On finite dual Cayley graphs

Problem 1

Characterizing finite dual Cayley graphs.

Lemma 2.1

[3, Lemma 2.1] Let Γ = Cay ( G , S ) . Then, the normalizer N Aut Γ ( G ˆ ) = G ˆ : Aut ( G , S ) .

Theorem 2.2

Let Γ = Cay ( G , S ) . Then, the following statements are equivalent:

1. Γ is a dual Cayley graph on G.

2. G ˜ ≤ Aut Γ .

3. τ ∈ Aut Γ .

Proof

Let g , h , x ∈ G .

( 1 ) ⇒ ( 2 ) . Since Γ is a dual Cayley graph on G, G ˇ ≤ Aut Γ , then as g ˜ = g ˇ g ˆ , we obtain G ˜ ≤ G ˇ G ˆ ≤ Aut Γ .

( 2 ) ⇒ ( 3 ) . Since h ˜ ∈ G ˜ ≤ Aut Γ fixes 1 and Γ ( 1 ) = S , h ˜ fixes S setwise. Then, we have the following equivalences.

Therefore, τ ∈ Aut Γ .

( 3 ) ⇒ ( 1 ) . Since x g ˇ = g − 1 x = ( x − 1 g ) − 1 = x τ g ˆ τ , we have g ˇ = τ g ˆ τ ∈ Aut Γ . Hence, G ˇ ≤ Aut Γ , and Γ is a dual Cayley graph on G.□

Lemma 2.3

1. G ˆ ∩ G ˇ ≅ Z ( G ) ; in particular, G ˆ ≠ G ˇ if and only if G is a nonabelian group.

2. G ˆ τ = G ˇ and G ˇ τ = G ˆ .

3. 〈 Aut ( G ) , τ 〉 = Aut ( G ) if G is abelian ; Aut ( G ) × 〈 τ 〉 if G is nonabelian .

Proof

1. If g ˆ = h ˇ for some g , h ∈ G , then x g = x g ˆ = x h ˇ = h − 1 x for each x ∈ G , and by choosing x = 1 we obtain h − 1 = g . So x g = g x , namely, g ∈ Z ( G ) . Consequently, G ˆ ∩ G ˇ = { g ˆ | g ∈ Z ( G ) } ≅ Z ( G ) .

2. Obviously, o ( τ ) = 1 if G is an elementary abelian 2-group, and o ( τ ) = 2 otherwise. For each g ∈ G , as discussed above, g ˆ τ = τ g ˆ τ = g ˇ , thus G ˆ τ = G ˇ . One similarly shows G ˇ τ = G ˆ .

3. Let σ ∈ Aut ( G ) . For each x ∈ G , we have x σ τ = ( σ ( x ) ) − 1 = σ ( x − 1 ) = x τ σ , hence σ τ = τ σ . Now note that o ( τ ) ≤ 2 , and τ ∈ Aut ( G ) if and only if G is abelian, part (iii) follows.□

Lemma 2.4

Let Γ = Cay ( G , S ) be a dual Cayley graph on a group G. Then,

or equivalently, S = S − 1 is a union of some G-conjugate classes.

Conversely, each Cayley graph constructed as above is a dual Cayley graph.

Proof

Let Γ = Cay ( G , S ) be a dual Cayley graph. Then, 〈 G ˜ , τ 〉 ≤ Aut Γ by Theorem 2.2 and fixes 1, hence it preserves S = Γ ( 1 ) setwise. Noting that each orbit of G ˜ × 〈 τ 〉 on S has the form { s , s − 1 } G with s ∈ G \ { 1 } , we obtain S = s 1 , s 1 − 1 G ∪ ⋯ ∪ s m , s m − 1 G for some s 1 , … , s m ∈ G \ { 1 } .

Conversely, assume Γ = Cay ( G , S ) with S = s 1 , s 1 − 1 G ∪ ⋯ ∪ s m , s m − 1 G . Since G ˜ ≤ Aut ( G ) fixes S setwise, we have G ˜ ≤ Aut ( G , S ) ≤ Aut Γ , hence Γ is a dual Cayley graph by Theorem 2.2.□

Lemma 2.5

1. The complete graphs K n with n ≥ 6 not a prime are nontrivial dual Cayley graphs.

2. The complete bipartite graphs K n , n with n ≥ 3 are nontrivial dual Cayley graphs.

3. The graphs K p , p − p K 2 with p an odd prime are not nontrivial dual Cayley graphs.

Proof

Write Γ = K n , Σ = K n , n and Δ = K p , p − p K 2 .

1. Let G be a nonabelian group of order n. Then, Γ ≅ Cay ( G , G \ { 1 } ) . Since G \ { 1 } = ( G \ { 1 } ) − 1 is a union of some G-conjugate classes, by Lemma 2.4, K n is a nontrivial dual Cayley graph on G.

2. Let H = 〈 a , b | a n = b 2 = 1 , a b = a − 1 〉 ≅ D 2 n be a dihedral group of order 2 n . It is easy to show that Σ ≅ Cay ( H , { b , a b , … , a n − 1 b } ) . Since b a i = a − i ( b a i b ) b = a − 2 i b , and ( a b ) a i = a b a i = a − 2 i + 1 b , we conclude that { b , a b , … , a n − 1 b } = b G ∪ ( a b ) G . It then follows from Lemma 2.4 that K n , n is a nontrivial dual Cayley graph on H.

3. Suppose for a contrary that Δ = Cay ( R , S ) is a dual Cayley graph on a nonabelian group R. Since Δ is of valency p − 1 and order 2 p , we have | S | = p − 1 and | R | = 2 p . Since R is nonabelian of order twice a prime, it is easy to show R ≅ 〈 x , y | x p = y 2 = 1 , x y = x − 1 〉 ≅ D 2 p . If some x i y ∈ S , by Lemma 2.4, ( x i y ) R ⊆ S ; however, as p is an odd prime, a simple computation shows ( x i y ) R = { x , x y , … , x p − 1 y } , contradicting | S | = p − 1 . Hence, S ⊆ 〈 x 〉 . It follows that 〈 S 〉 ≤ 〈 x 〉 < R , and so Δ is disconnected, also a contradiction.□

Lemma 2.6

Let Γ = Cay ( G , S ) be a dual Cayley graph on a nonabelian group G. Then, the following statements hold.

1. If Γ is X-vertex-transitive and X-edge-transitive, then Γ is 〈 X , τ 〉 -arc-transitive. In particular, each edge-transitive dual Cayley graph is arc-transitive.

2. 〈 G ˆ , G ˇ , τ 〉 = ( G ˆ ∘ g ˇ ) : 〈 τ 〉 ≤ Aut Γ .

Proof

1. Since Γ = Cay ( G , S ) is a dual Cayley graph, by Theorem 2.2, τ ∈ Aut Γ , and as τ maps s to s − 1 for s ∈ S , we conclude that Γ is 〈 X , τ 〉 -arc-transitive.

1. By Theorem 2.2, 〈 G ˆ , G ˇ , τ 〉 ≤ Aut Γ . Since G ˇ and G ˆ centralize each other, we have 〈 G ˆ , G ˇ 〉 = G ˆ G ˇ = G ˆ ∘ G ˇ . For x , g , h ∈ G , since x τ − 1 g ˆ h ˇ τ = ( h − 1 x − 1 g ) τ = g − 1 x h = x g ˇ h ˆ , we have ( g ˆ h ˇ ) τ = g ˇ h ˆ = h ˆ g ˇ ∈ G ˆ G ˇ , namely, τ normalizes G ˆ G ˇ . Thus to finish the proof, we only need to prove τ ∉ G ˆ G ˇ . If not, then τ = g ˆ h ˇ for some g , h ∈ G . Hence for each x ∈ G , we have x − 1 = x τ = x g ˆ h ˇ = h − 1 x g . Choosing x = 1 , we obtain g = h and x τ = g − 1 x g , namely, τ is an inner automorphism of G induced by g, which is a contradiction as G is a nonabelian group.

From Lemma 2.6, we have the next corollary. Recall that a Cayley graph Γ on a group G is called a normal Cayley graph if G ˆ ⊲ Aut Γ , see [6].

Corollary 2.7

The dual Cayley graphs on nonabelian groups G are never normal Cayley graphs on G.

Proof

Suppose for a contradiction that Γ is a dual Cayley graph and is normal on a nonabelian group G. Then, G ˆ ⊲ Aut Γ and by Theorem 2.2, τ ∈ Aut Γ ; however, as G is nonabelian, by Lemma 2.3, G ˆ τ = G ˇ ≠ G ˆ , yielding a contradiction.□

Problem 2

1. Which nonabelian groups (especially nonabelian simple groups) admit DGRSAs?

2. Which nonabelian groups (especially nonabelian simple groups) admit arc-transitive DGRSAs?

Lemma 2.8

Let Γ = Cay ( G , S ) be a DGRSA of a nonabelian group G, and suppose Γ is arc-transitive. Then, S = { g , g − 1 } G for some g ∈ G \ { 1 } .

Proof

By assumption, Aut Γ = ( G ˇ ∘ G ˆ ) : 〈 τ 〉 , hence ( Aut Γ ) 1 = G ˜ × 〈 τ 〉 by Lemma 2.3. By the arc-transitivity, the vertex stabilizer ( Aut Γ ) 1 is transitive on Γ ( 1 ) = S , thus S = g ( Aut Γ ) 1 = g G ˜ × 〈 τ 〉 = g , g − 1 G for some g ∈ G \ { 1 } .□

Lemma 2.9

Let Γ be a connected arc-transitive DGRSA of a dihedral group G ≅ D 2 n . Then, n = 3 and Γ ≅ K 3 , 3 .

Proof

Let G = a , b | a n = b 2 = 1 , a b = a − 1 and let Γ = Cay ( G , S ) . By Lemma 2.8, S = { g , g − 1 } G for some g ∈ G , and by the connectivity, 〈 g G 〉 = G . Set g = a i b j with 0 ≤ i ≤ n − 1 and 0 ≤ j ≤ 1 . If j = 0 , then g ∈ 〈 a 〉 , and as 〈 a 〉 is a characteristic subgroup of G, we deduce 〈 S 〉 ≤ 〈 a 〉 < G , a contradiction. Therefore, g = a i b is of order 2 and S = g G .

Assume n is even. Since b a = a − 1 a b b = a − 2 b , we easily deduce S = { b , a 2 b , … , a n − 2 b } if g is conjugate to b in G, or S = { a b , a 3 b , … , a n − 1 b } if g is not conjugate to b in G. For the former case, 〈 S 〉 = 〈 a 2 , b 〉 = 〈 a 2 〉 : 〈 b 〉 ≅ D n , a contradiction. For the latter case, as ( a 2 ) a b = ( a 2 ) b = a − 2 ∈ 〈 a 2 〉 , one easily obtains 〈 S 〉 = 〈 a b , a 2 〉 = 〈 a 2 〉 : 〈 a b 〉 ≅ D n , also a contradiction.

Therefore, n is odd. Then, it is easy to see that each involution of G is conjugate to b, and a direct computation shows that S = b G = { b , a b , … , a n − 1 b } is of size n. Note that Γ is a bipartite graph with bipartitions 〈 a 〉 and S, and we further conclude that Γ ≅ K n , n and Aut Γ = ( S n × S n ) : Z 2 . On the other hand, since Γ is a DGRSA of D 2 n , we have Aut Γ = ( G ˆ ∘ G ˇ ) : 〈 τ 〉 ≅ D ˆ 2 n × D ˇ 2 n : 〈 τ 〉 as n is odd. It follows ( n ! ) 2 ⋅ 2 = ( 2 n ) 2 ⋅ 2 , which implies that n = 3 and Γ ≅ K 3 , 3 is an arc-transitive DGRSA of D 6 .□

Theorem 3.1

Let Γ be an arc-transitive dual Cayley graph on a nonabelian simple group T. Then, Γ is connected and s-transitive with s = 1 or 2. Furthermore, either

1. s = 2 and Γ is a complete graph; or

2. s = 1 , and there is a group H with Inn ( T ) ≤ H ≤ Aut ( T ) and an element g ∈ T \ { 1 } , such that Aut Γ = ( T ˆ : H ) : 〈 τ 〉 and

Proof

Suppose Γ = Cay ( T , S ) and set A = Aut Γ . Since T is nonabelian simple, T ˇ ∘ T ˆ = T ˇ × T ˆ , and by Lemma 2.6, A ≥ ( T ˇ × T ˆ ) : 〈 τ 〉 . It follows that A 1 ≥ T ˜ and g T ⊆ S for g ∈ S . Noting that 〈 g T 〉 is a normal subgroup of T, we conclude that 〈 S 〉 = T , so Γ is connected.

If Γ is a complete graph, then Γ is 2-transitive, as in part (i).

Now suppose Γ is not a complete graph. Then, Aut Γ is not the alternating or symmetric group on V Γ . Note that ( T ˇ × T ˆ ) : 〈 τ 〉 ≅ T ≀ Z 2 is a primitive group of simple diagonal type on V Γ = T , by [10, Proposition 8.1], its overgroup A is also a primitive group of simple diagonal type with unique minimal normal subgroup T 2 . Hence, A = ( T ˇ × T ˆ ) . ( o × 〈 τ 〉 ) with o ≤ Out ( T ) . Consequently, A 1 = T ˜ . o × 〈 τ 〉 = ( Inn ( T ) . o ) × 〈 τ 〉 because 〈 Aut ( T ) , τ 〉 = Aut ( T ) × 〈 τ 〉 by Lemma 2.3. Set H = Inn ( T ) . o . By the arc-transitivity of Γ , A 1 = H × 〈 τ 〉 is transitive on Γ ( 1 ) = S , hence S = { g , g − 1 } H for some g ∈ T \ { 1 } . Furthermore, if Γ is 2-arc-transitive, as A is primitive on V Γ , by [12, Theorem 2], A should be an affine group, an almost simple group, or is of twisted wreath product type or of product action type, which is a contradiction as A is of simple diagonal type. Therefore, Γ is 1-transitive.□

Theorem 3.1 Particularly says that dual Cayley graphs on nonabelian simple groups are at most 2-transitive, and Lemma 2.9 shows that there is a 3-transitive nontrivial dual Cayley graph on soluble group. We have not found examples with higher transitivity on insoluble and nonabelian soluble groups.

Problem 3

1. Are there 3-transitive dual Cayley graphs on insoluble (not nonabelian simple) groups?

2. Are there 4-transitive dual Cayley graphs on nonabelian soluble groups?

We finally answer Problem 2(2) for alternating groups and PSL (2, p ) with p a prime. Denote by C G ( g ) the centralizer of an element g in a group G.

Lemma 3.2

Let g be an element of A n . Then, C S n ( g ) = C A n ( g ) if and only if g may be written as a product of disjoint cycles with different odd lengths.

Proof

(Sufficiency) Suppose g = ( α 1 … α m 1 ) ( β 1 … β m 2 ) … ( γ 1 … γ m s ) is a product of disjoint cycles, where m 1 , … , m s are different odd positive integers. If h ∈ C S n ( g ) , then h − 1 g h = g , and one may deduce

for some positive integers k 1 , … , k m . Note that a cycle of odd length is an even permutation, so is h. Hence, C S n ( g ) = C A n ( g ) .

(Necessity) Assume on the contrary that C S n ( g ) = C A n ( g ) but g is not a product of disjoint odd cycles with different lengths. Write g = ( a 1 ⋯ a n 1 ) ( b 1 ⋯ b n 2 ) ⋯ ( c 1 ⋯ c n t ) , a product of disjoint cycles. Then, either at least one of n 1 , … , n t , say n 1 , is an even integer, or there are at least two equal odd integers in n 1 , … , n t , say n 1 and n 2 . For the former case, ( a 1 ⋯ a n 1 ) is an odd permutation and centralizes h, and for the latter case, ( a 1 b 1 ) ⋯ ( a n 1 b n 1 ) is an odd permutation and centralizes g, hence we always have C S n ( g ) > C A n ( g ) , which is a contradiction.□

Theorem 3.3

Each alternating simple group A n with n ≥ 5 has an arc-transitive DGRSA.

Proof

Let g = ( 123 ) ∈ A n and let

Set T = A n and S = { g , g − 1 } A n . Clearly, 〈 T ˆ , T ˇ 〉 = T ˆ × T ˇ . By Lemma 2.4, Γ is a dual Cayley graph on T. Since n ≥ 5 , and g = ( 123 ) and g − 1 = ( 132 ) are conjugate in S 3 , it is easy to see that g and g − 1 are conjugate in A n ; we further deduce S = g A n and Γ is T ˆ × T ˇ -arc-transitive. Also, since each element in S is of order 3, { g , g − 1 } A n ≠ A n \ { 1 } , so Γ is not a complete graph. It then follows from Theorem 3.1 that Aut Γ = ( A ˆ n : H ) : 〈 τ 〉 where Inn ( T ) ≤ H ≤ Aut ( T ) , hence ( Aut Γ ) 1 = H × 〈 τ 〉 by Lemma 2.3. By the arc-transitivity, S = g H × 〈 τ 〉 = { g , g − 1 } H = g H .

Assume first n ≠ 6 . Then, H ≤ Aut ( A n ) = S n . If H = S n , then g A n = g S n , so A n : C A n ( g ) = g A n = g S n = S n : C S n , implying C S n ( g ) = 2 C A n ( g ) ; however, as g = ( 123 ) is a 3-cycle, Lemma 3.2 implies C S n ( g ) = C A n ( g ) , yielding a contradiction. Hence, H = Inn ( T ) = T ˜ and Aut ( Γ ) = ( T ˆ : T ˜ ) : 〈 τ 〉 = ( T ˇ × T ˆ ) : 〈 τ 〉 , namely, Γ is a DGRSA of A n .

Assume now n = 6 . Then, Inn ( A 6 ) ≤ H ≤ Aut ( A 6 ) ≅ A 6 : Z 2 2 , by Atlas [13, Page 4], we have C H ( g ) = C A 6 ( g ) ≅ Z 3 2 , so | H | / 9 = g H = g A 6 = A 6 / 9 . It follows H = Inn ( A 6 ) = A ˜ 6 and Aut ( Γ ) = ( A ˇ 6 × A ˆ 6 ) : 〈 τ 〉 , hence Γ is a DGRSA of A 6 .□

Lemma 3.4

Let T = PSL (2, p ) with p ≥ 5 a prime, and let g ∈ T . Then, the following statements hold.

1. Elements of order p of T form two conjugate classes of T and are conjugate in Aut ( T ) .

2. Either o ( g ) = p or o ( g ) divides ( p − 1 ) / 2 or ( p + 1 ) / 2 .

3. If o ( g ) ≠ p , then g and g − 1 are conjugate in T.

4. If o ( g ) = p , then C T ( g ) = C Aut ( T ) ( g ) ≅ Z p , and g is conjugate to g − 1 in T if and only if p ≡ 1 ( mod 4 ) .

Theorem 3.5

Let T = PSL ( 2 , p ) with p ≥ 5 a prime. Then, T has an arc-transitive DGRSA if and only if p ≡ 1 ( mod 4 ) .

Proof

(Sufficiency) Let

By Lemma 2.4, Γ is a dual Cayley graph on T. Set S = { g , g − 1 } T . Since elements in S have the same order, S ≠ T \ { 1 } , and hence Γ is not a complete graph.

Since p ≡ 1 ( mod 4 ) , by Lemma 3.4(d), g is conjugate to g − 1 in T, so S = g T and Γ is T ˆ × T ˇ -arc-transitive. Since Γ is not a complete graph, by Theorem 3.1, Aut Γ = ( T ˆ : H ) : 〈 τ 〉 , where T ˜ ≤ H ≤ Aut ( T ) ≅ PGL (2, p ) , hence ( Aut Γ ) 1 = H × 〈 τ 〉 by Lemma 2.3(iii). By the arc-transitivity, S = g H × 〈 τ 〉 = { g , g − 1 } H = g H . If H = Aut ( T ) , then T : C T ( g ) = g T = S = g Aut ( T ) = Aut ( T ) : C Aut ( T ) ( g ) , it follows C Aut ( T ) ( g ) = 2 C T ( g ) , contradicting to Lemma 3.4(d). Hence H = T ˜ , Aut Γ = ( T ˆ : T ˇ ) : 〈 τ 〉 and Γ is an arc-transitive DGRSA of T.

(Necessity) Suppose for a contradiction that p ≡ 3 ( mod 4 ) and there exists an arc-transitive DGRSA Σ of T. By Lemma 2.8, we have

and by Lemma 3.4(b), either o ( h ) = p or o ( h ) divides ( p − 1 ) / 2 or ( p + 1 ) / 2 . Write S ′ = { h , h − 1 } T .

Assume first o ( h ) = p . Since p ≡ 3 ( mod 4 ) , by Lemma 3.4(d), S ′ consists of two conjugate classes of elements of order p of T, then by Lemma 3.4(a), we conclude S ′ = h Aut ( T ) . It then follows from Lemma 2.1 that Aut ( T ) = Aut ( T , S ′ ) ≤ Aut Σ . Hence, Aut Σ ≥ ( T ˆ : Aut ( T ) ) : 〈 τ 〉 , a contradiction.

Assume now o ( h ) divides ( p − 1 ) / 2 or ( p + 1 ) / 2 . Then, S ′ = h T by Lemma 3.4(c). Moreover, by [14, p. 417] and [15, Theorem 2], we have: if o ( h ) p − 1 2 , then o ( h ) is odd, and C Aut ( T ) ( h ) ≅ Z p − 1 ; if o ( h ) = 2 , then C T ( h ) ≅ D p + 1 and C Aut ( T ) ( h ) ≅ D 2 ( p + 1 ) ; if o ( h ) p + 1 2 with o ( h ) ≠ 2 , then C T ( h ) ≅ Z p + 1 2 and C Aut ( T ) ( h ) ≅ Z p + 1 . Thus, we always have h T = T : C T ( h ) = Aut ( T ) : C Aut ( T ) ( h ) = h Aut ( T ) , hence S ′ = h Aut ( T ) . Now, Lemma 2.1 implies Aut ( T ) = Aut ( T , S ′ ) ≤ Aut Σ , and hence Aut Σ ≥ ( T ˆ : Aut ( T ) ) : 〈 τ 〉 , also a contradiction.□