Conjugacy classes of torsion in GL_n(Z)

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1 Electronic Journal of Linear Algebra Volume 30 Volume 30 (2015) Article Conjugacy classes of torsion in GL_n(Z) Qingjie Yang Renmin University of China Follow this and additional works at: Recommended Citation Yang Qingjie (2015) "Conjugacy classes of torsion in GL_n(Z)" Electronic Journal of Linear Algebra Volume 30 DOI: This Article is brought to you for free and open access by Wyoming Scholars Repository It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository For more information please contact

2 CONJUGACY CLASSES OF TORSION IN GL N (Z) QINGJIE YANG Abstract The problem of integral similarity of block-triangular matrices over the ring of integers is connected to that of finding representatives of the classes of an equivalence relation on general integer matrices A complete list of representatives of conjugacy classes of torsion in the 4 4 general linear group over ring of integers is given There are 45 distinct such classes and each torsion element has order of or 12 Key words General linear group Ring of integers Integral similarity Direct sum Torsion Cyclotomic polynomial AMS subject classifications 53D30 15A36 1 Introduction The problem that we consider in this paper is the determination of the conjugacy classes of torsion matrices in the n n general linear group over Z the ring of integers Let M n m (Z) be the set of n m matrices over Z For a matrix A M n m (Z) the transpose of A is denoted by A T When n = m we simply denote M n m (Z) by M n (Z) Let I n be the identity matrix in M n (Z) A unimodular matrix of size n is an n n integer matrix having determinant +1 or 1 The general linear group of size n over Z denoted by GL n (Z) is the set of unimodular matrices in M n (Z) together with the operation of ordinary matrix multiplication That is GL n (Z) = {A M n (Z) A = ±1} where A is the determinant of A An element T of GL n (Z) is a torsion element if it has finite order ie if there is a positive integer m such that A m = I A d-torsion element is a torsion element that has order d Two matrices A B of M n (Z) are conjugates or integrally similar denoted by A B if there is a matrix Q GL n (Z) such that B = Q 1 AQ Finding finite groups or torsion of integral matrices up to conjugation has a long history see [5] Received by the editors on April Accepted for publication on June Handling Editor: Bryan L Shader School of Information Renmin University of China Beijing China (yangqj@ruceducn) 478

3 Conjugacy Classes of Torsion in GL n(z) 479 Given a matrix A M n (Z) we denote the characteristic polynomial of A by f A (x) = xi A If A GL n (Z) then f A (x) is a monic polynomial with constant term f(0) = ±1 Let f(x) = x n +a n 1 x n 1 + +a 1 x+a 0 Z[x] the polynomial ring over Z with f(0) = ±1 The set of all integral matrices with characteristic polynomial f(x) is denoted by M f That is M f = {A GL n (Z) f A (x) = f(x)} Let M f be the set of all conjugacy classes of matrices in M f The size of M f is denoted by M f The matrix C f given by a a 1 C f = a a n 1 is known as the companion matrix of f(x) = x n +a n 1 x n 1 + +a 1 x+a 0 It is known that C f M f and thus M f For any A GL n (Z) we use C(A) to denote its centralizer in GL n (Z) If A is similar to the companion matrix of a polynomial over Z then its centralizer is C(A) = {g(a) GL n (Z) g(x) Z[x] is of degree less than n} For A M n m (Z) and B M s t (Z) the direct sum of A and B is A B = A 0 M 0 B (n+s) (m+t) (Z) Obviously A B is a unimodular matrix if and only if both A and B are unimodular matrices A matrix A GL n (Z) is decomposable if it is conjugate to a direct sum of two matrices which have smaller sizes; otherwise A is said to be indecomposable The characteristic polynomial of a decomposable matrix is reducible over Z but the converse is not true

4 480 Q Yang In this paper we mainly consider the integral similarity problem for upper blocktriangular matrices of the form A X (11) 0 B where A B are unimodular matrices with coprime minimal polynomials Our results are based on the following lemmas We state them without proof Lemma 11 Each A in M n (Z) is integrally similar to a block-triangular matrix A 11 A 12 A 1r 0 A 22 A 2r 0 0 A rr where the characteristic polynomial of A ii is irreducible 1 i r The blocktriangularization can be attained with the diagonal blocks in any prescribed order See [6 9] for a proof Lemma 12 Let A GL n (Z) have irreducible minimal polynomial p(x) with M p = 1 Then A is integrally similar to C p C p C p where C p is the companion matrix of p(x) That is M p k = 1 See also [6] Consider two monic polynomials f(x) = x n +a n 1 x n 1 + +a 0 and g(x) = x m +b m 1 x m 1 + +b 0 in Z[x] Recall that the resultant of f(x) and g(x) is the determinant 1 a n 1 a a n 1 a a n 1 a 0 R(fg) = 1 b m 1 b b m 1 b b m 1 b 0 m rows n rows

5 Conjugacy Classes of Torsion in GL n(z) 481 It is known that f(x) and g(x) are coprime if and only if R(fg) 0 The following theorem which is a corollary of Lemma 31 gives a sufficient condition for decomposability Theorem 13 Let A M n (Z) with its characteristic polynomial a product of two coprime polynomials whose resultant is ±1 Then A is decomposable To explain our results we need to develop some notation For any A GL n (Z) A e A e we use A + A to denote the block matrices respectively where e = (100) T M n 1 (Z) Clearly A + A GL n+1 (Z) Also we let C n denote the companion matrix of Φ n (x) the nth cyclotomic polynomial of degree ϕ(n) where ϕ is the Euler s totient function Our results are given in following theorems We will prove them in Section 3 Theorem 14 Let n > 1 and A = C n C n C n the direct sum of s-copies of C n Let A X M = where X M 0 I sϕ(n) m (Z) m 1 If n = p k where p is a prime number and k 1 then (12) M C n + C+ n C n C n I }{{}}{{} m t t s t where the number t of C + n satisfies 0 t min(sm) and is uniquely determined by M 2 If n is not a power of a prime then M A I m The special case n = 2 was established by Hua and Reiner [4] Similarly we have the following Theorem 15 Let n > 2 and A = C n C n C n the direct sum of s-copies of C n Let A X M = where X M 0 I sϕ(n) m (Z) m 1 If n = 2p k where p is a prime and k 1 then M Cn Cn C n C n ( I }{{}}{{} m t ) t s t

6 482 Q Yang where the number t of Cn satisfies 0 t min(sm) and is uniquely determined by M 2 If n 2p k then M A ( I m ) A complete conjugacy list of torsion in GL 2 (Z) is already known Lemma 16 All torsion in GL 2 (Z) up to conjugation are given in the following table together with the centralizers and minimal polynomials of the conjugacy class representatives where I = order A C(A) m A (x) 1 I GL 2 (Z) Φ 1 (x) = (x 1) I GL 2 (Z) Φ 2 (x) = (x+1) 2 K ±I±K (x 1)(x+1) U ±I±U (x 1)(x+1) 3 W ±I±W±(I +W) Φ 3 (x) = x 2 +x+1 4 J ±I±J Φ 4 (x) = x W ±I±W±(I +W) Φ 6 (x) = x 2 x K = U = W = C 3 = J = C 4 = 1 1 For a proof see [8] Although the maximal finite subgroups of GL 4 (Z) up to conjugation have been determined by Dade [1] a complete set of non-conjugate classes of torsion in GL 4 (Z) is of value We have solved the closely related problem of classifying the conjugacy classes of elements of finite order in the 4 4 symplectic group over Z see [12] If A is a d-torsion element in GL 4 (Z) then its minimal polynomial m A (x) is a factor of x d 1 ie m A (x) is a product of cyclotomic polynomials It is easy to check that any torsion element A in GL 4 (Z) has order or 12 Note that ϕ(5) = ϕ(8) = ϕ(10) = ϕ(12) = 4 and then Φ n (x) is a quartic polynomial for n = or 12 According to Latimer MacDuffee and Taussky [11] if the characteristic polynomial of A is Φ 5 (x) Φ 8 (x) Φ 10 (x) or Φ 12 (x) then A is conjugate to C 5 C 8 C 10 or C 12 respectively since Q(ζ m ) where ζ m is a primitive mth root of unity has class number one for all positive integers m less than 12 We reduce the problem to the case that the characteristic polynomial of A is reducible The cases where m A (x) is an irreducible quadratic polynomial that is m A (x) = Φ n (x) n = 34

7 Conjugacy Classes of Torsion in GL n(z) 483 or 6 can be solved by Lemma 12 and Lemma 16 Furthermore the case where A 2 = I was solved in complete generality by Hua and Reiner [4] We only need to consider the cases where m A (x) is one of the following: (x 2 +1)(x 1) (x 2 +1)(x+1) (x 2 ±x+1)(x 1) (x 2 ±x+1)(x+1) (x 2 ±x+1)(x 2 +1) (x 2 +x+1)(x 2 x+1) (x 2 1)(x 2 + 1) and (x 2 1)(x 2 ± x + 1) As a consequence by Lemma 11 A is integrally similar to a block upper triangular matrix with different diagonal 2 2 blocks chosen from Lemma 16 By applying Theorem 14 or Theorem 15 we can solvetheproblemforthe firstfourcaseswhereoneandonlyoneof±1isaneigenvalue The case where m A (x) = (x 2 ±x+1)(x 2 +1) can be solved by Theorem 13 For the remaining three cases we have the following result Theorem 17 All elements in GL 4 (Z) with some given reducible characteristic polynomials f(x) up to conjugation are listed below: M f = 1 When f(x) = (x 2 1)(x 2 +λx+1) where λ = ±1 M f = { K 0 λ λ 0 W 2 When f(x) = (x 2 1)(x 2 +1) { K 0 K E K I K E U 0 λ λ 0 W 0 W K I E U 0 } U E ; 0 W U E } U I ; 3 When f(x) = (x 2 +x+1)(x 2 x+1) where E = M f = { W 0 0 W } W E 0 W To prove our results we need to develop some new tools The idea we use in this paper comes from Roth s Theorem [10] For a general form of Roth s Theorem see [2] or [3] We shall generalize Roth s Theorem to the integral similarity problem for upper block-triangular matrices of the form (11) with the diagonal blocks have coprime characteristic polynomials In Section 2 we shall study (A B)-equivalence in M n m (Z) Then we transform our similarity problem to the problem finding (AB)- equivalent classes and prove our results in Section 3 We use the program Mathematica to calculate some of results in this paper 2 (AB)-equivalence Let A GL n (Z) B GL m (Z) and suppose that their respective characteristic polynomials f(x) and g(x) are coprime We define a linear

8 484 Q Yang transformation ψ on M n m (Z) by ψ : M n m (Z) M n m (Z) T AT TB Since f(x) g(x) are coprime ψ is injective Let AB be the image of ψ that is AB = {AT TB T M n m (Z)} By choosing a suitable basis the matrix of ψ is A I m I n B T where is the Kronecker product of matrices Then the determinant of ψ is equal to R(fg) the resultant of f(x) and g(x) Let r = R(fg) the absolute value of R(fg) The quotient module M n m (Z)/ AB called the cokernel of ψ and denoted by cokerψ is of order r Let X M n m (Z) Then an equivalent condition for X AB is that the Sylvester equation (21) AT TB = X has a unique integral solution for matrix T Clearly if X 0 (mod r) then X AB Lemma 21 Let C f be the companion matrix of f(x) of degree n and α M n 1 (Z) be an integral column vector Then α C f I 1 if and only if the integer number f(1) divides l(α) the sum of components of α Proof Let f(x) = x n +a n 1 x n 1 + +a 1 x+a 0 and α = (c 1 c 2 c n ) T In this case C f I 1 = {(C f I)X X = (x 1 x 2 x n ) T M n 1 (Z)} So α C f I 1 if and only if the system of linear equations x 1 a 0 x n =c 1 x 1 x 2 a 1 x n =c 2 x 2 x 3 a 2 x n =c 3 x n 1 (1+a n 1 )x n =c n has an integral solution This system is equivalent to the following system x 1 a 0 x n =c 1 x 2 (a 0 +a 1 )x n =c 1 +c 2 x n 1 (a 0 +a 1 + +a n 1 )x n =c 1 +c 2 + +c n 1 (1+a 0 +a 1 + +a n 1 +a n )x n =c 1 +c 2 + +c n 1 +c n

9 Conjugacy Classes of Torsion in GL n(z) 485 which has an integer solution for x 1 x 2 x n if and only if f(1) = (1+a 0 +a 1 + +a n 1 +a n ) (c 1 +c 2 + +c n 1 +c n ) = l(α) Thus α C f I 1 if and only if f(1) divides l(α) Similarly α C f I 1 if and only if f( 1) divides c 1 c 2 + +( 1) n 1 c n the alternating sum of components of α We now define an equivalence relation on M n m (Z) Definition 22 Let XY M n m (Z) be any two matrices X and Y are said to be (AB)-equivalent denoted by X = Y (mod AB) or X = Y for short if there exist P C(A) and Q C(B) such that XQ PY AB The set of (A B)-equivalent classes is denoted by S(A B) It is obvious that if X Y AB then X = Y (mod AB) But the converse is not necessarily true Lemma 23 Let XY M n m (Z) Then 1 X = Y (mod AB) if and only if X PYQ 1 AB for some P C(A) Q C(B); 2 X = PXQ (mod AB) where P C(A) and Q C(B) In particular X = X; 3 If X Y (mod r) then X = Y (mod AB) where r = R(fg) Proof By definition X = Y if and only if there exist P C(A) and Q C(B) such that (22) XQ PY = AT TB for some T M n m (Z) Since Q commutes B so does Q 1 and then (22) is equivalent to Therefore Part 1 is true X PYQ 1 = A(TQ 1 ) (TQ 1 )B Part 2 is obtained by PXQ (PXQ) = 0 and ( I n ) C(A) Part 3 is true since X Y AB whenever X Y (mod r) Suppose that the cokernel of ψ has a set of representative cokerψ = {S 1 S 2 S r S i M n m (Z)}

10 486 Q Yang where S i = S i + AB i = 1r are all cosets of AB in M n m (Z) We define a group action of C(A) C(B) on cokerψ given by PS i Q = PS i Q for any (PQ) C(A) C(B) The action is well defined since P AB = AB = AB Q The set of orbits is denoted by cokerψ/c(a) C(B) Lemma 24 Let XY M n m (Z) Then a necessary and sufficient condition for X = Y is that X and Y are in the same orbit of the action That is S(AB) = cokerψ/c(a) C(B) Proof Note that X = Y if and only if X PYQ AB for some (PQ) C(A) C(B) which is equivalent to X = PYQ Therefore X = Y if and only if X and Y are in the same orbit From Lemma 24 when r = 1 S(AB) the class number of (AB)-equivalence is equal to 1 If r > 1 then 1 < S(AB) r because AB is a fixed point for all elements in C(A) C(B) In particular if r = 2 S(AB) = 2 LetM = A A Abethedirectsumofs-copiesofA andn = B B B be the direct sum of t-copies of B Let X 11 X 12 X 1t X 21 X 22 X 2t X = M sn tm(z) X s1 X s2 X st where X ij M n m Then we have the following Lemma 25 X MN if and only if X ij AB for i = 1s; j = 1t It is also easy to prove that the following block row/column elementary operations on X preserve the equivalence: 1 Row/column switching R i R j : switch the i-th row blocks and the j-th row blocks; C i C j : switch the i-th column blocks and the j-th column blocks 2 Row/column multiplication R i PR i : left-multiply the i-th row blocks by P where P C(A); C i C i Q: right-multiply the i-th column blocks by Q where Q C(B)

11 Conjugacy Classes of Torsion in GL n(z) Row/column addition R i R i +PR j : add the j-th row blocks left-multiplied by P to the i-th row; C i C i +C j Q: add the j-th column blocks right-multiplied by Q to the i-th column where P M n (Z) commutes A and Q M m (Z) commutes B 3 Proofs Before the proof we need to give a connection of (A B)-equivalence with integral similarity of matrices of the form (11) Lemma 31 Let A M n (Z) B M m (Z) and suppose that they have coprime characteristic polynomials Then if and only if X A X A Y 0 B 0 B = Y (mod AB) Proof First suppose that GL n+m (Z) such that We get that A X 0 B A Y There is Q = 0 B A X Q11 Q 12 Q11 Q 12 A Y = 0 B Q 21 Q 22 Q 21 Q 22 0 B Q11 Q 12 Q 21 Q 22 (31) (32) (33) (34) AQ 11 +XQ 21 = Q 11 A AQ 12 +XQ 22 = Q 11 Y +Q 12 B BQ 21 = Q 21 A BQ 22 = Q 21 Y +Q 22 B Q11 Q 12 By hypothesis (33) implies Q 21 = 0 Then Q = and (31) (34) say 0 Q 22 that Q 11 C(A) and Q 22 C(B) Thus (32) means X = Y Conversely if X = Y by some (PQ) C(A) C(B) and T M n m (Z) then A X A Y P T via the similarity 0 B 0 B 0 Q According to Lemma 31 integral similarity problem for block-triangular matrices of the form (11) can be transformed to the problem of finding (AB)-equivalent classes Now we can prove our theorems Proof of Theorem 14 For any matrix X M sϕ(n) m (Z) we write X as a block

12 488 Q Yang matrix α 11 α 12 α 1m α 21 α 22 α 2m X = α s1 α s2 α sm where α ij M ϕ(n) 1 (Z) Then by Lemma 25 X AI m if and only if α ij C n I 1 which is equivalent to that Φ n (1) is a factor of l(α ij ) by Lemma 21 for all α ij Note that for n > 2 see [7] { p n = p k p prime k 1 (35) Φ n (1) = 1 otherwise Case 1 n = p k is a power of prime p then Φ n (1) = p We first show that S(C n I 1 ) = {0e} For any α M ϕ(n) 1 (Z) let b = l(α) l(α be) = l(α) bl(e) = b b 1 = 0 So α = be (mod C n I 1 ) We only need to show be = e (mod C n I 1 ) provided p b Without loss of generality we assume 0 < b p 1 Let P = I+C n +Cn+ +C 2 n b 1 Then P(I C n ) = I Cn b and thus the determinant of P satisfies P I C n = I C b n Since (bp) = 1 C n and C b n have the same characteristic polynomial Φ n (x) So I C n = I C b n = Φ n (1) = p Therefore P = 1 and P GL ϕ(n) (Z) Also P commutes with C n It is easy to verify that l(pe) = b and then l(be Pe) = 0 So be = e (mod C n I 1 ) and hence S(C n I 1 ) = {0e} Now suppose that there is α ij / C n I 1 We can use row or column switchings move it to the left-top position So we may assume that α 11 / C n I 1 There is P C n such that Pα 11 e C n I 1 Then by a row multiplication and Lemma 25 X R1 PR1 Pα 11 β 12 β 1m e β 12 β 1m β 21 β 22 β 2m β 21 β 22 β 2m = β s1 β s2 β sm β s1 β s2 β sm

13 Conjugacy Classes of Torsion in GL n(z) 489 By row additions R i R i l(β i1 )R 1 and column additions C j C j l(β 1j )C 1 we get e β 12 β 1m e γ 12 γ 1m e 0 0 β 21 β 22 β 2m γ 21 γ 22 γ 2m 0 γ 22 γ 2m = β s1 β s2 β sm γ s1 γ s2 γ sm 0 γ s2 γ sm where γ i1 = β i1 l(β i1 )e γ 1j = β 1j l(β 1j )e C n I 1 Continue this process to the submatrix obtained by deleting first row block and first column and so on we obtain X Y 0 = where Y = e e e 0 0 }{{} t for some 1 t min(sm) Therefore by Lemma 31 C n e C n C n A X 0 I m C n 1 e 1 I m t where the number of C n is s and the number of e is t By some pairs of row and column switchings we get the matrix on the right is conjugate to the matrix (12) For the uniqueness let X i = e e e i = 12 with t }{{} 1 > t 2 and suppose t i that X1 0 X2 0 = P11 P 12 Q11 Q 12 There are P = C(A) Q = C(B) where P 11 is t 1 t 2 P 21 P 22 Q 21 Q 22 matrix Q 11 is t 1 t 2 matrix such that X1 0 Q P 0 0 X2 0 = 0 0 X1 Q 11 P 11 X 2 X 1 Q 12 AI m P 21 X 2 0

14 490 Q Yang We get X 1 Q 12 0 (mod p) so Q 12 0 (mod p) Note that the block Q 12 is a t 1 (m t 2 ) matrix and t 1 +(m t 2 ) > m the size of Q Therefore the determinant of Q satisfies Q 0 (mod p) This is impossible since Q is an unimodular matrix This completes the proof of uniqueness Case 2 n is not a powerofprime From (35) Φ n (1) = 1 and then M ϕ(n) 1 (Z) = C n I 1 There is only one (AI m )-equivalent class Therefore M A I m The proof of Theorem 15 is similar In this case we use the fact that { p n = 2p k p prime k 1 (36) Φ n ( 1) = 1 otherwise see [7] Proof of Theorem 17 By Lemma 11 and Lemma 31 we only need to calculate (AB)-equivalent classes for some special pairs of 2 2 matrices in Lemma 16 When A = K and B = W Clearly r = 3 The linear transformation ψ is given by ψ(t) = KT TW Then the Sylvester equation (21) becomes 1 0 T T = 1 1 It is equivalent to the system of linear equations t 11 t 12 =a t 11 +2t 12 =b t 21 t 22 =c =d t 21 a b c d which has integral solutions if and only if 3 a b Thus the submodule KW the image of ψ is { } a b KW = M 2 (Z) a b(mod 3) c d It isobviousthat E2E / KW Therefore cokerψ = M 2 (Z)/ KW = {0E2E} By choosing P = I C(K) Q = I C(W) we see that EQ P(2E) = 3 0 KW 0 0 and thus E = 2E (mod KW) Note that S(KW) > 1 hence S(KW) = {0E}

15 Conjugacy Classes of Torsion in GL n(z) 491 When A = U and B = W We also have r = 3 This time the Sylvester equation is equivalent to t 11 t 12 +t 21 =a t 11 +2t 12 +t 22 =b t 21 t 22 =c t 21 =d It is clear that UW = { a b c d } M 2 (Z) a+d b+c(mod 3) So cokerψ = {0E2E} and then S(UW) = {0E} Since A B = A B for any A and B we obtain that S( K W) = S( U W) = {0E} Similarly we have { } a b KJ = M 2 (Z) a+b c+d 0(mod 2) S(KJ) = {0EII E} c d { } a b UJ = M 2 (Z) a+b+c c+d 0(mod 2) S(UJ) = {0EI} c d { } a b W W = M 2 (Z) a+b+c a d(mod 2) S(W W) = {0E} c d In summary we have the following table A B S(A B) K or U W 0E K or U W 0E K J 0EII E U J 0EI W W 0 E We can use the results in this table Lemma 11 and Lemma 31 to complete the proof From above theorems and some simple calculations all torsion in GL 4 (Z) up to

16 492 Q Yang conjugation are listed as follows: d = 1 I 4 ; d = 2 I 4 ; K ( I) U ( I); I ( I) K U U U; I K I U; I E d = 3 W W; I W ; 0 W I E I E d = 4 J J; I J ; ( I) J ; K E K I K I E K J ; U E U I U J ; d = 5 C 5 ; d = 6 (W W); I ( W); ( I) W; (I W) d = 8 C 8 ; K W (K W) W ( W) d = 10 C 5 ; [ K E 0 W [ K E 0 W W E ; 0 W ] ; U W ] ; (U W) d = 12 C 12 ; J W; J ( W) U E ; 0 W U E ; 0 W I E ; 0 W Acknowledgment The author would like to thank the referee for helpful comments REFERENCES [1] EC Dade The maximal finite groups of 4 4 integral matrices Illinois J Math 9: [2] R Guralnick Roth s theorems and decomposition of modules Linear Algebra Appl 39: [3] R Guralnick Roth s Theorems for sets of matrices Linear Algebra Appl 71: [4] LK Hua and I Reiner Automorphisms of the unimodular group Trans Amer Math Soc 71(3):

17 Conjugacy Classes of Torsion in GL n(z) 493 [5] J Kuzmanovich and A Pavlichenkov Finite groups of matrices whose entries are integers Amer Math Monthly 109(2): [6] TJ Laffey Lectures on Integer Matrices Unpublished lecture notes 1997 [7] S Lang Algebra Graduate Texts in Mathematics Vol 211 Springer-Verlag New York 2002 [8] G Mackiw Finite groups of 2 2 integer matrices Math Mag 69(5): [9] M Newman Integral Matrices Academic Press New York 1972 [10] W Roth The equations AX YB = C and AX XB = C in matrices Proc Amer Math Soc 3(3): [11] O Taussky On a theorem of Latimer and Macduffee Canadian J Math 1: [12] Q Yang Conjugacy classes of Torsion in 4 4 integral symplectic group JMath Res Exposition 28(1):

Dihedral groups of automorphisms of compact Riemann surfaces of genus two

Electronic Journal of Linear Algebra Volume 26 Volume 26 (2013) Article 36 2013 Dihedral groups of automorphisms of compact Riemann surfaces of genus two Qingje Yang yangqj@ruc.edu.com Dan Yang Follow