Counting Centralizers of a Finite Group with an Application in Constructing the Commuting Conjugacy Class Graph
aa r X i v : . [ m a t h . G R ] J a n COUNTING CENTRALIZERS OF A FINITE GROUP WITH ANAPPLICATION IN CONSTRUCTING THE COMMUTINGCONJUGACY CLASS GRAPH
A. R. ASHRAFI AND M. A. SALAHSHOUR ⋆ Abstract.
The set of all centralizers of elements in a finite group G is denoted by Cent ( G ) and G is called n − centralizer if | Cent ( G ) | = n . In this paper, the structure ofcentralizers in a non-abelian finite group G with this property that GZ ( G ) ∼ = Z p ⋊ Z p is obtained. As a consequence, it is proved that such a group has exactly [( p + 1) + 1]element centralizers and the structure of the commuting conjugacy class graph of G iscompletely determined. Keywords:
Commuting graph, commuting conjugacy class graph, element centralizer.
Primary: 20 C
15; Secondary: 20 D Introduction
Throughout this paper all groups are assumed to be finite and C G ( x ) denotes thecentralizer of an element x in G . If G is a group containing two subgroups H and N suchthat N ⊳ G , H ∩ N = 1 and G = HN then G is said to be the semi-direct product of H by N and we write G = N ⋊ H . The group G is called capable if there exists anothergroup H such that G ∼ = HZ ( H ) . Our other notations are standard and taken mainly from[9] and our calculations are done with the aid of GAP [12].The center of G is denoted by Z ( G ) and Cent ( G ) = { C G ( x ) | x ∈ G } . The group G is called n − centralizer if n = | Cent ( G ) | [1]. The study of finite groups with respectto the number of distinct element centralizers was started by Belcastro and Sherman [3].It is clear that | Cent ( G ) | = 1 if and only if G is abelian and there is no finite groupwith exactly two or three element centralizers. They proved that if GZ ( G ) ∼ = Z p × Z p then | Cent ( G ) | = p + 2 [3, Theorem 5].Let H be a graph with vertex set { , , · · · , k } and G i ’s, 1 ≤ i ≤ k , are disjoint graphsof order n i . Following Sabidussi [10], the graph H [ G , G , · · · , G k ] is formed by takingthe graphs G , G , · · · , G k and connect a vertex of G i to another vertex in G j whenever ⋆ Corresponding author (Email: [email protected]). is adjacent to j in H . The graph H [ G , G , · · · , G k ] is called the H − join of the graphs G , · · · , G k .The notion of commuting conjugacy class graph of a non-abelian group G , Γ( G ),was introduced by Mohammadian et al. [8]. This is a simple graph with non-centralconjugacy classes of G as its vertex set and two distinct vertices A and B are adjacentif and only if there are x ∈ A and y ∈ B such that xy = yx . In such a case, we alsosay that two conjugacy classes commute to each other. The authors of the mentionedpaper obtained some interesting properties of this graph among them a classification oftriangle-free commuting conjugacy class graph of a finite groups is given.A CA-group is a group in which every noncentral element has an abelian centralizer.The present authors [11], obtained the structure of the commuting conjugacy class graphof finite CA-groups. They proved that this graph is a union of some complete graphs. As aconsequence of this work, the commuting conjugacy class graph of dihedral, semi-dihedral,dicyclic and three other families of meta-cyclic groups were constructed.The following two theorems are the main results of this paper:
Theorem 1.1.
Suppose p is a prime number and G is a group with center Z such that GZ ∼ = Z p ⋊ Z p . Then G has exactly [( p + 1) + 1] element centralizers. Theorem 1.2.
Suppose p is a prime number and G is a group with center Z such that GZ ∼ = Z p ⋊ Z p . The commuting conjugacy class graph of G has one of the following types: (1) GZ is abelian. In this case, GZ ∼ = Z p × Z p and the commuting conjugacy class graphof G is isomorphic to M [ K n ( p − , · · · , K n ( p − , K m ( p − p ) , · · · , K m ( p − p ) ] , where thereare p + 1 copies of K n ( p − , p + p copies of K m ( p − p ) and M is the graph depicted inFigure 1a. Here, m = | Z | p and n = | Z | p . (2) GZ is not abelian. In this case, the commuting conjugacy class graph of G is isomorphicto M [ K n ( p − , · · · , K n ( p − , K n ( p − p ) ] , where there are p + p + 1 copies of K n ( p − ,a copy of K n ( p − p ) and M is depicted in Figure 1b. Here, p is an odd prime and n = | Z | p . Preliminary Results
The aim of this section is to prove some preliminary results which are crucial in nextsections. The greatest common divisor of positive integers r and s is denoted by ( r, s ).We start this section by the following simple lemma: p p p p + 1( p + 1) p ( p + 1) ( p + 1) (a) The Graph M . p p p + 1( p + 1) p ( p + 1) ( p + 1) (b) The Graph M . Figure 1.
The Graphs of M and M . Lemma 2.1.
Let p be a prime number and G be a group with center Z such that x p , y p ∈ Z . Then, for each positive integer m with this condition that m mod p ) , we have C G ( x m ) = C G ( x ) and C G ( y m ) = C G ( y ) .Proof. By our assumption, ( m, p ) = 1 and there are n and k such that mn + kp = 1.Since x p ∈ Z , C G ( x m ) ⊆ C G ( x mn ) = C G ( x − kp ) = C G ( xz ) = C G ( x ), as desired. Theproof of second one is similar and so it is omitted. (cid:3) Lemma 2.2.
Let G be a finite group with center Z , p | | G | and a, b ∈ G \ Z such that o ( aZ ) = o ( bZ ) = p and G = { a i b j z | z ∈ Z } . Then the following hold: ( a ) Suppose ≤ i, j ≤ p − and at least one of i, j is not divisible by p . Then a i b j = b j a i . ( b ) C G ( a s b ) = C G ( a t b ) , ≤ s, t ≤ p − , if and only if s = t .Proof. Our argument consists in showing:( a ) Suppose ≤ i, j ≤ p − , a i b j = b j a i and at least one of i, j is not divisible by p .Without loss of generality we can assume that j mod p ). Then ( j, p ) = 1 andso there are integers m and n such that mj + np = 1. Since b p ∈ Z , a i b = ba i whichmeans that a i ∈ Z . This implies that p = o ( aZ ) | i , contradict by our condition that1 ≤ i ≤ p − b ) If s = t and C G ( a s b ) = C G ( a t b ) then a s b ∈ C G ( a t b ) and so a s − t b = ba s − t . Again p = o ( aZ ) | s − t which is impossible. ence the result. (cid:3) The following result is a particular case of [7, Theorem 8.1(c)].
Theorem 2.3.
The non-abelian split extension Z ⋊ Z is not capable. Suppose p is a prime number. Then it is easy to see that there are only two groupsof order p that can be written as a semidirect product of two cyclic groups of order p .One of these groups is the abelian group L = Z p × Z p and another one is a grouppresented as L = h x, y | x p = y p = 1 , yx = x p +1 y i . , see [5, pp. 145,146] for details.For simplicity of our argument, it is useful to write the presentations of L and L in theform h x, y | x p = y p = 1 , yx = x rp +1 y i , where r = 0 , Lemma 2.4.
Suppose G = L or L . Then for every i, j , ≤ i, j ≤ p − , which are notsimultaneously zero, we have: o ( x i y j ) = p p | ( i, j ) p Otherwise . Proof.
The result is trivial for L . So, it is enough to assume that r = 1. By definition,for every integers i and j , 0 ≤ i, j ≤ p −
1, such that i, j are not simultaneously zero, wehave yx = yxx = ( x p +1 y ) x = x p +1 ( yx )= x p +1 ( x p +1 y )= x p +1) y. By the same way for each i , yx i = x i ( p +1) y and so y x i = y ( yx i )= y ( x i ( p +1) y ) = ( yx i ( p +1) ) y = ( x i ( p +1) y ) y = x i ( p +1) y . On the other hand, y j x i = x i ( p +1) j y j . Since x p = 1 and ( p + 1) j = p j + jp j − + · · · + j ( j − p + jp + 1. So,(2.1) y j x i = x ijp + i y j . his shows that ( x i y j ) = ( x i y j )( x i y j ) = x i ( y j x i ) y j = x i ( x ijp + i y j ) y j = x ijp x i y j . An inductive argument now proves that, for each k ,(2.2) ( x i y j ) k = x k ( k − ijp x ki y kj . Therefore, for every i, j with 0 ≤ i, j ≤ p − i, j are not simultaneously zero, wehave: o ( x i y j ) = p p | ( i, j ) p otherwise . This completes the proof. (cid:3) Proof of Theorem 1.1
In this section, the proof of our first main result will be presented. To do this, we needsome information about a group G with this property that | GZ ( G ) | = p , p is prime, and GZ ( G ) can be generated by two elements of order p . Lemma 3.1.
Let G be a finite group with center Z , p be prime and a, b ∈ G \ Z . Moreover,we assume that a p , b p ∈ Z and ( a i b j ) k = a k ( k − ijrp a ki b kj z where z ∈ Z , r = 0 , and i, j, k are positive integers. The following statements hold: i ) If ≤ i, j ≤ p − , then there are n j and s such that a ip b jp = ( a sp b p ) n j z , where n j mid p ) , ≤ s ≤ p − and z ∈ Z . ii ) If ≤ i ≤ p − , ≤ j ≤ p − and j mod p ) , then there are n j and s suchthat a ip b j = ( a sp b ) n j z , where n j mod p ) , ≤ s ≤ p − and z ∈ Z . iii ) If ≤ j ≤ p − , ≤ i ≤ p − and i mod p ) , then there are n i and s such that a i b jp = ( ab sp ) n i z , where n i mod p ) , ≤ s ≤ p − and z ∈ Z . iv ) If ≤ i, j ≤ p − , i mod p ) and j mod p ) , then for there are n j and s such that a i b j = ( a s b ) n j z , where n j mod p ) , s mod p ) , ≤ s ≤ p − and z ∈ Z . v ) If ≤ i, j ≤ p − , i mod p ) and j mod p ) , then there are n i and s suchthat a i b j = ( ab s ) n i z , where n i mod p ) , s mod p ) , ≤ s ≤ p − and z ∈ Z . roof. By assumption, ( a i b j ) k = a k ( k − ijrp a ki b kj z . In this equality, we assume that i = k = p and j = 1. Since a p ∈ Z , ( a p b ) p = b p z ′ in which z ′ ∈ Z . Also ( a p b ) p ( a p b ) =( a p b )( a p b ) p then ( b p z ′ )( a p b ) = ( a p b )( b p z ′ ) so a p b p = b p a p . Since a p , a p ∈ Z ,( a ip b jp ) p = a p ( p − ijrp a ip b jp z ∈ Z then ( a ip b jp ) p ∈ Z . Also we can show that ( a ip b j ) p , ( a i b j ) p ∈ Z . Now, each part of thelemma will be proved separately. i ) In this case ( j, p ) = 1 and so there are integers m j and k such that m j j + kp = 1.By our assumption, a p b p = b p a p and b p ∈ Z and hence,( a ip b jp ) m j = a im j p b jm j p = a im j p b (1 − kp ) p = a im j p b p z . Since ( a ip b jp ) p ∈ Z , a ip b jp = ( a im j p b p ) n j z . It is clear that p ∤ im j and so there areintegers s and k such that im j = kp + s . It is now easy to see that for a fixed positiveinteger j , i = i if and only if s = s . Since 1 ≤ i ≤ p −
1, 1 ≤ s ≤ p −
1. So, byabove discussion and this fact that a p ∈ Z , a ip b jp = ( a im j p b p ) n j z = ( a kp + sp b p ) n j z = ( a sp b p ) n j z, as desired. ii ) Since p ∤ j , ( j, p ) = 1. So, there are m j and k such that m j j + kp = 1 in which( m j , p ) = 1. By our assumption a p , b p ∈ Z and ( a i b j ) k = a k ( k − ijtp a ki b kj . Thus,( a ip b j ) m j = a mj ( mj − ijtp a im j p b jm j z = a im j p b − kp z = a im j p bz . Again since ( m j , p ) = 1, ( m j , p ) = 1 and so there are n j and t such that n j m j + tp =1 in which p ∤ n j . We now apply above discussion and this fact that ( a ip b j ) p ∈ Z toderive the equality a ip b j = ( a im j p b ) n j z . A similar argument shows that there existsan integer s such that 1 ≤ s ≤ p − a ip b j = ( a sp b ) n j z which completes ourargument. iii ) The proof is similar to ( ii ) and so it is omitted. v ) Since p ∤ j , ( j, p ) = 1. So there are m j and k such that m j j + kp = 1 in which( m j , p ) = 1. By our assumption b p ∈ Z and ( a i b j ) k = a k ( k − ijtp a ki b kj . Therefore,( a i b j ) m j = a mj ( mj − ijtp a im j b jm j z = a mj ( mj − ijtp + im j b − kp z = a u bz , where(3.1) u = m j ( m j − ijtp + im j = i (cid:18) m j ( m j − jtp + m j (cid:19) . Since ( m j , p ) = 1, ( m j , p ) = 1 and again there are n j and t such that m j n j + tp = 1in which p ∤ n j . By above discussion and this fact that ( a i b j ) p ∈ Z , a i b j = ( a u b ) n j z .It is now clear that p ∤ u and so p ∤ u . Choose integers s and k such that u = kp + s and p ∤ s . By Equation 3.1, j is fixed, one can see that i = i if and only if u = u ifand only if s = s . Since 1 ≤ i ≤ p −
1, 1 ≤ s ≤ p − a p ∈ Z , a i b j = ( a u b ) n j z = ( a kp + s b ) n j z = ( a s b ) n j z which completesthe proof of this part. v ) The proof is similar to iv and so it is omitted.Hence the result. (cid:3) Suppose p is a prime number. By a result of Baer [2], the abelian group Z p × Z p iscapable. If p is an odd prime, then the non-abelian group Z p ⋊ Z p is also capable [13].This means that there exists a group G such that GZ ( G ) ∼ = Z p × Z p or Z p ⋊ Z p . Inthe next theorem, we obtain the number and structure of the centralizers of G . We alsoproved in Theorem 2.3 that the non-abelian group Z ⋊ Z is not capable. In the nexttheorem, if p = 2, then we will assume that GZ ∼ = Z × Z . Theorem 3.2.
Suppose p is a prime number and G is a group with center Z such that GZ ∼ = Z p ⋊ Z p . Then the number of centralizers of G is ( p + 1) + 1 .Proof. By our assumption, it is enough to assume that G is non-abelian and Z = 1. Byour discussion before Lemma 2.4, we can see that GZ ∼ = Z p ⋊ Z p = h x, y | x p = y p = 1 , yx = x rp +1 y i . Thus, there are a, b ∈ G \ Z such that GZ = h aZ, bZ | ( aZ ) p = ( bZ ) p = 1 , ( bZ )( aZ ) = ( aZ ) rp +1 ( bZ ) i nd so G = { a i b j z | ≤ i, j ≤ p − , ba = a rp +1 bz ′ , z, z ′ , a p , b p ∈ Z } . We now apply Equation 2.2 to prove that(3.2) ( a i b j ) k = a k ( k − ijrp a ki b kj z, where z ∈ Z . In Equation 3.2, set i = k = p and j = 1. Since a p ∈ Z , there exists z ∈ Z such that ( a p b ) p = b p z . Therefore, a p b p = b p a p . (3.3)So, for every 0 ≤ i, j ≤ p − { a sp b tp z | ≤ s, t ≤ p − , z ∈ Z } ⊆ C G ( a ip b jp z ) . Note that | G | = p | Z | and for each x ∈ G \ Z , Z (cid:8) C G ( x ) (cid:8) G . So,(3.5) | C G ( x ) | = p | Z | or p | Z | or p | Z | . Suppose i and j are fixed and 0 ≤ i, j ≤ p −
1. By Equation 3.2, ( a i b j ) p ∈ Z . Thus,(3.6) { ( a i b j ) k z | ≤ k ≤ p − , z ∈ Z } ⊆ C G ( a i b j ) . Next we partition the elements of G into the following eight parts: V = { a i z | p | i & z ∈ Z } ,V = { a i z | p ∤ i & z ∈ Z } ,V = { b j z | p | j & z ∈ Z } ,V = { b j z | p ∤ j & z ∈ Z } ,V = { a i b j z | p | ( i, j ) & z ∈ Z } ,V = { a i b j z | p | i, p ∤ j & z ∈ Z } ,V = { a i b j z | p ∤ i, p | j & z ∈ Z } ,V = { a i b j z | p ∤ i, p ∤ j & z ∈ Z } . To compute the centralizer of non-central elements of G , the following cases will beconsidered:a) i = 0 and j = 0. Then for every 1 ≤ i ≤ p − { a k z | ≤ k ≤ p − , z ∈ Z } ⊆ C ( a i ) . ) p | i . Suppose i = sp such that 1 ≤ s ≤ p −
1. By Equation 3.4, A = { a k b tp z | ≤ k ≤ p − , ≤ t ≤ p − , z ∈ Z } ⊆ C G ( a i ) = C G ( a sp ) . Since | A | = p | Z | , by Equation 3.5 and Lemma 2.1, C G ( a p ) = C G ( a sp ) = { a k b tp z | ≤ k ≤ p − , ≤ t ≤ p − , z ∈ Z } . p ∤ i . By Lemma 2.2( a ), the element a commutes only with elements of V ∪ V andby Equations 3.7, 3.5 and Lemma 2.1, C G ( a ) = C G ( a i ) = { a k z | ≤ k ≤ p − , z ∈ Z } ≤ C ( a p ) . b) j = 0 and i = 0. Apply a similar argument as the case of (1). To do this, we considertwo cases that p | j or p ∤ j .3) p | j . Suppose j = tp , 1 ≤ t ≤ p −
1. Then, C G ( b p ) = C G ( b tp ) = { a sp b k z | ≤ k ≤ p − , ≤ s ≤ p − , z ∈ Z } . p ∤ j . In this case, we can easily see that, C G ( b ) = C G ( b j ) = { b k z | ≤ k ≤ p − , z ∈ Z } ≤ C ( b p ) . c) i = 0 and j = 0. The following four subcases are considered into account.5) p | i and p | j . Suppose i = i p and j = j p in which 1 ≤ i , j ≤ p −
1. ByLemma 3.1(I), there are n j and s such that a i p b j p = ( a sp b p ) n j z , where p ∤ n j ,1 ≤ s ≤ p − z ∈ Z . By Equation 3.2, ( a sp b p ) p ∈ Z and by Lemma (2.1), C G ( a i p b j p ) = C G (( a sp b p ) n j z ) = C G (( a sp b p ) n j ) = C G ( a sp b p ) . Since 1 ≤ s ≤ p −
1, the number of distinct proper centralizers constructed from theelements of V is at most p −
1. We now obtain the structures of these centralizers.In Equation 3.2, we put i = s , j = 1 and k = p . Then ( a s b ) p = a sp b p z in which z ∈ Z . Thus, ( a sp b p )( a s b ) = ( a s b )( a sp b p ) and by Equation 3.4, A = { a ip ( a s b ) k b jp z | ≤ i, j, k ≤ p − , z ∈ Z } ⊆ C G ( a sp b p ) . It can easily see that | A | = p | Z | and by applying Equations 3.2 and 3.5,(3.8) C G ( a sp b p ) = { a ip a sk b k b jp z | ≤ i, j, k ≤ p − , z ∈ Z } . Suppose 1 ≤ t ≤ p −
1. By Equation 3.8,(3.9) a t b ∈ C G ( a sp b p ) if and only if t = s. Therefore, C G ( a tp b p ) = C G ( a sp b p ), 1 ≤ s, t ≤ p −
1, if and only if t = s . This showsthat the number of distinct proper centralizers is exactly p − ) p | i and p ∤ j . Suppose i = i p , where 1 ≤ i ≤ p −
1. By Lemma 3.1(II), there are n j and s such that a i p b j = ( a sp b ) n j z in which p ∤ n j , 1 ≤ s ≤ p − z ∈ Z . Nowby Equation 3.2, ( a sp b ) p ∈ Z . Hence by Lemma 2.1, C G ( a i p b j ) = C G (( a sp b ) n j z )= C G (( a sp b ) n j ) = C G ( a sp b ). Since 1 ≤ s ≤ p −
1, by Lemma 2.2(b), the numberof distinct proper centralizers is exactly p −
1. To obtain the structures of thesecentralizers, it is enough to apply Equation 3.6. This implies that A = { ( a sp b ) k z | ≤ k ≤ p − , z ∈ Z } ⊆ C ( a sp b ) . On the other hand, by Equation 3.2, ( a sp b ) k = a spk b k z ′ in which z ′ ∈ Z . We assumethat k = tp + k in which 0 ≤ t, k ≤ p −
1. Thus, ( a sp b ) k = a spk b k z ′ = a spk b k b tp z ′′ ,where z ′′ ∈ Z . So, A = { a spk b k b tp z | ≤ k , t ≤ p − , z ∈ Z } ⊆ C ( a sp b ) . By Lemma 2.2(a), a sp b cannot commute with an element of V ∪ V ∪ V ∪ V ∪ V .Furthermore, it is easy to see that V ⊆ A . We claim that there are no ele-ments x ∈ V and y ∈ V such that xy = yx . To see this, we assume that x = a i p b j ∈ V and y = a i b j ∈ V are arbitrary. By Lemma 3.1( iii , iv ), a i p b j = ( a sp b ) n j z and a i b j = ( a t b ) n j z , where p ∤ n j , p ∤ n j , p ∤ s and p ∤ t .By Equation 3.2, ( a sp b ) p , ( a t b ) p ∈ Z and by Lemma 2.1, C G ( a i p b j ) = C G ( a sp b )and C G ( a i b j ) = C G ( a t b ). Thus, ( a i p b j )( a i b j ) = ( a i b j )( a i p b j ) if and only if( a sp b )( a t b ) = ( a t b )( a sp b ) if and only if a sp ba t = a t ba sp if and only if a sp − t b = ba sp − t .We now apply Lemma 2.2(a) to deduce that p | t which is impossible. Next weprove that C ( a sp b ) = A . Choose x ∈ C ( a sp b ) \ A . By above discussion, x ∈ V andso we can write x = a ip b j , where 1 ≤ i ≤ p −
1. By Lemma 3.1( ii ), there are n j and m such that a ip b j = ( a mp b ) n j z with this condition that p ∤ n j , 1 ≤ m ≤ p − z ∈ Z . On the other hand, By Equation 3.2, ( a mp b ) p ∈ Z . So, by Lemma 2.1, C ( a ip b j ) = C ( a mp b ). Since x ∈ C ( a sp b ), a sp b ∈ C G ( x ) = C ( a mp b ). Hence, a ( s − m ) p b = ba ( s − m ) p . Again by Lemma 2.2(a), p | s − m and since 1 ≤ s, m ≤ p − s = m .This proves that C ( a sp b ) = A . Therefore, C ( a sp b ) = { a spk b k b tp z | ≤ k, t ≤ p − , z ∈ Z } . p | j and p ∤ i . A similar argument as Case (6) shows that the number of distinctproper centralizers of G is equal to p − C G ( ab tp ) = { a sp a k b tpk z | ≤ k, s ≤ p − , z ∈ Z } . p ∤ i and p ∤ j . By Lemma 3.1( iv ), there are n j and s such that a i b j = ( a s b ) n j z in which p ∤ n j , p ∤ s , 1 ≤ s ≤ p − z ∈ Z . By Equation 3.2, ( a s b ) p ∈ Z . hus, by Lemma 2.1, C G ( a i b j ) = C G (( a s b ) n j z ) = C G (( a s b ) n j ) = C G ( a s b ). Since p ∤ s , 1 ≤ s ≤ p − V is equal to p − p . We now obtain the structure of centralizers. ByEquations 3.6 and (3.2), A = { ( a s b ) k z | ≤ k ≤ p − , z ∈ Z } ⊆ C ( a s b )= { a k ( k − sp + sk b k z | ≤ k ≤ p − , z ∈ Z } . (3.10)By Lemma 2.2(a) and definition of V , one can see that a s b cannot be commutedwith an elements in V ∪ V ∪ V ∪ V ∪ V ∪ V . We claim that C ( a sp b ) = A . To prove,we assume that x ∈ C G ( a s b ) \ A . By above discussion, x ∈ V ∪ V . If x ∈ V , then x = a i ′ b j ′ and by Lemma 3.1( iv ), there are n j ′ and t such that a i ′ b j ′ = ( a t b ) n j ′ z inwhich p ∤ n j ′ , p ∤ t and z ∈ Z . Hence by Equation 3.2, ( a t b ) p ∈ Z and by Lemma2.1, we have C G ( a i ′ b j ′ ) = C G ( a t b ). Since x ∈ C G ( a s b ), a s b ∈ C G ( x ) = C G ( a t b )and so a s − t b = ba s − t . Apply Lemma 2.2(a) to deduce that p | s − t . Since1 ≤ s, t ≤ p − s = t and x = a i ′ b j ′ = ( a s b ) n j ′ z ∈ A . If x ∈ V , then x = a i ′ p b j ′ p and by Lemma 3.1( i ), there are n j ′ and t such that a i ′ p b j ′ p = ( a tp b p ) n j ′ z in which p ∤ n j ′ , p ∤ t and z ∈ Z . Next by Equation 3.2, ( a tp b p ) p ∈ Z and Lemma 2.1 impliesthat C G ( a i ′ p b j ′ p ) = C G ( a tp b p ). Since x ∈ C G ( a s b ), a s b ∈ C G ( x ) = C G ( a tp b p ) and byEquation 3.9, one can see that t = s . Hence Equation 3.2 shows that x = a i ′ p b j ′ p =( a sp b p ) n j ′ z = ( a s b ) n j ′ p z ∈ A . Therefore, in both cases x ∈ A which means that C G ( a s b ) = A = { ( a s b ) k z | ≤ k ≤ p − , z ∈ Z } . In addition, by Equations 3.8 and 3.10, we can see that C G ( a s b ) ≤ C G ( a sp b p ).Therefore, the structure of non-trivial proper centralizers of G is as follows: C G ( a p ) = { a k b jp z | ≤ j ≤ p − , ≤ k ≤ p − , z ∈ Z } ,C G ( a ) = { a k z | ≤ k ≤ p − } ≤ C G ( a p ) ,C G ( b p ) = { a ip b k z | ≤ i ≤ p − , ≤ k ≤ p − , z ∈ Z } ,C G ( b ) = { b k z | ≤ k ≤ p − } ≤ C G ( b p ) ,C G ( a sp b p ) = { a ip a sk b k b jp z | ≤ i, j, k ≤ p − , z ∈ Z } , ≤ s ≤ p − C G ( ab sp ) = { a ip a k b spk z | ≤ i, k ≤ p − , z ∈ Z } ≤ C G ( a p ) , ≤ s ≤ p − C G ( a sp b ) = { a spk b k b jp z | ≤ j, k ≤ p − , z ∈ Z } ≤ C G ( b p ) , ≤ s ≤ p − C G ( a s b ) = { ( a s b ) k a isp b ip z | ≤ i, k ≤ p − , z ∈ Z } ≤ C G ( a sp b p ) , ≤ s ≤ p − , p ∤ s o, the number of distinct proper centralizers of G is ( p − p )+3( p − p +2 p +1 =( p + 1) and therefore | Cent ( G ) | = ( p + 1) + 1. This completes the proof. (cid:3) We end this section by the following conjecture:
Conjecture 3.3.
Suppose p is a prime number, n is a positive integer and G is a groupwith center Z such that GZ ∼ = Z p n ⋊ Z p n . Then | Cent ( G ) | = ( p + 1) n + 1 . Proof of Theorem 1.2
In this section, we apply the results of Section 3 to obtain the structure of the commutingconjugacy class graph of G when GZ ( G ) ∼ = Z p ⋊ Z p . Theorem 4.1.
Suppose p is a prime and G is a group with center Z such that GZ ∼ = Z p × Z p . Then the commuting conjugacy classes graph of G has the graph structure M [ p +1 z }| { K n ( p − , · · · , K n ( p − , p + p z }| { K m ( p − p ) , · · · , K m ( p − p ) ] , where M is the graph depicted in Figure 2. Here, m = | Z | p and n = | Z | p . p p p p + 1( p + 1) p ( p + 1) ( p + 1) Figure 2.
The Graph M . Proof.
Since GZ is abelian, xgZ = xZgZ = gZxZ = gxZ , where x, g ∈ G \ Z are arbitrary.Hence there exists z ∈ Z such xg = gxz and so for every x ∈ G \ Z ,(4.1) x G = { g − xg | g ∈ G } = { g − gxz | z ∈ Z } = { xz | z ∈ Z } . We now apply Theorem 3.2 to prove that the group G has the following structure: G = { a i b j z | ≤ i, j ≤ p − , ba = abz t , z, z t ∈ Z } . ence for each x ∈ G \ Z , x = a i b j z , where 0 ≤ i, j ≤ p − i, j are not simulta-neously zero. To obtain the non-central conjugacy classes of G , the following have to beinvestigated:a) i = 0 and j = 0 . We will consider two cases that p | i and p ∤ i .1) p | i . In this case, i = sp , where 1 ≤ s ≤ p −
1. By Theorem 3.2, | C G ( a sp ) | = p | Z | and so | ( a sp ) G | = p . By Equation 4.1, ( a sp ) G = { a sp z , a sp z , · · · , a sp z p } = a sp H ,where H = { z , z , · · · , z p } ⊆ Z . Suppose | Z | > p and choose z r ∈ Z \ H . It is easyto see that a sp z r ( a sp ) G and so ( a sp z r ) G = ( a sp ) G . Since C G ( a sp z r ) = C G ( a sp ), | ( a sp z r ) G | = | ( a sp ) G | = p . Also, ( a sp z r ) G = ( a sp ) G z r = a sp Hz r . Since ( a sp ) G ∩ ( a sp z r ) G = ∅ , H ∩ Hz r = ∅ and H ∪ Hz r ⊆ Z . We now choose an element z r ∈ Z \ ( H ∪ Hz r ) and by above method we will see that p | | Z | . Suppose n = | Z | p . Thus,there are n distinct conjugacy classes of the form ( a sp z r ) G , ( a sp z r ) G , . . . , ( a sp z r n ) G .Since 1 ≤ s ≤ p −
1, there are n ( p −
1) distinct conjugacy classes with n = | Z | p andeach of which has p elements.2) p ∤ i . Then by Theorem 3.2, | C G ( a i ) | = p | Z | and so | ( a i ) G | = p . Also, byEquation 4.1, ( a i ) G = { a i z , a i z , · · · , a i z p } = a i H , where H = { z , z , · · · , z p } ⊆ Z . Suppose that | Z | > p and choose z r ∈ Z \ H . It is easy to see that a i z r ( a i ) G and hence ( a i z r ) G = ( a i ) G . Since C G ( a i z r ) = C G ( a i ), | ( a i z r ) G | = | ( a i ) G | = p . Also, ( a i z r ) G = ( a i ) G z r = a i Hz r . Note that ( a i ) G ∩ ( a i z r ) G = ∅ . Thus, H ∩ Hz r = ∅ and H ∪ Hz r ⊆ Z . Choose the element z r ∈ Z \ ( H ∪ Hz r ).By repeated applications of above method we can see that p | | Z | . Define m = | Z | p . Then for constant i , there are m distinct conjugacy classes in the form of( a i z r ) G , ( a i z r ) G , . . . , ( a i z r m ) G . Since 1 ≤ i ≤ p − p ∤ i , there are m ( p − p )distinct conjugacy classes, where m = | Z | p and each of which has p elements.b) i = 0 and j = 0 .
3) If p | j , then j = tp , where 1 ≤ t ≤ p −
1. By Theorem 3.2, | C G ( b tp ) | = p | Z | and so | ( b tp ) G | = p . Then by Equation 4.1 and a similar argument as (1), for each t thereare n distinct conjugacy classes of the form ( b tp z r ) G , ( b tp z r ) G , . . . , ( b tp z r n ) G , where n = | Z | p . Since 1 ≤ t ≤ p −
1, there are n ( p −
1) distinct conjugacy classes with n = | Z | p and each class has p elements.4) If p ∤ j , then by Theorem 3.2, | C G ( b j ) | = p | Z | and so | ( b j ) G | = p . Apply againEquation 4.1 and a similar argument as (2) to result that for each constant j , thereare m distinct conjugacy classes of the form ( b j z r ) G , ( b j z r ) G , . . . , ( b j z r m ) G , where m = | Z | p . Since 1 ≤ j ≤ p − p ∤ j , there are m ( p − p ) distinct conjugacyclasses with m = | Z | p and each class has p elements. ) i = 0 and j = 0. This case can be separated into the following four cases:5) p | i and p | j . In this case, i = sp and j = tp in which 1 ≤ s, t ≤ p −
1. ByTheorem 3.2, | C G ( a sp b tp ) | = p | Z | and so | ( a sp b tp ) G | = p . We now apply Equation4.1 and a similar argument as (1) for constants s and t to deduce that there are n distinct conjugacy classes of the form ( a sp b tp z r ) G , ( a sp b tp z r ) G , . . . , ( a sp b tp z r n ) G ,where n = | Z | p . Since 1 ≤ s, t ≤ p −
1, there are n ( p − distinct conjugacy classeswith n = | Z | p and each class has p members.6) p ∤ i and p | j . Clearly j = tp , 1 ≤ t ≤ p −
1. By Theorem 3.2, | C G ( a i b tp ) | = p | Z | and hence | ( a i b tp ) G | = p . On the other hand, Equation 4.1 and a similar argumentas (2) for constants i and t , show that there are m distinct conjugacy classes of theform ( a i b tp z r ) G , ( a i b tp z r ) G , . . . , ( a i b tp z r m ) G , where m = | Z | p . Since 1 ≤ t ≤ p − ≤ i ≤ p − p ∤ i , there are mp ( p − distinct conjugacy classes with m = | Z | p and each class has p members.7) p | i and p ∤ j . In this case i = sp , for some integer s such that 1 ≤ s ≤ p − | C G ( a sp b j ) | = p | Z | and so | ( a sp b j ) G | = p . Then by Equation4.1 and a similar argument as (2) for constants s and j , one can prove that thereare m distinct conjugacy classes of the form ( a sp b j z r ) G , ( a sp b j z r ) G , . . . , ( a sp b j z r m ) G ,where m = | Z | p . Since 1 ≤ s ≤ p −
1, 1 ≤ j ≤ p − p ∤ j , there are mp ( p − distinct conjugacy classes with m = | Z | p and each class has p members.8) p ∤ i and p ∤ j . By Theorem 3.2, | C G ( a i b j ) | = p | Z | and so | ( a i b j ) G | = p . Onthe other hand, by Equation 4.1 and applying a similar argument as (2) for con-stants i and j , one can see that there are m distinct conjugacy classes of the form( a i b j z r ) G , ( a i b j z r ) G , . . . , ( a i b j z r m ) G , where m = | Z | p . Since 1 ≤ i, j ≤ p − p ∤ i, j , there are mp ( p − distinct conjugacy classes with m = | Z | p and each classhas length p .The above discussion are summarized in Table 1.We now obtain the commuting conjugacy class graph of G . To do this, we consider thefollowing cases:(1) Suppose a s p and a s p are representatives of two conjugacy classes of Type 1 inTable 1. It is clear that a s p a s p = a s p a s p and so all such classes are com-muting together. Hence, the commuting conjugacy class graph has a subgraphisomorphic to the complete graph K n ( p − . We now assume that a sp is a repre-sentative of a conjugacy class of Type 1 and a u b v z is an arbitrary element of G such that ( a sp )( a u b v z ) = ( a u b v z )( a sp ). Thus a sp b v = b v a sp and by Lemma 2.2(a), ype The Representatives of Conjugacy Classes | x G | a sp ≤ s ≤ p − p n ( p − a i ≤ i ≤ p − , i p p mp ( p − b tp ≤ t ≤ p − p n ( p − b j ≤ j ≤ p − , j p p mp ( p − a sp b tp ≤ s, t ≤ p − p n ( p − a i b tp ≤ t ≤ p − , ≤ i ≤ p − , i p p mp ( p − a sp b j ≤ s ≤ p − , ≤ j ≤ p − , j p p mp ( p − a i b j ≤ i, j ≤ p − , i p , j p p mp ( p − Table 1.
Conjugacy classes of G with n = | Z | p and m = | Z | p . v ≡ mod p ) or v = tp in which 0 ≤ t ≤ p −
1. Hence a sp commutes with allelements in the form a u b tp z . Now the following cases can be occurred:(a) If t = 0, then the conjugacy classes of Types 1 and 2 in Table 1 are commutingto each other.(b) If t = 0 and u = 0, then the conjugacy classes of Types 1 and 3 in Table 1are commuting to each other.(c) If u, t = 0 and p | u , then the conjugacy classes of Types 1 and 5 in Table 1can be commuted to each other.(d) If t = 0 and p ∤ u , then the conjugacy classes of Types 1 and 6 in Table 1 arecommuting together.(2) It is clear that two conjugacy classes of Type 2 are commuted to each other andso the commuting conjugacy class graph has a subgraph isomorphic to K mp ( p − .We now determine the relationship between conjugacy classes of this and othertypes. To do this, we assume that a i is a representatives of a conjugacy class ofType 2 and a u b v z is an arbitrary element of G such that ( a i )( a u b v z ) = ( a u b v z )( a i ).So a i b v = b v a i and since p ∤ i , by Lemma 2.2(a) the latter is true if and only if v = 0. Therefore, a i commutes with all elements in the form a u z . This shows thatthe conjugacy classes of Type 2 are commuted only with the conjugacy classes ofType 1.(3) Suppose b t p and b t p are the representative of two conjugacy classes of Type 3. Itis clear that b t p b t p = b t p b t p and so all conjugacy classes of Type 3 are commutingtogether in the graph. Hence by Table 1, the graph has a subgraph isomorphic o K n ( p − . To determine the relationship between conjugacy classes of this andother types, we choose b tp to be a representative of a conjugacy class of Type 3 and a u b v z as an arbitrary element of G such that ( b tp )( a u b v z ) = ( a u b v z )( b tp ). Thus, b tp a u = a u b tp and by Lemma 2.2(a), the later is true if and only if u ≡ mod p ) or u = sp in which 0 ≤ s ≤ p −
1. Therefore, b tp can be commuted with all elementsin the form a sp b v z . Hence, the following are satisfied:(a) If s = 0, then the conjugacy classes of Types 3 and 4 are commuting together.(b) If s = 0 and v = 0, then the conjugacy classes of Types 3 and 1 are commutingto each other.(c) If v, s = 0 and p | v , then the conjugacy classes of Types 3 and 5 are com-muting together.(d) If s = 0 and p ∤ v , then the conjugacy classes of Types 3 and 7 are commutingto each other.(4) Suppose b j and b j are the representatives of two conjugacy classes of Type 4.It is obvious that b j b j = b j b j which shows that all conjugacy classes of Type4 are commuted to each other. So, the commuting conjugacy class graph has asubgraph isomorphic to K mp ( p − . To determine the relationship between this andother types in Table 1, we assume that b j is a representative of a conjugacy class oftype 4 and a u b v z is an arbitrary element of G such that ( b j )( a u b v z ) = ( a u b v z )( b j ).Thus, b j a u = a u b j and since p ∤ j , by Lemma 2.2(a), the latter is true if and onlyif u = 0. Therefore, b j commutes with all elements in the form b v z and hence theconjugacy classes of Type 4 can be commuted only with the conjugacy classes ofType 3.(5) Suppose a s p b t p and a s p b t p are the representatives of two conjugacy classes ofType 5. By Equation 3.3, ( a s p b t p )( a s p b t p ) = ( a s p b t p )( a s p b t p ) and so all con-jugacy classes of Type 5 are commuted to each other. This gives us the completegraph K n ( p − as a subgraph of the commuting conjugacy class graph. Hence, it isenough to determine the relationship between this and conjugacy classes of Types6, 7 and 8. Suppose a sp b tp and a u b vp are the representatives of the conjugacyclasses of Types 5 and 6, respectively, such that they are commuting together.By Lemma 3.1, a sp b tp = ( a ip b p ) n t z and a u b vp = ( ab jp ) n u z and so by Lemma 2.1, C G ( a sp b tp ) = C G ( a ip b p ) and C G ( a u b vp ) = C G ( ab jp ). It is now easy to prove that( a sp b tp )( a u b vp ) = ( a u b vp )( a sp b tp ) if and only if ( a ip b p )( ab jp ) = ( ab jp )( a ip b p ) if andonly if b p a = ab p which is a contradiction with Lemma 2.2(a). Therefore, a conju-gacy class of Type 5 is not adjacent with a conjugacy class of Type 6. Similarly, conjugacy class of Type 5 is not adjacent to another one of Type 7. We nowassume that a sp isb tp and a u b v are representatives of the conjugacy classes of Types5 and 8, respectively. By Lemma 3.1, a sp b tp = ( a ip b p ) n t z and a u b v = ( a j b ) n v z . Onthe other hand, By Lemma 2.1, C G ( a sp b tp ) = C G ( a ip b p ) and C G ( a u b v ) = C G ( a j b )and by Equation 3.9, we can see ( a sp b tp )( a u b v ) = ( a u b v )( a sp b tp ) if and only if( a ip b p )( a j b ) = ( a j b )( a ip b p ) if and only if j ≡ i ( mod p ). Since 1 ≤ i ≤ p −
1, theconjugacy classes of Types 5 and 8 can be divided into p − n ( p −
1) in the commutingconjugacy class graph.(6) Suppose a i b t p and a i b t p are the representative of two conjugacy classes of Types6 such that they are commute to each other. By Lemma 3.1, a i b t p = ( ab s p ) n i z and a i b t p = ( ab s p ) n i z . On the other hand, by Lemma (2.1), C G ( a i b t p ) = C G ( ab s p ) and C G ( a i b t p ) = C G ( ab s p ). Therefore, ( a i b t p )( a i b t p ) = ( a i b t p )( a i b t p )if and only if ( ab s p )( ab s p ) = ( ab s p )( ab s p ) if and only if ab ( s − s ) p = b ( s − s ) p a .By Lemma 2.2(a), the last equality is satisfied if and only if s = s . This provesthat in the commuting conjugacy class graph, two conjugacy classes of Type 6 areadjacent, when the centralizers of their representatives is equal. By Theorem 3.2,the number of centralizer of Types 6 is p − p − mp ( p − . Hence, each part of Type 6 is aclique of size mp ( p − a i b tp and a sp b j are the representatives of conjugacy classesof Types 6 and 7, respectively, such that they are commute to each other. ByEquation 3.3, ( a i b tp )( a sp b j ) = ( a sp b j )( a i b tp ) if and only if a i b j = b j a i . Since p ∤ i, j ,Lemma 2.2(a) implies that the last equality cannot be occurred. This means thata conjugacy class of Type 6 is not adjacent with another one of Type 7 in thecommuting conjugacy class graph. Next, we assume that a i b tp and a u b v are therepresentatives of two conjugacy classes of Types 6 and 8, respectively and they arecommute to each other. By Lemma 3.1, a i b tp = ( ab sp ) n i z and a u b v = ( ab j ) n u z and by Lemma 2.1, C G ( a i b tp ) = C G ( ab sp ) and C G ( a u b v ) = C G ( ab j ). Therefore,( a i b tp )( a u b v ) = ( a u b v )( a i b tp ) if an only if ( ab sp )( ab j ) = ( ab j )( ab sp ) if and only if b sp − j a = ab sp − j . We now apply Lemma 2.2(a) to prove that the last equality issatisfied if and only if p | ( sp − j ) or p | j which is a contradiction. Since p ∤ j , he conjugacy classes of Type 6 is not adjacent with another one of Type 8 in thecommuting conjugacy class graph.(7) Suppose that a s p b j and a s p b j are representatives of two conjugacy classes ofType 7 such that they are commuting to each other. By Lemma 3.1, a s p b j =( a t p b ) n j z and a s p b j = ( a t p b ) n j z and by Lemma 2.1, C G ( a s p b j ) = C G ( a t p b )and C G ( a s p b j ) = C G ( a t p b ). Therefore, ( a s p b j )( a s p b j ) = ( a s p b j )( a s p b j ) ifand only if ( a t p b )( a t p b ) = ( a t p b )( a t p b ) if and only if ba ( t − t ) p = a ( t − t ) p b . ByLemma 2.2(a), the last equality is satisfied if and only if t = t . This meansthat two conjugacy classes of Types 7 are adjacent in the commuting conjugacyclass graph, when their centralizers is equal. Also, by Theorem 3.2, the numberof centralizers of Type 7 is p −
1. So, the conjugacy classes of this type can bedivided into p − mp ( p − and the graph structure of each part of Type 7 gives acomplete subgraph of size mp ( p − a sp b j and a u b v are representatives of two conjugacy classes of Types 7 and 8, respectivelysuch that they are commute to each other. By Lemma 3.1, a sp b j = ( a tp b ) n j z and a u b v = ( a i b ) n v z . On the other hand, by Lemma 2.1, C G ( a sp b j ) = C G ( a tp b )and C G ( a u b v ) = C G ( a i b ). Therefore, ( a sp b j )( a u b v ) = ( a u b v )( a sp b j ) if and only if( a tp b )( a i b ) = ( a i b )( a tp b ) if and only if a tp − i b = ba tp − i . By Lemma 2.2(a), the lastequality is satisfied if and only if p | ( tp − i ) or p | i which is a contradiction.Thus, the conjugacy classes of Type 7 is not adjacent with a conjugacy class ofType 8 in the commuting conjugacy class graph.(8) Suppose a i b j and a i b j are the representatives of two conjugacy classes of Type8 such that they are commuting together. By Lemma 3.1, a i b j = ( a s b ) n j z and a i b j = ( a t b ) n j z and by Lemma 2.1, C G ( a i b j ) = C G ( a s b ) and C G ( a i b j ) = C G ( a t b ). Therefore, ( a i b j )( a i b j ) = ( a i b j )( a i b j ) if and only if ( a s b )( a t b ) =( a t b )( a s b ) if and only if a s − t b = ba s − t . By Lemma 2.2(a), the last equality issatisfied if and only if s = t . This means that two conjugacy classes of Type 8 areadjacent in the commuting conjugacy class graph, when their centralizers is equal.Also, by Theorem 3.2, the number of centralizers of Type 8 is p ( p −
1) and so theconjugacy classes of this type can be divided into p ( p −
1) parts. On the otherhand, by Table 1, the number of conjugacy classes of Type 8 is mp ( p − whichproves that each part of Type 8 gives a clique of size mp ( p −
1) in the commutingconjugacy class graph. By our discussion in Case 5, all conjugacy classes of Type are adjacent only with conjugacy classes of Type 5. Since the conjugacy classesof Types 5 and 8 can be divided into p − p ( p −
1) parts, every part of Type5 is adjacent with p parts of Type 8.By above discussion in eight parts, the commuting conjugacy class graph of G is aconnected graph with n ( p −
1) + m ( p − p ) vertices. Suppose M is a graph depictedin Figure 2. Therefore, the commuting conjugacy class graph of G is a M -join of graphsas follows: Γ( G ) = M [ p +1 z }| { K n ( p − , · · · , K n ( p − , p + p z }| { K m ( p − p ) , · · · , K m ( p − p ) ] , where n = | Z | p and m = | Z | p . The graph Γ( G ) is depicted in Figure 3. (cid:3) p −
34 7 p − p p p − ( p − ( p − ( p − p Figure 3.
The Commuting Conjugacy Class Graph of G with GZ ( G ) ∼ = Z p × Z p .In the following theorem, we study the commuting conjugacy class graph of G with thiscondition that GZ ( G ) ∼ = Z p ⋊ Z p , when the semidirect product is non-abelian. We recallthat by Theorem 2.3, p is an odd prime number. Theorem 4.2.
Suppose p is an odd prime number and G is a group with center Z suchthat GZ ∼ = Z p ⋊ Z p and GZ is non-abelian. Then the commuting conjugacy classes of G isas follows: M [ p + p +1 z }| { K n ( p − , · · · , K n ( p − , K np ( p − ] , here M is depicted in Figure 4. Here, n = | Z | p . p p p + 1( p + 1) p ( p + 1) ( p + 1) Figure 4.
The Graph M . Proof.
We first apply Equation 2.1 to deduce that(4.2) b j a i Z = a ijp + i b j Z which implies that b j Za i Z = ( a p Z ) ij a i Zb j Z, (4.3) a i Zb j Z = ( a p Z ) − ij b j Za i Z. (4.4)Choose elements x and g in G \ Z . By the proof of Theorem 3.2, x = a i b j z and g = a u b v z , where z , z ∈ Z , 0 ≤ i, j, u, v ≤ p − i, j , as well as u, v , are notsimultaneously zero. In Equation 4.2, we put i = p and j = 1. Since a p ∈ Z , bZa p Z = a p ZbZ . Thus a p Z ∈ Z (cid:18) GZ (cid:19) and by Equations 4.3 and 4.4, − xgZ = g − ZxZgZ = ( a u b v ) − Z ( a i b j ) Z ( a u b v ) Z = ( a u b v ) − Za i Zb j Za u Zb v Z = ( a u b v ) − Za i Z ( a p Z ) uj a u Zb j Zb v Z = ( a p Z ) uj ( a u b v ) − Za u Za i Zb v Zb j Z = ( a p Z ) uj ( a u b v ) − Za u Z ( a p Z ) − vi b v Za i Zb j Z = ( a p Z ) uj ( a p Z ) − vi ( a u b v ) − Za u Zb v Za i Zb j Z = ( a p Z ) uj − vi ( a u b v ) − Z ( a u b v ) Z ( a i b j ) Z = ( a ( uj − vi ) p Z )( a i b j ) Z = ( a ( uj − vi ) p a i b j ) Z. So, there exists z ∈ Z such that(4.5) ( a u b v ) − ( a i b j )( a u b v ) = a ( uj − vi ) p a i b j z. Suppose that x = a i b j z is an arbitrary element of G such that 0 ≤ i, j ≤ p − i, j are not simultaneously zero. To compute the number of non-central conjugacy classes of G and obtain their structure, we will consider three separate cases that all together canbe divided into eight subcases:a) i = 0 and j = 0 . This case is separated into two subcases that p | i and p ∤ i as follows:1) p | i . Suppose i = sp , where 1 ≤ s ≤ p −
1. By Equation 4.5 and this fact that a p ∈ Z , ( a u b v ) − ( a sp )( a u b v ) = a − vsp a sp z = a sp z and so ( a sp ) G = { a sp z | z ∈ Z } .On the other hand, by Theorem 3.2, | C G ( a sp ) | = p | Z | and | ( a sp ) G | = p . Thus,( a sp ) G = { a sp z , · · · , a sp z p } = a sp H in which H = { z , · · · , z p } ⊆ Z . Suppose | Z | > p and choose the elements z r ∈ Z \ H and an element in the form a sp z r .It is easy to see that a sp z r ( a sp ) G and hence ( a sp z r ) G = ( a sp ) G . Furthermore, C ( a sp z r ) = C ( a sp ) and so | ( a sp z r ) G | = | ( a sp ) G | = p . Also, ( a sp z r ) G = ( a sp ) G z r = a sp Hz r . Since ( a sp ) G ∩ ( a sp z r ) G = ∅ , H ∩ Hz r = ∅ and H ∪ Hz r ⊆ Z . Wenow choose the element z r ∈ Z \ ( H ∪ Hz r ) and continue this process to provethat p | | Z | . If n = | Z | p then there are n distinct conjugacy classes in the form of( a sp z r ) G , ( a sp z r ) G · · · ( a sp z r n ) G . Since 1 ≤ s ≤ p −
1, there are n ( p −
1) distinctconjugacy classes with n = | Z | p and each conjugacy class has p elements. ) p ∤ i . By Equation 4.5,(4.6) ( a u b v ) − ( a i )( a u b v ) = a − vip a i z. By Theorem 3.2, | C G ( a i ) | = p | Z | and so | ( a i ) G | = p . Note that 0 ≤ v ≤ p − v ′ and k such that − vi = v ′ p + k and 0 ≤ k ≤ p −
1. By Equation4.6 and the fact that a p ∈ Z ,(4.7) ( a i ) G = { a kp a i z ( k,l ) | ≤ k, l ≤ p − , z ( k,l ) ∈ Z } . Set A i = { a kp a i z | ≤ k ≤ p − , z ∈ Z } and H = { z (0 ,l ) | ≤ l ≤ p − } .Obviously, | A i | = p | Z | , ( a i ) G ⊆ A i and H ⊆ Z . If | Z | > p , then we choosean element z r ∈ Z \ H and an element a i z r ∈ A i . It can be easily seen that a i z r ( a i ) G and ( a i z r ) G = ( a i ) G . Since C ( a i z r ) = C ( a i ), | ( a i z r ) G | = | ( a i ) G | = p and ( a i z r ) G = ( a i ) G z r . On the other hand, since ( a i ) G ∩ ( a i z r ) G = ∅ , H ∩ Hz r = ∅ and H ∪ Hz r ⊆ Z . We now choose an element z r ∈ Z \ ( H ∪ Hz r ) and continue thisprocess to prove that ( a i z r ) G ⊆ A i . This shows that p | | A i | . Put n = | A i | p = | Z | p .Then, for each i there are n distinct conjugacy classes as ( a i z r ) G , · · · , ( a i z r n ) G .Again since 1 ≤ i ≤ p − p ∤ i , there are non-negative integers k ′ and i ′ suchthat i = k ′ p + i ′ and 1 ≤ i ′ ≤ p −
1. By Equation 4.7,( a i ) G = ( a k ′ p + i ′ ) G = { a kp a k ′ p + i ′ z ( k,l ) | ≤ k, l ≤ p − , z ( k,l ) ∈ Z } = { a ( k + k ′ ) p a i ′ z ( k,l ) | ≤ k, l ≤ p − , z ( k,l ) ∈ Z } . Therefore, there exists z i ′ ∈ Z such that ( a i ) G = ( a i ′ z i ′ ) G . This proves that thenumber of distinct conjugacy classes is equal to n ( p −
1) in which n = | Z | p and eachclass has p elements.b) i = 0 and j = 0 . Again there are two cases that p | j and p ∤ j .3) p | j . Suppose j = tp , where 1 ≤ t ≤ p −
1. By Equation 4.5 and the fact that a p is an element of Z , we can conclude that ( a u b v ) − ( b tp )( a u b v ) = a utp b tp z = b tp z and so ( b tp ) G = { b tp z | z ∈ Z } . By Theorem 3.2, | C G ( b tp ) | = p | Z | and hence | ( b tp ) G | = p . Therefore, similar to the Case 1, there are n distinct conjugacy classes( b tp z r ) G , · · · , ( b tp z r n ) G , where n = | Z | p . So, the number of distinct conjugacy classesis equal to n ( p −
1) and each class has p elements.4) p ∤ j . Then by Equation 4.5, ( a u b v ) − ( b j )( a u b v ) = a ujp b j z . By Theorem 3.2, | C G ( b j ) | = p | Z | and so | ( b j ) G | = p . A similar argument as in the Case 2 showsthat ( b j ) G = { a kp b j z ( k,l ) | ≤ k, l ≤ p − , z ( k,l ) ∈ Z } . Thus, there are n distinctconjugacy classes ( b j z r ) G , · · · , ( b j z r n ) G , when j is a fixed number and n = | Z | p . ince 1 ≤ j ≤ p − p ∤ j , the number of distinct conjugacy classes is equal to n ( p − p ) and each class has p elements.c) i = 0 and j = 0 . We have four subcases as follows:5) p | i and p | j . Suppose i = sp and j = tp such that 1 ≤ s, t ≤ p −
1. ByEquation 4.5 and the fact that a p ∈ Z , ( a u b v ) − ( a sp b tp )( a u b v ) = a ( utp − vsp ) p a sp b tp z = a ( ut − vs ) p a sp b tp z = a sp b tp z . Thus, ( a sp b tp ) G = { a sp b tp z | z ∈ Z } . Apply Theorem3.2 to deduce that | C G ( a sp b tp ) | = p | Z | which implies that | ( a sp b tp ) G | = p . Similarto the Case 1, for fixed non-negative integers s and t , there are n distinct conjugacyclasses as ( a sp b tp z r ) G , . . . , ( a sp b tp z r n ) G in which n = | Z | p . By above discussion, weresult that the number of the distinct conjugacy classes is equal to n ( p − andeach class has p elements.6) p ∤ i and p | j . Suppose j = tp , where 1 ≤ t ≤ p −
1. By Equation 4.5 and the factthat a p ∈ Z , ( a u b v ) − ( a i b tp )( a u b v ) = a ( utp − vi ) p a i b tp z = a − vip a i b tp z . By Theorem3.2, | C G ( a i b tp ) | = p | Z | and so | ( a i b tp ) G | = p . Now a similar argument as the Case2,(4.8) ( a i b tp ) G = { a kp a i b tp z ( k,l ) | ≤ k, l ≤ p − , z ( k,l ) ∈ Z } . Hence, for fixed positive integers i and t , there are n distinct conjugacy classes as( a i b tp z r ) G , . . . , ( a i b tp z r n ) G in which n = | Z | p . Since 1 ≤ i ≤ p − p ∤ i , thereare integers k ′ and i ′ such that i = k ′ p + i ′ and 1 ≤ i ′ ≤ p −
1. By Equation 4.8,( a i b tp ) G = ( a k ′ p + i ′ b tp ) G = { a kp a k ′ p + i ′ b tp z ( k,l ) | ≤ k, l ≤ p − , z ( k,l ) ∈ Z } = { a ( k + k ′ ) p a i ′ b tp z ( k,l ) | ≤ k, l ≤ p − , z ( k,l ) ∈ Z } . This shows that there exists z ( i ′ ,t ) ∈ Z such that ( a i b tp ) G = ( a i ′ b tp z ( i ′ ,t ) ) G . Since1 ≤ i ′ , t ≤ p −
1, the number of the distinct conjugacy classes is equal to n ( p − and each class has p elements.7) p | i and p ∤ j . Suppose i = sp such that 1 ≤ s ≤ p −
1. By Equation 4.5 and thefact that a p ∈ Z , we have ( a u b v ) − ( a sp b j )( a u b v ) = a ( uj − vsp ) p a sp b j z = a ujp a sp b j z = a ( uj + s ) p b j z . In this case, there is no new conjugacy class and the argument issimilar to the Case 4.8) p ∤ i and p ∤ j . By Equation 4.5, ( a u b v ) − ( a i b j )( a u b v ) = a ( uj − vi ) p a i b j z . By Theorem3.2, | C G ( a i b j ) | = p | Z | and so | ( a i b j ) G | = p . Now by a similar argument as in theCase 2, we can see that(4.9) ( a i b j ) G = { a kp a i b j z ( k,l ) | ≤ k, l ≤ p − , z ( k,l ) ∈ Z } . ype The Representative of Conjugacy Classes | x G | a sp ≤ s ≤ p − p n ( p − a i ≤ i ≤ p − p n ( p − b tp ≤ t ≤ p − p n ( p − b j ≤ j ≤ p − , j p p np ( p − a sp b tp ≤ s, t ≤ p − p n ( p − a i b tp ≤ i, t ≤ p − p n ( p − a sp b j ≤ s ≤ p − , ≤ j ≤ p − , j p a i b j ≤ i ≤ p − , ≤ j ≤ p − , j p p np ( p − Table 2.
Conjugacy Classes of G such that n = | Z | p .Hence, for fixed positive integers i and j , there are n distinct conjugacy classes( a i b j z r ) G , · · · , ( a i b j z r n ) G , where n = | Z | p . Since 1 ≤ i ≤ p − p ∤ i , there areintegers k ′ and i ′ such that i = k ′ p + i ′ and 1 ≤ i ′ ≤ p −
1. Therefore, by Equation4.9, ( a i b j ) G = ( a k ′ p + i ′ b j ) G = { a kp a k ′ p + i ′ b j z ( k,l ) | ≤ k, l ≤ p − , z ( k,l ) ∈ Z } = { a ( k + k ′ ) p a i ′ b j z ( k,l ) | ≤ k, l ≤ p − , z ( k,l ) ∈ Z } . Thus, there exists z ( i ′ ,j ) ∈ Z such that ( a i b j ) G = ( a i ′ b j z ( i ′ ,j ) ) G . Since 1 ≤ i ′ ≤ p − ≤ j ≤ p − p ∤ j , the number of distinct conjugacy classes is equal to np ( p − and each class has p elements.We now investigate the commuting conjugacy class graph of G . To do this, it is enoughto determine the relationship between conjugacy classes in Table 2. We consider thefollowing cases:(1) Suppose a s p and a s p are the representatives of two classes of Type 1 in Table2. It is clear that a s p a s p = a s p a s p and so all conjugacy classes of Type 1are commuting together. Hence, the commuting conjugacy class graph has asubgraph isomorphic to K n ( p − . We now determine the relationship betweenconjugacy classes of this and another types. For this purpose, we suppose a sp isa representative of a conjugacy class of Type 1 and a u b v z is an arbitrary elementof G such that they are commuted to each other. This means that ( a sp )( a u b v z ) = a u b v z )( a sp ) and hence a sp b v = b v a sp . By Lemma 2.2(a), the last equality is trueif and only if p | v or v = tp in which 0 ≤ t ≤ p −
1. Then a sp commutes with allelements of the form a u b tp z . Therefore, the following cases can be occurred:(a) t = 0. Then the conjugacy classes of Types 1 and 2 in Table 2 are commutingtogether.(b) t = 0 and u = 0. In this case, the conjugacy classes of Types 1 and 3 in Table2 are commuting to each other.(c) u, t = 0 and p | u . Then the conjugacy classes of Types 1 and 5 in Table 2are commuting together.(d) t = 0 and p ∤ u . By this condition, the conjugacy classes of Types 1 and 6are commuting together.(2) Suppose a i and a i are the representatives of two conjugacy classes of Type 2.Obviously, a i a i = a i a i and so all conjugacy classes of Type 2 are commuted toeach other in the group. Hence, the commuting conjugacy class graph has a cliqueof size n ( p − a i is a representative of a conjugacy class of Type 2and a u b v z is an arbitrary element of G such that ( a i )( a u b v z ) = ( a u b v z )( a i ). Thus, a i b v = b v a i . Since p ∤ i , then By Lemma (2.2)(a) the last equality is satisfied if andonly if v = 0. Then a i commutes with all elements of the form a u z which showsthat the conjugacy classes of Type 2 are commuting only with conjugacy classesof Type 1.(3) Suppose b t p and b t p are representatives of two conjugacy classes of Type 3. Itis clear that b t p b t p = b t p b t p . This shows that all conjugacy classes of Type 3are commuting together and hence the commuting conjugacy class graph has asubgraph isomorphic to K n ( p − . We now determine the relation between this andanother types. Suppose b tp is a representatives of a conjugacy class of Type 3 and a u b v z is an arbitrary element of G such that ( b tp )( a u b v z ) = ( a u b v z )( b tp ). Thus, b tp a u = a u b tp . By Lemma 2.2(a), the last equality is satisfied if and only if p | u or u = sp in which 0 ≤ s ≤ p −
1. Then b tp commutes with all elements in the formof a sp b v z . Again, the following cases can be occurred:(a) s = 0. Then the conjugacy classes of Types 3 and 4 in Table 2 are commutingtogether.(b) s = 0 and v = 0. Under this condition, the conjugacy classes of Types 3 and1 in Table 2 are commuting together. c) v, s = 0 and p | v . In this case, the conjugacy classes of Types 3 and 5 inTable 2 are commuting to each other.(d) s = 0 and p ∤ v . The classes classes of Types 3 and 7 in Table 2 are commutingtogether.(4) Suppose b j and b j are representatives of two conjugacy classes of Type 4. Since b j b j = b j b j , all conjugacy classes of Type 4 are commuting together and hencethe commuting conjugacy class graph has a clique of size np ( p − b j is a representative of a conjugacy class of Type 4 and a u b v z is an arbitraryelement of G such that ( b j )( a u b v z ) = ( a u b v z )( b j ). Thus, b j a u = a u b j . Since p ∤ j ,by Lemma 2.2(a) the last equality is satisfied if and only if u = 0. Then b j commutes with all elements in the form of b v z . So, the conjugacy classes of Type4 are commuting only with the conjugacy classes of Type 3.(5) Suppose a s p b t p and a s p b t p are representatives of two conjugacy classes of Type5. By Equation 3.3, ( a s p b t p )( a s p b t p ) = ( a s p b t p )( a s p b t p ). This proves that allconjugacy classes of Type 5 are commuting to each other and hence the commutingconjugacy class graph has a clique of size n ( p − . It is now enough to determinethe relationship between this and conjugacy classes of types 6, 7 and 8. Suppose a sp b tp and a u b vp are representatives of conjugacy classes of Types 5 and 6, respec-tively, which are commuting together. By Lemma 3.1, a sp b tp = ( a ip b p ) n t z and a u b vp = ( ab jp ) n u z . On the other hand, By Lemma 2.1, C G ( a sp b tp ) = C G ( a ip b p )and C G ( a u b vp ) = C G ( ab jp ). We note that ( a sp b tp )( a u b vp ) = ( a u b vp )( a sp b tp ) if andonly if ( a ip b p )( ab jp ) = ( ab jp )( a ip b p ) if and only if b p a = ab p , contradicts by Lemma2.2(a). Therefore, a conjugacy class of Type 5 is not adjacent with a conjugacyclass of Type 6. We now assume that a sp b tp and a u b v are representatives of theconjugacy classes of Types 5 and 8, respectively, which are commute to each other.By Lemma 3.1, a sp b tp = ( a ip b p ) n t z and a u b v = ( a j b ) n v z . On the other hand, ByLemma 2.1, C G ( a sp b tp ) = C G ( a ip b p ) and C G ( a u b v ) = C G ( a j b ). By Equation 3.9,( a sp b tp )( a u b v ) = ( a u b v )( a sp b tp ) if and only if ( a ip b p )( a j b ) = ( a j b )( a ip b p ) if and onlyif j ≡ i ( mod p ). Since 1 ≤ i ≤ p −
1, the conjugacy classes of Types 5 and 8 can bedivided into p − n ( p − a i b t p and a i b t p are representatives of two commuting conjugacy classesof Type 6. By Lemma 3.1, a i b t p = ( ab s p ) n i z and a i b t p = ( ab s p ) n i z . On theother hand, by Lemma 2.1, C G ( a i b t p ) = C G ( ab s p ) and C G ( a i b t p ) = C G ( ab s p ). his proves that ( a i b t p )( a i b t p ) = ( a i b t p )( a i b t p ) if and only if ( ab s p )( ab s p ) =( ab s p )( ab s p ) if and only if ab ( s − s ) p = b ( s − s ) p a . By Lemma 2.2(a), the lastequality is satisfied if and only if s = s and so in the commuting conjugacyclass graph two conjugacy classes of Type 6 are adjacent, when their centralizersis equal. Also, by Theorem 3.2, the number of centralizers of Type 6 is p −
1. So,the conjugacy classes of this type can be divided into p − n ( p − . Hence,each part of Type 6 gives a complete subgraph isomorphic to K n ( p − . It is nowenough to find the relationship between conjugacy classes of Types 6 and 8. Tosee this, we assume that a i b tp and a u b v are representatives of conjugacy classes ofTypes 6 and 8, respectively, such that they are commuting together. By Lemma3.1, a i b tp = ( ab sp ) n i z and a u b v = ( ab j ) n u z . On the other hand, by Lemma2.1, C G ( a i b tp ) = C G ( ab sp ) and C G ( a u b v ) = C G ( ab j ). Therefore, ( a i b tp )( a u b v ) =( a u b v )( a i b tp ) if and only if ( ab sp )( ab j ) = ( ab j )( ab sp ) if and only if b sp − j a = ab sp − j .By Lemma 2.2(a), the last equality is satisfied if and only if p | ( sp − j ) or p | j which is impossible. This shows that in the commuting conjugacy class graph, theconjugacy classes of Type 6 is not adjacent with any conjugacy classes of Type 8.(7) Suppose a i b j and a i b j are representatives of two commuting conjugacy classes ofType 8 such that they are commuting together. By Lemma 3.1, a i b j = ( a s b ) n j z and a i b j = ( a t b ) n j z . On the other hand, by Lemma 2.1, C G ( a i b j ) = C G ( a s b )and C G ( a i b j ) = C G ( a t b ). Therefore, ( a i b j )( a i b j ) = ( a i b j )( a i b j ) if and onlyif ( a s b )( a t b ) = ( a t b )( a s b ) if and only if a s − t b = ba s − t . By Lemma 2.2(a), the lastequality is satisfied if and only if s = t and so in the commuting conjugacy classgraph two conjugacy classes of Type 8 are adjacent, when their centralizers isequal. Also, by Theorem 3.2, the number of centralizers of Type 8 is equal to p ( p −
1) and so the conjugacy classes of this type can be divided into p ( p − np ( p − andhence each part of Type 8 gives a complete subgraph of size n ( p − G , the conjugacy classesof Type 8 are adjacent only with conjugacy classes of Type 5. Finally, we knowthat the conjugacy classes of Types 5 and 8 can be divided into p − p ( p − p conjugacy classes of Type 8.By above discussion, the commuting conjugacy class graph of G is a connected graphwith n ( p − p + 1) vertices. Suppose M is the graph depicted in Figure 3. Then the ommuting conjugacy class graph of G can be written as a M -join, i.e.Γ( G ) = M [ p + p +1 z }| { K n ( p − , · · · , K n ( p − , K np ( p − ] . The graph Γ( G ) is depicted in Figure 5. (cid:3) p −
34 5 p p p − ( p − ( p − ( p − p Figure 5.
The Graph Γ( G ), when GZ ( G ) is non-abelian and GZ ∼ = Z p ⋊ Z p . Acknowledgement.
The research of the first author is partially supported by the Uni-versity of Kashan under grant no. 364988/64.
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