A note on the automorphism group of the Hamming graph
aa r X i v : . [ m a t h . G R ] J a n A note on the automorphism group ofthe Hamming graph
S.Morteza Mirafzal and Meysam Ziaee
Abstract.
Let Ω be a m -set, where m >
1, is an integer. The Hamminggraph H ( n, m ), has Ω n as its vertex-set, with two vertices are adjacentif and only if they differ in exactly one coordinate. In this paper, weprovide a proof on the automorphism group of the Hamming graph H ( n, m ), by using elementary facts of group theory and graph theory. Mathematics Subject Classification (2010).
Primary 05C25 Secondary94C15.
Keywords.
Hamming graph, automorphism group, wreath product.
1. Introduction
Let Ω be a m -set, where m >
1, is an integer. The
Hamming graphH ( n, m ), has Ω n as its vertex-set, with two vertices are adjacent if and onlyif they differ in exactly one coordinate. This graph is very famous and muchis known about it, for instance this graph is actually the Cartesian productof n complete graphs K m , that is, K m (cid:3) · · · (cid:3) K m . In general, the connectionbetween Hamming graphs and coding theory is of major importance. If m =2, then H ( n, m ) = Q n , where Q n is the hypercube of dimension n . Since, theautomorphism group of the hypercube Q n has been already determined [10],in the sequel, we assume that m ≥
3. Figure 1. displays H (2 ,
3) in the plane.Note that in this figure, we denote the vertex ( x, y ) by xy .It follows from the definition of the Hamming graph H ( n, m ) that if θ ∈ Sym([ n ]), where Ω = [ n ] = { , · · · , n } , then f θ : V ( H ( n, m )) −→ V ( H ( n, m )) , f θ ( x , ..., x n ) = ( x θ (1) , ..., x θ ( n ) ) , is an automorphism of the Hamming graph H ( n, m ) , and the mapping ψ :Sym([ n ]) −→ Aut ( H ( n, m )), defined by this rule, ψ ( θ ) = f θ , is an injection.Therefore, the set H = { f θ | θ ∈ Sym([ n ]) } , is a subgroup of Aut ( H ( n, m )),which is isomorphic with Sym([ n ]). Hence, we have Sym([ n ]) ≤ Aut ( H ( n, m )). . S.Morteza Mirafzal and Meysam Ziaee00 01 0210 11 1220 21 22Figure 1. The Hamming graph H(2,3)Let
A, B , be non-empty sets. Let
F un ( A, B ), be the set of functionsfrom A to B , in other words, F un ( A, B ) = { f | f : A → B } . If B is agroup, then we can turn F un ( A, B ) into a group by defining a product,( f g )( a ) = f ( a ) g ( a ) , f, g ∈ F un ( A, B ) , a ∈ A, where the product on theright of the equation is in B . If f ∈ F un ([ n ] , Sym([ m ])), then we define themapping, A f : V (Γ) → V (Γ) , by this rule , A f ( x , · · · , x n ) = ( f (1)( x ) , · · · , f ( n )( x n )) . It is easy to show that the mapping A f is an automorphism of the Hamminggraph Γ = H ( n, m ), and hence the group, F = { A f | f ∈ F un ([ n ] , Sym([ m ])) } ,is also a subgroup of the Hamming graph Γ = H ( n, m ). Therefore, thesubgroup which is generated by H and F in the group Aut (Γ), namely, W = < H, F > is a subgroup of Aut (Γ). In this paper, we want to showthat;
Aut ( H ( n, m )) = W = < H, F > = Sym(Ω) wr I Sym([ n ])There are various important families of graphs Γ, in which we know that fora particular group G , we have G ≤ Aut (Γ), but showing that in fact we have G = Aut (Γ), is a difficult task. For example note to the following cases.(1) The
Boolean lattice BL n , n ≥
1, is the graph whose vertex set isthe set of all subsets of [ n ] = { , , ..., n } , where two subsets x and y are adja-cent if and only if their symmetric difference has precisely one element. The hypercube Q n is the graph whose vertex set is { , } n , where two n -tuples areadjacent if they differ in precisely one coordinates. It is an easy task to showthat Q n ∼ = BL n , and Q n ∼ = Cay ( Z n , S ), where Z is the cyclic group of order2, and S = { e i | ≤ i ≤ n } , where e i = (0 , ..., , , , ..., i thposition. It is an easy task to show that the set H = { f θ | θ ∈ Sym([ n ]) } , f θ ( { x , ..., x n } ) = { θ ( x ) , ..., θ ( x n ) } is a subgroup of Aut ( BL n ), and hence H is a subgroup of the group Aut ( Q n ). We know that in every Cayley note on the automorphism group of the Hamming graph 3graph Γ = Cay ( G, S ), the group
Aut (Γ) contains a subgroup isomorphicwith the group G . Therefore, Z n is a subgroup of Aut ( Q n ). Now, showingthat Aut ( Q n ) = < Z n , Sym([ n ]) > ( ∼ = Z n ⋊ Sym([ n ])), is not an easy task [10].(2) Let n, k ∈ N with k < n and Let [ n ] = { , ..., n } . The Kneser graphK ( n, k ) is defined as the graph whose vertex set is V = { v | v ⊆ [ n ] , | v | = k } and two vertices v and w are adjacent if and only if | v ∩ w | =0. The Knesergraph K ( n, k ) is a vertex-transitive graph [6]. It is an easy task to show thatthe set H = { f θ | θ ∈ Sym([ n ]) } , f θ ( { x , ..., x k } ) = { θ ( x ) , ..., θ ( x k ) } , is asubgroup of Aut ( K ( n, k )) [6]. But, showing that H = { f θ | θ ∈ Sym([ n ]) } = Aut ( K ( n, k ))is not very easy [6 chapter 7, 13].(3) Let n, k ∈ N with k < n, and let [ n ] = { , ..., n } . The Johnson graphJ ( n, k ) is defined as the graph whose vertex set is V = { v | v ⊆ [ n ] , | v | = k } and two vertices v and w are adjacent if and only if | v ∩ w | = k −
1. The John-son graph J ( n, k ) is a vertex-transitive graph [6]. It is an easy task to showthat the set H = { f θ | θ ∈ Sym([ n ]) } , f θ ( { x , ..., x k } ) = { θ ( x ) , ..., θ ( x k ) } ,is a subgroup of Aut ( J ( n, k )) [6]. It has been shown that Aut ( J ( n, k )) ∼ =Sym([ n ]), if n = 2 k, and Aut ( J ( n, k )) ∼ = Sym([ n ]) × Z , if n = 2 k , where Z is the cyclic group of order 2 [3,7,12].
2. Preliminaries
In this paper, a graph Γ = Γ(
V, E ) is considered as a simple undirected graphwith vertex-set V (Γ) = V , and edge-set E (Γ) = E . For all the terminologyand notation not defined here, we follow [1,2,5,6].The group of all permutations of a set V is denoted by Sym( V ) orjust Sym( n ) when | V | = n . A permutation group G on V is a subgroup ofSym( V ). In this case we say that G act on V . If Γ is a graph with vertexset V , then we can view each automorphism as a permutation of V , and so Aut (Γ) is a permutation group. Let G act on V , we say that G is transitive (or G acts transitively on V ), if there is just one orbit. This means thatgiven any two elements u and v of V , there is an element β of G such that β ( u ) = v. Let Γ , Λ be arbitrary graphs with vertex-set V , V , respectively. Anisomorphism from Γ to Λ is a bijection ψ : V −→ V such that { x, y } isan edge in Γ if and only if { ψ ( x ) , ψ ( y ) } is an edge in Λ. An isomorphismfrom a graph Γ to itself is called an automorphism of the graph Γ. The set ofautomorphisms of graph Γ with the operation of composition of functions is agroup, called the automorphism group of Γ and denoted by Aut(Γ). In mostsituations, it is difficult to determine the automorphism group of a graph,but there are various in the literature and some of the recent works appearin the references [7,8,9,11,13,14,15,16,17]. S.Morteza Mirafzal and Meysam ZiaeeThe graph Γ is called vertex - transitive, if Aut(Γ) acts transitively on V (Γ). In other words, given any vertices u, v of Γ, there is an f ∈ Aut(Γ)such that f ( u ) = v .For v ∈ V (Γ) and G = Aut (Γ), the stabilizer subgroup G v is thesubgroup of G containing of all automorphisms which fix v . In the vertex-transitive case all stabilizer subgroups G v are conjugate in G , and conse-quently isomorphic, in this case, the index of G v in G is given by the equa-tion, | G : G v | = | G || G v | = | V (Γ) | . If each stabilizer G v is the identity group,then every element of G , except the identity, does not fix any vertex and wesay that G act semiregularly on V . We say that G act regularly on V if andonly if G acts transitively and semiregularly on V, and in this case we have | V | = | G | . Let N and H be groups, and let φ : H → Aut ( N ) be a group homomor-phism. In other words, the group H acts on the group N , by this rule n h = φ ( h )( n ), n ∈ N, h ∈ H . Note that in this case we have ( n n ) h = n h n h , n , n ∈ N . The semidirect product N by H which is denoted by N ⋊ H is a group on the set N × H = { ( n, h ) | n ∈ N, h ∈ H } , with the multipli-cation ( n, h )( n , h ) = ( n ( n ) − h , hh ). Note that the identity element of thegroup N ⋊ H is (1 N , H ), and the inverse of the element ( n, h ) is the element(( n − ) h , h − ).
3. Main Results
Let Γ be a connected graph with diameter d . Then we can partition thevertex-set V (Γ) with respect to the distances of vertices from a fixed vertex.Let v be a fixed vertex of the graph Γ. We denote the set of vertices at distance i from v , by Γ i ( v ). Thus it is obvious that { v } = Γ ( v ) and Γ ( v ) = N ( v ), theset of adjacent vertices to vertex v , and V (Γ) is partitioned into the disjointsubsets Γ ( x ) , ..., Γ D ( x ). If Γ = H ( n, m ), then it is clear that two verticesare at distance k if and only if they differ in exactly k coordinates. Thenthe maximum distance occurs when the two vertices (regarded as ordered n -tuples) differ in all n coordinates. Thus the diameter of H ( n, m ) is equal to n . Lemma 3.1.
Let m ≥ and Γ = H ( n, m ) . Let x ∈ V (Γ) , Γ i = Γ i ( x ) and v ∈ Γ i . Then we have; \ w ∈ Γ i − ∩ N ( v ) ( N ( w ) ∩ Γ i ) = { v } .Proof. It is obvious that v ∈ \ w ∈ Γ i − ∩ N ( v ) ( N ( w ) ∩ Γ i ).Let x = ( x , · · · , x n ). Since the Hamming graph H ( n, m ) is a distance-transitive graph [3], then we can assume that, v = ( x , · · · , x n − i , y n − i +1 , · · · , y n ),where y j ∈ Z m − { x j } for all j = n − i + 1 , · · · , n . note on the automorphism group of the Hamming graph 5Let w ∈ Γ i − ∩ N ( v ). Then w, x differ in exactly i − w, v differ in exactly one coordinate. Note that, if in v we change one of x j s,where j = 1 , · · · , n − i , then we obtain a vertex u such that d ( u, x ) ≥ i + 1.Thus, w has a form such as; w = w r = ( x , · · · , x n − i , y n − i +1 , · · · , y r − , x r , y r +1 , · · · , y n )We show that if u ∈ Γ i and u = v and u is adjacent to some w r , then thereis some w p such that u is not adjacent to w p . If v = u ∈ Γ i is adjacent to w r then u has one of the following forms;(i) u = ( x , · · · , x n − i , y n − i +1 , · · · , y r − , y, y r +1 , · · · , y n ), where y ∈ Z m , and y = y r , x r (note that since m ≥
3, hence there is such a y ).(ii) u = ( x · · · , x j − , y, x j +1 , · · · , x n − i , y n − i +1 , · · · , y r − , x r , y r +1 , · · · , y n ),where y ∈ Z m , and y = x j .In the case (i), u is not adjacent to w t , for all possible t , t = r .In the case (ii), it is obvious that u is also not adjacent to w t for all possible t , t = r .Our argument shows that if u ∈ Γ i , and u = v , then there is some w r suchthat u is not adjacent to w r , in other words u / ∈ N ( w r ). Thus we have; \ w ∈ N ( v ) ∩ Γ i − ( N ( w ) ∩ Γ i ) = { v } . (cid:3) Let I = { γ , ..., γ n } be a set and K be a group. Let F un ( I, K ) be theset of all functions from I into K . We can turn F un ( I, K ) into a group bydefining a product:( f g )( γ ) = f ( γ ) g ( γ ) , f, g ∈ F un ( I, K ) , γ ∈ I, where the product on the right of the equation is in K . Since I is finite, thegroup F un ( I, K ) is isomorphic to K n (the direct product of n copies of K ),by the isomorphism f ( f ( γ ) , ..., f ( γ n )). Let H be a group and assumethat H acts on the nonempty set I . Then, the wreath product of K by H with respect to this action is the semidirect product F un ( I, K ) ⋊ H where H acts on the group F un ( I, K ), by the following rule, f x ( γ ) = f ( γ x − ) , f ∈ F un ( I, K ) , γ ∈ I, x ∈ H. We denote this group by
Kwr I H . Consider the wreath product G = Kwr I H .If K acts on a set ∆ then we can define an action of G on ∆ × I by the followingrule, ( δ, γ ) ( f,h ) = ( δ f ( γ ) , γ h ) , ( δ, γ ) ∈ ∆ × I, where ( f, h ) ∈ F un ( I, K ) ⋊ H = Kwr I H . It is clear that if I, K and H, arefinite sets, then G = Kwr I H , is a finite group, and we have | G | = | K | | I | | H | .We have the following theorem [4]. S.Morteza Mirafzal and Meysam Ziaee Theorem 3.2.
Let Γ be a graph with n connected components Γ , Γ , · · · , Γ n ,where Γ i is isomorphic to Γ for all i ∈ [ n ] = { , · · · , n } = I . Then we have, Aut (Γ) =
Aut (Γ ) wr I Sym ([ n ]) . Lemma 3.3.
Let n ≥ , m ≥ . Let v be a vertex of the Hamming graph H ( n, m ) . Then, Γ = < N ( v ) > , the induced subgraph of N ( v ) in H ( n, m ) , isisomorphic with nK m − , where nK m − is the disjoint union of n copies ofthe complete graph K m − .Proof. Let v = ( v , · · · , v n ). Then, for all i , i = 1 , · · · , n , there are m − w j , w j ∈ Z m − { v i } . Let x ij = ( v , · · · , v i − , w j , v i +1 , · · · , v n ), 1 ≤ i ≤ n, ≤ j ≤ n −
1. Then, N ( v ) = { x ij : 1 ≤ i ≤ n, ≤ j ≤ m − } .Let x ij , x rs , be two vertices in Γ = < N ( v ) > , then x ij , x rs are adjacent inΓ if and only if i = r . Note that two vertices, ( v , ..., v i − , w j , v i +1 , ..., v n )and ( v , ..., v i − , w s , v i +1 , ..., v n ) differ in only one coordinate. Therefore, foreach i = 1 , · · · , n , there are m − w ir in Γ which are adjacent tothe vertex w ij , where r = j . Now, it is obvious that the subgraph inducedby the set { x ij : 1 ≤ j ≤ m − } , is isomorphic with K m − , the completegraph of order m −
1. Now, it is easy to see that, the subgraph induced bythe set { x ij : i = 1 , · · · , n, j = 1 , · · · , m − } , is isomorphic with nK m − ,the disjoint union of n copies of the complete graph K m − . (cid:3) We now are ready to prove the main result of this paper.
Theorem 3.4.
Let n ≥ , m ≥ , and Γ = H ( n, m ) be a Hamming graph.Then Aut (Γ) ∼ = Sym ([ n ]) wr I Sym ([ m ]) , where I = [ n ] = { , , · · · n } .Proof. Let G = Aut(Γ). Let x ∈ V = V (Γ), and G x = { f ∈ G | f ( x ) = x } be the stabilizer subgroup of the vertex x in Aut(Γ). Let < N ( x ) > = Γ bethe induced subgroup of N ( x ) in Γ. If f ∈ G x then f | N ( x ) , the restrictionof f to N ( x ) is an automorphism of the graph Γ . We define the mapping ψ : G x −→ Aut(Γ ) by this rule, ψ ( f ) = f | N ( x ) . It is an easy task to showthat ψ is a group homomorphism. We show that Ker ( ψ ) is the identity group.If f ∈ Ker ( ψ ), then f ( x ) = x and f ( w ) = w for every w ∈ N ( x ). Let Γ i bethe set of vertices of Γ which are at distance i from the vertex x . Since, thediameter of the graph Γ = H ( n, m ), is n , then V = V (Γ) = n [ i =0 Γ i . We proveby induction on i , that f ( u ) = u for every u ∈ Γ i . Let d ( u, x ) be the distanceof the vertex u from x . If d ( u, x ) = 1, then u ∈ Γ and we have f ( u ) = u .Assume that f ( u ) = u , when d ( u, x ) = i −
1. If d ( u, x ) = i , then by Lemma 1. { u } = \ w ∈ Γ i − ∩ N ( u ) ( N ( w ) ∩ Γ i ). Note that if w ∈ Γ i − , then d ( w, x ) = i − f ( w ) = w . Therefore, { f ( u ) } = \ w ∈ Γ i − ∩ N ( u ) ( N ( f ( w )) ∩ Γ i ) = \ w ∈ Γ i − ∩ N ( u ) ( N ( w ) ∩ Γ i ) = u . note on the automorphism group of the Hamming graph 7Thus, f ( u ) = u for all u ∈ V (Γ), hence we have Ker ( ψ ) = { } . On the otherhand, G v Ker ( ψ ) ∼ = ψ ( G v ) ≤ Aut(Γ ), hence G v ∼ = ψ ( G v ) ≤ Aut(Γ ).Thus, | G v | ≤ | Aut(Γ ) | .We know by Lemma 3. that Γ ∼ = nK m − . We know that, Aut( K m − ) ∼ =Sym([ m − | G v | ≤ | Aut(Γ ) | = | Sym([ m − wr I Sym([ n ]) | = (( m − n n !,where I = [ n ] = { , · · · , n } .Since Γ = H ( n, m ) is a vertex-transitive graph, then we have | V (Γ) | = | G || G v | , and therefore; | G | = | G v || V (Γ) | ≤ | Aut( nK m − ) | m n = m n (( m − n n ! = ( m !) n n ! ( ∗ )We have seen (in the introduction section of this paper) that if θ ∈ Sym([ n ]),where Ω = [ n ] = { , · · · , n } , then f θ : V ( H ( n, m )) −→ V ( H ( n, m )) , f θ ( x , ..., x n ) = ( x θ (1) , ..., x θ ( n ) ) , is an automorphism of the Hamming graph H ( n, m ) , and the mapping ψ :Sym([ n ]) −→ Aut ( H ( n, m )), defined by this rule, ψ ( θ ) = f θ , is an injection.Therefore, the set H = { f θ | θ ∈ Sym([ n ]) } , is a subgroup of Aut (( H ( n, m ))),which is isomorphic with Sym([ n ]). Hence, we have Sym([ n ]) ≤ Aut ( H ( n, m )).On the other hand, if f ∈ F un ([ n ] , Sym([ m ])), then we define the map-ping; A f : V (Γ) → V (Γ) , by this rule, A f ( x , · · · , x n )=( f (1)( x ) , · · · , f ( n )( x n )) . It is an easy task to show that themapping A f is an automorphism of the Hamming graph Γ, and hence thegroup, F = { A f | f ∈ F un ([ n ] , ([ m ])) } , is a subgroup of the Hamming graphΓ = H ( n, m ). Therefore, the subgroup which is generated by H and F isin the group Aut (Γ), namely, W = < H, F > is a subgroup of Aut (Γ). Notethat W = Sym([ m ]) wr I Sym([ n ]), where I = [ n ] = { , , · · · , n } . Since, thesubgroup W has ( m !) n n ! elements, then by ( ∗ ), we conclude that; Aut (Γ) = W = Sym([ m ]) wr I Sym([ n ]) (cid:3) References [1] Biggs, NL. (1993). Algebraic Graph Theory (Second edition). Cambridge Math-ematical Library: Cambridge University Press.[2] Bondy, JA., Murty, USR. (2008). Graph Theory. New York: Springer-Verlag.[3] Brouwer, AE., Cohen, AM., Neumaier, A. Distance-Regular Graphs. (1998).New York: Springer-Verlag.[4] Cameron, PJ. (2005). Automorphisms of graphs, Topics in Algebraic GraphTheory. Cambridge Mathematical Library: Cambridge University Press.
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