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

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Lie Regular Generators of General Linear Group Gl(4,Z 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. jAj 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 2 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 2 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). 00 a 0 b1 00 a b 01 Bc 0 d 0C Bc 0 0 dC Proposition 3.3. Suppose A = B C and B = B C @0 e 0 fA @e 0 0 fA 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- 00 0 1 01 B0 0 1 1C ther the element C = B C is also a Lie regular element in @1 0 0 0A 0 1 0 0 GL(4; R). Proof. It follows once we observe that 200 0 0 01 00 −a 0 −b13 6B0 1 0 0C Bc 0 d 0 C7 A = 6B C ; B C7 : 4@0 0 0 0A @0 −e 0 −fA5 0 0 0 1 g 0 0 0 4 SWATI MAHESHWARI AND R. K. SHARMA 201 0 0 01 0 0 a b 0 13 6B0 0 0 0C B−c 0 0 −dC7 B = 6B C ; B C7 : 4@0 0 0 0A @−e 0 0 −fA5 0 0 0 1 0 g 0 0 200 0 0 01 00 0 −1 0 13 6B0 0 0 0C B0 0 −1 −1C7 C = 6B C ; B C7 : 4@0 0 1 0A @1 0 0 0 A5 0 0 0 1 0 1 0 0 Proposition 3.4. Suppose A 2 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 2 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 20 a b c d13 0αa αb αc αd1 6B e f g hC7 B e f g h C η 6B C7 = B C : 4@ i j k l A5 @ i j k l A m n o p m n o p We see that η is one-one. Thus jN(4; Zm)j ≥ jSL(4; Zm)j, which gives j [α2U(Zm) Nα(4; Zm)j ≥ φ(m)jSL(4; Zm)j. 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 jker(f)j = (pn−1)16. By using first isomorphism theorem, we get o(GL(4; Zpn )) = o(GL(4; Zp))jker(f)j = 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 00 0 1 01 00 0 1 01 0 0 1 0 01 B0 0 0 1C B0 0 1 1C B 0 0 1 0C a = B C ; b = B C and c = B C : @1 0 0 0A @1 0 0 0A @ 0 0 0 1A 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.
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