) satisfies | g | = g + · · · + g k . We have any element g in the subgroup h F ( G, n ) i has norm | g |∈ h g + · · · + g k i . Here we can suppose that h g + · · · + g k i ( G ,because in the other case G would be a finite cyclic group and for this case see Remark2.3. Therefore, we can choose g ′ ∈ G − h g + · · · + g k i and obtain (0 , . . . , , g ′ ) / ∈ h F ( G, n ) i .Therefore, F ( G, k ) is not a generating set of G k . (cid:3) Remark 2.5.
Non-abelian groups are pervasive in mathematics and physics. One of thesimplest examples of a non-abelian group is the dihedral group of order 6, D ∼ = S . Recallthat D = h r, a : r = 1 , a = 1 , ara = r − i . Then the group elements are given by D = { , r, r , a, ra, r a } . It is the smallest finite non-abelian group (up isomorphism).We have F ( D ,
6) = { σ (1 , r, r , a, ra, r a ) : σ ∈ S } . CESAR A. IPANAQUE ZAPATA
Then it is straightforward to check that all elements in F ( D , are of order . (cid:0) σ (1 , r, r , a, ra, r a ) (cid:1) = σ (1 , r , r, , , (cid:0) σ (1 , r, r , a, ra, r a ) (cid:1) = σ (1 , , , a, ra, r a ) (cid:0) σ (1 , r, r , a, ra, r a ) (cid:1) = σ (1 , r, r , , , (cid:0) σ (1 , r, r , a, ra, r a ) (cid:1) = σ (1 , r , r, a, ra, r a ) (cid:0) σ (1 , r, r , a, ra, r a ) (cid:1) = σ (1 , , , , , . We believe that F ( D , is not a generating set of D .There are several facts which seem to be relevant in studying non-abelian finite groupsof lower orders: • There exists no non-abelian finite group of odd orders. • For smaller order, i.e. Ord < , there exist only three types of non-abelian finitegroups (see [2]). Example 2.6.
For G = Z = { , , , } and k = 3 , the configuration F ( G, k ) = { σ (0 , , , σ (0 , , , σ (1 , , , σ (0 , ,
3) : σ i ∈ S , i = 1 , , } . Note that (1 , ,
0) = 3(2 , ,
1) + 3(0 , ,
3) + 3(1 , , , ,
0) = (2 , ,
1) + (3 , ,
1) + (3 , , , ,
1) = 3(0 , ,
2) + 3(1 , ,
0) + 3(3 , , and thus { (1 , , , (0 , , , (0 , , } ⊂ h F ( G, k ) i . Therefore, F ( G, k ) is a generating setof G k . Note that { (1 , , , (0 , , , (0 , , } is a generating set of G k . For k ≥ , note that the subset E k = { ( g , , · · · , , (1 , g , , · · · , . . . , (1 , · · · , , g k ) : g i ∈ G } is a generating set of G k . Proposition 2.7.
Let G be a group and k ≥ with | G |≥ k +1 . We have, E k ⊂ h F ( G, k ) i .In particular, the subgroup generated by F ( G, k ) is equal to G k . In particular, if G is aninfinite group, the subgroup generated by F ( G, k ) is equal to G k .Proof. We check (1 , , . . . , , g ) ∈ h F ( G, k ) i for any g ∈ G . The other cases are verysimilar to this one. If g = 1 , it holds. If g = 1 , choose h , . . . , h k − ∈ G − { g, } with h i = h j for i = j (it is possible because | G |≥ k + 1 and k ≥ ). Note that (1 , , . . . , , g ) = ( h , h , . . . , h k − , g ) · ( h − , h − , . . . , h − k − , , with ( h , h , . . . , h k − , g ) and ( h − , h − , . . . , h − k − , in F ( G, k ) . Thus, (1 , , . . . , , g ) be-longs to the subgroup h F ( G, k ) i . (cid:3) Proposition 2.2 together with Proposition 2.4 and Proposition 2.7 give the followingstatement.
Theorem 2.8.
Let G be a group. ONFIGURATION SETS FOR GROUPS 5 (1) If k = 2 or | G |≥ k + 1 ≥ , then F ( G, k ) is a generating set of G k . (2) If G is abelian and finite with | G | = k ≥ , then F ( G, k ) is not a generating setof G k . Next, we will characterize the set F ( G, k ) in algebraic terms as a subset of G k .Let G be a group and k ≥ . Set p : G k +1 → G k , p ( g , . . . , g k , g k +1 ) = ( g , . . . , g k ) be the natural projection. Definition 2.9.
Let X be a subset of G k +1 (with k ≥ ). We say that X has the configuration property if(P-1) p ( X ) ∩ F ( G, k ) = ∅ and(P-2) p ( x ) ∈ F ( G, k ) for x ∈ X implies x ∈ F ( G, k + 1) .For example, any subset of the configuration space F ( G, k + 1) has the configurationproperty. In contrast, the set X from Remark 2.3 has not the configuration property. Remark 2.10.
Let f : G → H be a monomorphism and f k : G k → H k , ( g , . . . , g k ) ( f ( g ) , . . . , f ( g k )) . If X has the configuration property in G k , then f k ( X ) has the config-uration property in H k .Note that if X is a subset of G , then X has the configuration property if and only if X ⊂ F ( G, . In particular, one has Proposition 2.11 (Characterization for k = 2 ) . Let C be a subset of G . One has that C is the maximal subset having the configuration property if and only if C = F ( G, . Example 2.12.
For G = Z = { , } and k = 2 . The configuration set F ( G, k ) = { (0 , , (1 , } . The set X = { (0 , , (1 , } is another generated system of G which hasnot the configuration property and the subset { (0 , } has the configuration property. Remark 2.13. If X is a subset of G k (for k ≥ ) which has the configuration property,then F ( G, k ) ∩ X = ∅ . Indeed, suppose X is a subset of G k having the configurationproperty. Recall that p : G k → G k − , p ( g , . . . , g k ) = ( g , . . . , g k − ) denote the natural projection. Since X has the configuration property, F ( G, k − ∩ p ( X ) = ∅ . Then there exists an element x ∈ X such that p ( x ) ∈ F ( G, k − with x ∈ F ( G, k ) and thus F ( G, k ) ∩ X = ∅ . Remark 2.14.
Let α ∈ G k \ F ( G, k ) with | G |≥ k + 1 . The set C ( α ) = F ( G, k ) ∪ { α } is a generated subset of G k . Note that C ( α ) has the configuration property if and only if p ( α ) / ∈ F ( G, k − .For k ≥ , note that the subset E = { ( g , , . . . , , (1 , g , , . . . , , . . . , (1 , . . . , , g k ) : g i ∈ G } ⊂ G k is a generating set of G k and F ( G, k ) ∩ E = ∅ . Here, note that E has not theconfiguration property. CESAR A. IPANAQUE ZAPATA
Lemma 2.15.
For k ≥ with | G |≥ k + 1 , suppose C ( G k be a subset such that C ∩ E = ∅ and C ∩ X = ∅ for all X = E generated system of G k . Then F ( G, k ) ⊂ C .In particular, C is a generated system of G k .Proof. Let x ∈ F ( G, k ) and consider X = E ∪ { x } , then X is a generated system of G k (because E is a generated system) and X = E ( because F ( G, k ) ∩ E = ∅ ). Then x ∈ C .Hence, we does. (cid:3) Example 2.16.
The set C ( α ) = F ( G, k ) ∪ { α } where α / ∈ E (as an example α =( g, g, , . . . , with g = 1 ) satisfies the hypothesis: C ( α ) ∩ E = ∅ and C ( α ) ∩ X = ∅ forall X = E generated system of G k having the configuration property. From Lemma 2.15,one has that the configuration space F ( G, k ) is the minimal subset (with respect to theinclusion) satisfying these hypothesis. Proposition 2.17 (Characterization for k ≥ ) . For k ≥ and G a group with | G |≥ k + 1 , let C ( G k be a proper subset. We have C = F ( G, k ) if and only if C is the minimalsubset satisfying the following conditions: (C-1) It is a generated system such that C ∩ E = ∅ , (C-2) C ∩ X = ∅ for all X = E generated system of G k having the configuration property. Configuration groups
In this section G will be denote a finite abelian group with order | G | = k ≥ . Werecall that the configuration set F ( G, k ) is not a generating set of G k (see Theorem 2.8.)Thus we have the following definition. Definition 3.1.
For G a finite abelian group with | G | = k ≥ . The configuration group is given by F ( G, k ) = h F ( G, k ) i . We are interesting to know the configuration group. Recall the norm map | · | : G k → G given by | ( g . . . , g k ) | = g + · · · + g k , and note that it is a homomorphism of groups.Let p be a prime number with p ≥ and Z p the field of integers module p . Note that inthis case norm map | · | : F ( Z p , p ) → Z p is trivial. Also, the configuration group F ( Z p , p ) is a Z p − vector subspace of Z pp = Z p ⊕ · · · ⊕ Z p , because ( λg , . . . , λg p ) ∈ F ( Z p , p ) for any ( g , . . . , g p ) ∈ F ( Z p , p ) and λ ∈ Z p . So we have the following diagram of Z p -vector spaces: / / Z p − p (cid:31) (cid:127) i / / Z pp |·| / / Z p / / / / F ( Z p , p ) ?(cid:31) O O = / / F ( Z p , p ) ?(cid:31) O O |·| / / ?(cid:31) O O / / where the inclusion i : Z p − p → Z pp is given by i ( g , . . . , g p − ) = ( − ( g + · · · + g p − ) , g , . . . , g p − ) .In particular, the dimension of the vector space F ( Z p , p ) over the field Z p is less or equalthan p − , dim Z p ( F ( Z p , p )) ≤ p − . ONFIGURATION SETS FOR GROUPS 7
Example 3.2.
For p = 3 , Z = { , , } , we have the configuration set F ( Z ,
3) = { (0 , , , (0 , , , (1 , , , (1 , , , (2 , , , (2 , , } . Note that the vectors (0 , , ∈ F ( Z , and (1 , , ∈ F ( Z , are linearly independent, thus dim Z ( F ( Z , and thus F ( Z , ∼ = Z ⊕ Z . Note that F ( Z p , p ) ⊂ { ( − ( x + . . . + x p ) , x , . . . , x p ) ∈ Z pp : x , . . . , x p ∈ Z p } , and the vector spaces (over Z p ) { ( − ( x + . . . + x p ) , x , . . . , x p ) ∈ Z pp : x , . . . , x p ∈ Z p } and Z p − p = Z p ⊕ · · · ⊕ Z p are isomorphic. On the other hand, the set { e , . . . , e p − } is abasis to F ( Z p , p ) , where e i = (1 , , , . . . , p − , for i = 1 ; (0 , , , . . . , p − , for i = 2 ; (0 , . . . , , , i, i + 3 , . . . , p − i − , for ≤ i ≤ p − ; (0 , . . . , , , p − , for i = p − .Note that e i = (1 , , , . . . , i − , , i, i + 1 , . . . , p −
1) + ( p − , p − , p − , . . . , p − i +1 , , , , . . . , p − i ) for ≤ i ≤ p − . Hence, F ( Z p , p ) = { ( − ( x + . . . + x p ) , x , . . . , x p ) ∈ Z pp : x , . . . , x p ∈ Z p } ∼ = Z p − p . Proposition 3.3.
Let p be a prime ( p ≥ ). Then the vector spaces (over Z p ) F ( Z p , p ) and Z p − p = Z p ⊕ · · · ⊕ Z p are isomorphic. Corollary 3.4.
Let z , . . . , z p ∈ F ( Z p , p ) with z i = z j for i = j . Let A ∈ M p ( Z p ) whose i − column is z Ti . Then the matricial system Ax = 0 admits a nontrivial solution. Proof.
Suppose z i = ( x ,i , . . . , x p,i ) and x = ( λ , . . . , λ k ) T . The equation Ax = 0 is equalto λ z + · · · + λ p z p = 0 . Because, dim Z p ( F ( Z p , p )) = p − , then z , . . . , z p are linearly dependent and thus thereexists ( λ , . . . , λ p ) = 0 . (cid:3) Cayley graphs
We recall the definition of a Cayley graph from [3]. Let G be a group and let S be asubset of G \ { } . The Cayley graph of G with respect to S is the graph Cay ( G, S ) = (
G, E ) , where E = { ( x, y ) | x − y ∈ S } . CESAR A. IPANAQUE ZAPATA
A sequence of distinct vertices [ x , x , . . . , x k ] such that ( x i , x i +1 ) is an edge, ≤ i ≤ k − , is said to be a path from x to x k with length k − . Of course, we admit the case k = 1 , where the path consists of a single vertex.Let G be a finite group and let S be a subset of G \ { } such that S = S − (itimplies that x − y ∈ S if and only if y − x ∈ S ). A factorization of an element x ∈ G \ { } with respect to S determines a unique path with the same length from to x and vice versa. Let x = x x · · · x ℓ , with x i ∈ S , be a factorization of x . The path [1 , x , x x , . . . , x x · · · x ℓ − , x x · · · x ℓ − x ℓ ] is from to x and it has length equal to l .Suppose [1 , x , . . . , x ℓ ] is a path from to x ℓ = x , then x = x ′ x ′ · · · x ′ ℓ , where x ′ = x andfor i = 2 , . . . , ℓ , x ′ i = x − i − x i . Note that each x ′ i ∈ S . Hence, S is a generating set of G ifand only if the Cayley graph Cay ( G, S ) is path-connected.Theorem 2.8 implies the following statement. Theorem 4.1.
Let G be a group. (1) If k = 2 or | G |≥ k + 1 ≥ , then the Cayley graph Cay ( G k , F ( G, k )) is path-connected. (2) If G is abelian and finite with | G | = k ≥ , then the Cayley graph Cay ( G k , F ( G, k )) is not path-connected. The set F ( G − { } , k ) Let k ≥ . If G is a group, then it is well-known that the configuration set F ( G, k + 1) is bijective to the product G × F ( G − { } , k ) , under the bijection ( g , g , . . . , g k ) ( g , g g − , . . . , g k g − ) with its inverse ( g , g , . . . , g k ) ( g , g g , . . . , g k g ) .More general, we can consider the map ϕ : G k +1 → G k given by the formula(5.1) ϕ ( x , x , . . . , x k ) = ( x x − , . . . , x k x − ) . Remark 5.1.
We note ϕ is a homomorphism of groups if and only if G is abelian. Supposethat ϕ is a homomorphism of groups and take any x, y ∈ G , we have yx − = ϕ ( x, y ) = ϕ (( x, · (1 , y )) = ϕ ( x, · ϕ (1 , y ) = (1 · x − ) · ( y · − ) = x − y and thus xy = yx .It is easy to check that ϕ ( F ( G, k + 1)) = F ( G − { } , k ) . Thus we will consider therestriction map over F ( G, k + 1) ,(5.2) ϕ := ϕ | : F ( G, k + 1) → F ( G − { } , k ) Let p = ( p , . . . , p k ) ∈ F ( G − { } , k ) be a fixed point. Let K be the set ϕ − ( p ) , and wehave(5.3) K = { ( g, p g, . . . , p k g ) ∈ F ( G, k + 1) | g ∈ G } . We will consider the quotient set F ( G, k + 1) /K given by the classesi) [ x , . . . , x k ] = { ( x , . . . , x k ) } if ( x , . . . , x k ) ∈ F ( G, k + 1) − K , andii) [ x , . . . , x k ] = K if ( x , . . . , x k ) ∈ K . ONFIGURATION SETS FOR GROUPS 9
The map ϕ | induces a bijection ψ : F ( G, k +1) /K → F ( G −{ } , k ) given by the formula ψ [ x , . . . , x k ] = ϕ ( x , . . . , x k ) = ( x x − , . . . , x k x − ) . Thus we have the following statement.
Theorem 5.2.
For any k ≥ . There is a bijection ψ : F ( G, k + 1) /K → F ( G − { } , k ) given by the formula ψ [ x , . . . , x k ] = ( x x − , . . . , x k x − ) . References [1] E. Fadell and L. Neuwirth, ’Configuration spaces’ , Math. Scand. (1962), no. 4, 111-118.[2] J, Kubo, ’The dihedral group as a family group’ , Quantum field theory and beyond, World Sci. Publ.,Hackensack, NJ, (2008), 46–63.[3] Y. O. Hamidoune, ’On the subsets product in finite groups’ , Europ. J. Combinatorics. (1991),211–221. Deparatmento de Matemática,UNIVERSIDADE DE SÃO PAULO INSTITUTO DE CIÊN-CIAS MATEMÁTICAS E DE COMPUTAÇÃO - USP , Avenida Trabalhador São-carlense,400 - Centro CEP: 13566-590 - São Carlos - SP, Brasil
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