Enumerating Cayley (di-)graphs on dihedral groups
aa r X i v : . [ m a t h . C O ] D ec Enumerating Cayley (di-)graphs on dihedral groups ∗ Xueyi Huang, Qiongxiang Huang † College of Mathematics and Systems Science, Xinjiang University, Urumqi, Xinjiang 830046, P. R. China
Abstract
Let p be an odd prime, and D p = h τ, σ | τ p = σ = e, στ σ = τ − i thedihedral group of order 2 p . In this paper, we provide the number of (connected)Cayley (di-)graphs on D p up to isomorphism by using the P´olya enumeration the-orem. In the process, we also enumerate (connected) Cayley digraphs on D p ofout-degree k up to isomorphism for each k . Keywords:
Cayley (di-)graph; Dihedral group; Cayley isomorphism; P´olyaenumeration theorem.
AMS Classification:
Let G be a finite group, and let S be a subset of G such that 1 S . The Cayleydigraph on G with respect to S , denoted by Cay( G, S ), is the digraph with vertexset G and with an arc from g to h if hg − ∈ S . If S is symmetric, i.e., S − = { s − | s ∈ S } = S , then hg − ∈ S if and only if gh − ∈ S , and so Cay( G, S ) can be viewedas an undirected graph, which is called the
Cayley graph on G with respect to S .In particular, if G is a cyclic group, then the Cayley (di-)graph Cay( G, S ) is calleda circulant (di-)graph .Let Cay(
G, S ) be the Cayley digraph on G with respect to S . Suppose that σ ∈ Aut( G ), where Aut( G ) is the automorphism group of G . Let T = α ( S ). Thenit is easily shown that α induces an isomorphism from Cay( G, S ) to Cay(
G, T ).Such an isomorphism is called a
Cayley isomorphism . However, it is possible fortwo Cayley digraphs Cay(
G, S ) and Cay(
G, T ) to be isomorphic but no Cayleyisomorphisms mapping S to T . The Cayley digraph (resp. Cayley graph) Cay( G, S )is called a
DCI-graph (resp.
CI-graph ) of G if, for any Cayley digraph (resp. Cayleygraph) Cay( G, T ), whenever Cay(
G, S ) ∼ = Cay( G, T ) we have α ( S ) = T for some α ∈ Aut( G ). A group G is called a DCI-group (resp.
CI-group ) if all Cayleydigraphs (Cayley graphs) on G are DCI-graphs (CI-graphs). Clearly, a DCI-groupis always a CI-group, but the converse is not always right. The investigation of CI-graphs (DCI-graphs) or CI-groups (DCI-groups) stems from a conjecture proposed ∗ Supported by National Natural Science Foundation of China (Nos. 11671344 and 11531011). † Corresponding author.
E-mail: [email protected] (X. Huang), [email protected] (Q. Huang).
G, S ) is a CI-graph (DCI-graph), then to decide whether or not Cay(
G, S ) isisomorphic to Cay(
G, T ), we only need to decide whether or not there exists anautomorphism α ∈ Aut( G ) such that α ( S ) = T . For this reason, Mishna [19] (seealso [2]) applied the P´olya enumeration theorem to count the isomorphic classesof Cayley graphs (Cayley digraphs) on some CI-groups (DCI-groups). Also, theisomorphic classes of some families of Cayley graphs which are edge-transitive butnot arc-transitive were determined in [18,25,26]. For more results about CI-problem(DCI-problem) and determination for isomorphic classes of Cayley graphs, we referthe reader to the review paper [16] and references therein.Let p be an odd prime, and D p = h τ, σ | τ p = σ = e, στ σ = τ − i the dihedralgroup of order 2 p . In [3], Babai showed that D p is a DCI-group. This remind us theP´olya enumeration theorem could be used to enumerate the isomorphic classes ofCayley (di-)graphs of D p . In this paper, inspired by Mishna’s work [19], we obtainthe number of (connected) Cayley (di-)graphs on D p up to isomorphism. In theprocess, we also enumerate (connected) Cayley digraphs on D p of out-degree k upto isomorphism for each k . Let G be a finite group and X a finite set. A ( left ) action of G on X , denoted by( G, X ), is a map from G × X to X , with the image of ( g, x ) being denoted by gx ,which satisfies the following two conditions:(1) ex = x for all x ∈ X , where e is the identity of G ;(2) ( gh ) x = g ( hx ) for all g, h ∈ G and all x ∈ X .If there is an action of G on X , then we say that G acts on X or that X is a G -set .The action ( G, X ) is called faithful if the only element g ∈ G satisfying gx = x forevery x ∈ X is the identity element. Let X be a G -set. For each x ∈ X , we definethe orbit of x to be the subset Gx = { gx | g ∈ G } of X , and we define the stabilizer of x to be the subset G x = { g ∈ G | gx = x } of G . It is easy to see that all theorbits form a partition of X and that G x is a subgroup of G .Let G be a permutation group on X ( | X | = n ). Then G acts on X naturallyby defining gx = g ( x ). For g ∈ G , we denote by b k ( g ) the number of cycles oflength k in the disjoint cycle decomposition of g , where k = 1 , , . . . , n . Thenthe cycle type of g ∈ G is defined as type( g ) = ( b ( g ) , b ( g ) , . . . , b n ( g )). Clearly, b ( g ) + 2 b ( g ) + · · · + nb n ( g ) = n . The cycle index I ( G, X ) of the permutation group G acting on X is defined to be the polynomial I ( G, X ) = P G ( x , x , . . . , x n ) = 1 | G | X g ∈ G x b ( g )1 x b ( g )2 · · · x b n ( g ) n , (1)where x , x , . . . , x n are indeterminates.Let A and C be finite sets. Denote by C A = { f | f : A → C } (2)the set of all maps from A to C . Let G be a permutation group acting on A . Thenwe obtain a group action ( G, C A ) naturally by setting: gf = f ◦ g − for every g ∈ G and f ∈ C A , (3)where f ◦ g − denotes the composite of the maps f and g − . Under the group action( G, C A ), two maps in C A are said to be G -equivalent if they belong to the sameorbit. The P´olya Enumeration Theorem provides the number of orbits of the groupaction ( G, C A ). Lemma 2.1. (P´olya Enumeration Theorem, see [10], Chapter 2.) Let A and C befinite sets with | A | = n and | C | = m . Let G be a permutation group acting on A .Denote by F the set of all orbits of the group action ( G, C A ) . Then |F | = P G ( m, m, . . . , m ) , (4) where P G ( x , x , . . . , x n ) is the cycle index of ( G, A ) defined in (1). Let G be a permutation group acting on A . Two k -subsets S and T of A are saidto be G -equivalent if there exists some g ∈ G such that g ( S ) = T . The followingresult enumerates the G -equivalent k -subsets of A . Lemma 2.2. (See [10], Chapter 2.) Let A be a finite set with | A | = n , and let G be apermutation group acting on A . Then the number of G -equivalent classes of k -subsetsof A is equal to the coefficient of x k in the polynomial P G (1 + x, x , . . . , x n ) ,where P G ( x , x , . . . , x n ) is the cycle index of ( G, A ) defined in (1). Recall that a group G is a DCI-group if all Cayley digraphs on G are DCI-graphs.The following result due to Babai [3] shows that the dihedral group D p ( p prime)is a DCI-group. Lemma 2.3. (See [3].) The two Cayley digraphs
Cay( D p , S ) and Cay( D p , T ) on D p ( p prime) are isomorphic if and only if there exists some α ∈ Aut( D p ) suchthat α ( S ) = T . In [23], Rotmaler determined the automorphism group of the dihedral group D n . Lemma 2.4. (See [23].) Suppose n > is an integer and D n = h τ, σ | τ n = σ = e, στ σ = τ − i = { τ i , τ j σ | i, j ∈ Z n } is the dihedral group of order n .Then Aut( D n ) = { α s,t | s ∈ Z × n , t ∈ Z n } ∼ = Z × n ⋉ Z n , where α s,t ( τ i ) = τ si and α s,t ( τ j σ ) = τ sj + t σ for all i, j ∈ Z n . D p Let D p = h τ, σ | τ p = σ = e, στ σ = τ − i = { τ i , τ j σ | i, j ∈ Z p } be the dihedralgroup of order 2 p ( p is an odd prime). Take A = D p \{ e } = { τ i , τ j σ | i ∈ Z p \{ } , j ∈ Z p } and C = { , } . Then Aut( D p ) is a permutation group acting on A and C A .For S ⊆ A , let f S denote the characteristic function of S , that is, f S ( a ) = 1 if a ∈ S ,and f S ( a ) = 0 if a ∈ A \ S . Clearly, f S ∈ C A and C A consists of all characteristicfunctions on A . By Lemma 2.3, we know that two Cayley digraphs Cay( D p , S ) andCay( D p , T ) on D p are isomorphic if and only if there exists some α ∈ Aut( D p )such that α ( S ) = T , which is the case if and only if f S , f T ∈ C A are Aut( D p )-equivalent. Thus the number of Cayley digraphs on D p up to isomorphism is equalto the number of orbits of the group action (Aut( D p ) , C A ). Therefore, by Lemma2.1, in order to enumerate Cayley digraphs on D p , we first need to compute thecycle index of the permutation group Aut( D p ) acting A .By Lemma 2.4, we have Aut( D p ) = { α s,t | s ∈ Z × p , t ∈ Z p } , where α s,t ( τ i ) = τ si and α s,t ( τ j σ ) = τ sj + t σ for all i, j ∈ Z p . Putting A = h τ i \ { e } = { τ i | i ∈ Z p \ { }} and A = h τ i σ = { τ j σ | j ∈ Z p } . Then A = A ∪ A , and furthermore, we observethat α s,t ( A ) = A and α s,t ( A ) = A for each α s,t ∈ Aut( D p ).Also, since p is an odd prime, we know that Z × p = { z ∈ Z p | gcd( z, p ) = 1 } = Z p \ { } is a cyclic group of order p − p . Thus we can assume that Z × p = h z i for some integer z ∈ Z × p = Z p \ { } . Thenfor any s ∈ Z × p , there exists some i s ∈ Z p − such that s = z i s . For example, i s = 0when s = 1 and i s = p − when s = −
1. Furthermore, if s ranges over all elementsof Z × p then i s ranges over all elements of Z p − . The following lemma is crucial tothe calculation of cycle index. Lemma 3.1.
Let A = D p \ { e } and α s,t ∈ Aut( D p ) be defined as above. Let z be agenerating element of the cyclic group Z × p . Under the action of Aut( D p ) on A , thecycle type of α s,t is given by type ( α s,t ) = ( b ( α s,t ) , b ( α s,t ) , . . . , b p − ( α s,t )) , where b k ( α ,t ) = p − if k = 1 and t = 0 ,p − if k = 1 and t ∈ Z p \ { } , if k = p and t ∈ Z p \ { } , otherwise , (5) and for each = s = z i s ∈ Z × p (i.e., i s = 0 ) and t ∈ Z p , b k ( α s,t ) = b k ( α s, ) = if k = 1 , · gcd( i s , p − if k = p − i s , p − , otherwise . (6) Proof.
Let A and A be defined as above. Since α s,t ( A ) = A and α s,t ( A ) = A for each α s,t ∈ Aut( D p ), we must have b p +1 ( α s,t ) = · · · = b p − ( α s,t ) = 0. For α s,t ∈ Aut( D p ), we consider the following two situations. Case 1. s = 1;Note that α ,t ( τ i ) = τ i for each i ∈ Z p \ { } , so the permutation α ,t splits A into p − α ,t ( τ j σ ) = τ j + t σ for j ∈ Z p . If t = 0, then α , ( τ j σ ) = τ j σ for each j ∈ Z p , and so α ,t splits A into p cycles oflength 1. If t ∈ Z p \ { } , the order of t in Z p is equal to o ( t ) = p gcd( t,p ) = p . Then,for any j ∈ Z p , τ j σ ∈ A is in the cycle ( τ j σ, α ,t ( τ j σ ) , α ,t ( τ j σ ) , . . . , α o ( t ) − ,t ( τ j σ )) =( τ j σ, τ j + t σ, τ j +2 t σ, . . . , τ j +( p − t σ ). Thus the permutation α ,t splits A into exactlyone cycle, which is of length p . Therefore, we have obtain the cycle type of α ,t , asshown in (5). Case 2. s = 1, say s = z i s ( i s = 0).Note that α s,t ( τ i ) = τ si for i ∈ Z p \ { } and α s,t ( τ j σ ) = τ sj + t σ for j ∈ Z p .Firstly, we shall prove that α s,t has the same cycle type as α s, for each t ∈ Z p . As α s,t ( τ i ) = τ si = α s, ( τ i ), α s,t and α s, have the same cycle type in A . Now consider α s,t and α s, acting on A . As s = 1, (1 − s ) is invertible in Z × p , we define a bijectionin A , that is, β : τ j σ τ j +(1 − s ) − t σ for j ∈ Z p . Assume that ( τ a σ, τ a σ, . . . , τ a r σ )is a cycle of α s, , i.e., a ℓ = sa ℓ − for ℓ ∈ Z r . Then we have β ( τ a ℓ σ ) = τ a ℓ +(1 − s ) − t σ ,and α s,t ( β ( τ a ℓ − σ )) = α s,t ( τ a ℓ − +(1 − s ) − t σ ) = τ s ( a ℓ − +(1 − s ) − t )+ t σ = τ a ℓ + s (1 − s ) − t + t σ = τ a ℓ +( s − − s ) − t +(1 − s ) − t + t σ = τ a ℓ +(1 − s ) − t σ = β ( τ a ℓ σ ) for each ℓ ∈ Z r . Thus ( β ( τ a σ ) ,β ( τ a σ ) , . . . , β ( τ a r σ )) is a cycle of α s,t . Therefore, α s,t and α s, also have the same cy-cle type in A because β is a bijection. Hence, we just need to consider the cycle typeof α s, in A = A ∪ A . Since the order of s in Z × p is equal to o ( s ) = o ( z i s ) = p − i s ,p − ,for any i, j ∈ Z p \ { } , τ i ∈ A is in the cycle ( τ i , α s, ( τ i ) , α s, ( τ i ) , . . . , α o ( s ) − s, ( τ i )) =( τ i , τ si , τ s i , . . . , τ s o ( s ) − i ) and τ j σ ∈ A is in the cycle ( τ j σ, α s, ( τ j σ ) , α s, ( τ j σ ) , . . . ,α o ( s ) − s, ( τ j σ )) = ( τ j σ, τ sj σ, τ s j σ, . . . , τ s o ( s ) − j σ ). Also note that α s, ( τ σ ) = τ σ ,so τ σ ∈ A is in the cycle ( τ σ ). Thus the permutation α s, splits A into p − o ( s ) = gcd( i s , p −
1) cycles each of length o ( s ) = p − i s ,p − , and splits A into p − o ( s ) = gcd( i s , p −
1) cycles each of length o ( s ) = p − i s ,p − and one cycle of length 1.Then we have obtained the cycle type of α s,t , as shown in (6).We complete the proof.Let n be a positive integer. The Euler’s totient function Φ( n ) is the number ofintegers k in the range 1 ≤ k ≤ n for which the greatest common divisor gcd( n, k )is equal to 1. Let n = p k · · · p k s s be the prime factorization of n . Then the Euler’sproduct formula states that Φ( n ) = n Q ri =1 (1 − p i ). According to Lemma 3.1, wenow give the cycle index of Aut( D p ) acting on A = D p \ { e } . Lemma 3.2.
The cycle index of
Aut( D p ) acting on A = D p \ { e } is given by I (Aut( D p ) , A ) = 1 p x p − ( x p − x p ) + 1 p − x · X d | ( p − Φ( d ) x p − d d , (7) where Φ( · ) denotes the Euler’s totient function. Proof.
By Lemma 3.1, the cycle index of Aut( D p ) acting on A = D p \ { e } is I (Aut( D p ) , A )= 1 | Aut( D p ) | X α s,t ∈ Aut( D p ) x b ( α s,t )1 x b ( α s,t )2 · · · x b p − ( α s,t )2 p − = 1 p ( p − X s ∈ Z × p X t ∈ Z p x b ( α s,t )1 x b ( α s,t )2 · · · x b p − ( α s,t )2 p − = 1 p ( p − (cid:20) X t ∈ Z p x b ( α ,t )1 x b ( α ,t )2 · · · x b p − ( α ,t )2 p − + X s ∈ Z × p \{ } X t ∈ Z p x b ( α s,t )1 x b ( α s,t )2 · · · x b p − ( α s,t )2 p − (cid:21) = 1 p ( p − (cid:20) X t ∈ Z p x b ( α ,t )1 x b ( α ,t )2 · · · x b p − ( α ,t )2 p − + p · X s ∈ Z × p \{ } x b ( α s, )1 x b ( α s, )2 · · · x b p − ( α s, )2 p − (cid:21) = 1 p ( p − (cid:20) x p − + ( p − x p − x p + px · X s = z is ∈ Z × p \{ } x · gcd( i s ,p − p − is,p − (cid:21) = 1 p ( p − (cid:20) x p − + ( p − x p − x p + px · X i s ∈ Z p − \{ } x · gcd( i s ,p − p − is,p − (cid:21) = 1 p ( p − (cid:20) x p − + ( p − x p − x p + px · X d | ( p − d =1 Φ( d ) x p − d d (cid:21) = 1 p ( p − (cid:20) x p − + ( p − x p − x p + px · X d | ( p − Φ( d ) x p − d d − px p − (cid:21) = 1 p x p − ( x p − x p ) + 1 p − x · X d | ( p − Φ( d ) x p − d d , where Φ( · ) denotes the Euler’s totient function.According to Lemmas 3.2, 2.1 and the arguments at the begining of the section,we obtain the number of Cayley digraphs on D p up to isomorphism immediately. Theorem 3.1.
Let p be an odd prime. The number of Cayley digraphs on D p upto isomorphism is equal to N = 1 p p (1 − p − ) + 2 p − · X d | ( p − Φ( d )2 p − d , (8) where Φ( · ) is the Euler’s totient function. In [19], Mishna enumerated the circulant digraphs of order p up to isomorphism. Lemma 3.3. (See [19].) Let p be an odd prime. The number of circulant digraphsof order p up to isomorphism is given by N c = 1 p − X d | ( p − Φ( d )2 p − d , (9) where Φ( · ) is the Euler’s totient function. It is well known that a Cayley digraph Cay(
G, S ) is connected if and only if h S i = G . Thus, for S ⊆ A = D p \ { e } , the Cayley digraph Cay( D p , S ) is disconnected ifand only if S ⊆ A = h τ i \ { e } = { τ i | i ∈ Z p \ { }} or S = { τ j σ } ⊆ A = h τ i σ for j ∈ Z p because p is a prime. Also note that Cay( D p , { τ j σ } ) ∼ = Cay( D p , { σ } ) foreach j since α ,j ( σ ) = τ j σ . Hence, from Theorem 3.1 and Lemma 3.3 we have thefollowing result immediately. Theorem 3.2.
Let p be an odd prime. The number of connected Cayley digraphson D p up to isomorphism is equal to N = N − N c − , (10) where N and N c are presented in (8) and (9), respictively. Theorem 3.3.
Let p be an odd prime and let M k denote the number of Cayleydigraphs on D p of out-degree k up to isomorphism. Then M k = if k = 0 or p − , p (cid:20)(cid:18) p − k (cid:19) − (cid:18) p − k (cid:19)(cid:21) + M k if ≤ k ≤ p − , p (cid:20)(cid:18) p − k − p (cid:19) − (cid:18) p − k (cid:19)(cid:21) + M k if p ≤ k ≤ p − , (11) where M k = 1 p − (cid:20) X d | gcd( p − ,k ) Φ( d ) (cid:18) p − dkd (cid:19) + X d | gcd( p − ,k − Φ( d ) (cid:18) p − dk − d (cid:19)(cid:21) and Φ( · ) is the Euler’s totient function. Proof.
Clearly, if k = 0 (resp. k = 2 p − D p , ∅ ) (resp. Cay( D p , D p \{ e } )) is the unique Cayley digraph on D p with out-degree k . It suffices to consider1 ≤ k ≤ p −
2. By Lemma 3.1, the cycle index of Aut( D p ) acting on A = D p \ { } is I (Aut( D p ) , A ) = 1 p x p − ( x p − x p ) + 1 p − x X d | ( p − Φ( d ) x p − d d . Putting x i = 1 + x i in the above equation, we obtain the polynomial Q ( x ) = 1 p ((1 + x ) p − (1 + x p ) − (1 + x ) p − )) + 1 p − x ) X d | ( p − Φ( d )(1 + x d ) p − d = 1 p (cid:20) p − X i =0 (cid:18) p − i (cid:19) ( x i + x i + p ) − p − X i =0 (cid:18) p − i (cid:19) x i (cid:21) + 1 p − X d | ( p − Φ( d ) p − d X i =0 (cid:18) p − d i (cid:19) ( x di + x di +1 ) . Note that the two Cayley digraphs Cay( D p , S ) and Cay( D p , T ) are isomorphic ifand only if S and T are Aut( D p )-equivalent by Lemma 2.3. Thus the number ofCayley digraphs on D p of out-degree k up to isomorphism is equal to the numberof Aut( D p )-equivalent k -subsets of A , which is also the coefficient of x k in thepolynomial Q ( x ) by Lemma 2.2. If 1 ≤ k ≤ p −
1, the coefficient of x k in thepolynomial Q ( x ) is given by1 p (cid:20)(cid:18) p − k (cid:19) − (cid:18) p − k (cid:19)(cid:21) + 1 p − (cid:20) X d | gcd( p − ,k ) Φ( d ) (cid:18) p − dkd (cid:19) + X d | gcd( p − ,k − Φ( d ) (cid:18) p − dk − d (cid:19)(cid:21) . Similarly, if p ≤ k ≤ p −
2, the coefficient of x k in the polynomial Q ( x ) is equal to1 p (cid:20)(cid:18) p − k − p (cid:19) − (cid:18) p − k (cid:19)(cid:21) + 1 p − (cid:20) X d | gcd( p − ,k ) Φ( d ) (cid:18) p − dkd (cid:19) + X d | gcd( p − ,k − Φ( d ) (cid:18) p − dk − d (cid:19)(cid:21) . We obtain the result as required.
Lemma 3.4. (See [19].) Let p be an odd prime. The number of circulant digraphsof order p with out-degree k up to isomorphism is given by M c,k = 1 p − X d | gcd( p − ,k ) Φ( d ) (cid:18) p − dkd (cid:19) , (12) where Φ( · ) is the Euler’s totient function. Let Cay( D p , S ) be a Cayley graph on D p with | S | = k . If k = 0 or 1, thenCay( D p , S ) is obviously disconnected. Observe that for 2 ≤ k ≤ p −
1, Cay( D p , S )is disconnected if and only if S ⊆ A = h τ i \ { e } = { τ i | i ∈ Z p \ { }} , and for p ≤ k ≤ p −
1, Cay( D p , S ) must be connected. By Theorem 3.3 and Lemma 3.4,we obtain the following result. Theorem 3.4.
Let p be an odd prime. The number of connected Cayley digraphson D p of out-degree k up to isomorphism is M ′ k = if k = 0 , , M k − M c,k if ≤ k ≤ p − , M k if p ≤ k ≤ p − , (13) where M k and M c,k are presented in (11) and (12), respictively. Example 1.
Take p = 3 and consider the dihedral group D = h τ, σ | τ = σ = e i = { τ i , τ j σ | ≤ i, j ≤ } . Let A = D \{ e } . Then it is easy to see that all the rep-resentative elements of Aut( D )-equivalent classes of subsets of A are as follows: ∅ , { τ } , { σ } , { τ, τ } , { τ, σ } , { σ, τ σ } , { τ, σ, τ σ } , { τ, τ , σ } , { σ, τ σ, τ σ } , { τ, σ, τ σ, τ σ } , { τ, τ , σ, τ σ } , { τ, τ , σ, τ σ, τ σ } . Thus there are exactly twelve Cayley digraphs on D up to isomorphism in which eight are connected. By Theorems 3.1 and 3.2, N = 13 · (1 − ) + X d | Φ( d )2 d = 12 , N = N − N c − − X d | Φ( d )2 d − , as required. Also, we see that there are exactly three Cayley digraphs on D without-degree k = 2 up to isomorphism in which two are connected. By Theorems 3.3and 3.4, we obtain M = 13 (cid:20)(cid:18) (cid:19) − (cid:18) (cid:19)(cid:21) + 12 (cid:20) X d | gcd(2 , Φ( d ) (cid:18) d d (cid:19) + X d | gcd(2 , Φ( d ) (cid:18) d d (cid:19)(cid:21) = 3 , M ′ = M − M c, = 3 − X d | gcd(2 , Φ( d ) (cid:18) d d (cid:19) = 2 . Example 2.
Combining Theorems 3.3 and 3.4, we have p − X k =2 M ′ k = N . (14)In Tab. 1, we list the number of connected Cayley digraphs on D p with out-degree k up to isomorphism for each 2 ≤ k ≤ p −
1, where 3 ≤ p ≤
19. One can easilyverify that (14) holds for each p .Tab. 1: The number of connected Cayley digraphs on D p (3 ≤ p ≤ p ( M ′ , . . . , M ′ p − ) N , , ,
1) 85 (2 , , , , , , ,
1) 377 (2 , , , , , , , , , , ,
1) 23311 (2 , , , , , , , , , , , , , , , , , , ,
1) 1936313 (2 , , , , , , , , , , , , , , , , , , , , , , ,
1) 21616717 (2 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,
1) 3159250319 (2 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,
1) 401911341 D p Let D p = h τ, σ | τ p = σ = e, στ σ = τ − i = { τ i , τ j σ | i, j ∈ Z p } be the dihedralgroup of order 2 p ( p is an odd prime). Take A = A ∪ A , where A = n ¯ τ i = { τ i , τ − i } | ≤ i ≤ p − o and A = { τ j σ | j ∈ Z p } . (15)Then Aut( D p ) acts on A naturally by setting α s,t (¯ τ i ) = ¯ τ si and α s,t ( τ j σ ) = τ sj + t σ for each α s,t ∈ Aut( D p ). Furthermore, if α s,t ∈ Aut( D p ) fixes every element of A ,0then α s,t ( τ j σ ) = τ j σ for each j ∈ Z p . Take j = 0, we have α s,t ( τ σ ) = τ t σ = τ σ ,which implies that t = 0. Take some j = j ∈ Z p \{ } , then α s,t ( τ j σ ) = α s, ( τ j σ ) = τ sj σ = τ j σ , which gives that sj = j in Z p , thus we must have s = 1 due to j ∈ Z p \ { } = Z × p is invertible. Therefore, we obtain that α s,t = α , , which is theidentity element of Aut( D p ). This implies that Aut( D p ) acts on A faithfully, andso can be viewed as a permutation group on A . Also observe that α s,t ( A ) = A and α s,t ( A ) = A for each α s,t ∈ Aut( D p ). Let C = { , } . As explained at thebegining of Section 3, we may conclude that the number of Cayley graphs on D p upto isomorphism is equal to the number of orbits of the group action (Aut( D p ) , C A ).Thus the primary task is to calculate the cycle index of Aut( D p ) acting on A . Lemma 4.1.
Let A = A ∪ A be defined as in (15), and let Aut( D p ) act on A as above. Let z be a generating element of the cyclic group Z × p . Under theaction of Aut( D p ) on A , the cycle type of α s,t ∈ Aut( D p ) is given by type ( α s,t ) =( b ( α s,t ) , b ( α s,t ) , . . . , b p − ( α s,t )) , where b k ( α ,t ) = p − if k = 1 and t = 0 ,p − if k = 1 and t ∈ Z p \ { } , if k = p and t ∈ Z p \ { } , otherwise , (16) and for each = s = z i s ∈ Z × p (i.e., i s = 0 ) and t ∈ Z p , b k ( α s,t ) = b k ( α s, ) = if k = 1 , gcd(2 i s , p − if k = p − i s , p − , gcd( i s , p − if k = p − i s , p − , otherwise . (17) (Note that if s = − , i.e., i s = p − , then p − i s ,p − = 1 , and so b ( α − ,t ) = b ( α − , ) = 1 + gcd(2 i s ,p − = p +12 for each t ∈ Z p .) Proof.
Since α s,t ( A ) = A and α s,t ( A ) = A for each α s,t ∈ Aut( D p ), we musthave b p +1 ( α s,t ) = · · · = b p − ( α s,t ) = 0. For α s,t ∈ Aut( D p ), we consider thefollowing two situations. Case 1. s = 1;Note that α ,t (¯ τ i ) = ¯ τ i for 1 ≤ i ≤ p − , so the permutation α ,t splits A into p − cycles each of length 1. Also note that α ,t ( τ j σ ) = τ j + t σ for j ∈ Z p . As in the proofof Lemma 3.1, we may conclude that the permutation α ,t splits A into p cycleseach of length 1 when t = 0 and exactly one cycle of length p when t ∈ Z p \ { } .Therefore, we have obtained the cycle type of α ,t , as shown in (16). Case 2. s = 1, say s = z i s ( i s = 0);Note that α s,t (¯ τ i ) = ¯ τ si for i ∈ Z p \ { } and α s,t ( τ j σ ) = τ sj + t σ for j ∈ Z p .Firstly, we shall prove that α s,t has the same cycle type as α s, for each t ∈ Z p . As1 α s,t (¯ τ i ) = ¯ τ si = α s, (¯ τ i ), α s,t and α s, have the same cycle type in A . Now consider α s,t and α s, acting on A . By the same method as in the proof of Lemma 3.2, onecan deduce that α s,t and α s, also have the same cycle type in A . Hence, we justneed to consider the cycle type of α s, in A = A ∪ A . For any fixed ¯ τ i ∈ A , assumethat ¯ o ( s ) is the minimal positive integer such that α ¯ o ( s ) s, (¯ τ i ) = ¯ τ s ¯ o ( s ) · i = ¯ τ z is ¯ o ( s ) · i = ¯ τ i .Then we have τ z is ¯ o ( s ) · i = τ i and τ − z is ¯ o ( s ) · i = τ − i , or τ z is ¯ o ( s ) · i = τ − i and τ − z is ¯ o ( s ) · i = τ i .For the former, we obtain that z i s ¯ o ( s ) = 1 and so i s ¯ o ( s ) ≡ p − i s ¯ o ( s ) ≡ p − ); for the later, we get z i s ¯ o ( s ) = − z p − and so i s ¯ o ( s ) ≡ p − (mod p − i s ¯ o ( s ) ≡ p − ). By theminimality of ¯ o ( s ), we claim that ¯ o ( s ) = p − gcd( i s , p − ) = p − i s ,p − . Therefore, ¯ τ i ∈ A is in the cycle (¯ τ i , α s, (¯ τ i ) , α s, (¯ τ i ) , . . . , α ¯ o ( s ) − s, (¯ τ i )) = (¯ τ i , ¯ τ si , ¯ τ s i , . . . , ¯ τ s ¯ o ( s ) − i ).Thus the permutation α s, splits A into p − ¯ o ( s ) = gcd(2 i s ,p − cycles each of length¯ o ( s ) = p − i s ,p − . Also note that ¯ o ( s ) = 1 if and only if s = − s = 1.Since the order of s in Z × p is equal to o ( s ) = o ( z i s ) = p − i s ,p − , for any j ∈ Z p \ { } , τ j σ ∈ A is in the cycle ( τ j σ, α s, ( τ j σ ) , α s, ( τ j σ ) , . . . , α o ( s ) − s, ( τ j σ )) =( τ j σ, τ sj σ, τ s j σ, . . . , τ s o ( s ) − j σ ). Also note that α s, ( τ σ ) = τ σ , so τ σ ∈ A is inthe cycle ( τ σ ). Thus the permutation α s, splits A into p − o ( s ) = gcd( i s , p −
1) cycleseach of length o ( s ) = p − i s ,p − and one cycle of length 1. Then we have obtainedthe cycle type of α s,t , as shown in (17).We complete the proof. Lemma 4.2.
Let A = A ∪ A be defined as in (15). The cycle index of Aut( D p ) acting on A is given by I (Aut( D p ) , A ) = 1 p x p − ( x p − x p ) + 1 p − x · p − X i =0 x gcd(2 i,p − p − i,p − x gcd( i,p − p − i,p − . (18) Proof.
By Lemma 4.1, the cycle index of Aut( D p ) acting on A is I (Aut( D p ) , A )= 1 | Aut( D p ) | X α s,t ∈ Aut( D p ) x b ( α s,t )1 x b ( α s,t )2 · · · x b p − ( α s,t ) p − = 1 p ( p − X s ∈ Z × p X t ∈ Z p x b ( α s,t )1 x b ( α s,t )2 · · · x b p − ( α s,t ) p − = 1 p ( p − (cid:20) X t ∈ Z p x b ( α ,t )1 x b ( α ,t )2 · · · x b p − ( α ,t ) p − + X s ∈ Z × p \{ } X t ∈ Z p x b ( α s,t )1 x b ( α s,t )2 · · · x b p − ( α s,t ) p − (cid:21) = 1 p ( p − (cid:20) X t ∈ Z p x b ( α ,t )1 x b ( α ,t )2 · · · x b p − ( α ,t ) p − + p · X s ∈ Z × p \{ } x b ( α s, )1 x b ( α s, )2 · · · x b p − ( α s, ) p − (cid:21) = 1 p ( p − (cid:20) x p − + ( p − x p − x p + px · X s = z is ∈ Z × p \{ } x gcd(2 is,p − p − is,p − x gcd( i s ,p − p − is,p − (cid:21)
2= 1 p ( p − (cid:20) x p − + ( p − x p − x p + px · p − X i s =1 x gcd(2 is,p − p − is,p − x gcd( i s ,p − p − is,p − (cid:21) = 1 p ( p − (cid:20) x p − + ( p − x p − x p + px · p − X i s =0 x gcd(2 is,p − p − is,p − x gcd( i s ,p − p − is,p − − px p − (cid:21) = 1 p x p − ( x p − x p ) + 1 p − x · p − X i =0 x gcd(2 i,p − p − i,p − x gcd( i,p − p − i,p − . According to Lemmas 4.2 and 2.1, we obtain the number of Cayley graphs on D p up to isomorphism immediately. Theorem 4.1.
Let p be an odd prime. The number of Cayley graphs on D p up toisomorphism is equal to N ′ = 1 p (2 p +12 − p − ) + 2 p − p − X i =0 gcd(2 i,p − +gcd( i,p − . (19)In [19], Mishna also enumerated the circulant digraphs of order p up to isomor-phism. Lemma 4.3. (See [19].) Let p be an odd prime. The number of circulant graphs oforder p up to isomorphism is equal to N ′ c = 2 p − X d | p − Φ( d )2 p − d , (20) where Φ( · ) is the Euler’s totient function. As in Section 3, from Theorem 4.1 and Lemma 4.3 we also give the number ofconnected Cayley graphs on D p up to isomorphism. Theorem 4.2.
Let p be an odd prime. The number of connected Cayley graphs on D p up to isomorphism is equal to N ′ = N ′ − N ′ c − , (21) where N ′ and N ′ c are presented in (19) and (20), respictively. Example 3.
As in Example 1, we take p = 3 and consider the dihedral group D . Let A = D \ { e } . Then it is easy to see that all the representative elementsof Aut( D )-equivalent classes of inverse-closed subsets of A are as follows: ∅ , { σ } , { τ, τ } , { σ, τ σ } , { τ, τ , σ } , { σ, τ σ, τ σ } , { τ, τ , σ, τ σ } , { τ, τ , σ, τ σ, τ σ } . Thus thereare exactly eight Cayley graphs on D up to isomorphism in which five are connected.By Theorems 4.1 and 4.2, we have N ′ = 13 (2 − ) + X i =0 gcd(2 i, +gcd( i, = 8 , N ′ = N ′ − N ′ c − − X d | Φ( d )2 d − . D p ( p prime) up to isomorphism for 3 ≤ p ≤
41 by applying Theorem 4.2 (see Tab. 2).Tab. 2:
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