Permutation-like Matrix Groups with a Maximal Cycle of Prime Square Length
aa r X i v : . [ m a t h . G R ] N ov Permutation-like Matrix Groups with a MaximalCycle of Prime Square Length
Guodong Deng, Yun Fan
School of Mathematics and StatisticsCentral China Normal University, Wuhan 430079, China
Abstract
A matrix group is said to be permutation-like if any matrix of thegroup is similar to a permutation matrix. G. Cigler proved that, if apermutation-like matrix group contains a normal cyclic subgroup whichis generated by a maximal cycle and the matrix dimension is a prime, thenthe group is similar to a permutation matrix group. This paper extendsthe result to the case where the matrix dimension is a square of a prime.
Key words : matrix group, matrix similarity, permutation-like group,permutation matrix group.
Mathematics Subject Classification 2010 : 15A18, 15A30, 20H20.
A multiplicative group consisting of complex invertible matrices of size n × n is said to be a matrix group of dimension n . A matrix group G is said to bepermutation-like if any matrix of G is similar to a permutation matrix, see [2, 3].If there exists an invertible matrix Q such that Q − AQ is a permutation matrixfor all A ∈ G , then we say that G is similar to a permutation matrix group, or G is a permutation matrix group for short. A matrix is called a maximal cycle ifit is similar to a permutation matrix corresponding to a cycle permutation withcycle length equal to the dimension. G. Cigler in [3] showed that a permutation-like matrix group is not a permutation matrix group in general, and suggesteda conjecture as follows. Conjecture.
A permutation-like matrix group containing a maximal cycle issimilar to a permutation matrix group.
G. Cigler in [3] proved it affirmatively in two cases: the dimension ≤
5, or thedimension is a prime integer and the cyclic subgroup generated by the maximalcycle is normal.
Email address : [email protected] (Guodong Deng), [email protected] (Yun Fan).
1n this paper we extend the result of [3] to the case where the length of themaximal cycle is a square of a prime.
Theorem 1.1.
Let G be a permutation-like matrix group of dimension p where p is a prime. If G contains a maximal cycle C such that the subgroup h C i generated by C is normal in G , then G is a permutation matrix group. In Section 2 we state some preliminaries as a preparation. The theorem willbe proved in Section 3.
The complex field is denoted by C . For a positive integer n , by Z ∗ n we denotethe multiplicative group consisting of units of the residue ring Z n of the integerring Z modulo n . A diagonal blocked matrix B . . . B k is denoted by B ⊕ · · · ⊕ B k for short. The identity matrix of dimension n is denoted by I n × n .All complex invertible matrices of dimension n consist the so-called generallinear group, denoted by GL n ( C ). We denote the characteristic polynomial ofa complex matrix A by char A ( x ). Lemma 2.1.
The following two are equivalent to each other:(i) A is similar to a permutation matrix;(ii) A is diagonalizable and char A ( x ) = Q i ( x ℓ i − .If it is the case, then each factor x ℓ i − of char A ( x ) corresponds to exactlyone ℓ i -cycle of the cycle decomposition of the permutation of the permutationmatrix. Proof.
It is clear.We’ll apply the lemma to the case where A p = 1 with p being a prime,at that case it is easy to check the condition (ii) of the lemma, because thereare only three divisors 1 , p, p of p which form a chain with respect to thedivision relation, and x p − ( x )Φ p ( x )Φ p ( x ) where Φ p i ( x ) denotes the p i ’th cyclotomic polynomial.Let C ∈ GL n ( C ) be a maximal cycle of dimension n , and λ be a primitive n ’th root of unity. Then there is a basis of the vector space C n : e , e , · · · , e n − , (2.1)such that C i e j = λ ij e j for all i, j = 0 , , · · · , n −
1; with respect to this basis, C isa diagonal matrix C = 1 ⊕ λ ⊕· · ·⊕ λ n − ; see [3, § C e j , for j = 0 , · · · , n −
1, is2ust the eigen-subspace of the eigenvalue λ j , for j = 0 , · · · , n − C ; or, in representation theoretic notations, C e j for j = 0 , · · · , n − h C i , see [1, §
15 Example 1].Taking any non-zero complexes c , · · · , c n − and setting f = P n − j =0 c j e j , weobtain another basis of C n : f, Cf, · · · , C n − f, (2.2)with which C is a cycle permutation matrix, see [3, Lemma 4.1].Let B ∈ GL n ( C ) with ord( B ) = ℓ , where ord( B ) denotes the order of B .Assume that B normalizes the group h C i . Since the automorphism group of thecyclic group h C i is isomorphic to Z ∗ n , there is an r ∈ Z ∗ n such that B − C i B = C ri , ∀ i ∈ Z n ; (2.3)thus the action by conjugation of B on h C i is determined by the action of µ r on Z n , where µ r ( a ) = ra for all a ∈ Z n . Further, C i Be j = B · B − C i Be j = B · C ri e j = B · λ rij e j = λ rij Be j ;taking i = 1, we see that Be j is an eigenvector of the eigenvalue λ rj of C , i.e. B C e j = C e rj , j = 0 , , · · · , n − . (2.4)Thus, B permutes the eigen-subspaces (cid:8) C e , C e , · · · , C e n − (cid:9) of C in the sameway as µ r permutes Z n .Let Γ , Γ , · · · , Γ m be orbits of the action of the group h B i on the set of1-dimensional subspaces { C e , C e , · · · , C e n } . Assume that the length of Γ k is n k for k = 1 , · · · , m . Since |h B i| = ℓ where |h B i| denotes the order of thegroup h B i , we have that n k | ℓ for k = 1 , · · · , m . Lemma 2.2.
Let notation be as above.(1) For each k , take any one C e j k ∈ Γ k and any non-zero e ′ k ∈ C e j k , set V k = L n k − h =0 B h C e ′ k and E k = { e ′ k , Be ′ k , · · · , B n k − e ′ k } ; then B n k e ′ k = ω k e ′ k where ω k is an ( ℓ/n k ) ’th root of unity, V k is a B -invariant subspace of C n , and E k is a basis of V k , with which the matrix of B restricted to V k is B | V k = · · · ω k . . . . . . ... . . . . . . ... · · · n k × n k . (2.5) (2) C n = V ⊕ · · · ⊕ V m , the union E = E ∪ · · · ∪ E m is a basis of C n and,with respect to the basis E , the matrix of B is B = B | V ⊕ · · · ⊕ B | V m . (2.6)3 roof. (1). Since the length of Γ k is n k , Γ k = (cid:8) C e ′ k , B C e ′ k , · · · , B n k − C e ′ k (cid:9) and B n k C e ′ k = C e ′ k , hence there is an ω k ∈ C such that B n k e ′ k = ω k e ′ k . Then itis clear that V k is B -invariant, E k is a basis of V k and Eqn (2.5) is the matrixof B | V k . Since B ℓ = I n × n , we have ( B | V k ) ℓ = I n k × n k ; but by Eqn (2.5),( B | V k ) n k = ω k I n k × n k ; so ω k is an ( ℓ/n k )’th root of unity.(2). Applying (1) to all orbits Γ , · · · , Γ m , by Eqn (2.1) one can check theconclusions in (2) easily. Proposition 2.3.
Let notation be as in Lemma 2.2. Assume that the followingcondition is satisfied: (SC)
For any e j and B i , if B i C e j = C e j then B i e j = e j .Then the matrix group h C, B i generated by C and B is a permutation matrixgroup. Proof.
We keep the notations in Lemma 2.2 and its proof. We have seenthat B n k C e ′ k = C e ′ k ; by the condotion (SC) we have B n k e ′ k = e ′ k , i.e. ω k = 1and B | V k = · · · · · · n k × n k ; (2.7)hence B P n k − h =0 B h e ′ k = P n k − h =0 BB h e ′ k = P n k − h =0 B h e ′ k .Now we set f = P mk =1 P n k − h =0 B h e ′ k ; then Bf = B m X k =1 n k − X h =0 B h e ′ k = m X k =1 n k − X h =0 BB h e ′ k = m X k =1 n k − X h =0 B h e ′ k = f. By Eqn (2.2) the set of the vectors: f, Cf, · · · , C n − f, is a basis of C n ; and with respect to this basis C is a cycle permutation matrix.Further, by Eqn (2.3) we have BC i f = BC i B − Bf = C r − i f ;that is, with respect to the basis f, Cf, · · · , C n − f , the B is also a permutationmatrix. In conclusion, the matrix group h C, B i generated by C and B is apermutation matrix group.We’ll quote a result of [3] repeatedly, so state it as a lemma: Lemma 2.4. ([3, Proposition 4.2]) If h C, B i is an abelian permutation-likematrix group where C is a maximal cycle, then B ∈ h C i .
4e state some group-theoretic information as a remark for later quotations.We say that an action of a group G on a set X is free if the stabilizer of any X ∈ X in G is trivial; at that case, X is partitioned in to G -orbits such thateach orbit is a regular G -set (i.e. equivalent to the set G on which the group G acts by left translation). Remark 2.5.
Let p be an odd prime.(1) Let C = h C i be a cyclic group of order p . It is easy to see that C p = h C p i where C p = { X p | X ∈ C} ; and, mapping X ∈ C to X p ∈ C p is a surjectivehomomorphism from C onto C p with kernel C p .(2) Let G be a finite group containing a normal cyclic subgroup C = h C i oforder p such that C is self-centralized (i.e. the centralizer C G ( C ) = C ).Then G / C is isomorphic to a subgroup of the automorphism group of C ,hence to a subgroup of the multiplicative group Z ∗ p which is a cyclic groupof order p ( p − r ∈ Z ∗ p and a B ∈ G such that G = h B, C i and C B = C r where C B = B − CB denotes the conjugate of C by B ; andthe order ord( r ) = |G / C| .(3) If ord( r ) (cid:12)(cid:12) ( p − B ) = |G / C| and the action by conjugation of B on C is equivalent to the action of r by multiplication on Z p , the latter isdenoted by µ r , i.e. µ r ( a ) = ra for all a ∈ Z p ; it is easy to check that thegroup h B i acts freely by conjugation on the difference set C − { } (cid:0) this isa specific case of [4, Corollary 4.35] (cid:1) .(4) If ord( r ) = |G / C| = p , then we can choose B such that B p = 1 and r = ap + 1 with 0 < a < p ; further, replacing B by a suitable power of B ,we can get r = p + 1; see [1, § B p = 1 and r = p + 1 as in (4), then it is easy to check that:(i) h B i centralizes the subgroup C p = h C p i of C , and acts freely byconjugation on the difference set C − C p ; in particular, C − C p ispartitioned into p − h B i , the length of everyclass is p .(ii) For any X ∈ C the product Q p − j =0 ( X ) B j ∈ C p ; the mapping X to Q p − j =0 ( X ) B j is a surjective homomorphism from C onto C p ; hencethe homomorphism induces a bijection from the set of the conjugacyclasses in C − C p on to the set C p − { } . If p = 2 then p = 4 and the conclusion of Theorem 1.1 has been checked in [3].In the following, we always assume that p is an odd prime and G is apermutation-like matrix group of dimension p which contains a normal cyclic5ubgroup h C i generated by a maximal cycle C ; and prove that G is a permuta-tion matrix group. If G is abelian, by Lemma 2.4, G = C which is a permutationmatrix group. So we further assume that G is non-abelian.Let λ be a primitive p ’th root of unity, By Eqn (2.1) there is a basis e , e , · · · , e p − of C p such that C i e j = λ ij e j , i, j = 0 , , · · · , p − . Let C = h C i and q = |G / C| . By Lemma 2.4, C is self-centralized in G ; byRemark 2.5(2), there are a B ∈ G , an r ∈ Z ∗ p and integers s, t such that • q = p δ s where δ = 0 or 1, p − st ; • G = h B i · h C i and the quotient h B i (cid:14) h B i ∩ h C i ∼ = h r i ≤ Z ∗ p ; • B − CB = C r , i.e. the action by conjugation of h B i (cid:14) h B i ∩ h C i on h C i isequivalent to the action of h µ r i on the residual set Z p , where µ r ( a ) = ra for all a ∈ Z p .The residual set Z p is a disjoint union of two µ r -stable subsets: Z p = Γ ∪ Z ∗ p , where Γ = Z p − Z ∗ p = { , p, p, · · · , ( p − p } ; (3.1)in fact, Γ corresponds to the subgroup C p of C . We prove the theorem in threecases. Case 1. δ = 0, i.e. q = s | ( p − |h B i| = s and h B i ∼ = h r i ≤ Z ∗ p . The group h µ r i fixes 0 ∈ Z p , and acts freely on both Z ∗ p and Γ − { } . There are t orbits of h µ r i on Γ − { } ; taking representatives v , · · · , v t from the t orbits, we can write the orbits of h µ r i on Γ as follows:Γ = { } , Γ = { v , rv , · · · , r s − v } , · · · , Γ t = { v t , rv t , · · · , r s − v t } . There are m orbits of h µ r i on Z ∗ p where m = p ( p − s = pt ; taking representatives w , · · · , w m from these m orbits, we have the orbits of h µ r i on Z ∗ p as follows:Γ = { w , rw , · · · , r s − w } , · · · , Γ m = { w m , rw m , · · · , r s − w m } . Accordingly, we apply Lemma 2.2 and its notation to get the basis of C p : E = E ∪ E ∪ · · · ∪ E t ∪ E ∪ · · · ∪ E m , and write matrices with respect to this basis. So C = 1 ⊕ (cid:16) s − ⊕ i =0 λ r i v (cid:17) ⊕ · · · ⊕ (cid:16) s − ⊕ i =0 λ r i v t (cid:17) ⊕ (cid:16) s − ⊕ i =0 λ r i w (cid:17) ⊕ · · · ⊕ (cid:16) s − ⊕ i =0 λ r i w m (cid:17) , s ’th root ε of unity such that B = ε ⊕ t + m z }| { P ⊕ · · · ⊕ P , where P is the cycle matrix of dimension s (see Lemma 2.2 and Eqn (2.7)): P = · · · · · · s × s . Then the characteristic polynomial of B ischar B ( x ) = ( x − ε )( x s − t + m ;since B is similar to a permutation matrix, by Lemma 2.1, we have ε = 1. Sothe condition (SC) of Proposition 2.3 is satisfied: (SC) For any e j and B i , if B i C e j = C e j then B i e j = e j ; hence G is a permutation matrix group. Case 2. s = 1, i.e. q = p . By Remark 2.5(4), we can assume that B p = 1and the conjugation of B on C is equivalent to the action of µ p +1 on Z p ,where µ p +1 a = ( p + 1) a for a ∈ Z p ; further, µ p +1 centralizes the subset Γ inEqn (3.1), and h µ p +1 i partitions Z ∗ p into p − h µ p +1 i -orbits, see Remark 2.5(5.i);take representatives u , · · · , u p − from each h µ p +1 i -orbit, the p − = { u , ( p + 1) u , · · · , ( p + 1) p − u } , Γ = { u , ( p + 1) u , · · · , ( p + 1) ( p − u } , · · · Γ p − = { u p − , ( p + 1) u p − , · · · , ( p + 1) ( p − u p − } . Accordingly, we apply Lemma 2.2 and its notation to get the basis of C p : E = E ∪ E ∪ · · · ∪ E p − , where E k is corresponding to Γ k for k = 1 , · · · , p − E iscorresponding to Γ ; and we write matrices with respect to this basis. So C = D ⊕ D ⊕ · · · ⊕ D p − , where D = 1 ⊕ λ p ⊕ · · · ⊕ λ ( p − p , (3.2) D i = λ u i ⊕ λ ( p +1) u i ⊕ · · · ⊕ λ ( p +1) p − u i , i = 1 , · · · , p − B = B ⊕ p − z }| { P ⊕ · · · ⊕ P , where B = ε ⊕ ε ⊕ · · · ⊕ ε p − (3.3)with ε , ε , · · · , ε p − being p ’th roots of unity and P is the cycle matrix ofdimension p (see Lemma 2.2 and Eqn (2.7)): P = · · · · · · p × p . Any element of G has the form C k B h , 0 ≤ k ≤ p −
1, 0 ≤ h ≤ p −
1; and C k B h = D k B h ⊕ D k P h ⊕ · · · ⊕ D kp − P h . Obviously, D k B h = ε h ⊕ λ pk ε h ⊕ · · · ⊕ λ p ( p − k ε hp − . (3.4)It is easy to calculate the characteristic polynomials:char D ki P h ( x ) = x p − λ P j ∈ Γ i jk = x p − λ u i pmk , i = 1 , · · · , p − pm = 1 + ( p + 1) + · · · + ( p + 1) p − = ( p +1) p − p , hence m is an integercoprime to p . If k p ), by Remark 2.5(5.ii), λ u i pmk for i = 1 , · · · , p − p ’th root of unity. Thus the characteristic polynomial ofthe matrix C k B h ischar C k B h ( x ) = ( char D k B h ( x ) · ( x p − p − , k ≡ p );char D k B h ( x ) · Φ p ( x ) , k p );where Φ p ( x ) denotes the p ’th cyclotomic polynomial. Since C k B h is similarto a permutation matrix, by Lemma 2.1, for any k, h we obtain thatchar D k B h ( x ) = ( x p − x − p , k ≡ p ); x p − , k p ) . (3.5)By Eqn (3.2), we can view D as a maximal cycle of dimension p ; byEqn (3.3), the matrix B of dimension p commutes with D ; by Lemma 2.1,from Eqn (3.5) we see that the abelian matrix group h D , B i of dimension p isa permutation-like matrix group of dimension p ; so, by Lemma 2.4, we have aninteger 0 ≤ ℓ ≤ p − ε , ε , · · · , ε p − ) = (1 , λ pℓ , · · · , λ p ( p − ℓ ) . D k B h ( x ) = ( x − x − λ p ( k + ℓh ) ) · · · ( x − λ p ( p − k + ℓh ) ) , so char D k B h ( x ) = ( ( x − p , k + ℓh ≡ p ); x p − , k + ℓh p ) . (3.6)Suppose that 0 < ℓ ≤ p −
1; taking k p ), we have an h such that k + ℓh ≡ p ), then, by Eqns (3.5) and (3.6) we have x p − D k B h ( x ) = ( x − p , which is impossible. Thus ℓ = 0, i.e. ( ε , ε , · · · , ε p − ) = (1 , , · · · , B = I p × p ⊕ p − z }| { P ⊕ · · · ⊕ P .
Similar to Case 1, the group h B i satisfies the condition (SC) of Proposition 2.3: (SC) For any e j and B i , if B i C e j = C e j then B i e j = e j . Thus G is a permutation matrix group. Case 3. q = ps and s >
1. First we show that B ps = C ap with 0 ≤ a < p, hence ord( B ) = ( ps, a = 0; p s, < a < p. (3.7)For: otherwise B ps ∈ h C i − h C p i , then h C i = h B ps i , hence B centralizes h C i ,which contradicts to that G is non-abelian.It is easy to see that the group h µ r i of order q = ps acts freely on Z ∗ p , hencepartitions Z ∗ p in to t orbits of length q ; taking representatives w , · · · , w t fromthese t orbits, we have the orbits of h µ r i on Z ∗ p as follows:Γ = { w , rw , · · · , r q − w } , · · · , Γ t = { w t , rw t , · · · , r q − w t } . On the other hand, since the group h µ r s i of order p centralizes Γ , the group h µ r i fixes 0 and partitions Γ − { } in to t orbits of length s (cf. Case 2); takingrepresentatives v , · · · , v t from these t orbits, we have the orbits of h µ r i on Γ as follows:Γ = { } , Γ = { v , rv , · · · , r s − v } , · · · , Γ t = { v t , rv t , · · · , r s − v t } . According to the orbits Γ , Γ , · · · , Γ t , Γ , · · · , Γ t , we apply Lemma 2.2and its notation to get the basis of C p : E = E ∪ E ∪ · · · ∪ E t ∪ E ∪ · · · ∪ E t , C = 1 ⊕ (cid:16) s − ⊕ i =0 λ r i v (cid:17) ⊕ · · · ⊕ (cid:16) s − ⊕ i =0 λ r i v t (cid:17) ⊕ (cid:16) q − ⊕ i =0 λ r i w (cid:17) ⊕ · · · ⊕ (cid:16) q − ⊕ i =0 λ r i w t (cid:17) ;because p | v j for j = 1 , · · · , t (see Eqn (3.1)), λ r i v j p = 1 for 0 ≤ i ≤ s − ≤ j ≤ t , so C ap = I p × p ⊕ (cid:16) q − ⊕ i =0 λ r i w ap (cid:17) ⊕ · · · ⊕ (cid:16) q − ⊕ i =0 λ r i w t ap (cid:17) . And, there are complexes ε and ε j , ω j for j = 1 , · · · , t such that B = ε ⊕ P ⊕ · · · ⊕ P t ⊕ Q ⊕ · · · ⊕ Q t , where P j , Q j are as described in Eqn (2.5) (see Lemma 2.2) : P j = · · · ε j · · · s × s . Q j = · · · ω j · · · q × q ;hence B ps = ε ps ⊕ ε p I s × s ⊕ · · · ⊕ ε pt I s × s ⊕ ω I q × q ⊕ · · · ⊕ ω t I q × q . Since B ps = C ap (cid:0) see Eqn (3.7) (cid:1) , the collection of q z }| { ω , · · · , ω , · · · , q z }| { ω t , · · · , ω t (3.8)is coincide with the collection of λ w ap , λ rw ap , · · · , λ r q − w ap , · · · , λ w t ap , λ rw t ap , · · · , λ r q − w t ap . (3.9)Note that (cid:8) w , rw , · · · , r q − w , · · · , w t , rw t , · · · , r q − w t (cid:9) = Z ∗ p . If 0 < a < p (cf. Eqn (3.7)), then the collection (3.9) is just all primitive p ’throots of unity with multiplicity p for each one, see Remark 2.5(1); on the otherhand, the number of the elements appeared in the collection (3.8) is at most t ;but ts = p − s >
1, so t is less than the number of primitive p ’th roots;that is a contradiction to the coincidence of the collections (3.8) and (3.9).In conclusion, a = 0 and B ps = 1.Since p and s are coprime, we have h B i = h B p i × h B s i , |h B p i| = s, |h B s i| = p.
10e have considered h C, B p i and h C, B s i in Case 1 and Case 2 respectively,and have concluded that in both cases the condition (SC) in Proposition 2.3 issatisfied. Assume that B i C e j = C e j ; taking integers k, h such that ph + sk = 1,we have B i = B phi B ski , B phi ∈ h B p i , B phi C e j = C e j and B ski ∈ h B s i , B ski C e j = C e j ;by the conclusions in Case 1 and in Case 2, B phi e j = e j and B ski e j = e j ; hence B i e j = e j . Thus, by Proposition 2.3, G is a permutation matrix group.The proof of Theorem 1.1 is completed. Acknowledgements
The research of the authors is supported by NSFC with grant numbers 11171194and 11271005.