Lie Regular Generators of General Linear Group Gl(4,Z

Home , General linear group, Linear group, Multiplicative group

LIE REGULAR GENERATORS OF GENERAL LINEAR GROUP GL(4, Zn)

SWATI MAHESHWARI AND R. K. SHARMA

Abstract. In this article, we discuss the existence of Lie regular matrices in M(4, Zn). It is shown that the linear group GL(4, Zn) is generated by Lie regular matrices for all n > 1.

1. Introduction

Special linear group SL(n, Z) is the multiplicative group of all n × n matrices with integer entries having determinant 1. It is well known that SL(n, Z) is generated by transvections, the matrices Tij ( 1 ≤ i, j ≤ n, i 6= j) with 1’s on the diagonal and in the (i, j)-th position and 0’s elsewhere. Generators of unimodular group, the linear group GL(n, Z) of all n × n matrices with integer entries having determinant ±1, has been discussed in [1, p.85], [5]. In 2012, Sharma et al. intro- duced Lie regular elements and Lie regular units for non commutative rings (see [2, 3]). They have given generators of linear group GL(2, Zn), for some n > 1 in terms of Lie regular units. They proposed an open problem to determine rings that have Lie regular units and whether it is possible to generate the whole unit group of such rings using Lie regular units. We are considering this problem with the ring M(4, Zn). In this article, we will discuss the existence of Lie regular units in the ring M(4, Zn) and will show that Lie regular units generate GL(4, Zn) for n > 1.

Throughout this article, φ denotes the Euler’s totient function and U(R) denotes the unit group of the ring R. |A| denotes the cardinality of the set A. Suppose G is a group then o(G) denotes the order of G and o(g) denotes the order of an element g of G.

2. Preliminaries We shall frequently use the following well known results to prove our main results.

2010 Mathematics Subject Classiﬁcation. Primary 20H25; 16U60; 20F05. Key words and phrases. Linear Group; Lie regular elements. 1 2 SWATI MAHESHWARI AND R. K. SHARMA

Lemma 2.1. The linear group SL(n, Z) is generated by transvections, the matrices Tij (1 ≤ i 6= j ≤ n) with 1’s on the diagonal and in the (i, j)-th position and 0’s elsewhere.

Lemma 2.2. [4, p.21]The natural map from SL(n, Z) → SL(n, Zm) is onto.

Note that we denote image of a transvection Tij by Tij itself.

Corollary 2.1. Transvections generate SL(n, Zm). Lemma 2.3. Suppose m and n are two positive integers such that ∼ (m, n) = 1. Then U(Zmn) = U(Zm) × U(Zn). Deﬁnition 2.1. [3] An element ‘a’ of a ring R is said to be Lie regular if a = [e, u] = eu − ue, where e is an idempotent in R and u is a unit in R. Further, a unit in R is said to be Lie regular unit if it is Lie regular as an element of R.

Proposition 2.1. [3, Proposition 2.6] Let F be a ﬁeld then the inverse of a Lie regular unit in M(2, F) is again Lie regular.

Proposition 2.2. [3, Proposition 2.14] If R is a commutative ring then any element in M(2, R) of the form λy x −λy −y −λy y λy −x , , , , −y −λy x λy −x λy y −λy where λ, x and y belong to R and xy is invertible in R, is a Lie regular element.

3. Lie regular elements in M(4, Zm) Proposition 3.1. If R is a commutative ring with unity and A, B are Lie regular elements (units ) in M(2, R), that is, A = [e, u] and B = A 0 [e , u ]. Then the block matrix is also a Lie regular element 1 1 0 B (unit) in M(4, R).

Proof. Observe that, A 0 e 0 u 0 = , . 0 B 0 e1 0 u1 This completes the proof. A 0 Note that if R is a ﬁeld then the inverse of is also a Lie 0 B regular unit in M(4, R). LIE REGULAR GENERATORS OF GENERAL LINEAR GROUP GL(4, Zn) 3 Proposition 3.2. If R is a commutative ring with unity and A, B are Lie regular elements (units ) in M(2, R) such that A = [e, u] and B = 0 A [e, u ]. Then the block matrix is also a Lie regular element (unit) 1 B 0 in M(4, R).

Proof. Observe that, 0 A e 0 0 u = , . B 0 0 e u1 0 This completes the proof.

For example take A, B ∈ M(2, Zm) such that 0 1 1 0 0 1 A = = , 1 0 0 0 −1 0 and 0 1 1 0 0 1 B = = , , α 0 0 0 −α 0 0 A where α ∈ U( ). Then is a Lie regular unit in M(4, ). Zm B 0 Zm 0 A Note that If R is a ﬁeld then the inverse of is also a Lie B 0 regular unit in M(4, R).

0 a 0 b 0 a b 0 c 0 d 0 c 0 0 d Proposition 3.3. Suppose A =   and B =   0 e 0 f e 0 0 f g 0 0 0 0 g 0 0 are elements in GL(4, R). Then A, B are Lie regular in GL(4, R). Fur- 0 0 1 0 0 0 1 1 ther the element C =   is also a Lie regular element in 1 0 0 0 0 1 0 0 GL(4, R).

Proof. It follows once we observe that

0 0 0 0 0 −a 0 −b 0 1 0 0 c 0 d 0  A =   ,   . 0 0 0 0 0 −e 0 −f 0 0 0 1 g 0 0 0 4 SWATI MAHESHWARI AND R. K. SHARMA

1 0 0 0  0 a b 0  0 0 0 0 −c 0 0 −d B =   ,   . 0 0 0 0 −e 0 0 −f 0 0 0 1 0 g 0 0 0 0 0 0 0 0 −1 0  0 0 0 0 0 0 −1 −1 C =   ,   . 0 0 1 0 1 0 0 0  0 0 0 1 0 1 0 0 Proposition 3.4. Suppose A ∈ GL(2, R). Then the block matrix AA is a Lie regular element in GL(4, R). −A −A Proof. Observe that AA I I 0 A = d d , . −A −A 0 0 A 0 This completes the proof. o(GL(4,Zm)) Lemma 3.1. The order of the linear group SL(4, Zm) is φ(m) .

Proof. Consider a group homomorphism θ : GL(4, Zm) → U(Zm) by θ(A) = determinant(A).

The map is surjective, thus result holds. 4. Generators of Linear Groups

Lemma 4.1. For each invertible element α ∈ Zm, the number of matrices in M(4, Zm) having determinant α is equal to the order of SL(4, Zm).

Proof. Let α be an invertible element in Zm and Nα(4, Zm) is the set of all 4 × 4 matrices having determinant α. Deﬁne a map

η : SL(4, Zm) → Nα(4, Zm) by  a b c d αa αb αc αd  e f g h  e f g h  η   =   .  i j k l   i j k l  m n o p m n o p

We see that η is one-one. Thus |N(4, Zm)| ≥ |SL(4, Zm)|, which gives

| ∪α∈U(Zm) Nα(4, Zm)| ≥ φ(m)|SL(4, Zm)|. The result follows from the Lemma 3.1. LIE REGULAR GENERATORS OF GENERAL LINEAR GROUP GL(4, Zn) 5 Proposition 4.1. Let p be a prime. Then the order of the linear group 3(4n−2) 2 2 2 n 4 GL(4, Zpn ) is p (p + 1) (p + 1)(p + p + 1)(φ(p )) . Proof. This is well known that the natural homomorphism

f : GL(4, Zpn ) → GL(4, Zp) is onto. Observe that |ker(f)| = (pn−1)16. By using ﬁrst isomorphism theorem, we get

o(GL(4, Zpn )) = o(GL(4, Zp))|ker(f)| = p3(4n−2)(p + 1)2(p2 + 1)(p2 + p + 1)(φ(pn))4.

Proposition 4.2. Suppose m > 1 is a positive integer such that m = Qk ri i=1 pi , where pi’s are distinct primes. Then k Y r o(GL(4, Zm) = o(GL(4, Zpi i )). i=1 ∼ Proof. Since r1 rk = r1 × · · · × rk . Thus Zp1 ···pk Zp1 Zpk k Y r r r o(GL(4, 1 k )) = o(GL(4, pi i )). Zp1 ···pk Z i=1

Lemma 4.2. Any transvection Tij (1 ≤ i ≤ 4) in M(4, Zm) can be written as a combination of a, b and c, where 0 0 1 0 0 0 1 0  0 1 0 0 0 0 0 1 0 0 1 1  0 0 1 0 a =   , b =   and c =   . 1 0 0 0 1 0 0 0  0 0 0 1 0 1 0 0 0 1 0 0 −1 0 0 0 Proof. The proof is based on the following well known result −1 −1 [Tij,Tjk] = TijTjkTij Tjk = Tik, whenever i, j, k are distinct and 1 ≤ i, j, k ≤ 4. Observe that

Tij combination in a, b, c Tij combination in a, b, c T43 ab T21 ba −1 −1 −1 T14 c (ab) c T32 c bac −1 −1 −1 −1 T13 T14T43T14 T43 T12 T13T32T13 T32 −1 −1 −1 −1 T23 T21T13T21 T13 T24 T21T14T21 T14 −1 −1 −1 −1 T31 T32T21T32 T21 T34 T32T24T32 T24 −1 −1 −1 −1 T41 T43T31T43 T31 T42 T43T32T43 T32 6 SWATI MAHESHWARI AND R. K. SHARMA

This completes the proof.

Lemma 4.3. For m > 1, SL(4, Zm) is generated by Lie regular elements a, b and c, where 0 0 1 0 0 0 1 0  0 1 0 0 0 0 0 1 0 0 1 1  0 0 1 0 a =   , b =   and c =   . 1 0 0 0 1 0 0 0  0 0 0 1 0 1 0 0 0 1 0 0 −1 0 0 0

Proof. First, observe that a, b and c ∈ SL(4, Zm). Suppose H is a subgroup of SL(4, Zm) generated by a, b, c. Let K denotes a subgroup generated by transvections Tij. Then by above lemma, we have

K ≤ H ≤ SL(4, Zm).

By using Corollary 2.1, K = SL(4, Zm) and result follows. Qk ri Let n = i=1 pi be an odd positive integer. Then an element α ∈ ri U(Zn) is called a primitive element modulo pi in Zn if order of α ri modulo n is φ(pi ). For the forthcoming results, m denotes an odd positive integer such Qk ri that m = i=1 pi , where pi’s are distinct primes and ri ≥ 0.

ri Theorem 4.1. Let αi ∈ U(Z2m) be a primitive element modulo pi Qk ji ri in Z2m for each i and i=1 αi 6≡ 1 mod 2m, 0 ≤ ji < φ(pi ), where j1, j2 . . . jk are not simultaneously zero. Then the linear group GL(4, Z2m) is generated by Lie regular units a, b, c, d and ei, where 1 ≤ i ≤ k, 0 0 1 0 0 0 1 0  0 1 0 0 0 0 0 1 0 0 1 1  0 0 1 0 a =   , b =   , c =   , 1 0 0 0 1 0 0 0  0 0 0 1 0 1 0 0 0 1 0 0 −1 0 0 0 0 0 0 1  0 0 0 1 0 0 1 0  0 0 1 0 d =   , ei =   . 0 1 0 0  0 1 0 0 1 0 0 0 αi 0 0 0

Proof. By using Lemma 2.3, for each i there exists an element αi ∈ ri Qk ji U(Z2m) such that o(αi) = φ(pi ) and i=1 αi 6≡ 1 mod 2m for 0 ≤ ri ji < φ(pi ). First, observe that a, b, c, d and ei ∈ GL(4, Z2m). Let G be a subgroup of GL(4, Z2m) generated by a, b, c, d and ei. Set 1 0 0 0 

0 1 0 0  ri xi = eid =  . Then the order of xi is φ(p ). Consider a 0 0 1 0  i 0 0 0 αi LIE REGULAR GENERATORS OF GENERAL LINEAR GROUP GL(4, Zn) 7 group

ri φ(pi ) H = hx1, x2, . . . , xk| xi , xixj = xjxi, 1 ≤ i, j ≤ ki.

Qk ji Then H is a subgroup of G. Since i=1 αi 6≡ 1 mod 2m, clearly Qk ji i=1 xi 6≡ 1 mod 2m. Thus the canonical form of H is given by k Y ji ri h xi | 0 ≤ ji < φ(pi )i i=1

Qk ri and the order of H is i=1 φ(pi ). We can see that an arbitrary element of H is of the form 1 0 0 0 0 1 0 0   0 0 1 0 0 0 0 β

Qk ji with determinant β, where β = i=1 αi . We also have that a, b, c generate SL(4, Z2m) (by Lemma 4.3). Observe that

H ∩ SL(4, Z2m) = {I4}. Now we have k Y ri o(HSL(4, Z2m)) = φ(pi )o(SL(4, Z2m)). i=1

Since HSL(4, Z2m) ⊆ G ≤ GL(4, Z2m) and by using Lemma 3.1, k Y ri o(GL(4, Z2m)) = φ(pi )o(SL(4, Z2m)). i=1

Therefore HSL(4, Z2m) = GL(4, Z2m). Hence result follows.

Theorem 4.2. Let m > 1 and αi ∈ U(Zm) be a primitive element ri Qk ji ri modulo pi in Zm for each i and i=1 αi 6≡ 1 mod m, 0 ≤ ji < φ(pi ), where j1, j2 . . . jk are not simultaneously zero. Then the linear group GL(4, Zm) is generated by Lie regular units a, b, c, d and ei, where 1 ≤ i ≤ k, 0 0 1 0 0 0 1 0  0 1 0 0 0 0 0 1 0 0 1 1  0 0 1 0 a =   , b =   , c =   , 1 0 0 0 1 0 0 0  0 0 0 1 0 1 0 0 0 1 0 0 −1 0 0 0 0 0 0 1  0 0 0 1 0 0 1 0  0 0 1 0 d =   , ei =   . 0 1 0 0  0 1 0 0 1 0 0 0 αi 0 0 0 8 SWATI MAHESHWARI AND R. K. SHARMA

Proof. Proof of this theorem is similar to the previous theorem.

ri Theorem 4.3. Let αi ∈ U(Z4m) be a primitive element modulo pi in Qk ji Z4m for each i and β ∈ U(Z4m) with o(β) = 2 such that i=1 αi 6≡ j ri β mod 4m, 0 ≤ ji < φ(pi ) and j = 0, 1, where j, j1, j2 . . . jk are not simultaneously zero. Then the linear group GL(4, Z4m) is generated by a, b, c, d, ei and f, where 1 ≤ i ≤ k, 0 0 1 0 0 0 1 0  0 1 0 0 0 0 0 1 0 0 1 1  0 0 1 0 a =   , b =   , c =   , 1 0 0 0 1 0 0 0  0 0 0 1 0 1 0 0 0 1 0 0 −1 0 0 0 0 0 0 1  0 0 0 1 0 0 0 1 0 0 1 0  0 0 1 0 0 0 1 0 d =   , ei =   , f =   . 0 1 0 0  0 1 0 0 0 1 0 0 1 0 0 0 αi 0 0 0 β 0 0 0

Proof. By using Lemma 2.3, for 1 ≤ i ≤ k, we have αi, β ∈ U(Z4m) such Qk ji ri that i=1 αi 6≡ β mod 4m, where 0 ≤ ji < φ(pi ). First, observe that a, b, c, d, ei and f ∈ GL(4, Z4m). Let G be a subgroup of GL(4, Z4m) generated by a, b, c, d, ei and f. 1 0 0 0  1 0 0 0 0 1 0 0  0 1 0 0 Set xi = eid =   and y = fd =  . Then 0 0 1 0  0 0 1 0 0 0 0 αi 0 0 0 β ri the order of xi is φ(pi ) and order of y is 2. Consider a group ri φ(pi ) 2 H = hx1, x2, . . . , xk, y| xi , y , yxi = xiy, xixj = xjxi, 1 ≤ i, j ≤ ki.

Qk ji j Then H is a subgroup of G. Since i=1 αi 6≡ β mod 4m. Thus the canonical form of H is given by

j j1 jk ri hy x1 . . . xk | j = 0, 1, 0 ≤ ji < φ(pi )i and the order of H is 2φ(m). We can see that an arbitrary element of H is of the form 1 0 0 0 0 1 0 0   0 0 1 0 0 0 0 γ with determinant γ, where k Y ji j γ = ( αi )β . i=1 LIE REGULAR GENERATORS OF GENERAL LINEAR GROUP GL(4, Zn) 9

We also have that a, b, c generate SL(4, Z4m) (by Lemma 4.3). Observe that H ∩ SL(4, Z4m) = {I4}. Now we have k Y ri o(HSL(4, Z4m)) = 2 φ(pi )o(SL(4, Z4m)). i=1

Since HSL(4, Z4m) ⊆ G ≤ GL(4, Z4m) and by using Lemma 3.1

o(GL(4, Z4m)) = 2φ(m)o(SL(4, Z4m)).

Therefore HSL(4, Z4m) = GL(4, Z4m). Hence result follows.

Proposition 4.3. Consider the group U(Z2n ). For n > 2 the order of 3 is 2n−2.

Proof. For n = 3, we can directly see that o(3) = 2. To prove the result, we shall show that 32n−3 ≡ 1 + 2n−1(mod 2n). Observe that 32 ≡ 1 + 23(mod 24). Suppose result holds for any positive integer less than n. 32n−4 ≡ 1 + 2n−2(mod 2n−1). Hence there exist an integer r so that 32n−4 = 1 + 2n−2 + r2n−1 32n−3 = (1 + 2n−2 + r2n−1)2 32n−3 ≡ 1 + 2n−1(mod 2n) 6≡ 1(mod 2n)

n−2 Hence we get o(3) ≥ 2 and result follows, since U(Z2n ) is not cyclic.

Theorem 4.4. Let n > 2 and αi, β ∈ U(Z2nm), such that o(β) = n−2 ri 2 and αi is a primitive element modulo pi in Z2nm for each i with Qk ji j n ri n−2 ( i=1 αi )β 6≡ ±1 mod 2 m, 0 ≤ ji < φ(pi ) and 0 ≤ j < 2 , where ji, j2 . . . jk, j are not simultaneously zero. Then the linear group GL(4, Z2nm) is generated by Lie regular units a, b, c, d, ei, fand g, where 1 ≤ i ≤ k, 0 0 1 0 0 0 1 0  0 1 0 0 0 0 0 1 0 0 1 1  0 0 1 0 a =   , b =   , c =   , 1 0 0 0 1 0 0 0  0 0 0 1 0 1 0 0 0 1 0 0 −1 0 0 0 0 0 0 1  0 0 0 1 0 0 0 1 0 0 1 0  0 0 1 0 0 0 1 0 d =   , ei =   , f =   , 0 1 0 0  0 1 0 0 0 1 0 0 1 0 0 0 αi 0 0 0 β 0 0 0 10 SWATI MAHESHWARI AND R. K. SHARMA

 0 0 0 1  0 0 1 0 g =   .  0 1 0 0 −1 0 0 0 Proof. By using Lemma 2.3 and Proposition 4.3, we have elements n−2 ri β, αi ∈ U(Z2nm) such that o(β) = 2 and o(αi) = φ(pi ) with Qk ji j n ( i=1 αi )α 6≡ ±1 mod 2 m. Let G be a subgroup of GL(4, Z2nm) generated by a, b, c, d, ei, f and g. 1 0 0 0  1 0 0 0 0 1 0 0  0 1 0 0 Set xi = eid =  , y = fd =   and z = 0 0 1 0  0 0 1 0 0 0 0 αi 0 0 0 β 1 0 0 0 

0 1 0 0  ri n−2 gd =  . Then the order of xi is φ(p ), order of y is 2 0 0 1 0  i 0 0 0 −1 and order of z is 2. Consider the group r i n−2 φ(pi ) 2 2 H = hx1, x2, . . . , xk, y, z| xi , y , z , yxi = xiy, xixj = xjxi, zy = yz, zxi = xiz, 1 ≤ i, j ≤ ki.

Qk ji n Then H is a subgroup of G. Since ( i=1 αi )β 6≡ ±1 mod 2 m. Thus the canonical form of H is given by

j1 jk j l ri n−2 hx1 . . . xk , y , z | 0 ≤ ji < φ(pi ), 0 ≤ j < 2 , l = 0, 1i and the order of H is 2n−1φ(m). We can see that an arbitrary element of H is of the form 1 0 0 0 0 1 0 0   0 0 1 0 0 0 0 γ with determinant γ, where k Y ji j γ = ±( αi )β . i=1

We also have that a, b, c generate SL(4, Z2nm) (by Lemma 4.3). Observe that H ∩ SL(4, Z2nm) = {I4}. Now we have k Y ri o(HSL(4, Z2nm)) = 2 φ(pi )o(SL(4, Z2nm)). i=1

Since HSL(4, Z2nm) ⊆ G ≤ GL(4, Z2nm) and by using Lemma 3.1

o(GL(4, Z2nm)) = 2φ(m)o(SL(4, Z2nm)). LIE REGULAR GENERATORS OF GENERAL LINEAR GROUP GL(4, Zn) 11

Therefore HSL(4, Z2nm) = GL(4, Z2nm). Hence result follows. References [1] H. S. M. Coxeter and W. O. J. Moser. Generators and Relations for Discrete Groups. Springer-Verlag Berlin Heidelberg, New York, 1972. [2] P. Kanwar, R. K. Sharma, and P. Yadav. Lie regular generators of general linear groups. Int. Electron. J. Algebra, 13(91-108), 2013. [3] R. K. Sharma, P. Yadav, and P. Kanwar. Lie regular generators of general linear groups. Comm. Algebra, 40(4)(1304-1315), 2012. [4] G. Shimura. Introduction to the arithmetic theory of automorphic functions. Princeton university press, New Jersey, 1971. [5] S. M. Trott. A pair of generators for the unimodular group. Canad. Math. Bull., 5(3), 1962.

Department of Mathematics Indian Institute of Technology Delhi Delhi - 110016 India E-mail address: [email protected]