_{1}

An automorphism of a graph X is a permutation σ of vertex set of X with the property that, for any vertices u and v, we have { u σ , v σ } is an edge of X if and only if { u , v } is the edge of X. As usual, we use u σ to denote the image of the vertex u under the permutation σ and { u , v } to denote the edge joining vertices u and v. All automorphisms of graph X form a group under the composite operation of mapping. This group is called the full automorphism group of graph X, denoted by A in this paper.

For a graph X, we denote vertex set and edge set of X by V ( X ) and E ( X ) . A v is the stabilizer of vertex v in the automorphism group of X. X k ( v ) denotes the set of vertices at distance k from vertex v. D 2 n means the dihedral group of order 2n. A graph is called vertex-transitive if its automorphism group A is transitive on the vertex set V ( X ) . An s-arc in a graph is an ordered ( s + 1 ) -tuple ( v 0 , v 1 ,..., v s − 1 , v s ) of vertices of the graph such that v i − 1 is adjacent to v i for 1 ≤ i ≤ s and v i − 1 ≠ v i + 1 for 1 ≤ i ≤ s . A graph is said to be s-arc-transitive if the automorphism group A acts transitively on the set of all s-arcs in X. When s = 1 , 1-arc called arc and 1-arc transitive is called arc-transitive or symmetric.

Throughout this paper, graphs are finite, simple and undirected.

Let G be a finite group and S be a subset of G such that 1 ∉ S . The Cayley graph X = C a y ( G , S ) on G with respect to S is defined to have vertex set V ( X ) = G and edge set E ( X ) = { { g , s g } | g ∈ G and s ∈ S } . Let set S − 1 = { s − 1 | s ∈ S } . If S − 1 = S , C a y ( G , S ) is undirected. If S is a generating system of G, C a y ( G , S ) is connected. Two subsets S and T of group G are called equivalent if there exists a group automorphism of group G mapping S to T: S α = T for some α ∈ A u t ( G ) . Denote by S ≡ T . If S and T are equivalent, Cayley graphs C a y ( G , S ) and C a y ( G , S ) are isomorphic.

The right regular representation R ( G ) of group G is a subgroup of the the automorphism group A of the Cayley graph X. In particular by [

In [

Summarising theorem 4.1, 4.2, 4.3 in Part 4 gives the main results.

Theorem 1.1. Let G = D 2 n p m be a dihedral group where n ≥ 2 and p is an odd prime number. S is an inverse-closed generating system of three elements without identity element. Then Cayley graph C a y ( G , S ) is GRR except the following cases:

1) S ≡ { b , a b , a k b } where k 2 ≡ 1 ( mod 2 n − 1 p m ) and gcd ( k ,2 n − 1 p m ) = 1 , A u t ( X ) ≅ G : ℤ 2 .

2) S ≡ { b , a b , a 2 n − 1 p m b } , A u t ( X ) ≅ ℤ 2 2 n − 2 p m ⋊ D 2 n − 1 p m .

3) S ≡ { a , a − 1 , b } , A u t ( X ) = G : ℤ 2 .

4) S ≡ { b , a b , a 2 n − 2 p m } , A u t ( X ) = G : ℤ 2 .

Results used to prove main theorem are listed here.

Proposition 2.1. Suppose that G = < a , b | a n = b 2 = 1 , b − 1 a b = a − 1 > is a dihedral group, then the automorphism group A u t ( G ) of G has the following properties.

1) Any automorphism of G can be defined as a ↦ a i and b ↦ a j b where i ∈ ℤ n * and j ∈ ℤ n .

2) A u t ( G ) = < α > ⋊ < β > ≅ ℤ n ⋊ ℤ n * where α : a ↦ a , b ↦ a b ; β : a ↦ a i , b ↦ b , i ∈ ℤ n * .

Proposition 2.2. Suppose G is a finite group and subsets S ≡ T , then C a y ( G , S ) ≅ C a y ( G , T ) .

Proposition 2.3. Let G = < a , b | a n = b 2 = 1 , b − 1 a b = a − 1 > be the dihedral group of order 2n. Subsets { b , a b , a k b } ≡ { b , a b , a 1 − k b } .

Proof Let σ ∈ A u t ( G ) : a ↦ a − 1 , b ↦ a b then { b , a b , a k b } σ = { b , a b , a 1 − k b } .

The following sufficient and necessary condition of normality of Cayley graph is from paper [

Proposition 2.4. Let X = C a y ( G , S ) be connected. Then X is a normal Cayley graph of G if and only if the following conditions are satisfied:

1) For each φ ∈ A 1 there exists σ ∈ A u t ( G ) such that φ | X 1 ( 1 ) = σ | X 1 ( 1 ) ;

2) For each φ ∈ A 1 , φ | X 1 ( 1 ) = 1 X 1 ( 1 ) implies φ | X 2 ( 1 ) = 1 X 2 ( 1 ) .

A classification of locally primitive Cayley graphs of dihedral groups from paper [

Proposition 2.5. Let X be a locally-primitive Cayley graph of a dihedral group of order 2n. Then one of the following statements is true, where q is a prime power.

1) X is 2-arc-transitive, and one of the following holds:

a) X = K 2 n , K n , n or K n , n − n K 2 ;

b) X = H D ( 11,5,2 ) or H D ( 11,6,2 ) , the incidence or non-incidence graph of the Hadamard design on 11 points;

c) X = P H ( d , q ) or P H ′ ( d , q ) , the point-hyperplane incidence or non-incidence graph of ( d − 1 ) -dimension projective geometry P G ( d − 1, q ) , where d ≥ 3 ;

d) X = K q + 1 2 d , where d is a divisor of q − 1 2 if q ≡ 1 (mod 4), and a divisor of q − 1 if q ≡ 3 (mod 4) respectively.

2) X = N D 2 n , r , k is a normal Cayley graph and is not 2-arc-transitive, where n = r t p 1 e 1 p 2 e 2 ⋯ p s e s ≥ 13 with r , p 1 , p 2 , ⋯ , p s distinct odd primes, t ≤ 1 , s ≥ 1 and r | ( p i − 1 ) for each i. There are exactly ( r − 1 ) s − 1 non-isomorphism such graphs for a given order 2n.

In the following, group G means that G = < a , b | a 2 n − 1 p m = b 2 = 1 , b − 1 a b = a − 1 > be dihedral group of order 2 n p m where n ≥ 2 and p is an odd prime number.

Proposition 3.1. If S = { a i b , a j b , a r b } is a generating system of G of three elements, then S ≡ { b , a b , a k b } for some 2 ≤ k ≤ 2 n − 1 p m − 1 .

There are two types of S classified by the number of subsets of two elements generating G.

Type 1: S has only one subset of two elements generating G.

Type 2: S has exactly two subsets of two elements generating G. In this case, S ≡ { b , a b , a k b } where gcd ( k ,2 n − 1 p m ) = 1 .

The proof of Proposition 3.1 will be done by the following three lemmas.

Lemma 3.1. If S = { a i b , a j b , a r b } is a generating system of G of three elements, then S is equivalent to a subset of type { b , a b , a k b } for some 2 ≤ k ≤ 2 n − 1 p m − 1 .

Proof By proposition 2.1 in preliminary, automorphism group Aut(G) of dihedral group G is transitive on the set of involutions { a i b | 0 ≤ i ≤ 2 n − 1 p m − 1 } . One may assume that b ∈ S and S = { b , a i b , a j b } be a generating system of G of three elements. S has three subsets of two elements: { b , a i b } , { b , a j b } and { a i b , a j b } ..

Note that, subset T ⊂ G is a generating system of G if and only if T α is a generating system of G for any α ∈ A u t ( G ) .

Suppose that subset { b , a x b } ( x = i or j ) generates G. Let α ∈ A u t ( G ) : a ↦ a x , b ↦ b , then { b , a b , a k b } α = { b , a i b , a j b } for some k ≠ 0,1 . Hence S ≡ { b , a b , a k b } .

Assume that both subset { b , a i b } and { b , a j b } do not generate G. Next will show that { a i b , a j b } must be able to generate G.

G = < S > = < b , a i b , a j b > = < a i , a j > < b > = < a gcd ( i , j ) > < b > . Hence gcd ( i , j ) and 2 n − 1 p m are mutually prime.

G ≠ < b , a i b > = < a i > < b > . Hence i and 2 n − 1 p m are not mutually prime.

Similarly, G ≠ < b , a j b > implies that j and 2 n − 1 p m are also not mutually prime.

( gcd ( i , j ) , 2 n − 1 p m ) = 1 , ( i ,2 n − 1 p m ) ≠ 1 and ( j ,2 n − 1 p m ) ≠ 1 imply that, for i and j , one number is power of 2 and the other one is power of p. Thus i − j and 2 n − 1 p m are mutually prime.

Hence, { a i b , a j b } is a generating system of G since < a i b , a j b > = < a i − j > < a i b > = G .

Let α ∈ A u t ( G ) : a ↦ a i − j , b ↦ a j b . Then { b , a b , a k b } α = { b , a i b , a j b } for some k. S ≡ { b , a b , a k b } .

Corollary 3.1. If S = { a i b , a j b , a r b } is a generating system of G of three elements, there exists at least one subset of two elements generating G.

Lemma 3.2. If S = { a i b , a j b , a r b } is a generating system of G of three elements, there are only one or two subsets of two elements of S generating G.

Proof By Lemma 3.1, we assume that S = { b , a b , a k b } where k ≠ 0,1 . S has three subsets of two elements: { b , a b } , { b , a k b } and { a b , a k b } . Next we will show that it is impossible that all three subsets of two elements generating G.

< b , a k b > = < a k > < b > is a dihedral subgroup of G. < a b , a k b > = < a k − 1 > < a b > is also a dihedral subgroup of G.

For k and k − 1 , one is an even number and the other one is an odd number. The orders of elements a k and a k − 1 are different: ∘ ( a k ) ≠ ∘ ( a k − 1 ) . This implies that at least one subset of { b , a k b } and { a b , a k b } does not generate G.

Hence there are only one or two subsets of two elements of S generating G.

Lemma 3.3. Let S = { a i b , a j b , a r b } be a generating system of G of three elements and S has two subsets of two elements generating G. If S ≡ { b , a b , a k b } , either gcd ( k ,2 n − 1 p m ) = 1 or gcd ( 1 − k ,2 n − 1 p m ) = 1 .

Proposition 3.2. Suppose that S = { a i b , a j b , a r b } is a generating system of G of three elements and S ≡ { b , a b , a k b } .

(1) If S has only one subset of two elements generating G, then A u t ( G , S ) = 1 .

(2) If S has two subsets of two elements generating G, then A u t ( G , S ) = 1 except the following two cases. A u t ( G , S ) ≅ ℤ 2 if k 2 ≡ 1 ( mod 2 n − 1 p m ) and gcd ( k ,2 n − 1 p m ) = 1 ; A u t ( G , S ) ≅ ℤ 2 if ( 1 − k ) 2 ≡ 1 ( mod 2 n − 1 p m ) and gcd ( 1 − k ,2 n − 1 p m ) = 1 .

Proof (1) If there is only one subset of two elements in S = { b , a b , a k b } generating G, then G ≠ < b , a k b > , G ≠ < a b , a k b > and G = < b , a b > . For any σ ∈ A u t ( G , S ) , { b , a b } σ is also a generating system of G. { b , a b } σ = { b , a b } . Since S σ = S . Hence a k b = S − { b , a b } is fixed by σ . ( a k b ) σ = a k b .

If b σ = b and ( a b ) σ = a b then a σ = ( a b b ) σ = ( a b ) σ b σ = a b b = a , hence σ = 1 .

If b σ = a b and ( a b ) σ = b , then a σ = ( a b b ) σ = ( a b ) σ b σ = b a b = a − 1 . This implies that a k b = ( a k b ) σ = ( a k ) σ b σ = a − k a b = a 1 − k b . Thus a k = a 1 − k . This is a contradiction. For k and 1 − k , one is an even number and the other one is an odd number. This implies that the orders of the element a k and a 1 − k are not equal: ∘ ( a k ) ≠ ∘ ( a 1 − k ) .

Hence A u t ( G , S ) = 1 .

(2) If there are two subsets of two elements of S generating G, we assume that gcd ( k , 2 n − 1 p m ) = 1 . G = < b , a b > = < b , a k b > and G ≠ < a b , a k b > .

Since subset { a b , a k b } is the only subset of two elements not generating G, { a b , a k b } σ = { a b , a k b } for any σ ∈ A u t ( G , S ) . b = S − { a b , a k b } is fixed by σ . ( a b ) σ = a b or a k b .

If ( a b ) σ = a b , then σ = 1 .

If ( a b ) σ = a k b , then a σ = ( a b b ) σ = ( a b ) σ b σ = a k b b = a k . ( a k b ) σ = ( a k ) σ b σ = ( a k ) k b = a k 2 b = a b . So k 2 ≡ 1 ( mod 2 n − 1 p m ) .

Hence A u t ( G , S ) = 1 if k 2 ≡ 1 ( mod 2 n − 1 p m ) . A u t ( G , S ) ≅ ℤ 2 if k 2 ≡ 1 ( mod 2 n − 1 p m ) .

Similarly, when gcd ( 1 − k , 2 n − 1 p m ) = 1 , A u t ( G , S ) = 1 if ( 1 − k ) 2 ≡ 1 ( mod 2 n − 1 p m ) . A u t ( G , S ) ≅ ℤ 2 if ( 1 − k ) 2 ≡ 1 ( mod 2 n − 1 p m ) .

Proposition 3.3. Suppose that S is inverse-closed generating system of three elements of G, then S ≡ { a , a − 1 , b } , { b , a b , a 2 n − 2 p m } or { b , a b , a k b } ( k ≠ 0,1 ) .

Proof Since S contains three elements and inverse-closed, there must be an involution in S. There are two orbits of involutions in G under the action of group automorphism A u t ( G ) : { a 2 n − 2 p m } and { a i b | 0 ≤ i ≤ 2 n − 1 p m − 1 } .

Suppose that a 2 n − 2 p m ∈ S . S − { a 2 n − 2 p m } is also inverse-closed hence it is a set of two involutions from orbit { a i b | 0 ≤ i ≤ 2 n − 1 p m − 1 } . S generating G implies that S − { a 2 n − 2 p m } also generates G. We get S ≡ { b , a b , a 2 n − 2 p m } .

Suppose that S contains an involution from { a i b | 0 ≤ i ≤ 2 n − 1 p m − 1 } . A u t ( G ) is transitive on this orbit, we can assume that b ∈ S . If S − { b } contains an involution, S ≡ { b , a b , a k b } ( k ≠ 0,1 ) by Proposition 3.1 and 2.1. If S − { b } contains no involutions, S ≡ { b , a , a − 1 } by Proposition 2.1.

By Proposition 3.3, we only need to discuss X = C a y ( G , S ) for S = { a , a − 1 , b } , { b , a b , a 2 n − 2 p m } and { b , a b , a k b } ( k ≠ 0,1 ) .

Firstly, we discuss X = C a y ( G , { b , a b , a k b } ) ( k ≠ 0,1 ) .

Theorem 4.1. Suppose that S = { a i b , a j b , a m b } is a generating system of three involutions of G and S ≡ { b , a b , a k b } .

X is GRR except the following cases.

(1) When gcd ( k ,2 n − 2 p m ) = 1 , k 2 ≡ 1 ( mod 2 n − 1 p m ) and k ≠ 2 n − 2 p m + 1 then A u t ( X ) ≅ R ( G ) : ℤ 2 .

(2) When gcd ( 1 − k ,2 n − 2 p m ) = 1 , ( 1 − k ) 2 ≡ 1 ( mod 2 n − 1 p m ) and k ≠ 2 n − 2 p m then A u t ( X ) ≅ R ( G ) : ℤ 2 .

(3) If k = 2 n − 2 p m + 1 or k = 2 n − 2 p m , then A u t ( X ) ≅ ℤ 2 2 n − 2 p m ⋊ D 2 n − 1 p m .

Proof Let S = { b , a b , a k b } where 2 ≤ k ≤ 2 n − 1 p m − 1 and X = C a y ( G , S ) . Classify X in two cases: there are 4-cycles in X and there is no 4-cycle in X.

(1) Note that X 2 ( 1 ) = { a , a k , a − 1 , a k − 1 , a − k , a 1 − k } is the set of vertices at distance 2 from vertex 1.

If there are 4-cycles in X, some vertices in X 2 ( 1 ) are coincident. Solving a = a k − 1 and a − 1 = a 1 − k we get k = 2 . Solving a = a − k and a k = a − 1 we get k = − 1 . Solving a k = a − k we get k = 2 n − 2 p m . Solving a k − 1 = a 1 − k we get k = 2 n − 2 p m + 1 . There is no solution for other equations. Note that −1 and 2 n − 2 p m + 1 are two solutions of equation k 2 ≡ 1 ( mod 2 n − 1 p m ) . 2 and 2 n − 2 p m are two solutions of equation ( 1 − k ) 2 ≡ 1 ( mod 2 n − 1 p m ) . Since { b , a b , a 2 b } ≡ { b , a b , a − 1 b } and { b , a b , a 2 n − 2 p m b } ≡ { b , a b , a 2 n − 2 p m + 1 b } we only discuss k = 2 and k = 2 n − 2 p m .

(1.1) When k = 2 , X = C 2 n − 1 p m × K 2 is a cylinder as

(1.2) When k = 2 n − 2 p m , X is a thickened 2-cover of the cycle graph C 2 n − 1 p m as

kernel of the action of A on the imprimitive block system is isomorphic to ℤ 2 2 n − 2 p m . Thus A ≅ Z 2 2 n − 2 p m ⋊ D 2 n − 1 p m .

(2) Suppose that there is no 4-cycle in X. We will count 6-cycles passing through vertex 1.

X 3 ( 1 ) = { a − k b , a − 1 b , a 1 − k b , a 2 − k b , a k − 1 b , a k + 1 b , a 2 b , a 2 k − 1 b , a 2 k b } is the set of

vertices at distance 3 from vertex 1.

a) Solving a 2 k b = a 2 − k b and a 2 k − 1 b = a 1 − k b , we get 3 k ≡ 2 ( mod 2 n − 1 p m ) . Solving a 2 k b = a 1 − k b and a 2 k − 1 b = a − k b , we get 3 k ≡ 1 ( mod 2 n − 1 p m ) . Solving a k − 1 b = a 2 b and a − 1 b = a 2 − k b we get k = 3 . Solving a − k = a 2 and a k + 1 = a − 1 we get k = − 2 . There is no solution for other equations.

The induced subgraph of the set of vertices at distance less than or equal to 3 from vertex 1 in X are isomorphic in these four cases. The following uses C a y ( G , { b , a b , a 3 b } ) as representative to discuss. See

We count the number of 6-cycles passing through vertex 1. There are four 6-cycles through edge { 1, b } . There are five 6-cycles through edge { 1, a b } . There are three 6-cycles through edge { 1, a 3 b } . For any σ ∈ A 1 , A 1 fixes edged { 1, b } , { 1, a b } , { 1, a 3 b } and hence σ fixes vertices set X 1 ( 1 ) = { b , a b , a 3 b } pointwise. σ fixes all vertices on X by the connectivity of X and the transitivity of A on V ( X ) . Hence A 1 = 1 . X is GRR.

b) Suppose that k ≡ 3 , k ≡ − 2 , 3 k ≡ 2 , 3 k ≡ 1 (mod 2 n − 1 p m ). Then the induced subgraph of the set of vertices at distance less than or equal to 3 from vertex 1 in X is the as

Firstly, show that the action of A 1 on X 1 ( 1 ) is faithful.

Let σ ∈ A 1 and σ fixes X 1 ( 1 ) pointwise. Passing through vertices { 1, b , a b } ,

there is a unique 6-cycle [ 1 , b , a k , a 1 − k b , a k − 1 , a b ] ≜ C 1 . Passing through vertices { 1, b , a k b } , there is a unique 6-cycle [ 1 , b , a , a k − 1 b , a 1 − k , a k b ] ≜ C 2 . Passing through vertices { 1, a b , a k b } , there is a unique 6-cycle [ 1 , a b , a − 1 , a k + 1 b , a − k , a k b ] ≜ C 3 . For any α ∈ A , the image of a cycle of length l under α is also a cycle of length l. Note that σ ∈ A 1 fixes { 1, b , a b , a k b } pointwise, hence C 1 σ is also a 6-cycle passing through vertices 1, b , a b . Hence C 1 σ = C 1 . Follow the same argument, C 2 σ = C 2 , C 3 σ = C 3 . So σ fixes all vertices on cycles C 1 , C 2 , C 3 . In particular, σ fixes X 2 ( 1 ) pointwise. By the connectivity of X and the transitivity of A on V ( X ) , we get A 1 acts on X 1 ( 1 ) = S faithfully.

Next, show that X is normal.

A 1 acting on X 1 ( 1 ) faithfully implies that A 1 is isomorphic to a subgroup of symmetric group of degree 3. A 1 ≲ S 3 .

If A 1 ≅ A 3 or S 3 , then A 1 is transitive on X 1 ( 1 ) . Since | X 1 ( 1 ) | = 3 is prime, X is a locally-primitive Cayley graph. Theorem 1.5 in [

Since the order of G is 2 n p m where n ≥ 2 and p is odd, C a y ( G , S ) is not on the list of locally-primitive Cayley graphs. Thus, A 1 is not transitive on X 1 ( 1 ) . A 1 ≅ ℤ 1 or ℤ 2 . | A : R ( G ) | = | A 1 | = 1 or 2, R ( G ) ⊴ A . X is normal. A = R ( G ) ⋊ A u t ( G , S ) .

By Proposition 3.2 and part(1) of this proof, A = R ( G ) : ℤ 2 if k 2 ≡ 1 ( mod 2 n − 1 p m ) , k ≠ 2 n − 2 p m + 1 and gcd ( k ,2 n − 1 p m ) = 1 or ( 1 − k ) 2 ≡ 1 ( mod 2 n − 1 p m ) , k ≠ 2 n − 2 p m and gcd ( 1 − k ,2 n − 1 p m ) = 1 .

Theorem 4.2. Suppose that S ≡ { a , a − 1 , b } , then X is normal and A = G : ℤ 2 .

Proof Suppose that S ≡ { a , a − 1 , b } and X = C a y ( G , S ) . Cayley graph X is also a cylinder as

Theorem 4.3. Suppose that S ≡ { b , a b , a 2 n − 2 p m } , then X is normal and A = G : ℤ 2 .

Proof Suppose that S ≡ { b , a b , a 2 n − 2 p m } and X = C a y ( G , S ) . The Cayley graph is an Möbius ladder as

The author declares no conflicts of interest regarding the publication of this paper.

Kong, X.F. (2020) Automorphism Groups of Cubic Cayley Graphs of Dihedral Groups of Order 2^{n}p^{m} (n ≥ 2 and p Odd Prime). Journal of Applied Mathematics and Physics, 8, 3075-3084. https://doi.org/10.4236/jamp.2020.812226