IDENTIFYING CONGRUENCE SUBGROUPS OF THE MODULAR GROUP

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1 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 24, Number 5, May 996 IDENTIFYING CONGRUENCE SUBGROUPS OF THE MODULAR GROUP TIM HSU (Communicated by Ronald M. Solomon) Abstract. We exhibit a simle test (Theorem 2.4) for determining if a given (classical) modular subgrou is a congruence subgrou, and give a detailed descrition of its imlementation (Theorem 3.). In an aendix, we also describe a more invariant and arithmetic congruence test.. Notation We describe (conjugacy classes of) subgrous Γ PSL 2 (Z) intermsofermutation reresentations of PSL 2 (Z), following Millington [, 2] and Atkin and Swinnerton-Dyer []. We recall that a conjugacy class of subgrous of PSL 2 (Z) isequivalenttoa transitive ermutation reresention of PSL 2 (Z). Such a reresentation can be defined by transitive ermutations E and V which satisfy the relations (.) =E 2 =V 3. The relations (.) are fulfilled by 0 (.2) E =, V =. 0 0 Alternately, such a reresentation can be defined by transitive ermutations L and R which satisfy (.3) =(LR L) 2 =(R L) 3, with the relations being fulfilled by 0 (.4) L =, R =. 0 One can also use ermuations E and L such that (.5) =E 2 =(L E) 3, with E and L corresonding to the indicated matrices in (.2) and (.4), resectively. Received by the editors Setember, Mathematics Subject Classification. Primary 20H05; Secondary 20F05. Key words and hrases. Congruence subgrous, classical modular grou. The author was suorted by an NSF graduate fellowshi and DOE GAANN grant #P200A0022.A c 996 American Mathematical Society

2 352 T. HSU (.6) (.7) (.8) The various notations can be translated using the following conversion table: E = LR L, V = R L, L = EV, R = EV 2, R = E L E. Examle.. The ermutations E = ( 2)(3 4)(5 6)(7 8)(9 0), (.9) V = ( 3 5)(2 7 4)(6 8 9), or, alternately, L = ( 4)( )(3 7 6), (.0) R =(7906)(23)(458), describe a conjugacy class of subgrous of index 0 in PSL 2 (Z). Remark.2. Note that any concrete method of secifying a modular subgrou can easily be converted to ermutation form. For instance, one way in which a modular subgrou Γ might be secified is by a list of generators. Such a list can be converted into ermutations as follows: First, use the Euclidean algorithm to exress each generator matrix as a roduct of L s and R s, where L and R are the elements in (.4). Then enumerate the cosets of Γ in terms of these generators and resentation (.3). This coset enumeration is easily converted into aroriate ermutations L and R. Similarly, any reasonable membershi test for Γ can be used to enumerate the cosets of Γ, with the same results as before. 2. Congruence subgrous and the level We recall the following definitions. Definition 2.. Γ(N) is defined to be the grou (2.) {γ PSL 2 (Z) γ ±I (mod N)}. Γ(N) is the kernel of the natural rojection from PSL 2 (Z) tosl 2 (Z/N )/{±I}. We say that a modular subgrou Γ is a congruence subgrou if Γ contains Γ(N) for some integer N. Otherwise,wesayΓisanon-congruence subgrou. An imortant invariant of (conjugacy classes of) modular subgrous is the following. Definition 2.2. The level of a modular subgrou Γ, as secified by ermutations L and R, is defined to be the order of L (or the order of R, sincelis conjugate to R ). We need the following result, sometimes known as Wohlfahrt s Theorem (Wohlfahrt [3]). Theorem 2.3. Let N be the level of a modular subgrou Γ. subgrou if and only if it contains Γ(N). Γ is a congruence Proof. This amounts to roving that, for congruence subgrous, our definition of the level is the same as the classical definition of the level. See Wohlfahrt [3].

3 IDENTIFYING CONGRUENCE SUBGROUPS 353 Theorem 2.4. Let Γ be a modular subgrou of level N, andlet (2.2) L, R r,r 2,... be a resentation for SL 2 (Z/N )/{±I} which is comatible with (.4). Then Γ is a congruence subgrou if and only if the reresentation of PSL 2 (Z) induced by Γ resects the relations {r i }. Proof. From Theorem 2.3, we only need to check if Γ contains Γ(N). Now, since Γ(N) isnormalinpsl 2 (Z), Γ contains Γ(N) if and only if the normal kernel of Γ contains Γ(N). However, the normal kernel of Γ is exactly the kernel of the reresentation induced by Γ, and since the relations {r i } generate Γ(N) astheir normal closure, the theorem follows. Comare Magnus [9, Ch. III], Britto [4], Wohlfahrt [3], and Larcher [8]. Lang, Lim, and Tan [7] have also develoed a congruence test; see the related aer Chan, Lang, Lim, and Tan [5]. Examle 2.5. Suose Γ is the conjugacy class of subgrous secified by (.0). Since L has order 30, we need to use a resentation for SL 2 (Z/30)/{±I}. We find that SL 2 (Z/30)/{±I} has a resentation with defining relations (2.3) =L 30, (2.4) =[L 2,R 5 ]=[L 3,R 0 ]=[L 5,R 6 ] in addition to the relations in (.3). (The commutator [x, y] is defined to be x y xy, so = [x, y] means x commutes with y.) Only the commutator relations (2.4) need to be checked. However, (2.5) L 2 =(29850)(367), which does not commute with (2.6) R 5 =(23), so Γ is a non-congruence subgrou. (It is worth mentioning that Larcher s results also imly that Γ is non-congruence, since L does not contain a 30-cycle.) Remark 2.6. The results in this section extend essentially verbatim to the Bianchi grous SL 2 (O d ), where O d is the ring of algebraic integers of an imaginary quadratic field Q [ d ] with class number. (See Fine [6] for more on the Bianchi grous.) However, for ractical use, one needs a uniform resentation of SL 2 (O d ) /A for A any ideal of O d. 3. Imlementation To assure the reader that the rocedure described by Theorem 2.4 is ractical, we rovide the following detailed algorithm. Suose we are given a subgrou Γ of finite index in PSL 2 (Z).. Describe Γ in terms of ermutations L and R. If necessary, use conversion (.7), conversion (.8), or another similar conversion. (See also Remark.2.) 2. Let N be the order of L, andletn=em, whereeisaowerof2andmis odd. 3. We have three cases: (a) N is odd: Γ is a congruence subgrou if and only if the relation =(R 2 L 2 ) 3 (A)

4 354 T. HSU is satisfied, where 2 is the multilicative inverse of 2 mod N. (b) N is a ower of 2: Let S = L 20 R 5 L 4 R,where 5 is the multilicative inverse of 5 mod N. Γ is a congruence subgrou if and only if the relations (LR L) S(LR L)=S, (B) S RS = R 25, =(SR 5 LR L) 3 are satisfied. (c) Both e and m are greater than : (i) Let 2 be the multilicative inverse of 2 mod m, andlet 5 be the multilicative inverse of 5 mod e. (ii) Let c be the unique integer mod N such that c 0(mode)andc (mod m), and let d be the unique integer mod N such that d 0(modm) and d (mode). (iii) Let a = L c, b = R c, l = L d, r = R d,andlets=l 20 r 5 l 4 r. (iv) Γ is a congruence subgrou if and only if the relations =[a, r], =(ab a) 4, (ab a) 2 =(b a) 3, (C) (ab a) 2 =(b 2 a 2 ) 3, (lr l) s(lr l)=s, s rs = r 25, (lr l) 2 =(sr 5 lr l) 3 are satisfied. Theorem 3.. The above rocedure determines if Γ is a congruence subgrou. Before roving Theorem 3., we need an algebraic trick (Lemma 3.2) and some known results (Lemma 3.3, due to Behr and Mennicke [2]; and Lemma 3.4, due to Mennicke [0]). Lemma 3.2 (Braid trick). Let x and y be elements which generate a grou G and satisfy the relation (3.) (xyx) 2 =(yx) 3. Then the element (xyx) 2 =(yx) 3 is central in G. Furthermore, (3.2) xyx = yxy and (3.3) (xyx) x(xyx) =y. We call this the braid trick because (3.2) is the defining relation for the 3-string braid grou. Proof. The elements X = xyx and Y = yx also generate G, and the element Z =(xyx) 2 =(yx) 3 = X 2 = Y 3 commutes with both X and Y,soZis central. (3.2) and (3.3) follow from cancellation in xyxxyx = yxyxyx.

5 IDENTIFYING CONGRUENCE SUBGROUPS 355 Lemma 3.3. Let m be an odd integer, and let 2 be the multilicative inverse of 2 mod m. SL 2 (Z/m) is isomorhic to G = a, b (3.4) =a m, (3.5) =(ab a) 4, (3.6) (ab a) 2 =(b a) 3, (3.7) Relations (3.4) (3.7) are fulfilled by a = (ab a) 2 =(b 2 a 2 ) 3. and b = 0 0 in SL 2 (Z/m). Proof. G is equivalent to Behr and Mennicke s resentation [2, (2.2)] by the following Tietze transformations. Add generators A = b and B = ab a. Alying the braid trick to (3.6), we get that B 2 is central, and from (3.2), we also get that (3.8) BA = b a. (3.8) imlies that a = ABA, which means that we can eliminate a and b. Using (3.3), (3.8), and the centrality of B 2, we see that (3.4) (3.6) become (3.9) (3.0) =A m =B 4, B 2 =(AB) 3, so it remains to convert (3.7) to Behr and Mennicke s form. However, alying (3.3), we have (3.) B 2 =(b 2 a 2 ) 3 =(A 2 B A 2 B) 3, so, using = B 8 and the centrality of B 2, (3.2) =(A 2 B A 2 B) 3 B 6 =(A 2 BA 2 B) 3. Lemma 3.4. Let e =2 n,let 5 be the multilicative inverse of 5 mod e, andlet s=l 20 r 5 l 4 r. SL 2 (Z/e) is isomorhic to G = l, r (3.3) =l e, (3.4) =(lr l) 4, (3.5) (lr l) 2 =(r l) 3, (3.6) (lr l) s(lr l)=s, (3.7) s rs = r 25, (3.8) Relations (3.3) (3.8) are fulfilled by l = in SL 2 (Z/e). (lr l) 2 =(sr 5 lr l) 3., r = 0 0,ands=

6 356 T. HSU Proof. As the reader may verify, the relations (3.3) (3.8) and s = l 20 r 5 l 4 r are satisfied in SL 2 (Z/e), so it suffices to show that G is a homomorhic image of Mennicke s resentation [0,. 20]. Add generators A = r, B = lr l,andt=s. Alying the braid trick to (3.5), we get that B 2 is central, BA = r l,andlis conjugate to A. As in the roof of the revious lemma, we can then eliminate generators l and r. Then (3.3), (3.4), (3.5), (3.6), (3.7), and (3.8) become Mennicke s relations (X), (Y), (P), (Z), (Q), and (R), resectively. For Lemma 3.5, we consider the following relations: (3.9) =L N, (3.20) =[a, r], (3.2) =[b, l], (3.22) =(ab a) 4, (3.23) (ab a) 2 =(b a) 3, (3.24) (ab a) 2 =(b 2 a 2 ) 3, (3.25) =(lr l) 4, (3.26) (lr l) 2 =(r l) 3, (3.27) (lr l) s(lr l)=s, (3.28) s rs = r 25, (3.29) (lr l) 2 =(sr 5 lr l) 3. All notation is as described in (2) and (3c)(i iii) of the algorithm. Note that = L N imlies that L = al and R = br. Lemma 3.5. SL 2 (Z/N ) has a resentation with generators ( L and ) R, and ( defining ) 0 relations (3.9) (3.29). The relations are fulfilled by L = and R = 0 in SL 2 (Z/N ). Proof. The Chinese Remainder Theorem imlies that (3.30) SL 2 (Z/N ) = SL 2 (Z/m) SL 2 (Z/e). It also follows from the Chinese Remainder Theorem that, if L = and R = 0 0 in SL 2 (Z/N ), the SL 2 (Z/m) factor is recisely a, b and the SL 2 (Z/e) factor is recisely l, r. Therefore, the above relations are satisfied in SL 2 (Z/N ). On the other hand, since (3.9) imlies (3.4) and (3.3), comarison with Lemmas 3.3 and 3.4 shows that the above resentation is the direct roduct of SL 2 (Z/m) and SL 2 (Z/e). The lemma follows. Proof of Theorem 3.. After stes and 2 of the rocedure, we know that the relations (3.3) =L N, (3.32) =(LR L) 2, (3.33) =(R L) 3

7 IDENTIFYING CONGRUENCE SUBGROUPS 357 must be satisfied. From Theorem 2.4, we see that if (3.3) (3.33) and (A) (res. (B), (C)) are defining relations for SL 2 (Z/N )/{±I} when N is odd (res. N is a ower of 2, and e and m are greater than ), then Theorem 3. follows. Comaring (A) and Lemma 3.3, with a = L and b = R, and comaring (B) and Lemma 3.4, with l = L and r = R, the first two cases follow easily, so it remains to check the third. Comaring (C) and (3.9) (3.29), we see that it is enough to show that given (3.3) (3.33) and (3.9) (3.29), the relations (3.2), (3.25), and (3.26) are redundant. First, (3.3), (3.32), (3.20), and (3.2) give us =(LR L) 4 (3.34) =(alr b al) 4 =(ab a) 4 (lr l) 4, which means that (3.22) imlies (3.25). Similarly, (3.3), (3.32), (3.33), (3.20), and (3.2) imly (LR L) 2 =(R L) 3, (3.35) (ab a) 2 (lr l) 2 =(b a) 3 (r l) 3, which means that (3.23) imlies (3.26). Finally, since (3.32), (3.33), and the braid trick (3.3) imly that L is conjugate to R, we can eliminate (3.2), since it is imlied by (3.20). For hand calculations, and for further study, we note the following relations which occur in SL 2 (Z/N ): (SL Z =(LR L) 2 =(R L) 3,=Z 2 2 ), (level) =L N =R N, (ab 0 (ab (ab 2 (mod N)) (mod N)) (mod N)) =[L a,r b ], (L a R b ) 3 = Z, (L a R b ) 2 = Z. It has been verified by coset enumeration that the relations (SL 2 ), (level), and (ab 0(modN)) are defining relations when N 360. This means that if the level N divides 360, the congruence test reduces to checking that the relations (ab 0(modN)) are satisfied. Acknowledgements The author would like to thank J. H. Conway and the referee for many helful comments and suggestions. Aendix A. An arithmetic congruence test In this aendix, we resent an arithmetic [ ]) and invariant congruence test which uses the Ihara modular grou SL 2 (Z.

8 358 T. HSU We begin by quoting the following result (Theorem A.) of J. Mennicke [0]. (Note that Mennicke s Schur multilier calculation and subsequent argument require the reairs described in F.R. Beyl [3, 5], but the main result still holds.) Let N be an integer, let be a rime not dividing N, let R N be the kernel [ ]) in SL 2 (Z resulting from reduction mod N, andletq N be the normal closure of L N in SL 2 [ ]) (Z. Theorem A.. R N = Q N. Let Γ be a modular subgrou of level N and index m in SL 2 (Z). Consider the commutative diagram in Figure A.. SL 2 (Z[ ]) i r r SL 2 (Z/N) SL 2 (Z) f f 2 ρ S m Figure A.. Commutative diagram for Theorem A.2 Here, S m is the symmetric grou on m objects (the cosets of Γ in SL 2 (Z)), r is reduction mod N, i is inclusion, and ρ is the ermutation reresentation of SL 2 (Z) induced by Γ. Note that f 2 exists if and only if Γ is a congruence subgrou, and that such an f 2 is uniquely determined. The setu in Figure A. rovides us with an invariant congruence test. Theorem A.2. In the notation of Figure A., a ma f exists if and only if f 2 exists. In other words, [ ]) Γ is congruence if and only if ρ can be factored through inclusion in SL 2 (Z. Proof. If f 2 exists, let f = f 2 r. Conversely, if f exists, since L N is in the kernel of ρ, L N must be in the kernel of f,soinfact,f is well defined on [ ( [ ])/ (A.) SL 2 (Z ])/Q N = SL 2 Z R N = SL2 (Z/N ), which means that f defines an aroriate ma f 2. Corollary A.3. In Figure A., f is determined uniquely if it exists. One curious feature of Theorem A.2 is that if we know a given family of modular subgrous all have levels relatively rime to, then we can handle all of them in a uniform manner. This is the rincile behind Behr and Mennicke s resentation of SL 2 (Z/N )fornodd, as these cases can be handled in SL 2 ( Z [ 2]). We also note that if we fix the level N, then we can choose any not dividing N to use in Theorem A.2. This leads to the following idea: For a given family of modular subgrous of level N, it seems lausible that one might be able to reduce the extensibility of ρ to the question of whether there exists a which satisfies certain congruences mod N. Dirichlet s theorem might then be used to find a which satisfies those congruences.

9 IDENTIFYING CONGRUENCE SUBGROUPS 359 References [] A.O.L.AtkinandH.P.F.Swinnerton-Dyer,Modular forms on noncongruence subgrous, Proc. Sym. Pure Math., Combinatorics (Providence) (T. S. Motzkin, ed.), vol. 9, AMS, Providence, 97,. 26. MR 49:2550 [2] H. Behr and J. Mennicke, A resentation of the grous PSL(2,), Can. J. Math. 20 (968), MR 38:4566 [3] F. R. Beyl, The Schur multilicator of SL(2,Z/mZ) and the congruence subgrou roerty, Math. Z. 9 (986), MR 87b:2007 [4] J. Britto, On the construction of non-congruence subgrous, ActaArith. XXXIII (977), MR 56:242 [5] S.-P. Chan, M.-L. Lang, C.-H. Lim, and S.-P. Tan, Secial olygons for subgrous of the modular grou and alications, Internat. J. Math. 4 (993), no., 34. MR 94j:045 [6] B. Fine, Algebraic theory of the Bianchi grous, Marcel Dekker, Inc., New York, 989. MR 90h:20002 [7] M.-L. Lang, C.-H. Lim, and S.-P. Tan, An algorithm for determining if a subgrou of the modular grou is congruence, rerint, 992. [8] H. Larcher, The cus amlitudes of the congruence subgrous of the classical modular grou, Ill. J. Math. 26 (982), no., MR 83a:0040 [9] W. Magnus, Noneuclidean tesselations and their grous, Academic Press, 974. MR 50:4774 [0] J. Mennicke, On Ihara s modular grou, Invent. Math. 4 (967), MR 37:485 [] M. H. Millington, On cycloidal subgrous of the modular grou, Proc.Lon.Math.Soc.9 (969), MR 40:484 [2], Subgrous of the classical modular grou, J. Lon. Math. Soc. (969), MR 39:5477 [3] K. Wohlfahrt, An extension of F. Klein s level concet, Ill. J. Math. 8 (964), MR 29:4805 Deartment of Mathematics, Princeton University, Princeton, New Jersey address: timhsu@math.rinceton.edu Current address: Deartment of Mathematics, University of Michigan, Ann Arbor, Michigan address: timhsu@math.lsa.umich.edu