Iranian Journal of Science & Technology, Transaction A, Vol. 34, No. A4 Printed in the Islamic Reublic of Iran, Shiraz University SUBORBITAL GRAPHS FOR A SPECIAL SUBGROUP * m OF THE NORMALIZER OF S. KADER, B. O. GULER ** AND A. H. DEGER 3 Deartment of Mathematics, Nigde University, Nigde, Turkey Email: skader@nigde.edu.tr Deartment of Mathematics, Rize University, Rize, Turkey Email: bahadir.guler@rize.edu.tr 3 Deartment of Mathematics, Karadeniz Technical University, Trabzon, Turkey Email: ahikmetd@ktu.edu.tr Abstract ( m) In this aer, we find the number of sides of circuits in suborbital grah for the normalizer of in PSL(,), where m will be of the form, is a rime and mod4. In addition, we give a number theoretical result which says that the rime divisors of u uare of the form mod4. Keywords Normalizer, imrimitive action, suborbital grah, circuits. INTRODUCTION az b Let PSL(,) denote the grou of all linear fractional T : z, where a, b, c, d are real and cz d ad bc. The modular grou Γ is the subgrou of PSL(,) such that a, b, c and d are integers. For any natural number m, ( m) is the subgrou of Γ with m c. The elements of PSL(,) are reresented as a b, a,b,c,d and ad bc. c d We will omit the symbol and identify each matrix with its negative. ( m) will denote the normalizer of ( m) in PSL(,). The elements of ( m) by [] are of the form ae cm h b h de where all letters are integers, e m and h is the largest divisor of 4 for which h m with the h s understanding that the determinant is e >, and that r s means that r s and r,. r Here, m will be mod4. All circuits in suborbital grah, where is a rime such that for the normalizer of ( m) in PSL(,) where m is a square-free ositive integer was studied in [, 3]. Received by the editor February, 9 and in final revised form December 8, Corresonding author
36 B. O. Guler / et al Our main idea is that we investigate a case in which m is not square-free. Similar studies were done for the modular grou and some Hecke grous [4-6]. In this case, h will be and e is,, or.. THE ACTION OF ON ( ) Any element of can be given as a reduced fraction x, with, and ( x, y). is reresented as a b y. The action of c d on x y is a b x ax by :. c d y cx dy Therefore, the action of a matrix on x x and on is identical. If the determinant of the matrix a b y y is and ( x, y), then ( ax by, cx dy). A necessary and sufficient condition for c d ( m) to act transitively on is given in [7]. 3 r Lemma.. Let m be any integer and m.3. 3... r, the rime ower decomosition of m. Then ( m) is transitive on if and only if 7, 3 and i for i 3,..., r. Corollary.. The action of the normalizer ( ) is not transitive on. Since the action is not transitive on we now find a maximal subset of on which the normalizer acts transitively. First we start with Lemma.3. The orbits of the action of ( ) on are ; ;,,..., ; 3,,...,,, 4,..., ; ;, where,,,. Proof: It is well known that if k s is given, then there exists some T ( ) such that k k T with s s s. And furthermore, for a a d, if and only if d d a amod d,. So the result follows. d Lemma.4. The orbits of the action of ( ) are as follows. Let l,,...,. Then (a) If l is odd then (b) If l is even then l l l l l l l l Iranian Journal of Science & Technology, Trans. A, Volume 34, Number A4 Autumn
Suborbıtal grahs for a secıal subgrou of the normalızer of m 37 (c) Proof: We rove only (a). The rest are similar. ae b Let T be an arbitrary element in c de ( ). Then e must be,, or. Case. Let e =. Then T ( ). Therefore T fixes l. a b l al b Case. Let e =. Then c d cl d. a b l al b Since and ad bc, we conclude that ( al b, cl d) =. c d cl d Therefore, al b x, where x ( al b)( cl d)mod ( cl d). This shows that and must be in a single orbit of ( ). a b 4 Case 3. Let e. Then T, ad bc. c d l a l b al b T 3 cl d cl d and as in Case, ( al b, cl d ). Therefore, l T = x, where x ( al b)( cl d)mod or x bcl mod. l Since bc mod, x l mod. Therefore and l must be in a single orbit of ( ). l Case 4. Let e. Then we easily find that T sends to l. So we consequently have l l l l the orbit. Corollary.6. The action of ( ) on ( ) is transitive. Lemma.7. The stabilizer of a oint in ( ) is an infinite cyclic grou. Proof: Since the action is transitive, stabilizers of any two oints are conugate. Therefore, we can only look at the stabilizer of in ( ). ae b ae T c de c, Autumn Iranian Journal of Science & Technology, Trans. A, Volume 34, Number A4
38 B. O. Guler / et al b then c. In this case e and since ad, T. This shows that stabilizer ( ) of is. We know from [7] (see also [8]) that the orders of the ellitic elements of ( ) may be, 3, 4, or 6. Therefore, we give the following: Lemma.8. Let be a rime and mod4. Then the normalizer ( ) contains an ellitic a b element E of order 4 and that E is of the form, det E =. c ( a) Let ( G, X ) be transitive ermutation grou, and suose that R is an equivalence relation on X. R is said to be G-invariant if x, y R imlies, a G-invariant relation are called blocks. We give the following from [9]. g x g y R for all g G. The equivalence classes of Lemma.9. Suose that ( G, X ) is a transitive ermutation grou, and H is a subgrou of G such that, for some x X, Gx H. Then R g x, gh x : gg, h H is an equivalence relation. Lemma.. Let ( G, X ) be a transitive ermutation grou, and the G-invariant equivalence relation defined in Lemma.9; then g g if and only if g g H. Furthermore, the number of blocks is G: H. To aly the ideas, we take ( ), a b, ( ), and the stabilizer ( c ( a) ) of in ( ) instead of ( G, X ), H and G. In this case the number of blocks is and these blocks are [ ]: and []:. x 3. SUBORBITAL GRAPHS OF ON ( ) ( ) Let ( G, X ) be a transitive ermutation grou. Then G acts on X X by g, g, g, gg;, X. The orbits of this action are called suborbitals of the normalizer G. The orbit containing, is denoted by O,. From O, we can form a suborbital grah G, : its vertices are the elements of X, and there is a directed edge from to if, O,. A directed edge from to is denoted by. If, O,, then we will say that there exists an edge in G,. If, the corresonding suborbital grah G,, called the trivial suborbital grah, is selfaired: it consists of a loo based at each vertex x X. We will mainly be interested in the remaining non-trivial suborbital grahs. These ideas were first introduced by Sims []. We now investigate the suborbital grahs for the action of ( ) on ( ). Since the action of ( ) on ( ) is transitive, ( ) ermutes the blocks transitively; so the subgrahs are all isomorhic. Hence, it is sufficient to study with only one block. On the other hand, it is clear that each Iranian Journal of Science & Technology, Trans. A, Volume 34, Number A4 Autumn
Suborbıtal grahs for a secıal subgrou of the normalızer of m non-trivial suborbital grah contains a air,u for some u ( ). Therefore, we work on the following case: We denote by F, u the subgrah of G, u such that its vertices are in the block [ ]. Theorem 3.. Let rs and x y be in the block [ ]. Then there is an edge rs xy if and only if (i) If s but s, then x ur mod, y us mod, ry sx (ii) If x ur mod, y us mod, ry sx. s, then Proof: Assume first that rs xy is an edge in F, u and that there exists some T in the normalizer 39 in F, u s but s ( ) such that T sends the air,u. Therefore, to the air rsxy,, that is T( ) r / s T u x y. Since s, T must be of the form i a b c d. a ( ) r T ( ) i gives that ( ) i r a and s ( ) i c, for i =,. c ( ) s a b u au b ( ) x Tu for =,. c d cu d ( ) y a b Since the matrix c d has determinant and ( u, ), then ( au b, cu d ). And therefore, ( au b, cu d ). So and x au b y cu d ( ) ( ), ( ) ( ). i i That is, x ( ) au mod, y ( ) su mod. Finally, since a c i b u ( ) r ( ) x i, for i, =,, d ( ) s ( ) y we get ry sx. This roves (i). Secondly, let rs xy be an edge in F, u and s. In this case T must be of the form i a b c d, det T=. Therefore, since a ( ) r T ( ) i we get a r and s c, by c ( ) s taking i to be. Likewise, since a c we have x ur mod and mod minus sign holds. b u d = au b ( ) x, cu d ( ) y y us and that ry sx. In the case where i = and =, the In the oosite direction we do calculations only for (i) and the lus sign. The other are likewise x ur mod, y us mod, ry sx, done. So suose s and s. Therefore there exists b, d in such that x ur b and y us d. Since ry sx, we Autumn Iranian Journal of Science & Technology, Trans. A, Volume 34, Number A4
3 B. O. Guler / et al r b get rd bs, or 4rd bs. Hence the element T : is not only in the normalizer s d ( ), but also H. It is obvious that r u x T ( ) and T. s y Theorem 3.. If we resent edges of F, u edges of the subgrah F, u of ( ) as hyerbolic geodesics in the uer half-lane, no cross in. Proof: Without loss of generality, since the action on ( ) is transitive, suose that u, x y x y and x y u x y, where all letters are ositive integers. Since x y x y and x y u x y, then xy xy and x y u x y, resectively. Therefore Then x y x y u x y. xy xy yy uy x y. So y uy x, a contradiction []. 4. THE NUMBER OF SIDES OF CIRCUITS Let (, ) G, be a suborbital grah. By a directed circuit in G,, we mean a sequence v v... vm v, where m 3 ; an anti-directed circuit will denote a configuration like the above with at least one arrow (not all) reversed. If m,3 or 4 then the circuit, directed or not, is called a self-aired, a triangle or a rectangle, resectively. G X be a transitive ermutation grou and Theorem 4.. F, u has a self-aired edge if and only if u mod. Proof: Without loss of generality, from transitivity, we can suose that the self-aired edge be u. Alying Theorem 3., the roof then follows. Theorem 4.. F, u contains no triangles. contains a triangle. From transitivity and Theorem 3. the form of such a triangle u x. But, to be x gives a contradiction to Theorem 3.(ii). Proof: Suose contrary F, u Theorem 4.3. The normalizer ( ) does not contain eriod 3. Proof: Suose the converse that ( ) does have a eriod 3. Then it has an ellitic element T of a b order 3. T must be of the form, det T and a d. Take ad. Then c d ad mod, and since ad, then aa mod, or a a mod, which is a contradiction. Theorem 4.4. The subgrah F, u u u mod. contains a rectangle if and only if Iranian Journal of Science & Technology, Trans. A, Volume 34, Number A4 Autumn
Suborbıtal grahs for a secıal subgrou of the normalızer of m 3 k m s x k. It can be easily l n t y l transitively. Therefore we suose that the m x k above rectangle is transformed under H to the rectangle. y l Furthermore, without loss of generality, suose m x k. From the first edge and Theorem 3. y l m u mod x um mod and ym x ; and that from Proof: Assume first that F, u/ has a rectangle shown that H ermutes the vertices and edges of F, u we get. The second edge gives the third edge we have k ux mod and x ky u ym mod or u uy mod. If we combine these we obtain. Since x ymky, then ym ( k). This gives that y=. Therefore holds then we conclude that u u mod u u mod. m x k If y l u u mod then we get the rectangle u u u., and furthermore, if Secondly suose that u u mod. Then, using Theorem 3., we see that u u u is a rectangle. As an examle, 35 75 45 G,3 5. is a rectangle in Corollary 4.5. For some u in, F, u 4. Proof: Firstly suose F, u contains a rectangle if and only if the grou H has a eriod contains a rectangle. Then, Theorem 4.4 shows that u u u u mod u. So we have the ellitic element of order 4 in H. u Since the index of H is in ( ), the elements of this form must be in H. Conversely, suose that H has a eriod for order 4, so H contains an ellitic element of order 4. Let a b this element be a, det =. From this we get (u u ). Therefore F, u contains a rectangle. We redict from the above lemmas that the ellitic elements of ( ) corresond to the circuit in F, u. To suort this idea we have Theorem 4.6. The set H ( ) has a eriod for order if and only if there exists some, (, ) F, u has a self-aired edge. u such that Autumn Iranian Journal of Science & Technology, Trans. A, Volume 34, Number A4
3 B. O. Guler / et al a b Proof: First suose that the set has such an ellitic element T. Then T must be of the form, a det T =. Therefore we have a mod. So, Theorem 3. shows that a is a self- F, u. aired edge in Secondly, let F, u have a self-aired edge. Without loss of generality, from transitivity, we u can suose that the self-aired edge be. So we have, by Theorem 3., (u ) u mod. This showes that there exists some such that b. Therefore a b a is an ellitic element of order in the set H ( ). Notice that H (5 ) has no eriod for order, and therefore F, u 5 does not have a selfaired edge. Finally, as a finishing oint, we give a number theoretical result as follows: Theorem 4.7. The rime divisors of, for any, are of the form mod4 u u. Proof: Let u be any integer and a rime divisor of u u. Then, without any difficulty, it can be easily seen that the normalizer ( ), like ( ) u u, has the ellitic element u of u order 4. From Lemma.8 we get that mod4. REFERENCES. Conway, J. H. & Norton, S. P. (977). Montorous Moonshine. Bull. London Math. Soc., 38-339.. Akbas, M. & Baskan, T. (996). Suborbital grahs for the normalizer of N. Tr. J. of Mathematics, 379-387. 3. Keskin, R. (6). Suborbital grahs for the normalizer m. Euroean J. Combin. 7 (), 93-6. 4. Akbas, M. (). On suborbital grahs for the modular grou. Bull. London Math. Soc. 33(6), 647-65. 5. Jones, G. A., Singerman, D. & Wicks, K. (99). The Modular Grou and Generalized Farey Grahs. London Math. Soc. Lecture Note Series, 6, 36-338. 6. Keskin, R. (). On suborbital grahs for some Hecke grous. Discrete Math. 34(-3), 53-64. 7. Akbas, M. & Singerman, D. (99). The Signature of the normalizer of N. London Math. Soc. Lecture Note Series 65, 77-86. 8. Machlaclan, C. (98). Grous of units of zero ternary quadratic forms. Proceeding of the Royal Society of Edinburg, 88 A, 4-57. 9. Bigg, N. L. & White, A. T. (979). Permutation grous and combinatorial structures. London Mathematical Society Lecture Note Series 33, CUP.. Sims, C. C. (967). Grahs and Finite Permutation Grous. Math. Z., 95, 76-86.. Rose, H. E. (988). A Course in Number Theory. Oxford University Press. Iranian Journal of Science & Technology, Trans. A, Volume 34, Number A4 Autumn