How to find G-admissible abelian coverings of a graph?
aa r X i v : . [ m a t h . C O ] M a r How to find G -admissible abelianregular coverings of a graph? Haimiao Chen ∗ Department of mathematics, Peking University, Beijing, China
Hao Shen † Institute of mathematics, Chinese Academy of Sciences, Beijing, China
Abstract
Given a finite connected simple graph Γ, and a subgroup G of itsautomorphism group, a general method for finding all finite abelianregular coverings of Γ that admit a lift of each element of G is devel-oped. As an application, all connected arc-transitive abelian regularcoverings of the Petersen graph are classified up to isomorphism. Keywords: abelian, regular covering of a graph, G -admissible, automorphism,lift, Petersen graph MSC2010:
In this paper we develop an effective method to find all finite abelian regularcoverings (up to isomorphism) of a graph, admitting lifts of a given groupof automorphisms of the graph. We then apply the method to classify arc-transitive finite abelian regular coverings of the Petersen graph. This extendsthe work of [2] which classified arc-transitive elementary abelian regular cov-erings of the Petersen graph. ∗ Email: [email protected] † Email: [email protected] partially supported by NSFC project 61173009 Z p -coverings were found. A systematic method was developed in [11, 12] tofind isomorphism classes of elementary abelian coverings of a graph that ad-mit lifts of given automorphisms; the method was applied in [12] to classifyvertex-transitive elementary abelian coverings of the Petersen graph. Besidesthe ones mentioned above, there are many such results, see [3, 9, 13, 14], andso on.But classification of general finite abelian regular coverings of a graphis never seen. Theoretically, as pointed out in [4], classification of generalfinite abelian regular coverings can be reduced to that of elementary abelianones. But the difficulty is that, when elementary abelian regular coveringsare classified, one still needs to go on to classify elementary abelian regularcoverings of each of these covering graphs, and go on again. One is not ableto stop at a step and say that the classification of abelian regular coveringsof the initial graph is completed!Our paper makes a breakthrough. The method is based on the criterionfor lifting automorphisms proposed by the first author in [1]. We reduce theproblem (Problem 3.1) of finding abelian coverings to that (Problem 3.13) offinding numbers and matrices satisfying certain conditions.The content is organized as follows.Section 2 is a preliminary on topological graph theory. Section 3 is aframework suitable for general graphs and automorphisms; to determine allabelian regular coverings to which some given automorphisms can be lifted, apractical method is developed and written down as Algorithm 3.18. Section4 is devoted to classifying all arc-transitive finite abelian regular coverings ofthe Petersen graph, and the result is given in Theorem 4.4.Finally, some notational conventions.For a commutative ring R , let R × be the set of units of R . Let R n,m denotethe set of all n × m matrices with entries in R . Identify R ,m with R m . For amatrix U ∈ R n,m , let U i,j denote its ( i, j )-th entry and write U = ( U i,j ) n × m ;the row space of the matrix U is denoted as h U i . Let GL( m, R ) be the setof invertible m × m matrices in R . For U ∈ R n,m , C ∈ Z k,n , K ∈ Z m,l , it is2eaningful to talk about CU ∈ R k,m and U K ∈ R n,l .We use Σ m to denote the permutation group on m letters. Each τ ∈ Σ m is identified with a permutation matrix which is also denoted as τ and givenby τ i,j = δ i,τ ( j ) .For n ∈ Z with n >
1, denote the quotient group of Z modulo n as Z n .Actually it is a quotient ring of Z , and is a field when n is prime. Note thatthere is a standard bijection f n : Z n → { , · · · , n − } . We shall often identify λ ∈ Z n with f n ( λ ) ∈ Z . For λ, λ ′ ∈ Z n , we say λ > λ ′ if f n ( λ ) > f n ( λ ′ ). In this section we recall some terminologies on graph theory and give somenecessary definitions. For more one can refer to [2, 5–7, 10].In this paper all graphs are finite, connected, and simple (there is no loopor parallel edges). For a graph Γ, let V (Γ) , E (Γ) , Ar(Γ) be the set of vertices,edges, arcs of Γ respectively. For any v ∈ V (Γ), we denote its neighborhoodby N ( v ). Let { u, v } ∈ E (Γ) denote the edge connecting the vertices u, v , andlet ( u, v ) ∈ Ar(Γ) denote the arc from u to v . Each edge { u, v } gives rise totwo arcs ( u, v ) and ( v, u ) which are opposite to each other: ( v, u ) = ( u, v ) − .A walk W = ( v , · · · , v n ) is a tuple such that ( v i , v i +1 ) ∈ Ar(Γ), for 1 i < n .A covering π : Λ → Γ is a map of graphs such that π : V (Λ) → V (Γ) issurjective, and π | N (˜ v ) → π | N ( v ) is bijective for any v ∈ V (Γ) and ˜ v ∈ π − ( v ).The group A of automorphisms of Λ which fix each fiber setwise is called the covering transformation group . The covering is called regular if A acts oneach fiber transitively.For two coverings π : Λ → Γ , π ′ : Λ ′ → Γ, an isomorphism π → π ′ is a pairof isomorphisms ( α, ˜ α ) with α ∈ Aut(Γ), ˜ α : Λ → Λ ′ such that α ◦ π = π ′ ◦ ˜ α .Call ˜ α a lift of α along ( π, π ′ ), and say that α is lifted to Λ → Λ ′ . When α = id, call ( α, ˜ α ) an equivalence and say that π is equivalent to π ′ .When Λ = Λ ′ and π = π ′ , ˜ α is called a lift of α along π , and α is said tobe lifted to Λ. If G Aut(Γ) such that each α ∈ G can be lifted to Λ, then π is said to be G -admissible .Given a finite abelian group A , a voltage assignment of Γ in A is a map φ : Ar(Γ) → A such that φ ( x − ) = − φ ( x ) for all x ∈ Ar(Γ). Each voltageassignment φ determines a regular covering Γ × φ A of Γ, having vertex set V (Γ × φ A ) = V (Γ) × A and edge set E (Γ × φ A ) = {{ ( u, g ) , ( v, φ ( u, v ) +3 ) } : { u, v } ∈ E (Γ) } ; there is a canonical covering map given by the projec-tion onto the first coordinate.Let T be an arbitrarily chosen spanning tree of Γ. A voltage assignment φ is called T -reduced if φ ( x ) = 0 for all x ∈ Ar( T ) ⊂ Ar(Γ). It is known(see [5]) that every regular covering of Γ with covering transformation group A is isomorphic to Γ × φ A for some T -reduced voltage assignment φ .For any walk W = ( v , · · · , v n ), the voltage of W is by definition φ ( W ) = φ ( v , v ) + · · · + φ ( v n − , v n ).Choose a base vertex v . For each v ∈ V (Γ) there is a unique reduced walk W ( v ) in T from v to v . For each arc ( u, v ) let L ( u, v ) = W ( u ) · ( u, v ) · W ( v ) − .Let e , · · · , e b ( b is the first Betti number of the graph) be all the cotreeedges, and for each i , choose an arc x i from e i . Then the first homology group H (Γ; Z ) is the free abelian group generated by L ( x ) , · · · , L ( x b ). There is awell-defined homomorphism E ( φ ) : H (Γ; Z ) → A, (1)which is surjective because Γ × φ A is assumed to be connected.Each automorphism α ∈ Aut(Γ) induces an automorphism α ∗ : H (Γ; Z ) → H (Γ; Z ) . (2)The following proposition comes from Theorem 3 in [7] and its proof: Proposition 2.1.
Given two A -coverings Γ × φ A and Γ × ψ A . An automor-phism β ∈ Aut(Γ) can be lifted to Γ × φ A → Γ × ψ A if and only if thereexists a group automorphism σ : A → A such that E ( ψ ) ◦ β ∗ = σ ◦ E ( φ ) , orequivalently, β ∗ (ker( E ( φ ))) = ker( E ( ψ )) .In particular, Γ × φ A is equivalent to Γ × ψ A if and only if ker( E ( φ )) =ker( E ( ψ )) . As a corollary, we obtain a theoretical criteria for lifting automorphisms.See [2] Proposition 1.2.
Proposition 2.2.
An automorphism α ∈ Aut(Γ) can be lifted to Γ × φ A if andonly if there is an automorphism σ : A → A such that E ( φ ) ◦ α ∗ = σ ◦ E ( φ ) ,or equivalently, α ∗ (ker( E ( φ ))) = ker( E ( φ )) . General framework
We want to solve the following
Problem 3.1.
Given a graph Γ and a subgroup G of Aut(Γ), find all G -admissible finite abelian regular coverings of Γ, up to isomorphism. Definition 3.2.
Given voltage assignments φ i : Ar(Γ) → A i , i m ,there is a product voltage assignment φ = m Q i =1 φ i : Ar(Γ) → m Q i =1 A i , given by φ ( x ) = ( φ ( x ) , · · · , φ m ( x )) . Call Γ × φ ( m Q i =1 A i ) the fibered product of the coverings Γ × φ i A i .Suppose A = Q p ∈ I A ( p ) where I is a finite set of prime numbers, and foreach p ∈ I , A ( p ) = n p Q η =1 Z p k ( p,η ) with k ( p, > · · · > k ( p, n p ) > T is a spanning tree of Γ as in Section 2, and φ : Ar(Γ) → A is avoltage assignment which is assumed to be T -reduced as we always do. Foreach p ∈ I , let q ( p ) : A → A ( p ) denote the projection, and let φ ( p ) = q ( p ) ◦ φ ;call Γ × φ ( p ) A ( p ) the p -primary part of Γ × φ A .It is easy to see that E ( φ ) is surjective if and only if E ( φ ( p ) ) is for all p ,and α ∗ (ker( E ( φ ))) = ker( E ( φ )) if and only if α ∗ (ker( E ( φ ( p ) ))) = ker( E ( φ ( p ) ))for all p . Hence we have Lemma 3.3.
The regular covering Γ × φ A is the fibered product of its primaryparts; it admits a lift of α if and only if each primary part does. Without loss of generality, from now on we assume that A is an abelian p -group for some prime p : A = n Q η =1 Z p k ( η ) with k (1) > · · · > k ( n ) > φ : Ar(Γ) → A , it is clear that p k (1) Z b ker E ( φ ). Let ker( E ( φ )) = ker E ( φ ) /p k (1) Z b Z bp k (1) . (3)5 efinition 3.4. A G -admissible pair ( k, K ) consists of a positive integer k and a subgroup K Z bp k such that α ∗ ( K ) = K for all α ∈ G , where α ∗ isthe induced action of α ∈ Aut(Γ) on H (Γ; Z p k ) ∼ = Z bp k .Say two pairs ( k, K ) and ( k ′ , K ′ ) are isomorphic if k = k ′ and there exists β ∈ Aut(Γ) such that β ∗ ( K ) = K ′ . Remark 3.5.
From Proposition 2.1 and Proposition 2.2, we see that Problem3.1 can be reduced to finding G -admissible pairs up to isomorphism.Be careful that when ( k, K ) is G -admissible, it is not guaranteed that( k, β ∗ ( K )) also is, unless G is normal in Aut(Γ). Let p be a prime number, k >
1, and let R = Z p k . Definition 3.6.
For λ ∈ R , the p -degree of λ , denoted deg p ( λ ), is the uniqueinteger r , 0 r k such that λ = p r · χ , with χ ∈ R × .The following theorem, proved in [1], plays a fundamental role. We in-clude the proof here for the reader’s convenience. Theorem 3.7.
For each X ∈ R n,m , there exist Q ∈ GL( n , R ) , S ∈ GL( m , R ) such that X := QXS is “in normal form”, that is, ( X ) i,j = δ i,j · p r i , with r · · · r n k .Proof. If X = 0, there is nothing to show. So let us assume X = 0.Choose an entry X i ,j = 0 with smallest p -degree, and suppose X i ,j = p r · χ with χ invertible. Interchange the i -th row of X with the first row,and the j -th column with the first column, and divide the first row by χ .The matrix obtained has (1 , p r , dividing all entries. Then performrow-transformations to eliminate the ( i, < i n . Denote thenew matrix by X (1) .Do the same thing to the ( n − × ( m −
1) down-right minor of X (1) ,and then go on. At each step, we can take row transformations to eliminateelements in a column below the main diagonal, and rearrange the diagonalelements. At l -th step for some l min { m, n } , we get some matrix X ( l ) whose i -th diagonal entry is p r i for 1 i l , with r · · · r l , and theelements under the main diagonal all vanish.Finally, perform column transformations to X ( l ) , to eliminate all the “off-diagonal” entries. Then the resulting matrix is in normal form.6 emark 3.8. From the proof it is easy to see that S can be chosen to be ωS ′ for some upper-triangular matrix S ′ and some permutation matrix ω . Theorem 3.9.
Each subgroup C R m is of the form h P Qω i such that(a) P ∈ R l,m , l m , P i,j = δ i,j · p r i , r · · · r l < k ; (b) Q ∈ GL( m , R ) is upper-triangular with diagonal 1’s, and Q i,j < p r j − r i forall j > i , (by convention set r i = k for i > l ); (c) ω ∈ Σ m . Moreover, C ∼ = Z p k − rl × · · · × Z p k − r , hence P is uniquely determined by C .Proof. Choose a set of generators { X , · · · , X l ′ } of C and make the matrix X ∈ R l ′ ,m , with row vectors X , · · · , X l ′ . By Theorem 3.7 and Remark3.8, there exist Y ∈ GL( l ′ , R ) , S ′ ∈ GL( m, R ) , ω ′ ∈ Σ m such that S ′ isupper-triangular, and P ′ := Y XωS ′ takes the form P ′ i,j = δ i,j · p r i , with0 r · · · r l ′ k . Let l be the maximal i with r i < k , and let P bethe matrix obtained by taking the first l rows of P ′ . Note that via scalarmultiplication and division with remainder, we are able to take elementaryrow transformations to convert P S ′− into P Q for some Q ∈ GL( m, R )satisfying (b). Then C = h X i = h Y X i = h P Qω i with ω = ω ′− .Now verify the last assertion. For a general matrix B , let B i denoteits i -th row vector. Define a homomorphism f : C = h P Qω i → h P i ∼ = Z p k − rl × · · · × Z p k − r by sending ( P Qω ) i to P i , 1 i l . It is well-defined,because for all w = ( w , · · · , w l ) ∈ Z l , we have wP Qω = 0 if and only if wP = 0. Clearly f is an isomorphism. Definition 3.10.
Suppose l m , P ∈ R l,m with P i,j = δ i,j · p r i with 0 r · · · r l < k . Two matrices M, M ′ ∈ GL( m, R ) are called P -equivalent and denoted as M ∼ P M ′ , if h P M i = h P M ′ i . Lemma 3.11.
Let P be given in Definition 3.10. Then M ∼ P M ′ is equiv-alent to deg p ( M M ′− ) i,j > r j − r i for all j > i , (set r l +1 = · · · = r m = k byconvention).Proof. Define a homomorphism h P i → h P ( M M ′− ) i by w w ( M M ′− ).This is an isomorphism, hence the finite group h P ( M M ′− ) i contains as manyelements as h P i does. So M ∼ P M ′ , which is equivalent to h P ( M M ′− ) i = h P i , holds if and only if h P ( M M ′− ) i h P i . The later condition is equivalentto deg p ( M M ′− ) i,j > r j − r i for all j > i , using the characterization that( w , · · · , w m ) ∈ h P i if and only if deg p ( w j ) > r j for all j .7 .3 The method of finding admissible abelian cover-ings Recall Remark 3.5 that we are going to find G -admissible pairs ( k, K ) up toisomorphism. For each k , there is always an admissible pair ( k, ) where is the trivial subgroup. In the rest of this section, let us assume K = .For α ∈ Aut(Γ), let S α ∈ GL( b, Z ) denote the matrix corresponding tothe induced isomorphism (2) of α , so that α ∗ ( L ( x i )) = b X j =1 ( S α ) i,j L ( x j ) , i = 1 , · · · , b. (4)Abusing the notation, we denote the image of S α under the homomorphismGL( b, Z ) → GL( b, Z p k ) induced by the quotient map Z → Z p k also by S α .Now if K = h P Qω i Z bp k , then α ∗ ( K ) = h P QωS α i . Definition 3.12. A G -admissible solution is a tuple ( p ; k, l ; r , · · · , r l ; Q, ω )consisting of integers k >
1, 0 < l < b, r · · · r l < k , a prime num-bers p , matrices Q ∈ GL( b, Z p k ) , ω ∈ Σ b , satisfying the following condition(by convention set r i = k for l < i b ):( ⋆ ) Q is upper-triangular with diagonal 1 ′ s, Q i,j < p r j − r i for all i < j ;for all α ∈ G, QωS α ∼ P Qω, where P ∈ Z l,bp k , P i,j = δ i,j · p r i . Call two solutions ( p ; k, l ; r , · · · , r l ; Q, ω ) and ( p ′ ; k ′ , l ′ ; r ′ , · · · , r ′ l ; Q ′ , ω ′ ) iso-morphic if p = p ′ , k = k ′ , l = l ′ , r i = r ′ i , i l , and Q ′ ω ′ ∼ P QωS β forsome β ∈ Aut(Γ).By Theorem 3.9 and Lemma 3.11, G -admissible pairs are the same as G -admissible solutions. Moreover, two solutions are isomorphic if and only if thecorresponding pairs and the corresponding regular coverings are isomorphic.Thus Problem 3.1 is further reduced to Problem 3.13.
Find all G -admissible solutions up to isomorphism. Remark 3.14.
Each solution determines, up to equivalence, a unique con-nected G -admissible covering Γ × φ ( n Q η =1 Z p k ( η ) ) with n = b − i ( i being thelargest i with r i = 0), k ( η ) = r b +1 − η , η n , and φ ( x i ) = ((( Qω ) − ) i,b , · · · , (( Qω ) − ) i,i +1 ) ∈ Z p k (1) × · · · × Z p k ( n ) , i b ;(5)8his is because there are isomorphisms Z bp k (1) / h P Qω i ∼ = Z bp k (1) / h P i , w w ( Qω ) − , and Z bp k (1) / h P i ∼ = n Y η =1 Z p k ( η ) , ( u , · · · , u b ) ( u b (mod p k (1) ) , · · · , u i +1 (mod p k ( n ) )) , where by u (mod p r ) for u ∈ Z p k , r k , we mean the image of u under thecanonical homomorphism Z p k → Z p r . (6)The second part of ( ⋆ ) is, by Lemma 3.11, equivalent todeg p ( QωS α ( Qω ) − ) i,j > r j − r i for all j > i and α ∈ G. (7)It may be frustrating that there are so many variables, especially that ω may take b ! values. Nevertheless, we have some key observations which helpsimplify the calculations very much. Proposition 3.15.
Suppose that ( p ; k, l ; r , · · · , r l ; Q, ω ) is a solution.(a) If σ ∈ Σ b satisfies r σ ( i ) = r i for all i ∈ { , · · · , b } , then ( p ; k, l ; r , · · · , r l ; σQσ − , σω ) is a solution isomorphic to ( p ; k, l ; r , · · · , r l ; Q, ω ) .(b) If there exists β ∈ Aut(Γ) such that βGβ − = G , and S β = Dτ with τ ∈ Σ b and D diagonal, then ( p ; k, l ; r , · · · , r l ; Q ′ , ωτ ) is a solutionisomorphic to ( p ; k, l ; r , · · · , r l ; Q, ω ) , for some Q ′ .Proof. (a) It is easy to see that ( σQσ − , σω ) satisfies ( ⋆ ). Since r σ ( i ) = r i ,we have σ ∼ P id, hence ( σQσ − )( σω ) = σ ( Qω ) ∼ P Qω .(b) Suppose D i,j = δ i,j · d j . Let D ′ be the diagonal matrix with D ′ i,j = δ i,j · d ω − ( j ) . Similarly to the proof of Theorem 3.9, we may take row transfor-mations to convert P QD ′ into P Q ′ with Q ′ ∈ GL( b, Z p k ) satisfying the firstpart of ( ⋆ ), then Q ′ ∼ P QD ′ . Hence QωS β = QωDτ = QD ′ ωτ ∼ P Q ′ ωτ .This also shows that ( p ; k, j ; s , · · · , s j ; Q ′ , ωτ ) is a solution: for all α ∈ G ,we have QωS β S α = QωS βαβ − S β ∼ P QωS β , hence Q ′ ωτ S α ∼ P Q ′ ωτ .9et Aut G (Γ; T ) = { β ∈ Aut(Γ) : β ( T ) = T, βGβ − = G } . (8)Note that for each β ∈ Aut G (Γ; T ), S β = D β τ β , (9)where D β is some diagonal matrix and τ β ∈ Σ b is the induced permutationon cotree edges.Suppose r i η − +1 = · · · = r i η < r i η +1 , 0 η s , where i s = l , and byconvention we have set i − = 0 , i s +1 = b . Define an equivalence relation ∼ on Σ b by declaring ω ∼ ω ′ if there exists β ∈ Aut G (Γ; T ) such that { ( ωτ β ) − ( i ) : i η − < i i η } = { ω ′− ( i ) : i η − < i i η } , η = 0 , · · · , s + 1 . (10)Let Σ b ( i , · · · , i s ) be the set of equivalence classes.By Proposition 3.15, to find isomorphism classes of G -admissible solutionswith r i η − +1 = · · · = r i η < r i η +1 , it is sufficient to take a representative fromeach equivalence class in Σ b ( i , · · · , i s ).For r k and X ∈ Z n,mp k , let X (mod p r ) ∈ Z n,mp r denote the matrixobtained by replacing each entry by its image under (6). Proposition 3.16.
Suppose that ( p ; k, l ; r , · · · , r l ; Q, ω ) is a solution, r i The first half of ( ⋆ ) is clearly satisfied. For the second half, writing QωS α ω − Q − (mod p r ) = (cid:18) ∗ ∗ V α (cid:19) with V α ∈ GL( b − i, Z p r ), and re-garding ω, S α as in GL( b, Z p r ), we have QωS α ω − Q − = (cid:18) ∗ ∗ Q − V α Q (cid:19) .Thus ( ⋆ ) holds. 10 emark 3.17. The special case r = 1 implies that, if r i < r i +1 , then( p ; k, l ; r , · · · , r l ; Q, ω ) cannot be a G -admissible solution unless there ex-ists a G -admissible Z ip -covering of Γ. This is practically useful as long as theclassification of G -admissible elementary abelian coverings is known.Summarizing above discussions, now we go to solve Problem 3.13 by Algorithm 3.18. (a) Set s = 0 , k = 1 and run Step (c) and (d) belowto find elementary abelian regular coverings. That is to say, for each i ∈{ , , · · · , b − } and each class in Σ b ( i ), choose a representative ω , then findall upper-triangular matrices Q ∈ GL( b, Z p ) satisfying Q i,i = 1 , i b ,and ( QωS α ω − Q − ) i,j = 0 for all i i < j and α ∈ G .Let P be the set of pairs ( p, i ) such that ( p ; 1 , i ; 0 , · · · , Q, ω ) emergesas a solution, i.e., there is a G -admissible Z b − i p -covering of Γ.(b) Determine the set P := { ( p ; i , i , · · · , i s ) : 0 i < i < · · · < i s < b, ( p, i η ) ∈ P , η s } . (c) For each ( p ; i , · · · , i s ) ∈ P , determine Σ b ( i , · · · , i s ).For each equivalence class in Σ b ( i , · · · , i s ), choose a representative ω , andfind all ( c , · · · , c s , k ) ∈ Z s +1 and upper-triangular matrices Q ∈ GL( b, Z p k )satisfying(i) 1 c < · · · < c s < k ;(ii) Q i,i = 1 , i b ; 0 Q i,j < p c η − c η ′ for all i, j with i η ′ < i i η ′ +1 i η < j i η +1 ;(iii) ( QωS α ω − Q − ) i,j ≡ p c η − c η ′ ), for all i, j with i η ′ < i i η ′ +1 i η < j i η +1 and all α ∈ G .Then ( p ; k, l ; r , · · · , r l ; Q, ω ) is a solution, where l = i s , r i = 0 for i i ,and r i = c η for i η − < i i η .(d) Distinguish the isomorphism classes among all the solutions. Remark 3.19. The goal of Step (a) can be achieved alternatively by applyingthe method of [11, 12]. Remark 3.20. In Step (c), when checking condition (c)(iii), it is sufficientto do it for a set of generators of G .When s = 0, (in which case we call the solution “ pure ”), by (c)(ii), Q can be written in block form as (cid:18) I Θ0 I (cid:19) for Θ ∈ Z i ,b − i p k . Writing11 S α ω − = (cid:18) B α B α B α B α (cid:19) , condition (c)(iii) is equivalent to the equationsover Z p k : ( B α + Θ B α )Θ = B α + Θ B α . (12)One obtains a system of quadratic congruence equations from (12) for var-ious α . It is convenient to first get pure solutions and then apply Proposition3.16 to obtain some constraints on p, k, l, r , · · · , r l , Q, ω . Remark 3.21. If ( p ; k, l ; r , · · · , r l ; Q, ω ) is a solution as in (c) such that i = 0 (equivalently, r > p ; k ′ , l ; r ′ , · · · , r ′ l ; Q, ω ) with k ′ = k − c , r ′ i = r i − c is also a solution. Conversely, from any positive integer c and any solution ( p ; k, l ; r , · · · , r l ; Q, ω ), one can construct another solution( p ; k ′′ , l ; r ′′ , · · · , r ′′ l ; Q, ω ) with k ′′ = k + c , r ′′ i = r i + c , r ′′ > i > i = 0. Let X = { a, b, c, d, e } , and V = { , , , , , , , , , } , where 0 = { a, b } ,1 = { c, d } , 2 = { c, e } , 3 = { d, e } , 4 = { a, e } , 5 = { b, e } , 6 = { a, d } ,7 = { b, d } , 8 = { a, c } , 9 = { b, c } . The Petersen graph has V (Γ) = V and E (Γ) = {{ i, j }| i, j ∈ V, i ∩ j = ∅} . It is known that Aut(Γ) ∼ = Σ , the actionbeing induced from that on X .Fix a spanning tree T as in Figure 1(a); the induced subgraph is shownin (b). For each cotree edge { i, j } , choose the arc ( i, j ) with i < j . Label x = (5 , , x = (7 , , x = (4 , , x = (4 , , x = (6 , , x = (5 , α = (13)(67)(49)(58), α = (19)(56)(28)(03), α = (123)(468)(579), α = (45)(67)(89), induced by ( ab )( ce ) , ( ae )( bd ) , ( ced ) , ( ab ) respectively.According to [2], h α , α , α , α i = Aut(Γ), and the subgroup H := h α , α , α i is isomorphic to the alternating group on 5 elements, hence isnormal. A regular covering ˜Γ of Γ is arc-transitive if and only if H can belifted. Moreover, ˜Γ is 2-arc-transitive if H can be lifted, and ˜Γ is 3-arc-transitive if the whole Aut(Γ) can be lifted.12 igure 1: (a) The spanning tree T; (b) the induced subgraph The matrices determined by (4) can be calculated: S α = − − 10 0 0 0 − − − − , S α = − − − − − − − , S α = − − − − , S α = . Recall Theorem 3.2 of [2]: Theorem 4.1. Each connected arc-transitive elementary abelian covering of Γ is isomorphic to a unique one given in Table 1. Moreover, X (2 , , X (2 , ,X (5 , , X ( p, are 3-arc-transitive, and X ′ (2 , , X ± ( p, are 2-transitivebut not 3-arc-transitive. Remark 4.2. As pointed out in [2], X + ( p, 3) is not isomorphic to X − ( p, emark 4.3. In Table 1, λ ± are the two solutions of the equation λ − λ − Z p . In Table 2, the λ ± in 4th row are the two solutions of the equation λ − λ − Z p k , while the ones in 6th row are the two solutions of theequation λ − λ − Z p s and are identified with elements of Z p k viathe map Z p s f ps −−→ { , , · · · , p s − } ⊂ { , , · · · , p k − } f − pk −−→ Z p k , (for f n seethe end of Section 1, and it should be warned that ( f − p k f p s )( λ ± ) need not besolutions to the equation λ − λ − Z p k ).Here is the main result of this section: Theorem 4.4. Each connected primary arc-transitive finite abelian coveringof Γ is isomorphic to a unique one in Table 1 and Table 2. Moreover, inTable 2, X k, (2 , , X k, (2 , , X k, (5 , , X k, ( p, are 3-arc-transitive andthe others are 2-arc-transitive but not 3-arc-transitive.Each connected arc-transitive finite abelian covering of Γ is isomorphicto the fibered product of connected p -primary ones, with distinct primes p . Table 1 Elementary abelian coverings of the Petersen graphs Covering A φ t ( x ) φ t ( x ) φ t ( x ) φ t ( x ) φ t ( x ) φ t ( x ) Condition Admissiblefor X (2 , Z S X ′ (2 , Z A X (2 , Z 10 01 10 01 10 01 S X (5 , Z 100 134 001 311 010 143 S X ± ( p, Z p 100 1 λ ± − λ ± 11 010 1 − λ ± p ≡± A X ( p, Z p p arbitrary S able 2 New families of abelian coverings of the Petersen graphCovering A φ t φ t φ t φ t φ t φ t Condition admissiblefor X k, (2 , Z k − × Z k − − − − − k > S X ′ k, (2 , Z k − × Z k − − k > A X k, (2 , Z k − × Z k k > S X ± k, ( p, Z p k 001 010 1 − λ ∓ − λ ± λ ± λ ± − λ ± − λ ∓ − λ ± k > A X k, (5 , Z k − × Z k k > S X ± k,c, ( p, Z p c × Z p k − λ ∓ − λ ± λ ± λ ± − λ ± − λ ∓ − λ ± 100 100000 k > c > A X k, ( p, Z p k p arbitrary k> S .3 Proof of Theorem 4.4 The proof is a good illustration of the method developed in Section 3.3,although some modifications are taken in order to simplify the process.The trivial pair ( k, ) gives rise to X k, ( p, 6) for any p and any k .By Theorem 4.1, P = { (2 , , (2 , , (5 , } ∪ { ( p, 3) : p ≡ ± } , hence P = { (2; 5) , (2; 4) , (2; 4 , , (2; 0 , , (2; 0 , , (2; 0 , , }∪ { ( p ; 3) , ( p ; 0 , 3) : p = 5 or p ≡ ± } . (13)Note that α , α , α ∈ Aut G (Γ; T ); since α has order 2, we can deducethat Aut G (Γ; T ) = h α , α , α i . In the notation of (9), τ α = (26)(35) , τ α = (153)(264) , τ α = (14)(25)(36) . (14)Suppose ( p ; k, l ; r , · · · , r l ; Q, ω ) is a G -admissible solution. We shall dis-cuss various cases according to i (the largest i with r i = 0) and l .Note that, since Aut(Γ) /H ∼ = { , α } , two solutions ( p ; k, l ; r , · · · , r l ; Q, ω )and ( p ; k, l ; r , · · · , r l ; Q ′ , ω ′ ) are isomorphic if and only if h P Qω i = h P Q ′ ω ′ i or ( α ) ∗ h P Qω i = h P Q ′ ω i , where P ∈ Z l, p k , P i,j = δ i,j · p r i .Remark 3.14 tells us how to explicitly construct the corresponding cov-ering from a solution. And for each covering, it is easy to check whether itis S -admissible by testing whether it is possible to lift α , in a routine way. i = 5Then p = 2, l = 5, 0 = r = · · · = r < r = k .First, we show that Σ (5) = { [id] } . Each element of Σ is equivalent tosome ω with ω − (1) < · · · < ω − (5); there are 6 such elementsid , ω = (654321) , ω = (65432) , ω = (6543) , ω = (654) , ω = (65) . It is easy to verify that id α ∼ ω α ∼ ω α ∼ ω α ∼ ω α ∼ ω , where by ω α ∼ ω ′ wemean (10) holds for β = α .Thus Σ (5) = { [id] } and we can assume ω = id.16n the notation of Remark 3.20, suppose Θ = ( q , · · · , q ) t . The equation(12) for α = α , α , α can be written as, respectively, − − q − q − q − − q − − q − q q q q q = − , (15) q − − q q − − q q − − q q − − q − q − q q q q q q = − q q q q q , (16) q − q − q q − q q q q q q = − , (17)which hold in Z k .From (15)-(17) one can deduce that k = 1, q = q = 1, q = q = q .Hence there are two possibilities: Θ = (1 , , , , t or Θ = (0 , , , , t . Let Q = , Q ′ = . (18)It is clear that the solution (2; 1 , 5; 0 , , , , Q , id) gives rise to X (2 , , 5; 0 , , , , Q ′ , id) gives rise to a covering isomorphic to X ′ (2 , i = 4Then p = 2, l = 4 or 5, and 0 = r = · · · = r < r r = k .(a) When l = 4, r = r .Σ (4) = { [id] , [(45)] , [(354)] } . Let Θ = ( q ij ) × .(i) When ω = id. 17e have the following equations in Z k , obtained from (12): − − q − q − q − q − q − q − q − q − q q q q q q q q = − − , (19) q − q q − q − q − q − q − q − q q q q q q q q = − − q − q − q − q − q − q − q − q , (20) − q q − − q q − − q q − q q q q q q q q q q = − . (21)Use (20)-(1,1) to denote the equation obtained by comparing the (1,1)-entries of (20) and so on.We have the following deductions:(20)-(1,1),(20)-(2,2),(21)-(1,1), (21)-(1,2) ⇒ q − q = 1 + q ;(19)-(1,1), (19)-(1,2) ⇒ q (1 + q ) = 0;(20)-(1,1), (20)-(1,2) ⇒ q = 1 , q = − q , ( q − q ) = 1;(20)-(2,2) ⇒ q ( q − q ) = 0;(21)-(2,1), (21)-(2,2) ⇒ q − q = q q − q q − q q + q ;(20)-(4,2) ⇒ q ( q − q ) = q ( q q − q q − q q + q ) = q q ( q − q ) + q q ( q − q ) = q q ( q − q ) ⇒ q q = 0 , q q = 0;(21)-(1,1) ⇒ q = − ⇒ q = q ;(19)-(2,2) ⇒ q = 1, q = 0;(21)-(2,2) ⇒ q = 1;(21)-(3,2) ⇒ q = 0;(21)-(4,2) ⇒ q = − ⇒ q = 0;(20)-(2,1) ⇒ q = − ⇒ q = 0;(20)-(3,2) ⇒ ⇒ k = 1. 18hus the solution is (2; 1 , 4; 0 , , , Q , id) with Q = . (22)It can be checked that the S -admissible covering determined by this solutionis isomorphic to X (2 , ω = (45).The equation (12) over Z k for α = α , α reads respectively − − q − q − q − − q − q q q q q q q q = − q − q − − q − q , (23) q − − q q − − q q − q q − − q q q q q q q q q = − q − q − q − q − − q − q − q − q . (24)(23)-(2,2) ⇒ q = 1;(24)-(2,2) ⇒ q ( q − q ) = 0 ⇒ q = q ;(24)-(3,2) ⇒ q + 2 q = 0.This leads to contradiction. Hence in this case there is no solution.(iii) When ω = (354).Similarly, the equations (12) for α = α , α lead to contradiction.(b) When l = 5, r < r = k .It turns out that Σ (4 , 5) = { [id] , [(45)] , [(354)] } = Σ (4), so we still onlyneed to consider ω = id , (45) , (354).Suppose Q i,j = q ij , ( i = 1 , · · · , , j = 5 , Q , = q . From Proposition3.16 and the result of (2.1), it follows that ω = id, r = 1 and Q (mod 2) =19 ; by Proposition 3.16 and the result of Section 4.3.1, we have r = 1 and − q − q − q − q q q q q q = or . Hence q = q = q = 1, q = q = 0, q , q ∈ { , } , q , q ∈ { , } . Theequations ( QS α Q − ) i, ≡ i α = α , α , α imply q + q + q q − q − q (1 + q ) q (1 + q ) ≡ q + q + q q − q − q q − q q − − − q q ≡ − (1 − q ) q (2 − q ) q − q q − q − q + q − q q ≡ , which is impossible to hold, as can be checked easily. Thus there is nosolution. i = 3Then p = 5, or p ≡ ± l = 3, 0 = r = r = r < r = r = r = k , Σ (3) = { [(14253)] , [(34)] , [(2453)] } . Let Θ = ( q ij ) × .(i) ω = (14253).We have the following equations over Z p k : q q q q q q q q q = q q q q q q q q q , (25) − − q − q − − q − q − q − q q q q q q q q q q = q − q − q q − q − q q − − q − q , (26) − q − q − q − q − q − q q q q q q q q q q = q q − − q . (27)2025) ⇒ q = q , q = q , q = q , q = q ;(27)-(3,2),(3,3) ⇒ q ( q − q ) = q , q ( q + q ) = 0;(27)-(1,2) ⇒ q ( q + q ) = 0;(27)-(2,3) ⇒ q q + q = 1;(26)-(1,3) ⇒ q = q + q ( q + q ) ⇒ q or q is invertible ⇒ q = − q ;(26)-(2,2), (2,3) ⇒ ( q + q ) q = 0 = ( q + q ) q ⇒ q = − q = q ;(27)-(1,1) ⇒ q = 1 − q ;(27)-(2,3), (26)-(1,3) ⇒ q = − q − q − x = 5 , (28)with x = 2 q − p = 5, it is easy to see that (28) has a solution if and only if k = 1.If p = 5, by Proposition 5.1.1 of [8] (page 50), the equation (28) has asolution in Z p k if and only if the Legendre symbol (5 /p ) is equal to 1, whichturns out to be equivalent to p ≡ ± λ ± ∈ Z p k be the two solutions of λ − λ − , 3; 0 , , Q , (14253)) and ( p ; k, 3; 0 , , Q ± , (14253))( p ≡ ± , k > Q = ∈ Z , , (29) Q ± = λ ± λ ∓ − 10 1 0 λ ± − λ ± − λ ± λ ± − λ ∓ ∈ Z , p k . (30)The solution (5; 1 , 3; 0 , , Q , (14253)) gives rise to a covering isomorphicto X (5 , p ; k, 3; 0 , , Q ± , (14253)) gives rise to X ± k, ( p, X ± , ( p, 3) is isomorphicto X ± ( p, ω = (34). q q q − q q q q q q q q q = q q q q q q , (31) q − q − q q q − q − q − q q − q − q q + 1 q q q q q q q q q = − q q + 1 q − q q q − q q q , (32) − q q − q q − − q q q q q q q q q q q = − q q q − . (33)(31)-(2,3) ⇒ q = ± q = 1.(31)-(1,3) ⇒ q = 0;(31)-(1,2) ⇒ q = q ;(33)-(1,1),(1,2) ⇒ q = − ⇒ q = q = 1;(33)-(3,3) ⇒ − 1, a contradiction;if q = − ⇒ q = − q ;(31)-(3,3) ⇒ q = 0;(31)-(3,2) ⇒ q = − q ;(33)-(3,3) ⇒ q q = 1;(32)-(3,3) ⇒ q = 1;(32)-(3,1) ⇒ q = q − ⇒ q = − q ;(32)-(1,3) ⇒ ω = (2453). q q q q q q q q q = q q q q q q q q q , (34) − q q − q − q − q q q q q q q q q q q = q q q q − q q q q q , (35) − − q q q q q q q q q = q − q − q q − q − q q − q − q . (36)(35)-(1,3) ⇒ ( q + 1)( q − q ) = 0;(35)-(2,1) ⇒ q − q q − q q ;(34) ⇒ q = q , q = q , q = q , hence q = 1 / 2, ( q +1)( q − / 2) = 0.But (36)-(1,1) ⇒ q = − q = − / 2. So there is no solution. i = 0By Remark 3.21 and the previous results, the solutions are as follows:(2; k, k − , k − , k − , k − , k − Q , id), giving rise to X k, (2 , k, k − , k − , k − , k − , k − Q ′ , id) , k > 1, giving rise to X ′ k, (2 , k, k − , k − , k − , k − Q , id) , k > 1, giving rise to X k, (2 , k, k − , k − , k − Q , (14253)) , k > 1, giving rise to X k, (5 , p ; k, c, c, c ; Q ± , (14253)) , k > , p ≡ ± X ± k,c, ( p, References [1] H.M. Chen, Lifting automorphisms of abelian regular coverings of graphs,arXiv:1110.5038v3.[2] S.F. Du, J.H. Kwak, M.Y. Xu, Linear criteria for lifting automorphisms ofelementary abelian regular coverings, Linear Algebra and its Applications373 (2003) 101-119. 233] Y.Q. Feng, K.S. Wang, s -Regular cyclic coverings of the three-dimensional cube Q , European Journal of Combinatorics 24 (2003) 719-731.[4] S.F. Du, J.H. Kwak, M.Y. Xu, 2-Arc-transitive regular covers of completegraphs having the covering transformation group Z p , Journal of Combina-torial Theory, Series B 93 (2005) 73-93.[5] J.L. Gross, T.W. Tucker, Generating all graph coverings by permutationvoltage assignments, Discrete Math. 18 (1977) 273-283.[6] J.L. Gross, T.W. Tucker, Topological Graph Theory, Wiley-Interscience,New York, 1987.[7] S. Hong, J.H. Kwak, J. Lee, Regular graph coverings whose coveringtransformation groups have the isomorphism extension property, DiscreteMathematics 148 (1996) 85-105.[8] K. Ireland, M. Rosen, A Classical Introduction to Modern Number The-ory, Graduate Texts in Mathematics Vol 84, 2nd edition, Springer-Verlag,New York, 1990.[9] B. Kuzman, Arc-transitive elementary abelian covers of the completegraph K , Linear Algebra and its Applications, 433 (2010) 1909-1921.[10] A. Malniˇc, Group actions, coverings and lifts of automorphisms, DiscreteMath. 182 (1998) 203-218.[11] A. Malniˇc, D. Maruˇ s iˇ cc