Electronic Journal of Graph Theory and Applications 12 .
, 89Ae103 Tetravalent non-normal Cayley graphs of order 5p2 Soghra Khazaei.
Hesam Sharifi Department of Mathematics.
Faculty of Science.
Shahed University.
Tehran.
Iran Correspondence: hsharifi@shahed.
Abstract In this paper, we explore connected Cayley graphs on non-abelian groups of order 5p2 , where p is a prime greater than 5, and Sylow p-subgroup is cyclic with respect to tetravalent sets that encompass elements with different orders.
We prove that these graphs are normal.
however, they are not normal edge-transitive, arc-transitive, nor half-transitive.
Additionally, we establish that the group is a 5-CI-group.
Keywords: Cayley graph, normal edge-transitive, arc-transitive, graph automorphism, group automorphism.
Mathematics Subject Classification : 20D60, 05B25 DOI: 10.
5614/ejgta.
Introduction and Preliminary Suppose G is a group and S is a subset of G that does not include 1.
The Cayley graph associated with (G.
S), denoted by Cay(G.
S), is a directed graph with the vertex set G and the edge set consisting of .
, .
OO GyG when uv Oe1 OO S.
We notice that the Cayley graph Cay(G.
S), may depend on the choice of S, and is connected if and only if S generates G.
Also, we care that the edge set can be identified with set of ordered pairs {.
, s.
OO G, s OO S}.
The Cay(G.
S) can be considered as an undirected graph when S is closed under taking inverse i.
S = S Oe1 .
The degree of each vertex is easily seen to be S.
Following the definitions in .
, the graph e is called vertex-transitive, edge-transitive or arc-transitive if the automorphism group Aut.
, acts Received: 10 December 2022.
Revised: 20 December 2023.
Accepted: 25 February 2024.
Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi transitively on vertex-set, edge-set or arc-set of e, respectively.
A half-transitive graph is one that is vertex-set and edge-set transitive, but not arc-transitive.
Assume that e = Cay(G.
S).
Let A(G) = {Ag | g OO G}, where for g OO G, the map Ag : G Ie G is given by Ag .
= xg.
It is evident that.
Ag OO Aut.
The set A(G) forms a subgroup isomorphic to G) in Aut.
Therefore, since A(G) Aut.
, acting right regularly on the vertices of e, e is vertex-transitive.
However e is not edge-transitive in general.
Some notations used here are as follows.
In a graph e, the distance between two vertices u and v, denoted by d.
, .
, is the number of edge in a shortest path connecting them.
Let us introduce the Di .
= .
OO V .
| d.
, .
= .
For Semi-direct product of K by H in which H act on K, we write K o H.
Zn denotes a cyclic group of order n as well as the ring of integers modulo n, by Zy n we mean the multiplicative group consisting of the elements in Zn , which are coprime to By Aut(G.
S) = {E OO Aut(G)|E(S) = S}.
It is easy to see that Aut(G.
S) is a subgroup of the automorphisms group of Cay(G.
S).
For two subsets S and T of G such that 1 OO
/ S, 1 OO
/ T,
Oe1 Oe1 S = S and T = T , if there is a f OO Aut(G) such that f (S) = T , then S and T said to be equivalent and it can verified that in this case we have Cay(G.
S) O = Cay(G.
T ) and Cay(G.
S) is normal if and only if Cay(G.
T ) is normal.
LetAos review some fundamental facts about normal edge-transitive Cayley graphs.
A Cayley graph e = Cay(G.
S) is called normal if A(G) is a normal subgroup of Aut.
, e.
NAut.
(A(G)) = Aut.
and e is called normal edge-transitive or normal arc-transitive if NAut.
(A(G)) is transitive on the edges or arcs of e, respectively.
Lemma 1.
Lemma 2.
) For a Cayley graph e = Cay(G.
S), we have NAut.
(A(G)) = A(G) o Aut(G.
S).
Therefore, e is normal edge-transitive when A(G) o Aut(G.
S) is transitive on the edge-set of e.
Lemma 1.
Proposition 1.
]) Consider the Cayley graph e = Cay(G.
S).
Then the following are equivalent:
e is normal edge-transitive.
S = T O T Oe1 , where T is an Aut(G.
S)-orbit in G.
A(G) o Aut(G.
S) is transitive on the arcs of e if and only if Aut(G.
S) is transitive on Lemma 1.
Proposition 1.
) Let A = Aut(Cay(G.
S)).
The following are equivalent:
A(G) A.
Aut.
= A(G) o Aut(G.
S).
A1 Aut(G.
S), where A1 is the stabilizer of the identity 1 of G in A.
Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi Definition 1.
A Cayley graph Cay(G,S) is called a CI-graph, if whenever Cay(G.
S) O = Cay(G.
T ) for some subset T of G then S be equivalent to T, i.
T = (S) for some OO Aut(G).
The group G is called an m-CI-group if every Cayley graph over G of valency at most m is a CI-graph, and G is a CI-group if every Cayley graph over G is a CI-graph.
In .
, all the normal edge-transitive Cayley graphs of modular groups of order 8n, where n is a natural number, are determind and in .
, 5, .
all the tetravalent edge-transitive Cayley graphs on non-Abelian groups of order p2 , 3p2 , 4p2 are determined.
In .
all connected cubic non-normal Cayley graphs of order 2p2 are studied.
In this paper, motivated by .
, 4, 6, .
, we determine the structure of Cayley graphs of Frobenius group G of order 5p2 with cyclic kernel of order p2 , with respect to tetravalent sets, i.
|S| = 4, such that exactly two elements are same order.
Let G be a finite group of order 5p2 with cyclic Sylow p-subgroups of order p2 , where p is a prime number greater than 5.
It is not difficult to see, by the Sylow theorems, if the Sylow p-subgroup of G is cyclic then G is isomorphic to hx, .
x5 = y p = 1, xOe1 yx = y k i where 1 < k < p2 , p - k, xyxOe1 = y k , k 5 O 1 .
odp2 ) and o.
i y j ) = 5, for 1 O i O 4, 0 O j < p2 .
Lemma 1.
Lemma 2.
) Suppose G is a finite group of order 5p2 with cyclic Sylow psubgroups of order p2 , where p is a prime number greater than 5.
Then Aut(G) O = Zp2 o Zy where Zp2 denotes the group of the units .
nvertible element.
of the ring Zp2 .
Proof.
Suppose that f is an automorphism of G.
Therefore for the generators x and y of G, f .
and f .
must be of order 5 and p2 , respectively.
In fact, f .
OO .
i y j | 1 i < 5, 0 j < p2 } and f .
OO .
j | .
, .
= .
We claim that f .
= xy j , 0 j < p2 .
We shall prove this claim by the following steps:
Step 1.
6= x2 y j , 0 j < p2 .
Suppose that f .
= x2 y j and f .
= y j .
On the other hand xOe1 yx = y k .
Thus, we have f .
k ) = f .
Oe1 y.
= f .
Oe1 f .
= y Oej xOe2 y j x2 y j = y Oej xOe1 .
Oe1 y j .
xy j = y Oej xOe1 .
kj )xy j = y Oej y k j y j = y k j .
Also, f .
= y j Ne f .
k ) = y kj .
Therefore, y kj = y k j Ne p2 | .
2 Oe .
j 0 Ne p2 | .
2 Oe .
Since p - k, we have p2 | k Oe 1, contradicting to 1 < k < p2 .
Thus f .
6= x2 y j with 0 j p2 .
Step 2.
6= x3 y j , 0 j < p2 .
Suppose that f .
= x3 y j for some 0 j < p2 .
Similar to the first step, we conclude p2 | k.
2 Oe .
, since 1 < k < p2 , p2 | k 2 Oe 1.
That is a contradiction.
Because if p2 | k 2 Oe 1 Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi is true then there are three cases.
First, if .
k 1 and .
k Oe 1, then this is invalid because p is odd.
Secondly, if p2 .
1 then p2 = k 1, due to 1 < k < p2 .
Beside, k 5 O 1 .
ode p2 ).
So we have: .
2 Oe .
5 O 1 .
ode p2 ).
But this implies Oe1 O 1 .
ode p2 ), which is Third case i.
, p2 .
Oe 1 does not occur when 1 < k < p2 .
Step 3.
6= x4 y j , 0 j < p2 .
Suppose that f .
= x4 y j for some 0 j < p2 .
Again, similar to the first case, we have p2 | k.
3 Oe .
In the way, p2 | k 3 Oe 1, makes a contradiction, owing to if p2 .
3 Oe 1 then p2 .
5 Oe k 2 , also, k 5 O 1 .
ode p2 ), thus p2 will divide k 2 Oe 1.
But, in Step 2 we showed that this does not happen.
Therefore, there are p2 cases for the image of f on x and the image of f on y has I.
2 ) cases, where I is the Euler function.
Hence, all states are totally p3 .
Oe .
Elements of S are of the form xi y j , with 0 O i O 4 and 0 O j < p2 .
Since G = hSi and the inverses of x3 y j and x4 y j are x2 y j and xy j , respectively, we conclude that S = Si with i OO .
, 2, 3, 4, .
, where if xi y j OO S then .
i y j )Oe1 OO S.
S1 = .
y j , xy j , .
y j )Oe1 , .
y j )Oe1 }, j 6O j 0 .
od p2 ).
S2 = .
y j , x2 y j , .
y j )Oe1 , .
2 y j )Oe1 }, j 0 6O j.
od p2 ).
S3 = .
2 y j , .
2 y j )Oe1 , x2 y j , .
2 y j )Oe1 }, .
6O j 0 .
od p2 )).
S4 = .
y j , .
y j )Oe1 , y j , y Oej }, .
0 , .
= 1.
S5 = .
2 y j , .
2 y j )Oe1 , y j , y Oej }, .
0 , .
= 1 Lemma 1.
Main Theore.
) Let G be a finite group of order 5p2 with cyclic Sylow psubgroup where p is a prime number greater than 5.
There exists exactly three tetravalent subsets Si of G, 1 i 3, such that for each i.
G = hSi i, all elements of Si are of order 5 and one of the following holds.
S1 = .
, xy, xOe1 , .
Oe1 }, and each element of S1 has order 5.
e = Cay(G.
S1 ) is normal, normal edge transitive and edge transitive, but it is not arc-transitive.
Aut(G.
S1 ) O = Z2 and Aut.
O = A(G) o Z2 .
S2 = .
2 , xy, xOe2 , .
Oe1 }, and each element of S2 has order 5.
e = Cay(G.
S2 ) is normal but it is not normal edge transitive and arc-transitive.
Aut(G.
S2 ) is trivial and Aut.
O A(G).
S3 = .
2 , x2 y, xOe2 , .
Oe1 }, and each element of S3 has order 5.
e = Cay(G.
S3 ) is normal, normal edge transitive and edge transitive, but it is not arc-transitive.
Aut(G.
S1 ) O
Z2 and Aut.
= A(G) o Z2 .
Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi We are interested in the Cayley graph e = Cay(G.
S) with |S| = 4, 1 OO / S.
S = S Oe1 .
G = hSi and elements of S are of order 5 and p2 .
Our main result is the following.
Main Theorem.
Let G be a finite group of order 5p2 with cyclic Sylow subgroups, where p is a prime number greater than 5 and let S be a subset of G satisfying the following conditions .
|S| = 4, .
S = S Oe1 , 1 OO / S, .
S contains elements of order 5 and elements of order p2 .
Then one of the following holds.
S4 = .
, y, xOe1 , y Oe1 }, where x and y are of order 5 and p2 , respectively.
e4 = Cay(G.
S4 ) is normal but it is not normal edge transitive, nor edge transitive, nor arc-transitive, nor normal arc-transitive.
Aut(G.
S4 ) O = Z2 and Aut.
4 ) O = A(G) o Z2 .
S5 = .
2 , y, xOe2 , y Oe1 }, where x2 and y are of order 5 and p2 , respectively.
e5 = Cay(G.
S5 ) is normal but it is not normal edge transitive, nor edge transitive, nor arc-transitive, nor normal arc-transitive.
Aut(G.
S5 ) O = Z2 and Aut.
5 ) O = A(G) o Z2 .
e4 = Cay(G.
S4 ) In this section Cayley graph Cay(G.
S4 ) is denoted by e4 .
Lemma 2.
S4 is equivalent to .
, y, xOe1 , y Oe1 }.
Proof.
LetAos remind that S4 = .
y j , y j , .
y j )Oe1 , y Oej }, where 0 j < p and .
0 , .
= 1.
It is sufficient to consider f OO Aut(G) such that f .
= xy j and f .
= y j .
Since .
0 , p2 ) = 1, there exists such an automorphism f .
Theorem 2.
Aut(G.
S4 ) O = Z2 .
Proof.
Suppose that f OO Aut(G.
S4 ).
By order of elements of S4 and proof of Lemma 1.
4, clearly f .
= x and for f .
we have two cases.
= y or f .
= y Oe1 .
In the first case, f = id and in the second case o.
) = 2.
Lemma 2.
If i OO Aut.
4 )g and i.
= xg, then i.
= xi g for 1 i 4.
Proof.
Because x4 g OO D1 .
and i preserves distance, so:
i x4 g OO i(D1 .
) = D1 .
) = D1 .
= .
g, x4 g, yg, y Oe1 .
= .
, x4 g, yg, y Oe1 .
Since i is one-to-one, so i.
OO .
4 g, yg, y Oe1 .
On the other hand .
2 g, x3 .
is an edge and x2 g OO D1 .
, x3 g OO D1 .
, so there exists a member of i (D1 .
) that is adjacent with a member of i (D1 .
But if i .
= yg or y Oe1 g, these members donAot exist.
Thus i .
= x4 g.
x2 g and x3 g are only vertices D1 .
and D1 .
that are adjacent, so i will fix Similarly, if i.
= xOe1 g, then i.
= xOei g for 1 i 4 and if i.
Oe1 .
= xg, then i.
Oei .
= xi g for 1 i 4.
Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi Lemma 2.
If i OO Aut.
4 )g , then i .
OO / .
g, y Oe1 .
Proof.
If i .
= yg, then according to the above description, i .
Oe1 .
6= xg, thus i .
Oe1 .
= y Oe1 g.
2 g, x3 .
is an edge and x2 g OO D1 .
, x3 g OO D1 .
, so there exists a member of i (D1 .
) = D1 .
that is adjacent with a member of i (D1 .
) = D1 .
Oe1 .
, but these members do not exist.
Similarly i .
6= y Oe1 g.
Lemma 2.
If i OO Aut.
4 )g and i.
= yg, then i.
= y i g for 1 i < p2 .
Proof.
We prove that if i OO Aut.
4 )yiOe2 g and i OO Aut.
4 )yiOe1 g , then i OO Aut.
4 )yi g .
Since i is one-to-one, i (D1 .
iOe1 .
) = D1 .
iOe1 .
and the fact that D1 .
iOe1 .
= .
iOe2 g, xiOe1 yg, x4 y iOe1 g, y i .
= .
iOe2 .
, xiOe1 yg, x4 y iOe1 g, y i .
, we have i xiOe1 yg , i x4 y iOe1 g , i y i g OO xiOe1 yg, x4 y iOe1 g, y i g .
We know, x3 y iOe1 g OO D1 .
4 y iOe1 .
, x2 y iOe1 g OO D1 .
iOe1 y.
3 y iOe1 g, x2 y iOe1 .
is an edge, therefore we have i x3 y iOe1 g OO i D1 x4 y iOe1 g = D1 .
x4 y iOe1 g ), i x2 y iOe1 g OO i D1 xy iOe1 g = D1 .
xy iOe1 g ) and .
3 y iOe1 .
, i .
2 y iOe1 .
) is an edge.
Since there is no element of D1 .
adjacent to an element of D1 .
y iOe1 .
or an element of D1 .
4 y iOe1 .
Therefore, we have i xy iOe1 g , i x4 y iOe1 g OO xy iOe1 g, x4 y iOe1 g .
Thus i .
= y i g and the lemma is proved.
Lemma 2.
If i OO Aut.
4 )g and i.
= y Oe1 g, then i.
= y Oei g for 1 i < p2 .
Proof.
Since i.
= y Oe1 g, i(D1 .
) = D1 .
) = D1 .
Oe1 .
, we have .
, i.
, i.
4 y.
, i.
} = i .
, xy Oe1 g, x4 y Oe1 g, y Oe2 g .
On the other hand x2 yg OO D1 .
, x3 yg OO D1 .
4 y.
2 yg, x3 y.
is an edge, thus we have i x2 yg OO i (D1 .
) = D1 .
), i x3 yg OO i D1 x4 yg = D1 .
x4 yg ) and .
2 y.
, i .
3 y.
) is an edge.
Due to the fact that, do not exists any element of D1 .
Oe2 .
adjacent to an element of D1 .
y Oe1 .
or D1 .
4 y Oe1 .
, we see that i .
, i x4 yg OO xy Oe1 g, x4 y Oe1 g .
Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi y 2 g.
y Oe1 g y Oe2 g.
Figure 1.
The state of g with its neighboring vertices in the conditions of Lemma 2.
There for i .
= y Oe2 g.
Now, itAos sufficient that prove: if i .
iOe2 .
= y Oe.
Oe.
g and i .
iOe1 .
= y Oe.
Oe.
g, then i .
= y Oei g.
In Figure 1, replace AygAy by Ayy iOe1 gAy and Ayy Oe.
Oe.
gAy.
since i .
iOe1 .
= y Oe.
Oe.
g and i is automorphism, we have i xy iOe1 g , i x4 y iOe1 g = xy Oe.
Oe.
g, x4 y Oe.
Oe.
g i y i g , i y iOe2 g = y Oei g, y Oe.
Oe.
g = y Oei g, i y iOe2 g .
Thus i.
= y Oei g.
Lemma 2.
If i OO Aut.
4 )g , then i.
= xg.
Proof.
We know that xg OO D1 .
, so we have i .
OO i (D1 .
) = D1 .
) = D1 .
= xg, x4 g, yg, y Oe1 g .
On the contrary assume that i.
6= xg.
Thus by Lemma 2.
3 we have i.
= xOe1 g, so we encounter with two cases for i.
Case 1.
= yg.
by definition of group, xOe1 yx = y k .
so xg, xy k g and y k g, xy k g are two edges of graph.
since i preserves distance, so we have:
2 = d.
k g, x.
= d.
, i.
) = d.
k g, xOe1 .
> 2 which is a contradiction.
Case 2.
= y Oe1 g.
by Lemma 2.
5, we have:
2 = d.
k g, x.
= d.
, i.
) = d.
Oek g, xOe1 .
> 2 which is a contradiction.
Therefore this case does not happen.
Using Lemma 2.
6, if i OO Aut.
4 )1 , then one of the following two cases occurs: i.
= x, i.
= y or i.
= x, i.
= y Oe1 .
Lemma 2.
For Cayley graph e4 , if i OO Ag , and i fixes all the elements of D1 .
, then it fix all the elements of D2 .
Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi Proof.
We know that D2 .
= .
2 g, x3 g, xy k g, xy Oek g, x4 yg, xyg, y 2 g, y Oe2 g, x4 y Oe1 g, xy Oe1 g, x4 y Oek g, x4 y k .
In fact by Lemmas 2.
6 and 2.
4, we conclude that i keeps fixed them.
Immediately, the following result is obtained.
Corollary 2.
If i OO Aut.
4 )1 and i fixes all the elements of S4 , then i = id.
Proof.
Since graph is connected, it suffices to show that for every natural i 2, the statement g 0 OO Di .
Ne i .
0 ) = g 0 By Lemma 2.
7 the statement is true for i = 2.
Now assume that the statement is true for 1 i n, and we will show that the statement holds for n 1.
Let g 0 OO Dn 1 .
Hence, there is a sequence of adjacent vertices 1 = g 0 0 , g 0 1 , .
, g 0 nOe1 , g 0 n , g 0 n 1 .
OO DnOe1 .
and gn0 OO Dn .
Therefore by Clearly gnOe1 nOe1 = gnOe1 andi .
n ) = By applying Lemma 2.
7 for g := gn and i gnOe1 = gnOe1 , we conclude that i gn 1 = gn 1 or equivalently i .
) = g .
Lemma 2.
If i OO Aut.
4 )1 such that i.
= x, i .
= y Oe1 and f OO Aut(G.
S4 ) is non trivial, then i = f Proof.
Since i f .
= 1, i f .
= x and i f .
= y, by Corollary 2.
1, the statement is Theorem 2.
Aut.
4 )1 O = Z2 .
Proof.
By Lemmas 2.
6 and 2.
8 and Corollary 2.
1, the proof is straightforward.
Therefore, the first part of the main theorem is a consequence of Theorems 2.
1 and 2.
2 and Lemmas 1.
2 and 1.
e5 = Cay(G.
S5 ) We remind that S5 = .
2 y j , .
2 y j )Oe1 , y j , y Oej } where .
0 , .
= 1.
In this section Cayley graph Cay(G.
S5 ) is denoted by e5 .
Lemma 3.
S5 is equivalent to .
2 , xOe2 , y, y Oe1 }.
Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi Proof.
It is sufficient, consider f OO Aut(G) such that f .
= xy i and f .
= y , where i have two following cases if j is even.
i = 2j .
4 Oe k 3 k 2 Oe k .
, p2 j i = 2 .
Oe k k Oe k .
, if j is odd.
We also set = j 0 .
Therefore, f .
2 ) = x2 y j and f .
= y j .
From now on, we use the above mentioned equivalent for S5 .
Theorem 3.
Aut(G.
S5 ) O = Z2 .
Proof.
If we consider f OO Aut(G.
S5 ), then by attention to the orders of elements of S5 and proof of Lemma 1.
4, we have, f .
2 ) = x2 and f .
equals to y or y Oe1 .
Assume that f .
= y.
Then x2 = f .
2 ) = f .
2 = xy i xy i = x2 y i.
Hence, p2 | i.
We know, p - k 1, i.
, .
, k .
= 1, because otherwise, for some integer r, k = rp Oe 1, we will have and so, p2 | k 5 Oe 1, thus 1 O k 5 .
od p2 ) O 5rp Oe 1.
hence, p2 | 5rp Oe 2, which is a Thus p2 | i, then i = 0.
Therefore, f .
= x and f .
= y, i.
, f = id.
Now assume that f .
= y Oe1 .
Since f .
2 ) = x2 , similar to the previous case i = 0.
Thus, f .
= x.
It means that f is an element of order 2.
Lemma 3.
If i OO Aut.
5 )g and i.
= x2 g, then i.
= xi g for 1 i 4.
Proof.
Since x3 g OO D1 .
and i preserves distance, so we have i x3 g OO i (D1 .
) = D1 .
) = D1 .
= .
2 g, x3 g, yg, y Oe1 .
= .
, x3 g, yg, y Oe1 .
Since i is one-to-one, i .
OO .
3 g, yg, y Oe1 .
On the other hand .
4 g, x.
is an edge and x4 g OO D1 .
, xg OO D1 .
, so there exists a member of i (D1 .
) that is adjacent with a member of i (D1 .
But if i.
= yg or y Oe1 g, such members do not exist.
Thus i .
= x3 g.
Also x4 g and xg are the only vertices in D1 .
and D1 .
respectively, which adjacent, so i fixes them.
Similarly, if i.
= x3 g, then i.
Oei .
= xi g for 1 i 4 and if i.
= x2 g, then i.
Oei .
= xi g for 1 i 4.
Lemma 3.
If i OO Aut.
5 )g , then i .
OO / .
g, y Oe1 .
Proof.
If i .
= yg, then according to above description, i .
6= x2 g, thus i .
= y Oe1 g.
Also .
4 g, x.
is an edge and x4 g OO D1 .
and xg OO D1 .
, so there exists a member of i (D1 .
) = D1 .
that is adjacent with a member of i (D1 .
) = D1 .
Oe1 .
, but there are no such elements.
Similarly i .
6= y Oe1 g and the result now follows.
Lemma 3.
If i OO Aut.
5 )g and i.
= yg, then i.
= y i g for 1 i < p2 .
Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi Proof.
we prove that If i OO Aut.
5 )yiOe2 g and i OO Aut.
5 )yiOe1 g , then i OO Aut.
5 )yi g for i 2.
Since i is one-to-one, i (D1 .
iOe1 .
) = D1 .
iOe1 .
and D1 y iOe1 g = y iOe2 g, x2 y iOe1 g, x3 y iOe1 g, y i g = i y iOe2 g , x2 y iOe1 g, x3 y iOe1 g, y i g .
i x2 y iOe1 g , i x3 y iOe1 g , i y i g OO x2 y iOe1 g, x3 y iOe1 g, y i g .
We know xy iOe1 g OO D1 x3 y iOe1 g , x4 y iOe1 g OO D1 x2 y iOe1 g and .
y iOe1 g, x4 y iOe1 .
is an edge, therefore we have i xy iOe1 g OO i D1 x3 y iOe1 g = D1 .
x3 y iOe1 g ) i.
4 y iOe1 .
OO i(D1 .
2 y iOe1 .
= D1 .
2 y iOe1 .
y iOe1 .
, i .
4 y iOe1 .
) is and edge.
Because there is no element of D1 .
adjacent to an element of D1 .
2 y iOe1 .
or an element of D1 .
3 y iOe1 .
So we have i x2 y iOe1 g , i x3 y iOe1 g OO x2 y iOe1 g, x3 y iOe1 g .
Thus i .
= y i g and the lemma is proved.
Lemma 3.
If i OO Aut.
5 )g and i.
= y Oe1 g, then i.
= y Oei g for 1 i < p2 .
Proof.
Since i.
= y Oe1 g, i (D1 .
) = D1 .
) = D1 .
Oe1 .
, we have .
, i.
2 y.
, i.
3 y.
, i.
} = i .
, x2 y Oe1 g, x3 y Oe1 g, y Oe2 g .
On the other hand, since x4 yg OO D1 .
2 y.
, xyg OO D1 .
3 y.
4 yg, x3 y.
is an edge, thus we have i x4 yg OO i D1 x2 yg = D1 .
x2 yg ), i .
OO i D1 x3 yg = D1 .
x3 yg ) and .
4 y.
, i .
) is an edge.
Due to fact that, do not exists any element of D1 .
Oe2 .
adjacent to an element of D1 .
2 y Oe1 .
or D1 .
3 y Oe1 .
, we see that i x2 yg , i x3 yg OO x2 y Oe1 g, x3 y Oe1 g .
There for i .
= y Oe2 g.
Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi y 2 g.
x4 g y Oe1 g y Oe2 g.
Figure 2.
The state of g with its neighboring vertices in the conditions of Lemma 3.
Now, itAos sufficient that prove: if i .
iOe2 .
= y Oe.
Oe.
g and i .
iOe1 .
= y Oe.
Oe.
g, then i.
= y Oei g.
In Figure 2, replace AygAy by Ayy iOe1 gAy and Ayy Oe.
Oe.
gAy.
since i .
iOe1 .
= y Oe.
Oe.
g and i is automorphism, we have i x2 y iOe1 g , i x3 y iOe1 g = x2 y Oe.
Oe.
g, x3 y Oe.
Oe.
g i y i g , i y iOe2 g = y Oei g, y Oe.
Oe.
g = y Oei g, i y iOe2 g .
Thus i .
= y Oei g.
Lemma 3.
If i OO Aut.
5 )g , then i.
= x2 g.
Proof.
We know x2 g OO D1 .
, so we have i x2 g OO i (D1 .
) = D1 .
) = D1 .
= x2 g, x3 g, yg, y Oe1 g .
On the contrary assume that i.
6= x2 g.
Thus by Lemma 3.
3, we have i.
= x3 g.
Then we have two cases for i.
Case 1.
= yg.
by definition of group, xOe1 yx = y k .
2 g, x2 y k .
and y k g, x2 y k g are two edges of graph.
since i preserves distance, so we have 2 = d.
k g, x2 .
= d.
, i.
) = d.
k g, x3 .
> 2, which is a contradiction.
Case 2.
= y Oe1 g.
similar to the previous case, we conclude 2 = d.
k g, x2 .
= d.
, i.
) = d.
Oek g, x3 .
> 2, which is a contradiction, therefore this case does not happen.
Corollary 3.
If i OO Aut.
5 )g , then i.
= xg.
Proof.
By applying Lemma 3.
6 for g := x2 g and g := x4 g, it follows.
Similar to Lemma 2.
7, we have the following lemma.
Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi Lemma 3.
For Cayley graph e5 , if i OO Ag , and i fixes all the elements of D1 .
, then it fixes all the elements of D2 .
Proof.
We know D2 .
= .
4 g, xg, x2 y k g, x2 y Oek g, x3 y k g, x3 y Oek g, x2 y Oe1 g, x3 y Oe1 g, x2 yg, x3 yg, y 2 g, y Oe2 .
Already by Lemma 3.
4 and Corollary 3.
1, we know, i keeps fixed them.
Corollary 3.
If i OO Aut.
5 )1 and i fixes all the elements of D1 .
, then i = id.
Proof.
It follows immediately from Lemma 3.
7 and graph connectivity.
Lemma 3.
Let i OO Aut.
5 )1 .
If i.
= y, then i = id.
If i.
= y Oe1 , then i = f , where f is non-identity element of Aut(G.
S5 ).
Proof.
The first case is obtained using Corollaries 3.
1, 3.
2 and graph connectivity.
In the second case, since i f .
= x, i f .
= y and i f OO Aut.
5 )1 , by Corollary 3.
2, i f = id and because f is of order 2, so we have i = f .
Theorem 3.
Aut.
5 )1 O = Z2 .
Proof.
The statement is a consequence of Lemma 3.
8 and Corollary 3.
Therefore.
Lemmas 1.
2, 1.
3 and Theorems 3.
1, 3.
2 prove the second part of the Main Theorem.
Lemma 3.
e1 e3 .
Proof.
Suppose that there exists an automorphism i : e1 Ie e3 , so for some g OO G, we have i .
= g.
On other hand, since e3 is vertex-transitive, there exsits O OO Aut.
3 ) such that O .
= 1, therefore.
O i .
= 1.
Thus, without loss of generality, we assume i .
= 1.
Now, since n i is isomorphism oand .
, .
is an edge of e1 , .
, i .
} is an edge of e3 .
Therefore.
Oe1 i .
OO x2 , x3 , x2 y, .
In the following, we show that i .
canAot be equal to any of these 4 elements.
Oe1 Case 1.
6= x2 .
Because, if i .
= x2 , then i .
2 ) = x4 , and i .
OO x2 y, .
On the other hand.
D2 .
2 ) O D3 .
= y Oe1 , y Oek , x2 y k in e1 .
Since i is isomorphism.
D2 .
2 )) O D3 .
) has 3 elements in e3 .
But.
Oe1 n 4 k2 Oe1 o D2 x O D3 x y = D2 x O D3 x y = x y ,y which is a contradiction with isomorphism i.
2 Oe1 Case 2.
6= x .
Because, if i .
= x , then i .
) = x , and i .
OO x y, .
Oe1 Oe1 Oe1 If i .
= .
, then i .
= x2 y.
On the other hand.
D2 .
3 ) O D3 .
Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi 3 Oek3 x y ,y ,y in e1 , then.
D2 .
4 ) O D3 .
has 3 elements, but D2 .
4 ) O D3 .
= x4 y k , y Oe1 , which is a contrary to the fact that an isomorphism.
So we consider i .
= x2 y.
suppose that E 0 OO Aut(G.
S1 ), so E 0 .
= xy.
E 0 .
= y Oe1 and E OO Aut(G.
S3 ) are non trivial.
Let O = E i E 0 .
O defines an isomorphism from e1 to e3 , such that O .
= 1 and O .
= x2 .
But by case 1, there is no such isomorphism.
Case 3.
6= x2 y.
we assume that i .
= x2 y.
Let O = i E 0 .
here E 0 OO Aut(G.
S1 ) is non trivia.
Obviosly.
O .
= 1 and O .
OO .
2 , x3 }.
But by case 1 and case 2, there is no such Oe1 Oe1 Case 4.
6= .
If i .
= .
, then clearly i .
OO .
2 , x3 }.
Let O = i E 0 similar to the previous case.
O is a isomorphism that O .
= 1 and O .
OO .
2 , x3 }, which is a contradiction with cases 1 and 2.
Therefore, by these four cases the proof is complete.
Lemma 3.
e4 e5 .
Proof.
Suppose that i : e4 Ie e5 is an isomorphism.
Without loss of generality, we assume that i.
= 1.
Based on Figures 1 and 2, we have i.
OO .
2 , x3 } and i.
OO .
, y Oe1 }.
LetAos consider both possible cases.
Case 1.
If i.
= y, then i.
i ) = y i .
However, since d.
k , .
= 2 in e4 , if i.
= x2 , then d.
k , x2 ) = d.
k ), i.
) = 2 in e5 , which leads to a contradiction.
Similarly, if i.
= x3 , then d.
k , x3 ) = d.
k ), i.
) = 2 in e5 , resulting in a contradiction.
Therefore, this case is not Case 2.
If i.
= y Oe1 , then clearly i.
Oe1 ) = y.
According to Lemma 2.
8 and Theorem 2.
there exists a non-trivial element f belongs to Aut(G.
S4 ) such that f .
= x and f .
= y Oe1 .
Let O = i f .
Consequently.
O is an isomorphism between the graphs e4 and e5 with O.
= 1 and O.
= i.
) = i.
Oe1 ) = y.
Therefore.
O induse an isomorphic between e4 and e5 with the condition O.
= y, which is a contradiction similar to Case 1.
Lemma 3.
Cayley graphs Cay(G,S) with |S| = 4 are CI-graph.
Proof.
We know that there are five subsets non equivalent for S.
By Lemma 1.
Aut.
1 )1 Aut.
2 )1 .
Therefore e1 e2 .
Also by Lemma 3.
9, e1 e3 .
On the other hand, by Main Theorem e4 and e5 are not edge transitive, but by Lemma 1.
5, e1 is edge transitive.
Consequently, e4 and e5 canAot isomorphism with e1 .
According Lemma 1.
Theorem 2.
2 and Theorem 3.
2, since Aut.
2 )1 Aut.
3 )1 .
Aut.
2 )1 Aut.
4 )1 and Aut.
2 )1 Aut.
5 )1 , so e2 e3 , e2 e4 And e2 e5 respectively.
Morever, e3 e4 and e3 e5 , because, by Lemma 1.
5, e3 is normal edge-transitive, while e4 and e5 arenAot normal edge transitive.
Finally, by Lemma 3.
10, e4 e5 .
This completes the proof.
Lemma 3.
G is a 5-CI-graph.
Proof.
According to the order and relation between the generators of the group, the order of elements of S canAot be 1,2,3 and 5.
Therefore by Lemma 3.
11 and Definition 1.
1, the statement Tetravalent non-normal Cayley graphs of order 5p2 Khazaei and H.
Sharifi Conclusion In this paper, we consider Cayley graphs on Frobenius group of orders 5p2 , where p > 5 is prime, with cyclic Sylow p-subgroup and with respect to tetravalent sets.
In .
, we investigate graph automorphism and group automorphism determining all connected tetravalent normal edge transitive Cayley graphs on non-Abelian groups of order 5p2 with respect to tetravalent sets and same order elements.
the main result of which was the form of Lemma 1.
In this paper, we have focus on the tetravalent sets with different orders.
We prove that these graphs are normal.
but, they are not normal edge-transitive, arc-transitive, nor half- transitive.
Also, we show that the group is a 5-CI-group.
This can be an interesting research problem to investigate Cayley graphs on Frobenius groups of order qp2 .
Acknowledgement The authors would like to express their gratitude to the referee for his/her exact comments which made it possible to present the paper in this final form.
References