arxiv:1009.4619v1 [math.gr] 23 Sep 2010 G-subsets and G-orbits of Q ( n) under ation of the Modular Group. M. Aslam Malik and M. Riaz Department of Mathematis, University of the Punjab, Quaid-e-Azam Campus, Lahore, Pakistan. Abstrat It is well known that G = x,y : x 2 = y 3 = 1 represents the modular group PSL(2,Z), where x : z 1 z 1 z,y : z z are linear frational transformations. Let n = k 2 m, where k is any non zero integer and m is square free positive integer. Then the set Q ( n) := { a+ n : a,,b = a2 n Z and (a,b,) = 1} is a G-subset of the real quadrati field Q( m) [12]. We denote α = a+ n in Q ( n) by α(a,b,). For a fixed integer s > 1, we say that two elements α(a,b,), α (a,b, ) of Q ( n) are s-equivalent if and only if a a (mod s), b b (mod s) and (mod s). The lass [a,b,](mod s) ontains all s-equivalent elements of Q ( n) and Es n denotes theset onsisting ofall suhlasses of theform[a,b,](mod s). In this paper we investigate proper G-subsets and G-orbits of the set Q ( n) under the ation of Modular Group G. AMS Mathematis subjet lassifiation (2000): 05C25, 11E04, 20G15 Keywords: Real quadrati irrational number; Congruenes; Quadrati residues; Linear-frational transformations. malikpu@yahoo.om mriaz@math.pu.edu.pk 1
1 Introdution An integer m > 0 is said to be square free if its prime deomposition ontains no repeated fators. It is well known that every irrational member of Q( m) an be uniquely expressed as a+ n, where n = k 2 m for some integer k and a a, 2 n and are relatively prime integers. The set Q ( n) = { a+ n : a,,b = a2 n Z and(a,b,) = 1} is a proper G-subset of Q( m) [12]. If α = a+ n and ᾱ = a n have different signs, then α is alled an ambiguous number. These ambiguous numbers play an important role in the study of ation of G on Q( m) { }, as Stab α (G) are the only non-trivial stabilizers and in the orbit α G, there is only one (up to isomorphism). G. Higman(1978) introdued the onept of the oset diagrams for the modulargrouppsl(2,z)andq.mushtaq (1983)laiditsfoundation. Byusingthe oset diagrams for the orbit of the modular group G = x,y : x 2 = y 3 = 1 ating on the real quadrati fields Mushtaq [12] showed that for a fixed nonsquare positive integer n, there are only a finite number of ambiguous numbers in Q ( n), and that the ambiguous numbers in the oset diagram for the orbit α G form a losed path and it is the only losed path ontained in it. Let C = C {± } be the extended omplex plane. The ation of the modular group PSL(2,Z) on an imaginary quadrati field, subsets of C, has been disussed in[11]. The ationof the modular group onthe real quadrati fields, subsets of C, has been disussed in detail in [2], [12] and [13]. The exat number of ambiguous numbers in Q ( n) has been determined in [7], [15] as a funtion of n. The ambiguous length of an orbit α G is the number of ambiguous numbers in the same orbit [7], [15]. The Number of Subgroups of PSL(2,Z) when ating on F p { } has been disussed in [14] and the subgroups of the lassial modular group has been disussed in [10]. A lassifiation of the elements (a+ p)/, b = (a 2 p)/, of Q ( p), p an odd prime, with respet to odd-even nature of a,b, has been given in [3]. M. Aslam Malik et al. [8] proved, by using the notion of ongruene, that for eah non-square positive integer n > 2, the ation of the group G on a subset Q ( n) of the real Quadrati field Q( m) is intransitive. If p is an odd prime,then t 0(mod p) is said to be a quadrati residue of p if there exists an integer u suh that u 2 t(mod p). The quadrati residues of p form a subgroup Q of the group of nonzero inte- 2
gers modulo p under multipliation and Q = (p 1)/2. [1] Lemma 1.1 If v 1,v 2 Q, n 1,n 2 Q (v 1,v 2 are quadrati residues, and n 1,n 2 are quadrati non-residue, Then (a) n 1 v 1 is a quadrati non-residue. (b)n 1 n 2 is a quadrati residue. ()v 1 v 2 is a quadrati residue. In the sequel, q.r and q.nr will stand for quadrati residue and quadrati non-residue respetively. The Legendre symbol (a/p) is defined as 1 if a is a quadrati residue of p otherwise it is defined by 1. [1] We denote the element α = a+ n of Q ( n) by α(a,b,) and say that two elements α(a,b,) and α (a,b, ) of Q ( n) are s-equivalent (and write α(a,b,) s α (a,b, )orα s α )ifandonlyifa a (mods), b b (mods) and (mod s). Clearly the relation s is an equivalene relation, so for eah integer s > 1, we get different equivalene lasses [a,b,] modulo s of Q ( n). [8] Let E s denote the set onsisting of lasses of the form [a,b,] (mod s), n modulo s whereas if n i(mod s) for some fixed i {0,1,...,s 1} and the set onsisting of elements of the form [a,b,] with n i(mod s) is denoted by Ep i (or En s ). Obviously s 1 i=1 Ei s = E s and Es i Ej s = φ for i j. [6] The lassifiation of the real quadrati irrational numbers by taking prime modulus is very helpful in studying the modular group ation on the real quadrati fields. Thus it beomes interesting to determine the proper G- subsets of Q ( n) by taking the ation of G on the set Q ( n) and hene to find the G-orbits of Q ( n) for eah non square n. 2 Modular group G ating on Q ( n). In [8], it was shown that the ation of the group on Q ( 2) is transitive, whereas the ation of G on Q ( n), n 2 is intransitive. Speifially, it was proved with the help of lasses [a,b,](mod 2 2 ) of the elements of Q ( n) that Q ( n), n 2(mod 4), has two proper G-subsets. Q. Mushtaq [12], In the ase of PSL(2,13), showed one G-orbit of length 13 in the oset diagram for the natural ation of PSL(2,Z) on any subset of the real projetive line. In [6] it was proved that there exist two proper G-subsets of Q ( n) when n 0 (mod p) and four G-subsets of Q ( n) when n 0 (mod pq). In the present studies, with the help of the idea of 3
quadrati residues, we generalize this result and prove some ruial results whih provide us proper G-subsets and G-orbits of Q ( n). We extend this idea to determine four proper G-subsets of Q ( n) with n 0 (mod 2pq). Lemma 2.1 Let n 0(mod 2pq) where p and q are two distint odd primes, Let α = a+ n Q ( n) then the sets S 1 = {α Q ( n) : (/pq) = 1 or (b/pq) = 1 }, S 2 = {α Q ( n) : (/p) = 1 or (b/p) = 1 with (/q) = 1 or (b/q) = 1 }, S 3 = {α Q ( n) : (/p) = 1or(b/p) = 1with (/q) = 1or(b/q) = 1}, ands 4 = {α Q ( n) : (/p) = 1or(b/p) = 1with(/q) = 1or(b/q) = 1 } are four proper G-subsets of Q ( n). Q ( n) and n 0(mod 2pq), then a 2 n = b fores that Let a+ n a 2 b(mod 2pq) (1) where a,b, arebelonging totheomplete residue system {0,1,2,..., 2pq 1}. The ongruene (1) implies a 2 b(mod 2), a 2 b(mod p) and a 2 b(modq). Sine1istheonlyquadratiresidueof2andthereisnoquadrati non-residue of 2. Thus by Lemma 1.1 the quadrati residues and quadrati non residues of pq and 2pq are the same. We know that, if (t,m) = 1 and m = 2pq, then the ongruene x 2 t (mod m) is solvable and has four inongruent solutions if and only if t is quadrati residue of m [1], and in this ase ongruene (1) is solvable and has exatly four inongruent solutions. If a = b = and eah of a,b, are quadrati residue of pq then there exist four distint lasses [a,b,],[ a,b,],[a, b, ],[ a, b, ](mod pq) Thus for eah member [a,b,](mod pq), we have four ases. Case(i) The lasses [a,b,](mod pq) with (b/pq) = 1, Then all these lasses are ontained in S 1. ase(ii) The lasses [a,b,](mod pq) with (b/p) = 1, (b/q) = 1, Then all these lasses are ontained in S 2. ase(iii) The lasses [a,b,](mod pq) with (b/p) = 1, (b/q) = 1, Then all these lasses are ontained in S 3. 4
ase(iv) The lasses [a,b,](mod pq) with (b/p) = 1, (b/q) = 1, Then all these lasses are ontained in S 4. As x(α) = a+ n, where a b 1 = a, b 1 =, 1 = b and y( a+ n ) = a+b+ n b = a 2+ n 2 = a 1+ n 1, where a 2 = a+b, b 2 = 2a+b+, and 2 = b then by the ongruene (1) we have ( a+b) 2 ( 2a+b+)b(mod p) (2) Sine the modular group PSL(2,Z) has the representation G = x,y : x 2 = y 3 = 1 and every element of G is a word in the generators x,y of G, to prove that S 1 is invariant under the ation of G, it is enough to show that every element of S 1 is mapped onto an element of S 1 under x and y. Thus learly by the ongruenes (1) and (2) the sets S 1, S 2, S 3 and S 4 are G-subsets of Q ( n). Remark 2.2 Sine the quadrati residues and quadrati non residues of pq and 2pq are the same. Therefore the number of G-subsets of Q ( n) when n 0(mod pq) or n 0(mod 2pq) are same. Illustration 2.3 In the oset diagram for Q ( 15) there are four G-orbits namely ( 15) G,( 15 15 15) G,( 3 )G and ( 3 )G and similarly there are four G-orbits for Q ( 30) namely ( 30) G,( 30 30 30) G,( 2 )G and ( 2 )G. In the losed path lying in the orbit ( 15) G, the transformation (yx) 3 (y 2 x)(yx) 3 fixes k = 15, that is g 1 (k) = ((yx) 3 (y 2 x)(yx) 3 )(k) = k. Let k is an ambiguous number then x(k) is also ambiguous but one of the 5
number y(k) or y 2 (k) is ambiguous. The orientation of edges in the oset diagram is assoiated with the involution x and the small triangles with y whih hasorder 3. One of k and x(k) is positive andother is negative but one of k, y(k) or y 2 (k) is negative but other two are positive. We use an arrow head on an edge to indiate its diretion from negative to a positive vertex. The following table shows the details of the orbits α G, transformations whih fixes α, and the ambiguous lengths of eah orbit. Table 1: The Orbits of α Q ( n). G-orbits Transformations Ambiguous Length ( 15) G (yx) 3 (y 2 x)(yx) 3 14 ( 15) G (yx) 3 (y 2 x)(yx) 3 14 ( 15 3 )G (yx)(y 2 x) 3 (yx) 10 ( 15 3 )G (yx)(y 2 x) 3 (yx) 10 ( 30) G (yx) 5 (y 2 x) 2 (yx) 5 24 ( 30) G (yx) 5 (y 2 x) 2 (yx) 5 24 ( 30 2 )G (yx) 2 (y 2 x)(yx) 2 (y 2 x)(yx) 2 16 ( 30 2 )G (yx) 2 (y 2 x)(yx) 2 (y 2 x)(yx) 2 16 Now we extend this idea when n 0(mod p 1 p 2...p r ). Theorem 2.4 Let n 0(mod p 1 p 2...p r ), where p 1,p 2,...p r are distint odd primes, then there are exatly 2 r, G-subsets of Q ( n). 6
Let a+ n Q ( n) and n 0(mod p 1 p 2...p r ) where p 1,p 2,...p r are distint odd primes, then a 2 n = b gives The ongruene (3) implies a 2 b(mod p 1 ), a 2 b(mod p 1 p 2...p r ) (3) a 2 b(mod p 2 ),..., and a 2 b(mod p r ). We know that, if (t,m) = 1 and m = p 1 p 2...p r, then the ongruene x 2 t (mod m) is solvable if and only if t is quadrati residue of m [1], and in this ase ongruene (3) is solvable and has exatly 2 r inongruent solutions. Sine all values of b or whih are quadrati residues and quadrati nonresidues of m lie in the distint G-subsets and m is the produt of r distint primes, Thus onsequently we obtain 2 r, G-subsets of Q ( n). Corollary 2.5 Let n 0(mod 2p 1 p 2...p r ) where p 1,p 2,...p r are distint odd primes, then there are exatly 2 r, G-subsets of Q ( n). Let a+ n Q ( n) and n 0(mod 2p 1 p 2...p r ) where p 1,p 2,...p r are distint odd primes, then a 2 n = b gives a 2 b(mod 2p 1 p 2...p r ) (4) The ongruene (4) implies a 2 b(mod 2), a 2 b(mod p 1 ), a 2 b(mod p 2 ),..., and a 2 b(mod p r ) Sine 1 is the only quadrati residue of 2 and there is no quadrati nonresidue of 2. Thus by Lemma 1.1 the quadrati residues and quadrati non residues of p 1 p 2...p r and 2p 1 p 2...p r are same. Hene the result follows by the Theorem 2.4. Remark 2.6 The number of G-orbits of Q ( n) when n 0(mod p 1 p 2...p r ) or n 0(mod 2p 1 p 2...p r ) are same. 7
Illustration 2.7 Take n = 3.5.7.11 = 1155, Under the ation of G on Q ( 1155) there are sixteen G-orbits in the oset diagram for Q ( 1155) namely ( 1155 1155 1155 1155) G, ( 1 )G, ( ) G, ( 3 3 )G, 1155 1155 1155 1155 ( ) G, ( 5 5 )G, ( ) G, ( 7 7 )G, 1155 1155 1155 1155 ( ) G, ( 11 11 )G, ( ) G, ( 15 15 )G, 1155 1155 1155 1155 ( ) G, ( 21 21 )G, ( ) G, ( 33 33 )G. Similarly for n = 2.3.5.7.11 = 2310, Under the ation of G on Q ( 2310) there are sixteen G-orbits in the oset diagram for Q ( 2310) namely ( 2310 2310 2310 2310) G, ( 1 )G, ( ) G, ( 2 2 )G, 2310 2310 2310 2310 ( ) G, ( 5 5 )G, ( ) G, ( 7 7 )G, 2310 2310 2310 2310 ( ) G, ( 10 10 )G, ( ) G, ( 11 11 )G, 2310 2310 2310 2310 ( ) G, ( 14 14 )G, ( ) G, ( 22 22 )G. Theorem 2.8 Let h = 2k +1 3 then there are exatly two G-orbits of Q ( 2 h ) namely (2 k 2) G and ( 2k 2 1 )G. Let a+ n Q ( 2 h ), then a 2 n = b fores that a 2 2 h b(mod 2 h ) a 2 b(mod 2 h ) But the ongruene a 2 b(mod 2 h ) is solvable if and only if b 1(mod 8), Moreover the quadrati residue of 2 h,h 3 are those integers of the form 8l +1 whih are less than 2 h. Sine all values of b or whih are quadrati residues and quadrati non-residues of 2 h lie in the distint orbits. Thus the 8
lasses [a,b,] (modulo 2 h ) with b or quadrati residues of 2 h lie in the orbit (2 k 2) G and similarly the lasses [a,b,] (modulo 2 h ) with b or quadrati non-residues of 2 h lie in the orbit ( 2k 2 1 )G, This proves the result. Illustration 2.9 There are exatly two G-orbits of Q ( 2 7 ) namely (2 3 2) G and ( 23 2 1 )G, In the losed path lying in the orbit (2 3 2) G, the transformation (yx) 11 (y 2 x) 3 (yx) 5 (y 2 x) 3 (yx) 11 fixes 2 3 2. Similarly in the losed path lying in the orbit ( 23 2 1 )G, the transformation (yx) 11 (y 2 x) 3 (yx) 5 (y 2 x) 3 (yx) 11 fixes 2 3 2. 3 Ation of the subgroup G = yx and G = yx,y 2 x on Q ( n). Let us suppose that G = yx and G = yx,y 2 x are two subgroups of G. In this setion, we determine the G-subsets and G-orbits of Q ( n) by subgroup G and G ating on Q ( n). Let yx(α) = α+1 and y 2 x(α) = α α+1. Thus yx( a+ n ) = a 1+ n 1, with a 1 = a +, b 1 = 2a + b +, 1 = and y 2 x( a+ n ) = a 2+ n 2, with a 2 = a+b, b 2 = b, and 2 = 2a+b+. In the next Lemma we see that the transformation yxfixes the lasses [0,0,] (modulo p) and the hain of these lasses help us in finding G -subsets of Q ( n). Lemma 3.1 Let p be any prime, n 0(mod p), Then for any k 1, (yx) k [0,0,] = [k,k 2,] (mod p) and in partiular (yx) p [0,0,] = [0,0,] (mod p). Let α = [0,0,] (mod p) be a lass ontained in Ep 0. Applying the linear frational transformation yx on α suessively we see yx[0,0,] = [,,], (yx) 2 [0,0,] = [2,4,], (yx) 3 [0,0,] = [3,9,] ontinuing this proess k- times we obtain 9
(yx) k [0,0,] = [k,k 2,] (mod p) In partiular for k = p, kp 0(mod p) and k 2 p 0(mod p). Thus we get (yx) p [0,0,] = [0,0,] (mod p). Lemma 3.2 Let p be an odd prime, n 0(mod p) and G = yx, Then the sets A 1 = {α Q ( n) : (/p) = 1}, A 2 = {α Q ( n) : (/p) = 1}, C 1 = {α Q ( n) : 0(mod p) with (b/p) = 1}, C 2 = {α Q ( n) : 0(mod p) with (b/p) = 1}, are G -subsets of Q ( n). For any α = a+ n A 1 with n 0(mod p), a 2 n = b gives a 2 b(mod p) (5) we have two ases. (i) If a 0(mod p), The ongruene (5) fores that b 0(mod p), Then either b 0(mod p) or 0(mod p) but not both. So in this ase α belongs to the lass [0,b,0] or [0,0,] modulo p. (ii) If a 0(mod p), a 2 b(mod p), Then (5) fores that either both b, are quadrati residues of p or both quadrati non-residues of p. As yx : [a,b,] [a +,2a + b +,], Then it is lear that the set A 1 is invariant under the ation of the mapping yx, So the set A 1 is a G -subset Q ( n). Similarly the set A 2 is G -subset of Q ( n). Again for any α = a+ n C 1 by ongruene (5) 0(mod p) a 0(mod p), with b 0(mod p), so the lasses belonging to the set C 1 are of the form [0,b,0] with b quadrati residue of p. Sine the mapping yx fixes the lasses [0,b,0]. Thus learly the set C 1 is a G -subsets. Similarly the set C 2 is G -subsets of Q ( n). In the next theorem we determine two G-subsets of Q ( n) by using A 1, A 2, C 1 and C 2 as given in Lemma 3.2. Theorem 3.3 The sets S 1 = A 1 C 1 and S 2 = A 2 C 2 are two G-subsets of Q ( n). Let α = a+ n S 1 then either α A 1 or α C 1 with n 0(mod p). Thus it is lear that the lasses [a,b,] (mod p) with b or quadrati residues of p is ontained in A 1 C 1. By Lemma 3.1 yx fixes the lasses [0,0,] (modulo p). 10
Also the lasses belonging to A 1, A 2 are onneted to the lasses belonging to C 1, C 2, respetively under x. Sine the modular group PSL(2,Z) has the representation G = x,y : x 2 = y 3 = 1 and every element of G is a word in its generators x,y, to prove that S 1 is invariant under the ation of G, it is enough to show that every element of S 1 is mapped onto an element of S 1 under x and y. Thus learly we see that S 1 = A 1 C 1 and S 2 = A 2 C 2 are both G-subsets of Q ( n). In view of the above theorem we observe that for n = 2 the ation of G on Q ( n) is transitive. Sine 1 is the only quadrati residue of 2 and there is no quadrati non-residue of 2, Therefore the set S 2 beomes empty and S 1 is the only G-subset of Q ( 2). While the ation of G on Q ( n), n 2 is intransitive. Illustration 3.4 Let p = 5 and α = a+ n Q ( n), with n 0(mod 5), is of the form [a,b,](mod 5). In modulo 5, the squares of the integers 1,2,3,4 are 1 2 4 2 1 and 2 2 3 2 4 Consequently, the quadrati residues of 5 are 1,4, and the non residues are 2,3. Thus A 1 onsists of elements of Q ( n) of the form [0,0,1],[0,0,4],[1,1,1],[4,1,1],[2,4,1],[2,1,4],[3,1,4],[3,4,1],[1,4,4], [4,4,4]mod 5. Then A 1 is invariant under yx, Thus A 1 is G -subset of Q ( n). The elements of A 2 are of the form [0,0,2],[0,0,3],[2,2,2],[3,2,2],[2,3,3],[4,3,2], [4,2,3],[1,2,3),[1,3,2] and [3,3,3] mod 5 only. Again A 2 is invariant under yx, Thus A 2 is also G -subset of Q ( n). The elements of C 1 are of the form [0,1,0] and [0,4,0], and the elements of C 2 are of the form [0,2,0] and [0,3,0] Thus C 1 and C 2 are G -subsets. Then learly the sets S 1 = A 1 C 1 and S 2 = A 2 C 2 are two G-subsets of Q ( n). 11
In thenext lemma we find the onditions when n isquadrati residue of p. Lemma 3.5 For any α(a,b,) Q ( n), n is quadrati residue of p if and only if either or b 0(mod p). Let α = a+ n Q ( n), Then a 2 n = b fores that a 2 n b(mod p) (6) Let either b 0(mod p) or 0(mod p) then ongruene (6) implies that a 2 n(mod p) whih shows that n is quadrati residue of p Conversely let n be quadrati residue of p then learly a 2 n(mod p) that shows either b 0(mod p) or 0(mod p). Further we see the ation of G = yx,y 2 x on Q ( n) with n quadrati residue of p and determine four proper G -subsets of Q ( n). Theorem 3.6 Letpbeanoddprimeandnisquadratiresidueofp, letα = a+ n Q ( n) and G = yx,y 2 x, then the sets G 1 = {α Q ( n) : (/p) = 1}, G 2 = {α Q ( n) : (/p) = 1}, are two G -subsets of Q ( n). Let α = a+ n Q ( n), with a 2 n b(mod p), and a,b, modulo p are belonging to the set {0,1,2,...,p 1}. Let p is an odd prime with n quadrati residue of p, then either b 0(mod p) or 0(mod p). Sine yx : [a,b,] [a+,2a+b+,] and y 2 x : [a,b,] [a+b,b,2a+b+], Sine every element of G is a word in its generators yx,y 2 x, Then learly the sets G 1, G 2, are two G -subsets of Q ( n). Corollary 3.7 G,x = G We know that G = x,y : x 2 = y 3 = 1 and G = yx,y 2 x, the result follows from the fat that the generators x,y of G an be written as word by the elements of G,x. Finally we find G-orbits of Q ( 37) with help of G -subsets as given in 12
Theorem 3.6. It is important to note that 37 is the smallest prime whih have four G-orbits and all odd primes less than 37 has exatly two G-orbits. Illustration 3.9 IntheosetdiagramforQ ( 37), ThereareexatlyfourG-orbitsofQ ( 37) given by ( 37) G, ( 1+ 37 2 ) G, ( 1+ 37 3 ) G, ( 1+ 37 3 ) G. The G 1 ontains three orbits ( 37) G, ( 1+ 37 3 ) G, ( 1+ 37 3 ) G, While the set G 2 ontains only one orbit ( 1+ 37 2 ) G. In the losed path lying in the orbit ( 37) G, the transformation g 1 = (yx) 6 (y 2 x) 12 (yx) 6 fixes k = 37, that is g 1 (k) = ((yx) 6 (y 2 x) 12 (yx) 6 )(k) = k, and so gives the quadrati equation k 2 +37 = 0, the zeros, ± 37, of this equation are fixed points of the transformations g 1. In the losed path lying in the orbit ( 1+ 37 2 ) G, the transformations g 2 = (yx) 3 (y 2 x)(yx)(y 2 x) 5 (yx)(y 2 x)(yx) 2 fixes l = 1+ 37 2 and so gives the quadrati equation l 2 l 9 = 0, the zeros, 1± 37 2, of this equation are fixed points of g 2. Similarly in the losed path lying in the orbit ( 1+ 37 3 ) G the transformation g 3 = (yx) 2 (y 2 x) 2 (yx)(y 2 x) 3 (yx) 2 (y 2 x)(yx) fixes ( 1+ 37) and orresponding to the losed lying in the orbit ( 1+ 37) G, 3 3 the transformation g 4 = (yx)(y 2 x)(yx) 2 (y 2 x) 3 (yx)(y 2 x) 2 (yx) 2 fixes ( 1+ 37). 3 By [7], [15] we see that τ (37) = 124, That is there are 124 ambiguous numbers in the oset diagram for Q ( 37) while the ambiguous length of the orbits are 48, 28, 24 and 24 respetively. 13
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