On edge-transitive graphs of square-free order

We study the class of edge-transitive graphs of square-free order and valency at most k. It is shown that, except for a few special families of graphs, only finitely many members in this class are basic (namely, not a normal multicover of another member). Using this result, we determine the automorphism groups of locally primitive arc-transitive graphs with square-free order.


Introduction
For a graph Γ = (V, E), the number of vertices |V | is called the order of Γ.A graph Γ = (V, E) is called edge-transitive if its automorphism group AutΓ acts transitively on the edge set E. For convenience, denote by ETSQF(k) the class of connected edge-transitive graphs with square-free order and valency at most k.
The study of special subclasses of ETSQF(k) has a long history, see for example [1,4,5,17,18,21,22,23] for those graphs of order being a prime or a product of two primes.
Recently, several classification results about the class ETSQF(k) were given.Feng and Li [9] gave a classification of one-regular graphs of square-free order and prime valency.By Li et al. [12,14], one may obtain a classification of vertex-transitive and edge-transitive tetravalent graphs of square-free order.By Li et al. [13] and Liu and Lu [16], one may deduce an explicitly classification of ETSQF (3).In this paper, we give a characterization about the class ETSQF(k).
A typical method for analyzing edge-transitive graphs is to take normal quotient.Let Γ = (V, E) be a connected graph such that a subgroup G AutΓ acts transitively on E. Let N be a normal subgroup of G, denoted by N G. Then either N is transitive on V , or each N -orbit is an independent set of Γ.Let V N be the set of all N -orbits on V .The normal quotient Γ N (with respect to G and N ) is defined as the graph with vertex set V N such that distinct vertices B, B ∈ V N are adjacent in Γ N if and only if some α ∈ B and some α ∈ B are adjacent in Γ.We call Γ N non-trivial if N = 1 and |V N | 3. It is well-known and easily shown that Γ N is an edge-transitive graph.Moreover, if all N -orbits have the same length (which is obvious if G is transitive on V ), then Γ N is a regular graph of valency a divisor of the valency of Γ; in this case, Γ is called a normal multicover of Γ N .
A member in ETSQF(k) is called basic if it has no non-trivial normal quotients.Then every member in ETSQF(k) is a multicover of some basic member, or has a non-regular normal quotient (which might occur for vertex-intransitive graphs).Thus, to a great extent, basic members play an important role in characterizing the graphs in ETSQF(k).The first result of this paper shows that, except for a few special families of graphs, there are only finitely many basic members in ETSQF(k).
Theorem 1.Let Γ = (V, E) be a connected graph of square-free order and valency k 3. Assume that G AutΓ acts transitively on E and that each non-trivial normal subgroup of G has at most 2 orbits on V .Then one of the following holds: (1) Γ is a complete bipartite graph, and G is described in (1) and (5) of Lemma 13; (2) G is one of the Frobenius groups Z p :Z k and Z p :Z 2k , where p is a prime; (3) soc(G) = M 11 , M 12 , M 22 , M 23 , M 24 or J 1 ; (4) G = A n or S n with n < 3k; (5) G = PSL(2, p) or PGL(2, p); (ii) 2f < k, and soc(G) = G 2 (p f ), 3 D 4 (p f ), F 4 (p f ), 2 E 6 (p f ), or E 7 (p f ).
Remark 2 (Remarks on Theorem 1).For a finite group G, the socle soc(G) of G is the subgroup generated by all minimal normal subgroups of G.A finite group is called almost simple if soc(G) is a non-abelian simple group.
(a) The groups G in case (1) are known except for G being almost simple.
(b) The vertex-transitive graphs in case (5) are characterized in Theorem 27.
(c) Some properties about the graphs in cases ( 6)- (7) are given in Lemmas 14 and 15, respectively.
It would be interest to give further characterization for some special cases.
Problem 3. (i) Characterize edge-transitive graphs of square-free order which admits a group with socle PSL(2, q), Sz(q), A n or a sporadic simple group.
(ii) Classify edge-transitive graphs of square-free order of small valencies.
For a graph Γ = (V, E) and G AutΓ, the graph Γ is called G-locally primitive if, for each α ∈ V , the stabilizer of α in G induces a primitive permutation group on the neighbors of α in Γ.The second result of this paper determines, on the basis of Theorem 1, the automorphism groups of locally primitive arc-transitive graphs of square-free order.Theorem 4. Let Γ = (V, E) be a connected G-locally primitive graph of square-free order and valency k 3. Assume that G is transitive on V and that Γ is not a complete bipartite graph.Then one of the following statements is true.
(1) G = D 2n :Z k , 2nk is square-free, k is the smallest prime divisor of nk, and Γ is a bipartite Cayley graph of the dihedral group D 2n ; (2) G = M :X, where M is of square-free order, X is almost simple with socle T descried as in ( 3)-( 6) and ( 8) of Theorem 1 such that M T = M × T , T has at most two orbits on V and Γ is T -edge-transitive; in particular, if T = PSL(2, p), then M , T α and k are listed in Table 3, where α ∈ V .

Preliminaries
Let Γ = (V, E) be a graph without isolated vertices, and let G AutΓ.The graph Γ is said to be G-vertex-transitive or G-edge-transitive if G acts transitively on V or E, respectively.Recall that an arc in Γ is an ordered pair of adjacent vertices.The graph Γ is called G-arc-transitive if G acts transitively on the set of arcs of Γ.For a vertex α ∈ V , we denote by Γ(α) the set of neighbors of α in Γ, and by G α the stabilizer of α in G. Then it is easily shown that Γ is G-arc-transitive if and only if Γ is G-vertex-transitive and, for α ∈ V , the vertex-stabilizer G α acts transitively on Γ(α).
Let Γ = (V, E) be a connected G-edge-transitive graph.Note that each edge of Γ gives two arcs.Then either Γ is G-arc-transitive or G has exactly two orbits (of the same size |E|) on the arc set of Γ.If Γ is not G-vertex-transitive then Γ is a bipartite graph and, for α ∈ V , the stabilizer G α acts transitively on Γ(α).If Γ is G-arc-transitive, then there exists g ∈ G \ G α such that (α, β) g = (β, α) and, since Γ is connected, g, G α = G; obviously, this g can be chosen as a 2-element in (2) Γ is G-vertex-transitive, G α has exactly two orbits on Γ(α) of the same size |G α : Let Γ = (V, E) be a regular graph and G AutΓ.
α be the kernel of this action.Then G α .Considering the actions of Sylow subgroups of G [1] α on V , it is easily shown that the next lemma holds, see [7] for example.A permutation group G on a set Ω is semiregular if G α = 1 for each α ∈ Ω.A transitive permutation group is regular if further it is semiregular.
α acts trivially on Γ(β), and so AutΓ and α ∈ V .Assume that either N is regular on V , or Γ is a bipartite graph such that N is regular on both the bipartition subsets of Γ.Then G Thus our results follows from Lemma 7. Now assume that Γ is a bipartite graph with bipartition subsets U and W , and that N is regular on both U and W . Without loss of generality, we assume that α ∈ U .Then Γ(α) ⊆ W , and is a spanning subgraph of Γ, and X acts transitively on E 0 .Thus Σ is a regular graph, and X α is transitive on Σ(α).Noting Σ(α) ⊆ Γ(α), it follows that |Σ(α)| = 1, and hence Σ is a matching.In particular, Since all vertices in U are equivalent under X, we have X γ acts trivially on Γ(γ).Then a similar argument as above leads to G [1] α = X γ = X δ = X θ for any δ ∈ Γ(γ) and θ ∈ Γ(δ).Then, by the connectedness, we conclude that G [1] α fixes each vertex of Γ.Thus G We end this section by quoting a known result.Lemma 9 ([12]).Let Γ = (V, E) be a connected G-edge-transitive graph, N G AutΓ and α ∈ V .Then all N α -orbits on Γ(α) have the same length.

Complete bipartite graphs
We first list a well-known result in number theory.For integers a > 0 and n > 0, a prime divisor of a n − 1 is called primitive if it does not divide a i − 1 for any 0 < i < n.
Let G be a permutation group on V , and let x be a permutation on V which centralizes G.If x fixes some point α ∈ V , then x fixes α g for each g ∈ G. Thus the next simple result follows.
Lemma 11.Let G be a permutation group on V .Assume that N is a normal transitive subgroup of G. Then the centralizer C G (N ) is semiregular on V , and C G (N ) = N if further N is abelian.
Recall that a transitive permutation group G is quasiprimitive if each non-trivial normal subgroup of G is transitive.Let G be a quasiprimitive permutation group on V , and let B be a G-invariant partition on V .Then G induces a permutation group G B on B. Assume that |B| 2. Since G is quasiprimitive, G acts faithfully on B. Then G B ∼ = G, and so soc(G B ) ∼ = soc(G).
Lemma 12. Let G be a quasiprimitive permutation group of square-free degree.Then soc(G) is simple, so either G is almost simple or G AGL(1, p) for a prime p.
Proof.Let G be a quasiprimitive permutation group on V of square-free degree.Let B be a G-invariant partition on V such that |B| 2 and G B is primitive.Noting that |B| is the electronic journal of combinatorics 22(3) (2015), #P3.25 square-free, by [15], soc(G B ) is simple.Thus soc(G) ∼ = soc(G B ) is simple, and the result follows.
Let G be a permutation group on V .For a subset U ⊆ V , denote by G U and G (U ) the subgroups of G fixing U set-wise and point-wise, respectively.For X G and an X-invariant subset U of V , denote by X U the restriction of X on U .Then X U ∼ = X/X (U ) .
We now prove a reduction lemma for Theorem 1.
Lemma 13.Let Γ = (V, E) be a connected G-edge-transitive graph of square-free order and valency k 3, where G AutΓ. Assume that each minimal normal subgroup of G has at most two orbits on V .Then one of the following holds: (3) |V | = 2p with p 3 prime, and G is isomorphic to one of Z p :Z k and Z p :Z 2k ; (4) G is almost simple; for a nonabelian simple group T and, for α ∈ V , either Proof.Let N be a minimal normal subgroup of G. Then N is a directed product of isomorphic simple groups.Since Γ has valency k 3, we know that |V | > 3. Since |V | is square-free and N has at most two orbits on V , we conclude that N is not an elementary abelian 2-group.In particular, N has no a subgroup of index 2. Without loss of generality, we assume that N is intransitive on V .Then Γ is a bipartite graph, whose bipartition subsets are N -orbits, say U and V \ U .A similar argument as above paragraph yields that M has no subgroups of index 2.It follows that M fixes both U and V \ U set-wise, and hence U and V \ U are two M -orbits on V .
Let X = N M and ∆ = U or V \ U .By Lemma 11, both N ∆ and M ∆ are regular subgroups of X ∆ .Set N ∼ = T i , where T is a simple group.Then N (∆) ∼ = T j for some j < i, and so Since T is simple and |∆| is square-free, i − j = 1 and N ∆ ∼ = T ∼ = Z p , where p = |∆| is an odd prime.Similarly, M ∆ ∼ = Z p , and so M is abelian.In particular, X = N × M is abelian and |X| is a power of p.It implies that X ∆ ∼ = Z p .Then, by Lemma 11, Let α ∈ ∆.Then G α X (∆) .By Lemma 6, k = |Γ(α)| p, and so Γ ∼ = K p,p .Noting that N is regular on ∆ and V \ ∆, by Lemma 8, G α acts faithfully on Γ(α), and so G α is isomorphic to a subgroup of the symmetric group S p .Noting that G α has a normal subgroup X (∆) ∼ = Z p , it follows that G α is isomorphic to a subgroup of the Frobenius group Clearly, X (∆) has at least p + 1 orbits on V .Then, by the assumptions of this lemma, X (∆) is not normal in G. On the other hand, (X (∆) In the following we assume that N ∼ = T l for an integer l 2 and a simple group T .If N is transitive on V then G is quasiprimitive on V , and hence soc(G) = N is simple by Lemma 12, a contradiction.If G is intransitive on V , then G is faithful on each of its orbits, and then N is simple by Lemma 12, again a contradiction.Thus, in the following, we assume further that Γ is G-vertex-transitive and N has two orbits U and We claim that T is a nonabelian simple group.Suppose that T ∼ = Z p for some (odd) prime p.  Set N = T 1 ×T 2 , where T 1 ∼ = T 2 ∼ = T .Since T 1 and T 2 are isomorphic nonabelian simple groups, T 1 and T 2 are the only non-trivial normal subgroups of N .Thus or N is faithful on both U and W .The former case yields that N (U ) acts transitively on W , and so (i) of part ( 5) follows.
Assume that N is faithful on both U and W . Then neither T 1 nor T 2 is transitive on U .Let O be the set of T 1 -orbits on U , and let O ∈ O. Then T 2 is transitive on O. Thus T has two transitive permutation representations of degrees |O| and |O|, respectively.Then T has two primitive permutation representations of degrees n 1 and n 2 , where and n 2 are odd, square-free and coprime.Inspecting [15, Tables 1-4], we conclude that T is either an alternating group or a classical group of Lie type.
Suppose that T = PSL(d, q) with d 3.By the Atlas [8], neither PSL(3, 2) nor PSL(4, 2) has maximal subgroups of coprime indices.Thus we assume that (d, q) = (3, 2) or (4, 2).Then, by [15,Table 3], If q d − 1 has a primitive prime divisor r, then both n 1 and n 2 are divisible by r, which is not possible.Thus q d − 1 has no primitive prime divisor, and so (q, d) = (2, 6) by Theorem 10.Computation of n 1 and n 2 shows that this is not the case.Similarly, we exclude other classes of classical groups of Lie type except for PSL(2, p f ), where p is a prime.By the Atlas [8], we exclude PSL(2, p f ) while p f 31.Suppose that T = PSL(2, p f ) with p f 32.By [15, Table 3], one of n 1 and n 2 is p f + 1 and the other one is divisible by p.This is not possible since one of p f + 1 and p is even.Now let T = A c for some c 5. By the above argument, we may assume that A c is not isomorphic to a classical simple group of Lie type.Then c = 5, 6 or 8.Note that for c 5 and a < b < c 2 , the binomial coefficient . Thus ( c a ) and ( c b ) are not comprime, and so at most one of n 1 and n 2 equals to a binomial coefficient.Checking the actions listed in [15, Table 1] implies that either c = 7, or c = 2a for a ∈ {6, 9, 10, 12, 36}.Suppose the later case occurs.Then one of n 1 and n 2 is 1 2 ( 2a a ) and the other one is a binomial coefficient, say ( 2a b ).But computation shows that such two integers are not coprime, a contradiction.Therefore, T = A 7 .
Checking the subgroups of A 7 , we conclude that that T 2 is intransitive on V .Since T 2 N and N is transitive on U , we conclude that each T 2 -orbit on U has size 15.It follows that (T 2 ) O = (T 2 ) α .Then N α (T 1 ) α × (T 2 ) α , and so , and hence (ii) of part (5) occurs.
4 Graphs associated with PSL(2, p f ) and Sz(2 f ) Let Γ = (V, E) be a connected graph of square-free order and valency k.Assume that G AutΓ is almost simple with socle T .Assume further that G is transitive on E and that T has at most two orbits on V .Let {α, β} ∈ E.
Lemma 14.Let Γ = (V, E) be a connected G-edge-transitive graph of square-free order and valency k.Assume that soc(G) = PSL(2, p f ) with f 2 and p f > 9, and that soc(G) has at most two orbits on V .Then one of the following statements holds: Proof.Let T = soc(G).Take α ∈ V and a maximal subgroup M of T with T α M .Then both |T : M | and |M : T α | are square-free as |T : T α | is square-free.By [15], either and |T : Assume that T α is insoluble.Then f = 2 and [11,II.8.27], |T α : T αβ | is divisible by p or p + 1. Suppose that T α = PGL(2, p).Then T α is maximal in T , and so T = T α , T β .Thus |T β : T αβ | > 2 as T is simple; in particular, PSL(2, p) = T αβ .Checking the subgroups of T α which do not contain PSL(2, p) (refer to [3]), we conclude that |T α : T αβ | is divisible by p or p + 1.Thus part (i) occurs.
In the following, we assume that T α is soluble.Since p 2 is not a divisor of |T : T α |, each Sylow p-subgroup of T α has p f or p f −1 .Then, inspecting the subgroups of T , we conclude that T α ∼ = T β for β ∈ Γ(α), and that T α has a unique Sylow p-subgroup.
Let Q be a Sylow p-subgroup of T αβ .Then Q is normal in T αβ .Suppose that Q = 1.Let P 1 and P 2 be the Sylow p-subgroups of T α and T β , respectively.Then and P 2 are contained the same Sylow p-subgroup, say P of T .In particular, P 1 = P α and P 2 = P β .For γ ∈ Γ(β), since Γ is G-edge-transitive, we have |T αβ | = |T βγ |.A similar argument implies that P γ is the Sylow p-subgroup of T γ .It follows from the connectedness of Γ that P δ is the Sylow p-subgroup of T δ for any δ ∈ V .Thus P contains a normal subgroup P δ | δ ∈ V = 1 of G, a contradiction.Thus, T αβ is of order coprime to p, and so |T α : T αβ | is divisible by |P 1 | = p f −1 or p f .Thus, by Lemma 9, k is divisible by p f −1 or p f , respectively.
If M = PGL(2, p) then, inspecting the subgroups of M , we conclude that T α = Z p :Z l , where l is a divisor of p − 1 and divisible by 4. Assume that M = Z f p :Z p f −1 . Then Noting that M is a Frobenius group, T α is also a Frobenius group.It follows that l is a divisor of p f −1 − 1, and so l divides p − 1.
Assume further that Γ is G-locally primitive.Then T where l is a divisor of l.Since P 1 is the Sylow p-subgroup of T α , we have or p f .Thus one of (ii) and (iii) follows.
The following lemma gives a characterization of graphs admitting Suzuki groups.
Lemma 15.Let Γ = (V, E) be a connected G-edge-transitive graph of square-free order and valency k.Assume that soc(G) = Sz(2 f ) with odd f 3, and that soc(G) has at most two orbits on V .Then k is divisible by 2 2f −1 and Γ is not G-locally primitive.
Proof.Let α ∈ V and β ∈ Γ(α).Since |T : T α | is square-free, 4 does not divide |T : T α |, and hence 2 2f −1 divides |T α |.Then, inspecting the subgroups of T (see [20]), we get T α = [2 n ]:Z l , where n = 2f or 2f − 1, and l is a divisor of 2 f − 1.So T α has a unique Sylow 2-subgroup.By [20], for a Sylow 2-subgroup Q of T , all involutions of Q are contained in the center of Q.Noting that any two distinct conjugations of Q generate T , it follows any two distinct Sylow 2-subgroups of T intersect trivially.Thus, by a similar argument as in the above lemma, we know that T αβ has odd order.Thus Finally, suppose that G Γ(α) α is a primitive group.Let Q 1 be the Sylow 2-subgroup of T α , and Q be a Sylow 2-subgroup of By a similar argument as in the above lemma, we conclude that ).It follows that Q 1 is an elementary abelian 2-group.By [20], Q 1 lies in the center of Q, and so Q is abelian, which is impossible.Then this lemma follows.

Proof of Theorem 1
Let Γ = (V, E) be a connected graph of square-free order and valency k.Assume that a subgroup G AutΓ acts transitively on E and that each non-trivial normal subgroup of G has at most 2 orbits on V .By Lemma 13, to complete the proof of the theorem, we the electronic journal of combinatorics 22(3) (2015), #P3.25 may assume that G is almost simple.Let T = soc(G) and α ∈ V .Then T is transitive or has exactly two orbits on V , and every prime divisor of |T α | is at most k.
Let U be a T -orbit, and let B be a T -invariant partition on U such that |B| 2 and T B is primitive.Noting that |B| is square-free, T is listed in [15, Tables 1-4].In particular, if T is one of sporadic simple groups then part (3) of Theorem 1 follows.
Assume that T = A n , where n 5. Suppose that n 3k.By [15], there exists a prime p such that k < p < 3k/2, and thus p 2 divides |T |, and p divides |T α |.So p k, which is a contradiction.Therefore, n < 3k, as in part (4) of Theorem 1.
We next deal with the classical groups and the exceptional groups of Lie type.If T = PSL(2, p f ) or Sz(2 f ) then, by Lemmas 14 and 15, one of parts ( 5), ( 6) and ( 7) of Theorem 1 follows.Thus the following two lemmas will fulfill the proof of Theorem 1.
Finally we consider the exceptional simple groups of Lie type.
the electronic journal of combinatorics 22(3) (2015), #P3.25 Lemma 17.Let T be an exceptional simple group of Lie type defined over GF(p f ) with p prime.Then p k, and one of the following holds: Proof.Note that T has order divisible by p 2 .Then p divides |T α |, and so p k. k.If p 6f − 1 has no primitive prime divisor, then p = 2 and f = 1, and so |T : T α | is not square-free as it is divisible by 9, and hence T is described as in part(iii) of this lemma.

Graphs associated with PSL(2, p)
In this section, we investigate vertex-and edge-transitive graphs associated with PSL(2, p), and then give a characterization for such graphs.

Examples
It is well-known that vertex-and edge-transitive graphs can be described as coset graphs.Let G be a finite group and H be a core-free subgroup of G, where core-free means that ∩ g∈G H g = 1.Let [G : H] = {Hx | x ∈ G}, the set of right cosets of H in G.For an element g ∈ G \ H, define the coset graph Γ := Cos(G, H, H{g, g −1 }H) on [G : H] such that (Hx, Hy) is an arc of Γ if and only if yx −1 ∈ H{g, g −1 }H.Then Γ is a well-defined regular graph, and G induces a subgroup of AutΓ acting on [G : H] by right multiplication.The next lemma collects several basic facts on coset graphs.Lemma 18.Let G be a finite group and H a core-free subgroup of G. Take g ∈ G \ H and set Γ = Cos(G, H, H{g, g −1 }H).Then Γ is G-vertex-transitive and G-edge-transitive.Moreover, (2) Γ is connected if and only if H, g = G.
the electronic journal of combinatorics Note that the subgroups of T are known, refer to [11,II.8.27].We next analyze one by one the possible candidates for T α .
Lemma 24.Assume that T α is cyclic.Then T α ∼ = Z m for an even divisor m of p±1 2 , T is transitive on V , Γ is not G-locally-primitive, and one of the following holds: Proof.Note T α is a cyclic group of even order.By Lemma 7, T α is faithful and semiregular on Γ(α).It is easy to check that no primitive group contains a normal semiregular cyclic subgroup of even order.Thus Γ is not G-locally-primitive.By [11, II.8.5], T α is contained in a subgroup conjugate to Zp±1 Proof.By [11,II.8.27], recalling that T α has even order, T α ∼ = Z p :Z l for an even divisor l of p−1 2 .Since |T : 2l is even and square-free, p−1 2l is odd and T is transitive on V .By Lemma 7, noting that T α is a Frobenius group, T α acts faithfully on Γ(α).In particular, each T α -orbit on Γ(α) has size divisible by p.
Assume that Γ is G-locally primitive.Then T α is transitive on Γ(α) as T α G α .It implies that Γ has valency p and Γ is T -arc-transitive.Then Γ ∼ = Cos(T, T α , T α xT α ) for some x ∈ N T (T αβ ) with x 2 ∈ T αβ and x, T α = T , where β ∈ Γ(α).Note that , where i is odd and j is a power of 2. Then a = a i × a j .Since T αβ ∼ = Z l and p−1 2l is odd, we have a i ∈ T αβ T α .Since l is even, j = 1.It follows from x, T α = T that x = a si a tj b for some s and t.Then Proof.The first part follows from that Let {α, β} be an edge of Γ. Suppose that T αβ contains a cyclic subgroup C of order no less than 3. Then C is the unique subgroup of order |C| in both T α and T β .For an arbitrary edge {γ, δ}, since Γ is G-edge-transitive, {γ, δ} = {α, β} x for x ∈ G, so α is a normal cyclic subgroup of T α .By the argument in above paragraph, 2 contains the center of M .Without loss of generality, we choose b in the center of M , and so x = a i b for 1 i < p+ 2 .Thus Γ is isomorphic to a graph given in Example 20 (1).Now let If T is not transitive on V Γ, then G = PGL(2, p), Γ is a bipartite graph, and T α = G α .Thus we set X = PSL(2, p) or PGL(2, p) depending respectively on whether or not T is is transitive on V Γ.Then Γ ∼ = Cos(X, T α , T α xT α ) for some x ∈ N X (T αβ ) \ T αβ with x 2 ∈ T αβ ; in particular, N X (T αβ )/T αβ is of even order It implies that N T (T αβ ) ∼ = S 4 .Let M be the maximal subgroup of X with T α M .Then 8 divides |M |, and We write N X (T αβ ) = T αβ :( y : z ), where z ∈ N M (T αβ ) and y : z ∼ = S 3 .Noting that x ∈ N M (T αβ ) and x is of even order, we have x = x 1 y i z for some x 1 ∈ T αβ and i = 1 or 2. Noting that z normalizes T α and y z = y −1 , we have Cos(X, T α , T α xT α ) = Cos(X, T α , T α y i zT α ) ∼ = Cos(X, T α , T α yzT α ).Thus Γ is isomorphic to the graph given in Example 20 (2).
Theorem 27.Let Γ = (V, E) be a connected G-edge-transitive graph of square-free order and valency k 3, where G AutΓ. Assume that soc(G) = PSL(2, p) for a prime p 5, and that G is transitive on V .Then, for α ∈ V , the pair (soc(G) α , k) lies in Table 1.
Finally, if T α = A 5 then G α = T α and G αβ ∼ = A 4 , D 10 or S 3 , and thus Γ is isomorphic one of the graphs given in Example 23.

Locally primitive arc-transitive graphs
In this section we give a proof of Theorem 4. We first prove a technical lemma.
Lemma 28.Let G be a transitive permutation group on V of square-free degree and let M be a normal subgroup of G. Assume that M is semiregular on V and G/M acts faithfully on the M -orbits.Then there is X G such that G = M :X.
Proof.The result is trivial if M = 1.Thus we assume that M = 1.Note that M has square-free order.Let p be the largest prime divisor of |M | and P be the Sylow p-subgroup of M .Then P is cyclic and is normal in G. Let α ∈ V and B be the P -orbit with α ∈ B. Let V P be the set of P -orbits.Then |B| = p is coprime to 10.4] that the extension G = P.(G/P ) splits over P .Thus G = P :H for some H < G with H ∩ P = 1.If M = P , then the result follows.We assume M = P in the following.
Let K be the kernel of G acting on V P .Noting that each M -orbit on V consists of several P -orbits, we know that K fixes each M -orbits set-wise.It follows from the assumptions that K M .Then, considering the action of M on its an orbit, we conclude that K = P .Thus H is faithful and transitive on V P .Further, M = M ∩P H = P (M ∩H) the electronic journal of combinatorics 22(3) (2015), #P3.25 implies that M ∩H is semiregular on V P .It is easily shown that H/(M ∩H) acts faithfully on the (M ∩ H)-orbits on V P .Noting that |V P | < |V |, we may assume by induction that H = (M ∩ H)X with X ∩ (M ∩ H) = 1.Then G = P ((M ∩ H)X) = M X, and M ∩ X M ∩ H yielding M ∩ X M ∩ H ∩ X = 1, hence our result follows.
Let Γ = (V, E) be a connected G-locally primitive graph.Suppose that G has a normal subgroup N which has at least three orbits on V .Then either the quotient graph Γ N is a star, or Γ is a normal cover of Γ N , refer to [10,Theorem 1.1].Then following lemma is easily shown.
Lemma 29.Let Γ = (V, E) be a connected G-locally primitive and G-symmetric graph.Let N be a normal subgroup of G.If N is not semiregular on V , then N is transitive on E and has at most two orbits on V .
Theorem 30.Let Γ = (V, E) be a connected G-locally primitive graph of square-free order and valency k > 2. Let M G be maximal subject to that M has at least three orbits on V .Assume further that Γ M is not a star.Then one of the following holds.
Proof.Since Γ M is not a star, Γ is a normal cover of Γ M , hence M is semiregular on V ; in particular, |M | is coprime to |V M |.By the choice of M , we know that G/M is faithful on either V M or one of two G/M -orbits on V M .Then, by Lemma 28, we have G = M :X.Note that Γ M is G/M -locally primitive, and the pair G/M and Γ M satisfies the assumptions in Theorem 1.Let Y = soc(X).Then, by Lemma 13, Γ M ∼ = K k,k and Y ∼ = T 2 for a simple group T , or Y is a minimal normal subgroup of X.
, and hence M Y = M × Y , and so part (3) of this theorem occurs.We next complete the proof in two cases.
Case 1. Γ M ∼ = K k,k and Y ∼ = T 2 for a simple group T .In this case, by Lemma 13, X is transitive on V M , and so Γ M is X-arc-transitive.Then Γ is G-arc-transitive.Moreover, Y has exactly two orbits on V M of size k.Thus M Y has exactly two orbits U and W on V of length k|M |.Let U M and W M be the sets of M -orbits on U and W , respectively.Then U M and W M are Y -orbits on V M .
Assume first that T is a nonabelian simple group.Then part (5) of Lemma 13 holds for the pair (X, Γ M ).In particular, Y is the unique minimal normal subgroup of X.Let ∆ be an M -orbit on V .Suppose that T ∼ = A 7 .Then k = 105 and T ∆ ∼ = A 6 × PSL(3, 2).It is easily shown that Γ M is not X-locally primitive, which is not the case.Thus Y is unfaithful on both U M and W M .Let K be the kernel of Y acting on U Considering the action of M K on ∆, we conclude that K acts trivially on ∆.Then K acts trivially on U .Since K is transitive on W M , we conclude that Γ ∼ = K k,k .It follows that M = 1, and so (1) of this theorem occurs.Now let T ∼ = Z p for an odd prime p. Then k = p is coprime to |M |, and so |V | = 2k|M |.Noting that Γ M has odd valency k, it implies that Γ M has even order, and so |M | is odd.Moreover, by Lemma 13, X ∼ = G/M ∼ = (Z 2 k :Z l ).Z 2 is nonabelian, where l is a divisor of k − 1.Since |M | is square-free, M is soluble, and so G is soluble.Let F be the Fitting subgroup of G. Then C G (F ) F = 1.Suppose that F has at least three orbits on V .Since Γ is G-locally primitive and G-vertex-transitive, Γ is a normal cover of Γ F ; in particular, F has square-free order.Then G/F is isomorphic to a subgroup of AutΓ F acting transitively on the arcs of Γ F , and so G/F is not abelian.On other hand, since |F | is square-free, F is cyclic, and hence C G (F ) = F and Aut(F ) is abelian.Since G/F = N G (F )/C G (F ) Aut(F ), we know that G/F is abelian, a contradiction.Thus F has one or two orbits on V .Suppose that |F | is even.Let Q be the Sylow 2-subgroup of F .Then Q G. Consider the quotient Γ Q .Since |V | is square-free and Γ is G-vertextransitive, we get a graph of odd order k|M | and odd valency k, which is impossible.Then F has odd order, and hence F has exactly two orbits on V .
Assume |F | is divisible by k 2 .Let P be the Sylow k-subgroup of F .Then Z 2 k ∼ = Y = soc(X) = P G.By Lemma 29, we conclude that Γ ∼ = K k,k .This implies that M = 1, and Γ and G are described as in (1) of Lemma 13.Then (1) of this theorem occurs.
Assume that |F | is not divisible by k 2 .Then M = 1; otherwise Z 2 k ∼ = Y F , a contradiction.Since F has exactly two orbits on V , we know that |F | is divisible by k|M |.Let P be the Sylow k-subgroup of F .Then Z k ∼ = P G. Let q be the smallest prime divisor of |M |, and the let N be the q -Hall subgroup of M .Then N P is a normal subgroup of G.It is easy to see that N P is intransitive on both U and W . Then the quotient graph Γ N P is bipartite and of order 2q and valency k, and so q > k.Thus, since l is a divisor of k − 1, each possible prime divisor of l is less than q.Note that F M is a subgroup of G.  Assume that |L| has a prime divisor q such that either a Sylow q-subgroup of L is not normal in L or q is the smallest prime divisor of |L|.It is easily shown that L has a unique q -Hall subgroup N ; in particular, N is normal in L. Then N is normal in G, and N has q-orbits on each of U and W . Thus the quotient graph Γ N is bipartite and of order 2q and valency k.In particular, k q.Further, G/N = Z q :Z k is not abelain unless q = k.Since |N | is square-free, the outer automorphism group Out(N ) of N is abelian, refer to [12].Note that G/(N C G (N )) is isomorphic a quotient of a subgroup of Out(N ).Then G/(N C G (N )) is abelian.Thus either q = k, or N C G (N ) has order divisible by q.Suppose that q > k.Then q is not a divisor of |N | as N L and |L| is square-free.Note that N C G (N )/N ∼ = C G (N )/(N ∩ C G (N )).It follows that |C G (N )| is divisible by q.Let Q be a Sylow q-subgroup of C G (N ).Then Q is also a Sylow q-subgroup of G, and hence Q L.Moreover, N Q/N G/N , and so N Q G. Since N Q = N × Q, we know that Q G, which contradicts the choice of q.Therefore, q = k.This says that k is the smallest prime divisor of |G|, and either L ∼ = Z n or L ∼ = Z n k :Z k , where n = |L|.Thus G = Z n :Z k or Z n k :Z 2 k , and k is the smallest prime divisor of nk.Now let X ∼ = Z p :Z 2k .Then G has a normal regular subgroup R = L:Z 2 , and Γ is isomorphic a Cayley graph Cay(R, S), where S = {s τ i | 0 i k − 1} for an involution s ∈ R and an automorphism τ ∈ Aut(R) of order k such that S = R. Noting that |R| is square-free, it follows that R is a dihedral group, say D 2n .Then G = D 2n :Z k .Let q be the smallest prime divisor of n.Then G has a normal subgroup N with |G : N | = 2qk.It is easily shown that the quotient graph Γ N is bipartite and of valency k and order 2q.Then k q, and so k is the smallest prime divisor of nk.Thus part (2) follows.Now we are ready to give a proof of Theorem 4.
Proof of Theorem 4. Let Γ = (V, E) be a G-locally primitive arc-transitive graph, and let M G be maximal subject to that M has at least three orbits on V .Then Γ is a normal cover of Σ := Γ M .Note that Γ and Σ has even valency if |M | is even.
If G is soluble then, by Theorem 30, one of part (1) of Theorem 4 occurs.Thus we assume that G is insoluble.Then G = M :X, where T := soc(X) is a simple group descried in (3)-( 6) and (8) of Theorem 1.By Lemma 29, we conclude that either Γ is T -arc-transitive, or Γ is T -edge-transitive and T has exactly two orbits on V .We next consider the case where T = PSL(2, p) for a prime p 5.
Let ∆ be an M -orbit on V .Then either T ∆ is transitive on ∆; or T ∆ has exactly two orbits on ∆ and, in this case, T is intransitive on V and M × T is transitive on V .We take a normal subgroup N of G such that N = M if the first case occurs, or N is the 2 -Hall subgroup of M if the second case occurs.Let ∆ 1 be an N -orbit contained in ∆.Then T ∆ = T ∆ 1 is transitive on ∆ 1 and N is regular on ∆ 1 .Considering the action of N × T ∆ , we conclude that N ∼ = T ∆ /K, where K is the kernel of T ∆ on ∆ 1 .Note that T ∆ is known by Theorem 27, and that |V | = |T : T α | or 2|T : T α | is square-free.Then we get Table 3 by checking possible normal subgroups of T ∆ with square-free index.

( 6 )
soc(G) = PSL(2, p f ) with f 2 and p f > 9, and either k is divisible by p f −1 or f = 2 and k is divisible by p + 1;(7) soc(G) = Sz(2 f ) and k is divisible by 2 2f −1 ;(8) G is of Lie type defined over GF(p f ) with p k, and either(i) [ d 2 ]f < k,and G is a d-dimensional classical group with d 3; or the electronic journal of combinatorics 22(3) (2015), #P3.25

Case 1 .
Assume first that G has two distinct minimal normal subgroups N and M .Then N ∩ M = 1, and hence N M = N × M .Suppose that both N and M are transitive on V .By Lemma 11, N and M are regular on V ; in particular, |N | = |M | = |V |.Thus N and M are soluble, it implies that N ∼ = M ∼ = Z p for an odd prime p. Again by Lemma 11, N = M , a contradiction.

2 inT
. Thus T α ∼ = Z m for an even divisor m of p±1 2 .Then p(p ∓ 1) is a divisor of |T : T α |, and so |T : T α | is even.It follows that T is transitive on V .Note that |G α | = m or 2m.It follows that Γ has valency m, 2m or 4m.Then (i) or (ii) is associated with the case that T is transitive or intransitive on E, respectively.Lemma 25.Assume that |T α | is divisible by p. Then T α ∼ = Z p :Z l , T is transitive on V and Γ has valency divisible by p, where l is an even divisor of p−1 2 with p−1 2l odd.If Γ is G-locally primitive, then Γ is isomorphic to the graph in Example 19.
the electronic journal of combinatorics 22(3) (2015), #P3.25 Lemma 26.Assume that T α ∼ = D 2m with m > 1 coprime to p. Then m is a divisor of p±1 2 , and Γ has valency divisible by m 2 or m.If Γ is G-locally-primitive, then Γ has odd prime valency r, T α ∼ = D 2r or D 4r , and Γ is isomorphic to one of the graphs given in Example 20.

Case 2 .
soc(G/M ) ∼ = soc(X) = Y ∼ = Z p .Since Γ M is X-locally primitive, by Lemma 13, either X ∼ = Z p :Z k , or X ∼ = Z p :Z 2k and X is transitive on V M .Moreover,the electronic journal of combinatorics 22(3) (2015), #P3.25 |V M | = 2p, (p, |M |) = 1, p > k and k is an odd prime.Let L = M Y .Then L is a semiregular normal subgroup of G, and L has exactly two orbits U and W on V .Let X ∼ = Z p :Z k .Then |G| = kp|M | = k|L|.
and thus G can be viewed as a primitive subgroup of the affine group AGL(2, p).Since H U is an abelian normal subgroup of H, by [19, 2.5.10],H U is cyclic.It follows that H U Z p−1 .Since H U has index 2 in H, by [19, 2.5.7],H U is an irreducible subgroup of GL(2, p).Then, by [19, 2.3.2],|H U | is not a divisor of p − 1, a contradiction.Therefore, T is a nonabelian simple group.
the order |T | is divisible by (p f + 1) 2 and |T : T α | is divisible by p f + 1.If p 2f − 1 has a primitive prime divisor r, then |T | is divisible by r 2 , and |T α | is divisible by r, hence 2f < r m.Assume that p 2f − 1 has no primitive prime divisor.Then either f = 1 and 2f = 2 < k, or (p, 2f ) = (2, 6).For the latter, T = G 2 (8) or 3 D 4 (8), and so 9 is a divisor of |T : T α |, which contradicts that |T : T α | is square-free.Thus T is described as in part (ii) of this lemma.Assume that T is one of F 4 (p f ), 2 E 6 (p f ) and E 7 (p f ).Then |T | is divisible by (p 6f − 1) 2 and |T : T α | is divisible by p 6f −1 p f −1 .If p 6f − 1 has a primitive divisor r say, then r divides |T α |, and hence 6f < r