Smallest graphs with given automorphism group

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Theorem 1 Let G be a finite group of order n. Then one of the following is true:
(i) α(G) ≤ n, (ii) G is cyclic of order p k or 2 p, where p is prime and k is positive integer (n = 2), (iii) G is Q 2 r or Q 2 r × C 2 , where Q 2 r is the generalised quaternion group of order 2 r , r ≥ 3, (iv) G is one of the 17 exceptional groups of order at most 25 shown in Table 1.
As a consequence, we deduce when equality holds in Babai's bound.

Corollary 1.1 Let G be a finite group of order n. Then α(G) = 2n if and only if
(i) G is a generalised quaternion group of order 2 r , r ≥ 3, or (ii) G is cyclic of order p, where p is prime and p ≥ 7, or (iii) G is the abelian group C 3 × C 3 .
The main tool in the proof of Theorem 1 is the GRR-Theorem (Theorem 2.3). It states that, with some specified families of exceptions, every finite group G has a graphical regular representation (GRR), i.e., a Cayley graph having full automorphism group isomorphic to G. If a group G has a GRR, then α(G) ≤ |G|. Therefore, in order to prove Theorem 1, it suffices to study the exceptions in the GRR-Theorem.
Making use of the preliminary results presented in Sect. 2, we prove Theorem 1 across Sects. 3, 4, and 5. Section 3 concerns the case of abelian groups; the key fact is that α(G) has been determined for every abelian group G, in [1]. Sections 4 and 5 are devoted to the non-abelian exceptional groups of the GRR-Theorem. In Sect. 4, we address the non-abelian groups G for which the assertion of Theorem 1 is that α(G) ≤ |G|. For these groups, we construct a graph on at most |G| vertices having automorphism group isomorphic to G. In Sect. 5, we show that there exists no graph on at most |G| vertices with automorphism group isomorphic to G, for the non-abelian groups G for which the assertion of Theorem 1 is that α(G) > |G|. The values of α(Q 2 r ), α(Q 2 r × C 2 ) and α(G) for the non-abelian groups G in Table 1 are also justified in this section. Table 1 The groups G mentioned in Theorem 1,(iv), and the values α(G)

Background
Throughout the paper, all groups and graphs mentioned are assumed to be finite. Let us now present two families of groups that play an important role in our text.
Definition 2.1 [21] Let A be an abelian group that contains an element of order 2k for some k ≥ 2. A group G of the form is called generalised dicyclic and is denoted by Dic (A, b 2 ). If A is cyclic, G is simply called dicyclic. It is denoted by Dic m , where m = |G| 4 . A dicyclic group of order 2 r is called generalised quaternion (r ≥ 3). We denote it by Q 2 r .
Note that the existence of the element of order 2k in Definition 2.1 ensures that generalised dicyclic groups are non-abelian.

Definition 2.2 Let
A be an abelian group. A group G of the form is called generalised dihedral and is denoted by Dih(A).
If A is cyclic of order m, G is the dihedral group of order 2m, which we denote by D 2m .
A graph consists of a vertex set, which we denote by V ( ) and an edge set, denoted by E( ); we consider an edge to be an unordered pair of vertices of . We denote an edge between v, w ∈ V ( ) by v ∼ w or we say that [v, w] ∈ E( ). Moreover, if X is a subgraph of and v ∈ V (X ), we denote by ρ X (v) the valency of v in X and by ρ(v) the valency of v in the graph . If a group G acts on a graph and v ∈ V ( ), then we denote by O v the orbit containing v, O v = {gv | g ∈ G}, and by G v the stabilizer of v, G v = {g ∈ G | gv = v}.
Given a group G and a set S ⊂ G \ {1} that is inverse-closed, we define the Cayley graph Cay(G, S) to be the graph with vertex set G and edges {x, sx}, for all x ∈ G, s ∈ S.
A graph is called a Graphical Regular Representation (GRR) of a group G if there exists some S ⊂ G such that Cay(G, S) = and Aut( ) ∼ = G. The following theorem is known as the GRR-Theorem. It was proven by Godsil [6] for non-solvable groups and by Hetzel [11] for solvable groups using previous results of several authors including [12,[16][17][18][19][20][21]. Theorem 2.3 [6] A group admits a GRR if and only if it is not an abelian group of exponent greater than 2, a generalised dicyclic group, or one of the 13 exceptional groups shown in Table 2.

Corollary 2.4
If G is a non-abelian , non-generalised dicyclic group that is not one of the 13 groups shown in Table 2, then α(G) ≤ |G|.
Let us now state Babai's theorem.
The values of α(G) for cyclic groups G and graph constructions can be found in [1], which builds on the work of Sabidussi [18]. For the non-cyclic groups, we will use the following construction, given by Babai in [2]: Construction 2.6 Let G be a non-cyclic group of order |G| ≥ 6 and let H = {h 1 , . . . , h d } be a minimal generating set of G. Let G be an isomorphic copy of G with an isomorphism g −→ g from G to G . We define the graphs X 1 and X 3 to be such that Let ρ X s be the valency of the vertices of X s , s = 1, 3. We define the graph X 2 to be where X 3 is the complement graph of X 3 . Finally, let us define the graph X such that The map g : is a graph automorphism for every g ∈ G, and Aut(X ) ∼ = G; the proof appears in [2].
The inequality in Babai's Theorem 2.5 does not hold for the three cyclic groups excluded.
Example 2. 7 We will see shortly (Proposition 3.3) that α(C 4 ) = 10. A graph on 10 vertices that has automorphism group isomorphic to C 4 is shown in Fig. 1. In particular, the automorphism group of this graph can be realised as the subgroup b of S 10 , where b = (1 2)(3 4 5 6)(7 8

Proof of Theorem 1: abelian groups
The aim of this section is to prove that Theorem 1 holds for every abelian group G. Proposition 3.1 Let G be an abelian group. Then one of the following holds: G is cyclic of order p k or 2 p for some prime number p (|G| = 2), (iii) G is one of the 10 abelian groups shown in Table 1.

Proof of Theorem 1: the bound˛(G) ≤ |G|
In this section we prove the bound α(G) ≤ |G| for groups G that are non-abelian and do not satisfy (iii), (iv) in Theorem 1, as summarised in the following theorem.

Theorem 4.1 Let G be a non-abelian group such that
(i) G is not a generalised quaternion group, (ii) G is not a generalised dicyclic group of the form Q 2 r × C 2 , (iii) G is not one of the groups shown in Table 1.
By the GRR-Theorem (Theorem 2.3), in order to prove Theorem 4.1, we only need to consider the cases when G is generalised dicyclic and when G is one of the non-abelian groups that appear in Table 2 but not in Table 1. We will do this in Propositions 4.9 and 4.10. In particular, in Proposition 4.9, we consider the groups D 6 , D 8 , D 10 , and the remaining groups are addressed in Proposition 4.10. Let us start with two lemmas that will be used in the proof of Proposition 4.10.
Then there exists a GRR for G.
Proof By the GRR-Theorem, it suffices to prove that G is non-generalised dicyclic and not one of the groups appearing in Table 2.
Then the order of c is 4 and the order of b is 2, hence c ∈ X , b ∈ A. It follows from the properties of generalised dicyclic and generalised dihedral groups that The equalities given above imply that c 2 = 1, which is a contradiction. The restriction k = 9, implies that G is not the group of order 18 in Table 2. 10 or the group of order 27 in Table 2. On the other hand, the group A 4 has no abelian subgroup of index 2, hence it is not generalised dihedral. The remaining 4 suitable groups given in Table 2 contain a central element of order 3 or 4. However, if g is in the center of G and G is non-abelian then g ∈ X , hence bgb = g −1 . Furthermore, bgb = g, as g is central. Therefore, g has order 2.
The following lemma can be proven by elementary group theory arguments.

Lemma 4.3 Let G be an abelian 2-group and let c ∈ G be an element of order 2. Then there exists some y
Let us now define a collection of graphs, one for each group appearing in (7).

Construction 4.6 Let us also construct the graph 3 on 27 vertices and 171 edges.
Proof The n-cycle has full automorphism group D 2n .
The graphs i in Constructions 4.4−4.8 are designed to have at most |G i | vertices and automorphism groups Aut( i ) ∼ = G i , for each i ≥ 1. We omit the proof that Aut( i ) ∼ = G i for 1 ≤ i ≤ 4, which we verified using the mathematical software GAP [5].
We will now show that Theorem 4.1 also holds for the generalised dicyclic groups G that are different from Q 2 r × C 2 × C 2 × C 2 , r ≥ 3, completing the proof of Theorem 4.1.

Proposition 4.10 Let G be a generalised dicyclic group such that
iii) G is not one of the groups Dic 3 , Dic 5 , Dic 6 , G 16 that appear in Table 1.
The rest of this section concerns the proof of Proposition 4.10.
Let G = Dic(A, b 2 ) be a generalised dicyclic group as in Proposition 4.10 and let where A 2 is the Sylow 2-subgroup of A and A 2 is the Hall 2 -subgroup of A. Then, by Lemma 4.3, there exist y ∈ A 2 and B 2 < A 2 such that Let r , k be such that y ∼ = C 2 r , |X | = k. We note that the quotient group G/X is isomorphic to the generalised quaternion group Q 2 r +1 , for r > 1, and to the cyclic group C 4 , for r = 1. Moreover, the quotient group G/ y is isomorphic to the generalised dihedral group Dih(X ).

Construction 4.11
We will construct a graph such that Aut( ) ∼ = G. We start by defining two graphs, 1 For r ≥ 2, we let 1 be the graph with vertex set V ( 1 ) = G/X ∪(G/X ) that arises from Babai's Construction 2.6 for the generalised quaternion group G/X with respect to the minimal generating set H = {y X, bX}. Furthermore, we partition the set of vertices G/X of 1 into the sets T 1 , For r = 1, we define 1 to be the graph with automorphism group isomorphic to the cyclic group C 4 that was presented in Example 2.7. Likewise, we partition its vertex set into the sets T i , where Let us describe the graph 2 for all values of k. The conditions (i), (ii) in Proposition 4.10 ensure that k ≥ 3. For 3 ≤ k ≤ 5, we construct 2 with vertex set V ( 2 ) = G/ y ∪ (G/ y ) according to Babai's Construction 2.6 for the generalised dihedral group G/ y , with respect to some minimal generating set K = {k 1 , k 2 , . . . , k d } of G/ y such that k 1 = b y . For k ≥ 6, k = 9, we choose a GRR for G/ y , which exists by assumption (ii) in Proposition 4.10 and Lemma 4.2, and define the graph 2 to be either this GRR or its complement, with the additional property that the number k + ρ 2 (v) is even, for v ∈ V ( 2 ). Furthermore, for k ≥ 3, k = 9, we partition the set of vertices G/ y of 2 Finally, for k = 9, we let 2 be the graph on 18 vertices presented in Proposition 4.9, and S i = W i , where W i is as in Construction 4.5, Let us now define the graph such that Proof By construction, the groups G/X and G/ y act on V ( 1 ) and V ( 2 ), respectively. Using these actions, we associate a map g : V ( ) → V ( ) to every g ∈ G by setting In other words, the map g is defined to satisfy g V ( 1 ) = g X and g V ( 2 ) = g y .
Let g ∈ G. We will show that the map g is an automorphism of .
For the opposite inclusion, let φ ∈ Aut( ). Since, by assumption, φ fixes V ( 1 ), the restriction φ V ( 1 ) of φ is an automorphism of 1 . Since Aut( 1 ) ∼ = G/X , we have that φ V ( 1 ) = g 1 X , for some g 1 ∈ G. Then the automorphism g −1 1 φ acts trivially on V ( 1 ). As the set of vertices T 1 is fixed by g −1 1 φ, the set consisting of all neighbours of T 1 is also fixed by the same automorphism. However, g −1 1 φ fixes all neighbours of T 1 , except, possibly, from elements of the set S 1 . Therefore, S 1 is fixed by g −1 1 φ. Likewise, we consider the restriction of the automorphism g −1 1 φ on 2 to conclude that there exists some g 2 ∈ G such that the automorphism g −1 2 g −1 1 φ acts trivially on V ( 2 ); without loss of generality, we let g 2 ∈ X , b . Since the graph automorphisms g −1 1 φ and g −1 2 g −1 1 φ fix the set S 1 , we conclude that g 2 ∈ X . However, X acts trivially on V ( 1 ). Therefore, the graph automorphism g −1 2 g −1 1 φ is the identity automorphism of . Thus, and hence Aut( ) ≤ G.
We will examine all possible values of r , k in order to show that V ( 2 ) is fixed by φ, using the automorphisms' property to preserve the valency of the vertices permuting.
The valency of a vertex v ∈ V ( i ) in is where (8), which, together with the assumption that the number k + ρ 2 (y 1 ) is even, implies that We will show that the sets of vertices V (Y 2 ) and V (Y 1 ) are fixed by φ. Considering all abelian groups of order 3, 4 or 5, we conclude that the size of a minimal generating set of a generalised dihedral group of size 2k, such that 3 ≤ k ≤ 5, is between 2 and 3; hence 4 ≤ ρ(y 2 ) ≤ 5. On the other hand, by construction, min{ρ(x 1 ), ρ(x 2 ), ρ(y 1 )} > 5. As graph automorphisms preserve the valency of the vertices they permute, V (Y 2 ) is fixed by φ. It is implied that the set of neighbours of V (Y 2 ), which is V (Y 1 ), is also fixed by φ. Case 3 Assume now that r = 1, k ≥ 6, k = 9. First, we will compute the valency of each vertex of . By (8) . Without loss of generality, we assume that v ∈ S 1 . Graph automorphisms preserve the valency of the vertices they permute so ρ 2 (v) + 2 = k + j, for some j ∈ {0, 1, 2}.
The graph was constructed so that the number ρ 2 (y 1 ) + k is even, hence j is even. In other words, the set of vertices {1, 2} is fixed by φ, as is either the set {7, 8, 9, 10} or the set {3, 4, 5, 6}. The vertex 1 ∈ V ( 1 ) is connected to all other vertices in T 1 = {1, 3, 5, 7, 9} and no vertex in T 2 . Furthermore, the set of neighbours of v that lie in V ( 1 ) is T 1 . These properties combine to say that φ(v) is adjacent to either the vertices 1, 3 + 2 j, 5 + 2 j or the vertices 2, 4 + 2 j, 6 + 2 j. However, there exists no vertex in V ( 1 ) that is adjacent to any of these triplets of vertices. Thus, φ fixes V ( 2 ). Case 4 Finally, let r = 1, k = 9. We confirmed that the graph on 28 vertices constructed has the desired property using the mathematical software GAP [5].
The last step in the proof of Proposition 4.10 is to show that the order of the graph constructed is bounded by the order of the group G.
Lemma 4.14 The graph , defined in Construction 4.11, has at most |G| vertices.
Proof The graph was constructed so that the set V ( 1 ) has size 10, for r = 1, and 2|G/X |, for r ≥ 2. Moreover, the size of V ( 2 ) is 2|G/ y |, for 3 ≤ k ≤ 5, and |G/ y |, for k ≥ 6. Thus, where 5 k is the integer part of the real number 5 k . Let us now explain why |V ( )| ≤ |G|. If k = 3 then the assumption that r ≥ 3 (Proposition 4.10, (iii)) implies that

Proof of Theorem 1: the bound˛(G) > |G|
In this section we prove the bound α(G) > |G| and compute α(G) for groups G that are non-abelian and satisfy one of (iii), (iv) in Theorem 1.

Theorem 5.1 Let G be a group such that one of the following holds:
(i) G is a generalised quaternion group, (ii) G is a generalised dicyclic group of the form Q 2 r × C 2 , (iii) G is one of the non-abelian groups that appear in Table 1. Table 1.
The proof of Theorem 5.1 is the subject of this section. Specifically, we compute the value of α(G) when G is contained in one of the families of groups mentioned in Theorem 5.1, (i), (ii), in Propositions 5.4 and 5.5. Then, we calculate α(G) for the non-abelian groups G that are shown in Table 1 in Propositions 5.7, 5.10, 5.14 and 5.17.
Let us start by presenting two lemmas that will be used throughout the section.

Lemma 5.2 Let G be the dicyclic group of order 2 r +1 q, where q is an odd prime or q = 1. Let be a graph such that G ∼ = Aut( ) and consider the action of G on the vertex set V ( ).
If every orbit has size at most max{2 r +1 , 2 r q} then there exist at least 2 orbits of size 2 r +1 .
As the action of G on V ( ) is faithful, there exists a vertex w of such that b 2 / ∈ G w . Then, since b 2 is the only element of order 2, by Cauchy's Theorem, 2 does not divide |G w |. Therefore, by the orbit-stabilizer lemma, 2 r +1 divides |O w |. Thus |O w | = 2 r +1 , since |O w | ≤ max{2 r +1 , 2 r q}. Suppose that O w is the only orbit of size 2 r +1 .
Let B = {y k bw | k ∈ N}. We will show that the map φ : is an automorphism of . Indeed, the property v We have reached a contradiction since G w = G bw = x . Hence there exists a second orbit of size 2 r +1 .

Lemma 5.3
Let G = Dic q , q ∈ {3, 5}, and let be a graph on at most 4q + 4 vertices such that Aut( ) ∼ = G. Then, there is no orbit of size |G| = 4q in the action of G on V ( ).
Suppose that there is a vertex v ∈ V ( ) with stabilizer G v = {1}. If there exists an orbit of size 4, let u ∈ V ( ) be a vertex with stabilizer G u = x . By possibly replacing the graph with its complement, , we assume that v is adjacent to up to two vertices in the orbit O u , if it exists. Without loss of generality, we also assume We will show that φ ∈ Aut( ).
φ was constructed to preserve adjacency and non-adjacency between v 1 and v 2 , We have reached a contradiction, since φ fixes v but not bv and G v = G bv = {1}.
Using Lemma 5.2 we recover the following result, which was first proven in [8].

Proposition 5.5 The generalised dicyclic group Q
Let be a graph on at most 2 r +2 + 1 vertices with automorphism group isomorphic to G. The faithfulness of the action of G on V ( ) implies the existence of some if no such vertex exists, let u = w. Since |V ( )| ≤ 2 r +2 + 1, there exist at most two orbits of size 2 r +1 or one of size 2 r +2 . Therefore, if u = w then G u = G w , as the action of G on V ( ) is faithful.
Let B = y k x l bz | z ∈ {w, u}, k, l ∈ N and let φ : We will show that φ is an automorphism of by proving that The map φ fixes w but not bw and G w = G bw ; a contradiction. Thus, α(G) ≥ 2 r +2 +2.
Let us now construct a graph on 2 r +2 + 2 vertices such that Aut( ) ∼ = G.

Proposition 5.7
The generalised dicyclic group G 16 We will prove Proposition 5.7 using the following lemma.
Moreover, since the action is faithful, there is at most one i ∈ {1, . . . , s} such that G v i = x 2 b 2 . To sum up, we have s ≤ 2 or G v i = x 2 b 2 for a unique i ∈ {1, . . . , s}. If the latter is true, without loss of generality let v 1 have stabilizer G v 1 = x 2 b 2 ; alternatively, let v 1 be such that |O v 1 | ≥ |O v i |, for i ∈ {1, s}. Moreover, if s = 2, without loss of generality we assume that if v 1 is connected to O v 2 then v 1 ∼ v 2 . Finally, by possibly replacing with its complement, let v 1 be adjacent to at most half the vertices of O v 2 .
Let B = x k b l v 1 | k ∈ N, l ∈ {1, 3} . We will show that the map φ : is an automorphism of . Indeed, • If u 1 , u 2 are fixed by φ then clearly then φ was constructed to preserve adjacency and non-adjacency between u 1 , u 2 , We have reached a contradiction since G v 1 = G bv 1 and φ fixes v 1 but not bv 1 .

Proof of Proposition 5.7
Let G = G 16 . Suppose that is a graph on at most 17 vertices such that Aut( ) ∼ = G. As the action of G on V ( ) is faithful, there exists w 1 ∈ V ( ) such that x 2 / ∈ G w 1 ; by Lemma 5.8, which exist by Lemma 5.8. Since |V ( )| ≤ 17, there are up to two orbits, of total size at most eight, containing vertices that are not fixed by x 2 . If there are exactly eight such vertices, let w 2 ∈ V ( ) be such that x 2 / ∈ G w 2 and the vertices {w 1 , xw 1 , if only four vertices of are not fixed by x 2 , let w 2 = w 1 . By possibly replacing with its complement, we assume that the vertex w 1 is adjacent to up to two vertices of the set {w 2 , xw 2 , x 2 w 2 , x 3 w 2 }. Without loss of generality, if w 1 = w 2 , let these vertices be w 2 and x δ w 2 , δ ∈ {0, 1, 2, 3}; if w 1 is adjacent to exactly one vertex of the set {x k w 2 | k ∈ N} or w 1 = w 2 , let δ = 0. Then We will show that the map ψ : is an automorphism of . Indeed, • If u 1 = x k w 1 , u 2 = x l w 2 for k, l ∈ N and w 1 = w 2 then, by (9) We have reached a contradiction: ψ fixes v 1 and w 1 but We will complete the proof by constructing a graph on 18 vertices having Aut( ) ∼ = G.

Construction 5.9 Let be a graph with vertex set V
Using the mathematical software GAP [5] we verified that Aut( ) ∼ = G. The proofs of Propositions 5.10, 5.14, and 5.17 that follow are similar in nature to the proof of Proposition 5.7. Therefore, some of the technical details are omitted. Proof Let G = y, x, b be a generating set for G such that y 2 r = x q = 1, y 2 r −1 = b 2 , yx = x y, where r ∈ {1, 2}, q ∈ {3, 5}. Suppose that there exists a graph such that Aut( ) ∼ = G and |V ( )| < 3q + 2 r +2 .
Suppose, in addition, that less than 3q vertices are not fixed by x. The faithfulness of the action of G on V ( ) implies the existence of v ∈ V ( ) such that x / ∈ G v , hence q divides |O v |. Let V = x k z | z ∈ {v, u}, k ∈ N be the set of vertices of that are not fixed by x, where u / ∈ {x k v | k ∈ N}, if there exist two orbits of size q or one of size 2q, and u = v, if there is exactly one orbit of size q.
We may assume that v is adjacent to up to two vertices of the set {x k u | k ∈ N}. If v = u, let us consider these vertices to be u and Similar to the map ψ in Proposition 5.7, the map ψ : is an automorphism of . The faithfulness of the action implies the existence of w ∈ V ( ) such that b 2 / ∈ G w , hence G w = x (G w = {1}, since there exist less than 3q vertices that are not fixed by x). Then, ψ ∈ G w ∩ G v = {1} and ψ(xv) = x −1 v; a contradiction.
We will show that there is no orbit of size |G|. Indeed, if |G| = 24 then we assumed that |V ( )| ≤ 24 hence, by the GRR theorem, there exists no orbit of size 24. If |G| = 12, 20 then, by Lemma 5.3, every orbit has size at most 2 r q.
By Lemma 5.2, there exist at least two orbits of size 2 r +1 (a similar statement to Lemma 5.2 holds for the group G = Q 8 ×C 3 and the proof is analogous). Considering the number of vertices that are fixed or not fixed by x we conclude that |V ( )| ≥ 3q + 2 r +2 ; a contradiction. Hence, α(G) ≥ 3q + 2 r +2 .

Construction 5.11
Assume that G = Dic q for some q ∈ {3, 5}. Let be a graph

Construction 5.12
For G = Dic 6 , we let be the graph with vertex set V ( )

Construction 5.13
Finally, for G = Q 8 × C 3 , let 1 be a graph on 16 vertices constructed according to Babai's Construction 2.6 for the group Q 8 and let 2 be a graph on 9 vertices such that Aut( 2 ) ∼ = C 3 , which exists by Proposition 3.3. We let = 1 ∪ 2 .
Using GAP [5], we confirmed that each graph has the desired automorphism group.
Suppose that there exists graph with V ( ) ≤ 17 and Aut( ) ∼ = G. As G acts faithfully on V ( ), there exists w ∈ V ( ) such that a 4 / ∈ G w , hence G w ∈ { 1 , a 4 b , b }. If G w = 1 then the subgraph 1 of induced by O w has order 16 and Aut( 1 ) ∼ = G, contradicting the non-existence of a GRR for G. Since a 4 b , b are conjugate, we may assume that G w = b . Let u ∈ V ( ) such that u / ∈ O w and G u = b , if there exists a second orbit with elements not fixed by a 4 ; if no such orbit exists, let u = w.

Lemma 5.15
The vertex w is adjacent to exactly one of the vertices a 2 u, a 6 u in , with as above.
Proof Suppose, conversely, that Let B = {w, a 4 w, u, a 4 u}. Then, the map φ : is an automorphism of . The proof is similar to that of the other automorphisms defined in this section. However, φ fixes a 2 w but not w, contradicting the equality G w = G a 2 w .
Proof of Proposition 5.14 If is a graph on at most 17 vertices having Aut( ) ∼ = G and O w , O u are the orbits of size 8 listed above then by Lemma 5.15 either w ∼ a 2 u or w ∼ a 6 u (hence w = u). Arguing analogously, we can show that w ∼ u ⇐⇒ w a 4 u. Without loss of generality, we assume that w ∼ u, w ∼ a 2 u. Moreover, since b ∈ G w , b ∈ G u , we have that w ∼ a 3 u ⇐⇒ w ∼ ba 3 bu ⇐⇒ w ∼ a 7 u. Hence, it holds for every n ∈ N that w ∼ a n u ⇐⇒ w ∼ a 2−n u.
Since |V ( )| ≤ 17 and |O w ∪O u | = 16, there is at most one additional orbit, which has size 1. Similar to the map ψ in Proposition 5.7, the map ψ : is an automorphism. This is a contradiction as G u = G a 2 u and ψ fixes u but not a 2 u. We complete the proof by constructing a graph with V ( ) = 18 and Aut( ) ∼ = G.
Using GAP [5] we computed that Aut( ) ∼ = G. Since G has no subgroup of order 6, there is no orbit of size 2. Furthermore, there exists z ∈ V ( ) such that b / ∈ G z . Then |G z | ∈ {1, 2, 3}, hence the orbit O z has size 4, 6 or 12. We examine each of these cases.

Lemma 5.18 Let G = A 4 and let be as in the previous paragraph and consider the action of G on V ( ). Then, there exists no orbit of size 4.
Proof Suppose, in contrast, that there exists some orbit of size 4. If there also exists an orbit of size 6 as well as an orbit of size 3, then without loss of generality we let w ∈ V ( ) be such that |O w | = 3 and u ∼ aw ⇐⇒ u ∼ a 2 w, for every u ∈ V ( ) having G u = b ; this is possible since the bound |V ( )| ≤ 15 ensures that there is at most one orbit of size 6. Otherwise, let w be such that G w = aba . Then the map φ : is an automorphism. The proof is more technical but similar to others in this section. For a detailed justification, we refer the reader to the arXiv version of this article [3]. Practically, φ interchanges two pairs of vertices in the orbit of size 6, if it exists, one pair in the orbit of size 3, if both an orbit of size 3 and 6 exist, and one pair in every orbit of size 4. However, φ fixes two vertices, v 1 , v 2 , such that G v 1 = a , G v 2 = a 2 b , but G v 1 ∩ G v 2 = {1}; a contradiction. Proof Suppose that there exists an orbit of size 12. In [21,Proposition 3.7], Watkins proved that G has no GRR. The arguments in [21,Proposition 3.7] extend to the case that contains an additional orbit of size 3, or up to three additional orbits of size 1. For details on the extension we refer the reader to the arXiv version of this article [3].
Proof of Proposition 5. 17 We assumed that is a graph on at most 15 vertices having Aut( ) ∼ = G; then b / ∈ G z for some z ∈ V ( ). By Lemmas 5.18 and 5.19, there is no orbit of size 4 or 12, and |G z | = 6. Using the group structure of G we can prove that χ : V ( ) → V ( ), is an automorphism of . However, χ fixes two vertices, v 1 , v 2 ∈ O z such that G v 1 = aba 2 , G v 2 = a 2 ba , contradicting the property G v 1 ∩ G v 2 = {1}. Therefore, α(G) ≥ 16.
We will show that α(G) = 16 by constructing a graph on 16 vertices with Aut( ) ∼ = G.
Using the mathematical software GAP [5] we computed that Aut( ) ∼ = G.