J.
Indones.
Math.
Soc.
Vol.
No.
, pp.
117Ae126.
ON THE CUBIC EDGE-TRANSITIVE GRAPHS OF
ORDER 58p2 Mehdi Alaeiyan1 and Laleh Pourmokhtar2 Department of Mathematics.
Iran University of Science and Technology Narmak.
Tehran 16844 alaeiyan@iust.
Department of Mathematics.
Iran University of Science and Technology Narmak.
Tehran 16844 laleh pourmokhtar@iust.
Abstract.
A graph is called edge-transitive, if its full automorphism group acts transitively on its edge set.
In this paper, we inquire the existence of connected edge-transitive cubic graphs of order 58p2 for each prime p.
It is shown that only for p = 29, there exists a unique edge-transitive cubic graph of order 58p2 .
Key words: Edge-transitive graphs.
Symmetric graphs.
Semisymmetric graphs, sRegular graphs.
Regular coverings.
Abstrak.
Sebuah graf disebut transitif-sisi jika grup automorfisma penuhnya berlaku secara transitif pada himpunan sisinya.
Pada paper ini, kami meneliti keberadaan graf kubik transitif-sisi terhubung berorde 58p2 untuk setiap bilangan prima Kami tunjukkan bahwa hanya untuk p = 29 terdapat graf kubik transitif-sisi unik berorde 58p2 .
Kata kunci: Graf transitif-sisi, graf simetris, graf reguler-s, selimut reguler.
Introduction Throughout this paper, graphs are assumed to be finite, simple, undirected and connected.
For the group-theoretic concepts and notations not defined here we refer to Rose .
For a graph X, we denote its vertex set, edge set, arc set and full automorphism group of X by V (X).
E(X).
A(X) and Aut(X), respectively.
For u, v OO V (X), denote by .
, .
the edge incident to u and v in X.
Let G be a finite group and S a subset of G such that 1 OO / S and S = S Oe1 .
2010 Mathematics Subject Classification: 05C25, 20B25.
Received: 23-01-2015, revised: 28-09-2015, accepted: 29-09-2015.
Alaeiyan and L.
Pourmokhtar The Cayley graph X = Cay(G.
S) on G with respect to S is defined to have vertex set V (X) = G and edge set E(X) = {.
, s.
OO G, s OO S}.
Clearly.
Cay(G.
S) is connected if and only if S generates the group G.
The automorphism group Aut(X) of X contains the right regular representation GR of G, the acting group of G by right multiplication, as a subgroup, and GR is regular on V (X), that is.
GR is transitive on V (X) with trivial vertex stabilizers.
A graph X is isomorphic to a Cayley graph on a group G if and only if its automorphism group Aut(X) has a subgroup isomorphic to G, acting regularly on the vertex set.
An s-arc in a graph X is an ordered .
-tuple .
0 , v1 , .
, vsOe1 , vs ) of vertices of X such that viOe1 is adjacent to vi for 1 O i O s and viOe1 6= vi 1 for 1 O i < s.
A graph X is said to be s-arc-transitive if Aut(X) acts transitively on the set of its s-arcs.
In particular, 0-arc-transitive means vertex-transitive, and 1arc-transitive means arc-transitive or symmetric.
A graph X is said to be s-regular, if Aut(X) acts regularly on the set of its s-arcs.
Tutte .
showed that every finite connected cubic symmetric graph is s-regular for 1 O s O 5.
A subgroup of Aut(X) is said to be s-regular, if it acts regularly on the set of s-arcs of X.
If a subgroup G of Aut(X) acts transitively on V (X) and E(X), we say that X is G-vertex-transitive and G-edge-transitive, respectively.
In the special case, when G =Aut(X), we say that X is vertex-transitive and edge-transitive, respectively.
It can be shown that a G-edge-transitive but not G-vertex-transitive graph X is necessarily bipartite, where the two parts of the bipartition are orbits of G OAut(X).
Moreover, if X is regular then these two parts have the same cardinality.
A regular G-edge-transitive but not G-vertex-transitive graph will be referred to as a G-semisymmetric graph.
In particular, if G =Aut(X) the graph is said to be semisymmetric.
The classification of cubic symmetric graphs of different orders is given in many papers.
Ronald M.
Foster started collecting specimens of small cubic symmetric graphs prior to 1934, maintaining a census of all such graphs.
In 1988 the then current version of the census was published in a book entitled The Foster Census Foster .
, and contained data for the graphs on up to 512 vertices.
Conder .
, .
, the cubic s-regular graphs up to order 10000 are classified.
Throughout this paper, p and q are prime numbers.
The s-regular cubic graphs of some orders such as 2p2 , 4p2 , 6p2 , 10p2 were classified in Feng .
, 10, 11, .
Also, cubic s-regular graphs of order 2pq were classified in Zhou .
Also, we classified the cubic edge-transitive graphs of order 18p in Alaeiyan .
Furthermore, the study of semisymmetric graphs was initiated by Folkman .
For example, cubic semisymmetric graphs of orders 6p2 , 28p2 and 2pq are classified in Lu.
Alaeiyan and Du .
, 2, .
Now suppose that p is an odd prime.
Let N .
, p, .
= hxp = y p = z p = 1, .
, .
= z, .
, .
= .
, .
= 1i be a finite group of order p3 and G = ha, b, c, d | a2 = bp = cp = dp = .
, .
= .
, .
= .
, .
= 1, d = .
, .
, aba = bOe1 , aca = cOe1 i be a group of order 2p3 and S = .
, ab, a.
We write C(N .
, p, .
) = Cay(G.
S).
Feng .
Theorem 3.
C(N .
, p, .
) is a 2-regular graph of order 2p3 .
In this paper, we classify all the connected cubic edge-transitive .
ymmetric and also semisymmetri.
graphs of order 58p2 as follows.
Cubic symmetric graphs of order 58p2 Theorem 1.
Let p be a prime.
Then the only connected cubic edge-transitive graph of order 58p2 is the 2-regular graph C(N .
, 29, .
PRELIMINARIES
Let X be a graph and let N be a subgroup of Aut(X).
For u, v OO V (X), denote by .
, .
the edge incident to u and v in X, and by NX .
the set of vertices adjacent to u in X.
The quotient graph X/N or XN induced by N is defined as the graph such that the set of N -orbits in V (X) is the vertex set of X/N and B.
C OO are adjacent if and only if there exist u OO B and v OO C such that .
, .
OO E(X).
e is called a covering of a graph X with a projection Eo : X e IeX A graph X if there is a surjection Eo : V (X) Ie V (X) such that E.
NXf.
E) : NXe .
E) Ie NX .
is a bijection for any vertex v OO V (X) and vE OO EoOe1 .
The graph X is often called e of X with a projection Eo is said to be regular the base graph.
A covering graph X .
r K-coverin.
if there is a semiregular subgroup K of the automorphism group e such that graph X is isomorphic to the quotient graph X/K.
Aut(X) say by h, e Ie X/K and the quotient map X is the composition Eoh of Eo and h.
Proposition 2.
Lorimer .
Theorem .
Let X be a connected symmetric graph of prime valency and let G be an s-regular subgroup of Aut(X) for some s Ou 1.
a normal subgroup N of G has more than two orbits, then it is semiregular and G/N is an s-regular subgroup of Aut(XN ), where XN is the quotient graph of X corresponding to the orbits of N .
Furthermore.
X is an N -regular covering of XN .
The next proposition is a special case of Wang .
Proposition 2.
Proposition 2.
Let X be a G-semisymmetric cubic graph with bipartition sets U (X) and W (X), where G O A := Aut(X).
Moreover, suppose that N is a normal subgroup of G.
Then, .
If N is intransitive on bipartition sets, then N acts semiregularly on both U (X) and W (X), and X is an N -regular covering of a G/N -semisymmetric graph XN .
If 3 does not divide |A/N |, then N is semisymmetric on X.
Proposition 2.
DjokovicA .
Propositions 2-.
Let X be a connected cubic symmetric graph and G be an s-regular subgroup of Aut(X).
Then, the stabilizer Gv of v OO V (X) is isomorphic to Z3 .
S3 .
S3 y Z2 .
S4 , or S4 y Z2 for s = 1, 2, 3, 4 or 5.
Proposition 2.
MalnicU .
Proposition 2.
The vertex stabilizers of a connected G-semisymmetric cubic graph X have order 2r A 3, where 0 O r O 7.
Moreover, if u and v are two adjacent vertices, then the edge stabilizer Gu O Gv is a common Sylow 2-subgroup of Gu and Gv .
Alaeiyan and L.
Pourmokhtar Let G be a group.
If a, b OO G , then the commutator of a and b is the element abaOe1 bOe1 .
The commutator subgroup or derived subgroup of G is the subgroup generated by all the commutators of G and it is denoted by G0 or [G.
G].
Now, we have the following obvious facts in group theory.
Proposition 2.
Let G be a finite group and let p be a prime.
If G has an Abelian Sylow p-subgroup, then p does not divide |G0 O Z(G)|.
Proposition 2.
Wielandt .
Proposition 4.
Every transitive Abelian group G on a set E is regular and the centralizer of G in the symmetric group on E is G.
For a subgroup H of a group G, denote by CG (H) the centralizer of H in G and by NG (H) the normalizer of H in G.
Proposition 2.
Rose .
Lemme 4.
36 ] Let G be a finite group, and H G.
Then CG (H) is normal in NG (H), and NG (H)/CG (H) is isomorphic to a subgroup of AutH.
MAIN RESULTS
Let X be a cubic edge-transitive graph of order 58p2 .
By Tutte .
X is either symmetric or semisymmetric.
We now consider the symmetric case, and then we have the following lemma.
Lemma 3.
Let p be a prime and let X be a cubic symmetric graph of order Then X is isomorphic to the 2-regular graph C(N .
, 29, .
Proof.
By Conder .
, .
there is no symmetric graph of order 58p2 , where p < 7.
If p = 29, then by Feng .
Theorem 3.
X is isomorphic to the 2-regular graph C(N .
, 29, .
To prove the lemma, we only need to show that no cubic symmetric graph of order 58p2 exists, for p Ou 7 and p 6= 29.
We suppose to the contrary, that X is such a graph.
Set A := Aut(X).
Since X is symmetric, by Tutte .
X is at most 5-regular and by Proposition 2.
3, |Av | = 2sOe1 A 3 for some integer, 1 O s O 5 and hence |A| = 2s A 3 A 29 A p2 .
Let Q := Op (A) be the maximal normal p-subgroup of If |Q| = p2 , then by Proposition 2.
1, the quotient graph XQ of X corresponding to the orbits of Q is a cubic symmetric graph of order 58, which is impossible by Conder .
Thus |Q| = 1 or p.
First, suppose that |Q| = 1 and let N be a minimal normal subgroup of A.
N is unsolvable, then N O = T y T y A A A y T , where T is a non-Abelian simple group.
Since |A| = 2s A 3 A 29 A p2 , thus N O = T .
Suppose that N has more than two orbits in V (X).
By Proposition 2.
N is semiregular on V (X).
Thus |N | | 58p2 .
This forces that N is solvable, a contradiction.
It follows N has at most two orbits in Cubic symmetric graphs of order 58p2 V (X), implying 29p2 | |N |.
Since N is unsolvable, it is not a .
, .
-group.
Thus, |N | = 2t A 29 A p2 or 2t A 3 A 29 A p2 , where 1 O t O s.
Let q be a prime .
Then by Gorenstein .
, pp.
and Conway .
, a non-Abelian simple .
, p, .
-group is one of the following groups.
A5 .
A6 .
P SL.
, .
P SL.
, .
P SL.
, .
P SL.
, .
P SU .
, .
P SU .
, .
, .
with orders 22 A 3 A 5, 23 A 32 A 5, 23 A 3 A 7, 23 A 32 A 7, 24 A 32 A 17, 24 A 33 A 13, 25 A 33 A 7, 26 A 34 A 5.
This implies that for p Ou 7, there is no simple group of order 2t A29Ap2 .
Hence |N | = 2t A 3 A 29 A p2 .
Assume that T is a proper subgroup of N .
If T is unsolvable, then T has a non-Abelian simple composite factor T1 /T2 .
Since |T1 /T2 | | 2t .
p2 , by simple group listed in .
T1 /T2 cannot be a .
, 3, .
-, .
, 3, .
- or .
, 29, .
-group.
Thus.
T1 /T2 is a .
, 3, 29, .
-group.
One may assume that |T | = 2r A 3 A 29 A p2 or 2r A 3 A 29 A p, where r Ou 2.
Let |T | = 2r A 3 A 29 A p2 .
Then |N : T | O 8 because |N | = 2t A 3 A 29 A p2 .
Consider the action of N on the right cosets of by right multiplication, and the simplicity of N implies that this action is faithful.
It follows N O S8 and hence p O 7.
Since p Ou 7, one has p = 7 and hence N = 2t A 3 A 29 A 72 .
But by Conway .
, there is no non-Abelian simple group of order 2t A 3 A 29 A 72 , a Thus.
T is solvable and hence N is a minimal non-Abelian simple group, that is.
N is a non-Abelian simple group and every proper subgroup of N is solvable.
By Thompson .
Corollary .
N is one of the groups in Table 1.
can be easily verified that the order of the groups in Table 1 is not of the form 2r A 3 A 29 A p2 .
Thus |T | = 2r A 3 A 29 A p.
By the same argument as in the preceding paragraph .
eplacing N by T ).
T is one of the groups in Table 1.
Since |T | = 2r A 3 A 29 A p, the possible candidates for T is P SL.
, .
Clearly, m = p.
We show that |T | < 1025 .
If 29 - .
Oe .
/2, then .
Oe .
96, which implies that p O 193.
If p = 193, then 26 ||T |, a contradiction.
Thus p < 193 and hence p O 97 because .
Oe .
It follows that |T | O 96 A 29 A 97 = 270048.
If .
Oe .
/2, then p .
Consequently p O 47, implying |T | O 96 A 29 A 47 < 270048.
Thus, |T | O 214176.
Hence, by Conway .
, .
, is isomorphic to P SL.
, .
or P SL.
, .
Table I.
The possible for non-Abelian simple group N P SL.
, .
, m > 3 a prime and m2 6= 3 .
od p2 ) P SL.
, 2n ), n a prime P SL.
, 3n ), n an odd prime P SL.
, .
, n a prime Suzuki group Sz.
n ), n an odd prime
|N |
Oe .
n Oe .
1 n 2n 3 .
Oe .
A 33 A 24 22n .
n Oe .
It follows that p = 11 or 47 and hence |N | = 2t A 3 A 29 A 112 or 2t A 3 A 29 A 472 , which is impossible by Conway .
, pp.
Hence.
N is solvable and so elementary Abelian.
Since X has order 58p2 , by Proposition 2.
N is semiregular on V (X), implying |N | | 58p2 .
Consequently.
N O = Z2 , or Z29 , because |Q| = 1.
If N O = Z2 , then by Proposition 2.
XN is Alaeiyan and L.
Pourmokhtar O Z29 , then a cubic symmetric graph of odd order 29p2 , a contradiction.
If N = by Proposition 2.
1 the quotient graph XN is a cubic symmetric graph of order Let M/N be a minimal normal subgroup of A/N .
Since p Ou 7 and |A/N | = 2t A 3 A p2 , by the simple group listed in .
M/N is solvable and so elementary Abelian.
Again by Proposition 2.
M/N is semiregular on V (XN ), which implies that M/N O = Z2 .
Zp .
Zp yZp .
For the former by Proposition 2.
1, the quotient graph XM of X corresponding to the orbits of M is a cubic graph with an odd order p2 , a contradiction.
Thus M/N O = Zp .
Zp y Zp .
If p 6= 7, then, since p > 7.
M has a normal subgroup of order p or p2 , which is characteristic in M and hence is normal in A, because M is normal in A.
This contradicts our assumption that |Q| = 1.
Now, suppose that p = 7.
Consider the quotient graph XN .
Let T /N be a minimal normal subgroup of A/N .
Clearly.
T /N is solvable and so elementary Abelian.
By Proposition 2.
T /N is semiregular on V (XN ).
It implies |T /N | | 2 A 72 .
Consequently, |T /N | = 2, 7 or 72 .
If |T /N | = 2, then |T | = 58.
So, the quotient graph XT is a cubic symmetric graph with an odd order, a contradiction.
Now suppose that |T /N | = 7.
Thus |T | = 7 A 29.
If T be Abelian, then T O = Z7 y Z29 O Z203 and by Proposition 2.
X is a Z203 - covering of the Heawood graph.
But by Wang .
Theorem 1.
, there is no symmetric Z203 -covering of the Heawood graph, a contradiction.
Thus.
T is a non-Abelian group.
Let C = CA (N ) be the centralizer N in A.
Clearly C = N or C = T because T /N is a simple group.
If C = N , then by Proposition 2.
A/N Aut(N ) O = Z28 , a contradiction.
So C = T .
By Proposition 5, 7 - |T 0 O Z(T )| and hence T 0 O N = 1, where T 0 is the derived subgroup of T .
Also.
T 0 6= 1 and T 0 N .
Therefore.
T 0 O
= T 0 /(T 0 O N ) O
= T 0 N/N E T /N .
The simplicity of T /N implies T = T /N .
As T is characteristic in T and T A, we have T 0 A.
By Proposition 2.
1, the quotient graph XT 0 is a cubic symmetric graph of order 406.
But, by Conder .
there is no symmetric cubic graph of order 406, a contradiction.
Now, we show that |T /N | 6= 72 .
Let |T /N | = 72 , that is T /N O = Z7 y Z7 .
T = Z29 y Z7 y Z7 , then A has a normal subgroup of order 7 or 72 .
It contradicts with |Q| = 1.
So T is non-Abelian group.
Let C = CT (N ) be the centralizer N in T .
Then clearly N C.
Suppose that N = C.
Then by Proposition 2.
T /N is isomorphic to a subgroup of Aut(N ) O = Z28 , a contradiction.
Hence N C.
Since C/N T /N O = Z7 yZ7 .
So C/N O = T /N or Z7 .
If C/N = T /N , then by Proposition 5, 7 - |T 0 O Z(T )| and hence T 0 O N = 1, where T 0 is the derived subgroup of T .
Thus T 0 O = T 0 N/N T /N and so T 0 O = Z7 or T /N .
It implies |T 0 | = 7 or 72 .
If |T | = 7, then by a similar argument on the previous paragraph, we can get a Suppose now |T 0 | = 72 .
As T 0 is characteristic in T and T A.
Thus T 0 A.
Proposition 2.
1, the quotient graph XT 0 is a cubic symmetric graph of order 58.
But, by Conder .
there is no symmetric cubic graph of order 58, a contradiction.
Also.
If C/N O = Z7 , then by a similar argument on the case |T /N | = 7, we can get a contradiction.
Suppose now that Q O = Zp and let C = CA (Q) be the centralizer of Q in A.
By Proposition 2.
5, p - |C 0 O Z(C) and hence C 0 O Q = 1, where C 0 is the derived subgroup of C.
This force p2 - |C 0 |, because C 0 is normal in A.
It follows that C 0 Cubic symmetric graphs of order 58p2 has more than two orbits on V (X).
As C 0 is normal in A, by Proposition 2.
1, it is semiregular on V (X).
Moreover, the quotient graph XC 0 is a cubic graph and consequently, has even order.
Hence 2 - |C 0 | and since p2 - |C 0 |, the semiregularity C 0 implies |C 0 | | 29p.
Since the Sylow p-subgroups of A are Abelian, one has p2 | |C| and so |C/C 0 |.
Now let K/C 0 be a Sylow p-subgroup of the Abelian group C/C 0 .
As K/C 0 is characteristic in C/C 0 and C/C 0 A/C 0 , we have that K/C 0 A/C 0 .
Hence K is normal in A.
Clearly |K| = 29p2 because |Q| = p.
If p > 7, then K has a normal subgroup of order p2 , which is characteristic in K, hence is normal in A, contradicting the fact that Q O = Zp .
Hence p = 7.
Consider the quotient graph XQ .
By Proposition 2.
XQ is cubic symmetric graph and A/Q is an arc-transitive subgroup of Aut(XQ ).
Let T /Q be a minimal normal subgroup of A/Q.
If T /Q is unsolvable, then by Conway
T /Q O
= P SL.
, .
Let C = CT (Q) be the centralizer of Q in T .
Then C = Q or Q Z(T ).
If C = Q, then by Proposition 2.
T /Q is isomorphic to a subgroup of Aut(Q) O = Z6 , a contradiction.
Thus Q Z(C).
By Conway .
the Schur multiplier of P SL.
, .
is isomorphic to Z2 .
Thus, we have T O = T1 y Q where T1 is isomorphic to P SL.
, .
Since T1 is characteristic in T and T A.
has T1 is normal in A, implying A has an unsolvable minimal normal subgroup, a Again T /Q is solvable and so Abelian elementary.
By Proposition 1.
T /Q is semiregular on V (XQ ) an so |T /Q| | 2 A 29 A 7.
It implies |T /Q| = 2, 7 If |T /Q| = 2, then |T | = 14.
So the quotient graph XT is a cubic graph of odd order 29 A 7, a contradiction.
Also, if |T /Q| = 7, then the quotient graph XT is a cubic symmetric graph of order 58.
But, by Conway .
there is no symmetric cubic graph of order 58, a contradiction.
Therefore, |T /Q| = 29.
Let C = CT (Q) be the centralizer of T in Q.
Clearly Q C because Q is Abelian.
If Q = C, then by Proposition 2.
T /Q Aut(Q) O = Z6 , a contradiction.
Thus.
Q < C.
Since C/Q T /Q and T /Q O = Z29 , one has C/Q = T /Q, imply that Q Z(T ).
Let T 0 be the derived group of T .
Since the Schur multiplier of Z29 is trivial .
ee Atlas by Conway .
One has T 0 < T .
It follows that T = T 0 y Q, where T 0 O = Z29 .
Thus T O = Z29 y Z7 O = Z203 and by Proposition 2.
X is Z203 -covering of the Heawood But by Wang .
Theorem 1.
, we get a contradiction.
Hence, the result now follows.
Now, we consider the semisymmetric case, and we have the following result.
Lemma 3.
Let p be a prime.
Then, there is no cubic semisymmetric graph of order 58p2 .
Proof.
Let X be a cubic semisymmetric graph of order 58p2 .
Denote by U (X) and W (X) the bipartition sets of X, where |U (X)| = |W (X)| = 29p2 .
For p = 2, 3 by Conder .
there is no cubic semisymmetric graph of order 58p2 .
Thus we can assume that p Ou 5.
Set A := Aut(X) and also let Q := Op (A) be the maximal normal p-subgroup of A.
The automorphism group A acts transitively on the set U (X) .
nd also W (X)).
So by Proposition 2.
4, |A| = 2r A 3 A 29 A p2 , where 0 r 7.
Alaeiyan and L.
Pourmokhtar Let N be a minimal normal subgroup of A.
One can deduce that N is solvable.
Because if N is unsolvable, then N O = T y T y A A A y T = T k , where T is a non-Abelian .
, 3, .
, .
, 3, .
, 3, 29, .
-simple group .
ee Gorenstein .
For the two formers, since 32 - |N |, then k = 1.
So N O = T .
Since 3 - |A/N |, by Proposition 2.
N must be semisymmetric on X and then 29p2 | |N |, a contradiction.
For the latter, by a similar argument as in Lemma 3.
1, we get a contradiction.
Thus, we can assume that N is solvable, so elementary Abelian.
Clearly.
N acts intransitively on U (X) and W (X) and by Proposition 2.
2, it is semiregular on each Hence |N | | 29p2 .
So |N | = 29, p or p2 .
We show that |Q| = p2 as follows.
First.
Suppose that |Q| = 1.
It implies that N O = Z29 .
Now we consider XN be the quotient graph of X relative to N , where XN is a cubic A/N -semisymmetric graph of order 2p2 .
By Folkman .
XN is a vertex-transitive graph.
So XN is a cubic symmetric graph of order 2p2 .
Suppose that T /N is a minimal normal subgroup of A/N .
If T /N is not solvable, then by Conway .
T /N O = A5 or P SL.
, .
Thus, |T | = 22 A 3 A 5 A 29 or 23 A 3 A 7 A 29.
Since 3 does not divide A/T , then by Proposition 2.
T is semisymmetric on X.
Consequently, 52 or 72 | |T |, a contradiction.
Therefore.
T /N is solvable and so elementary Abelian.
First, suppose that p = 7, by Feng .
Lemma 3.
T /N is 7-subgroup Abelian elementary.
So |T /N | = 7 or 72 and hence |T | = 29 A 7 or 29 A 72 .
Now, let H be the Sylow 29-subgroup of T in Clearly H is normal in T , since T is characteristic in A.
So H is normal in A.
By Proposition 2.
2, the quotient graph XH is a cubic A/H- semisymmetric graph of order 98.
But, by Conder .
, there is no semisymmetric graph of order 98.
Therefore, we assume that p Ou 11.
If |T /N | = p2 , then |T | = 29p2 .
It is easily seen the Sylow p-subgroup of T is characteristic and consequently normal in A.
contradicts our assumption that |Q| = 1.
Therefore.
T /N acts intransitively on the bipartition sets of XN and by Proposition 2.
2, it is semiregular on each partition, which forces |T /N | | p2 .
Hence, |T /N | = p and so |T | = 29p.
We can deduce that A has a normal subgroup of order p, which is a contradiction.
Thus |Q| = We now suppose that |Q| = p.
Since |N | | 29p2 , then we have two cases:
= Z29 and N O = Zp .
Case I.
N O
= Z29 .
By Proposition 2.
XN is a cubic A/N -semisymmetric graph of order 2p2 .
Let T /N be a minimal normal subgroup of A/N .
Suppose first that p = 7.
By a similar argument as in the case |Q| = 1, we get a Now let p > 7.
Then by Feng .
Theorem 3.
, the Sylow p-subgroup of Aut(XN ) is normal, and also we know that A/N Aut(XN ).
Consequently, the Sylow psubgroup A/N is normal, say M/N .
It is easy to see that |M | = 29p2 .
Since p > 7, the Sylow p-subgroup of M is normal and hence characteristic in M .
Thus.
A has a normal subgroup of order p2 .
It contradicts our assumption that |Q| = p.
Case II.
N O
= Zp .
By Proposition 2.
XN is a cubic A/N -semisymmetric graph of order 58p.
Let T /N be a minimal normal subgroup of A/N .
By a similar way as above.
T /N is Cubic symmetric graphs of order 58p2 solvable and so elementary Abelian.
By Proposition 2.
T /N is semiregular.
implies that |T /N | | 29p.
If |T /N | = p, then |T | = p2 , a contrary to |Q| = p.
Hence |T /N | = 29 and so |T | = 29p.
By Proposition 2.
XT is a cubic A/T semisymmetric graph of order 2p.
Thus by a similar way as in Case I, we get a Therefore |Q| = p2 and so by Proposition 2.
X is a regular Q-covering of an A/Q-semisymmetric graph of order 58.
But, it is impossible because by Conder .
, .
there is no edge-transitive graph of order 58 The result now follows.
Proof of Theorem 1.
Now we complete the proof of the main theorem.
Let X be a connected cubic edge-transitive graph of order 58p2 , where p is a prime.
know that every cubic edge-transitive graph is either symmetric or semisymmetric.
Therefore, by Lemmas 3.
1 and 3.
2, the proof is completed.
REFERENCES