By Paula Ribenboim

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Now we multiply by T 3 to get AT- 2 ST 4 ST 3 = G -1)(1 -3 3) 0 1 = (0 1 -1)o = S. Solving for A we find A = ST- 3 ST- 4 ST 2 . At each stage there may be more than one power of T that makes 1d 1< 1c 1 so the process is not unique. To prove the theorem in general it suffices to consider those matrices A = (: ~) in r with c 2: O. We use induction on c. 29 2: The modular group and modular functions If c = 0 then ad = 1 so a = d = ± 1 and A=(±~ ±n=(~ ±~)=T±b. Thus, A is apower of T If c = 1 then ad - b = 1 so b = ad - 1 and A= G 1) ad; (~ ~)G -~)G ~) = TaST = d.

This number is called the dass number and is denoted by h(d). Hint: Show that 0< a s Jdi3 for each reduced form. 10. Determine all reduced forms with integer coefficients a, b, c and the c1ass number h(d) for each d in the interval 1 s d s 20. CONGRUENCESUBGROUPS The modular group r has many subgroups of special interest in number theory. The following exercises deal with a dass of subgroups called congruence subgroups. Let and 45 2: The modular group and modular functions be two unimodular matrices.

7. (a) If dis fixed, prove that there is a one-to-one correspondence between the set of forms with discriminant d and the set of complex numbers 1: with Im(1:) > O. (b) Prove that two quadratic forms with discriminant d are equivalent if and only if their representatives are equivalent under r. Note. A reduced form is one whose representative rE R r . Thus, two reduced forms are equivalent if and only if they are identical. Also, each dass of equivalent forms contains exactly one reduced form. 8.