202 IV Continued Fractions and Their Uses
Furthermore, if
g(z)=(a′z+b′)/(c′z+d′)is another modular transformation, then the composite maph=g◦fis again a mod-
ular transformation:
h(z)=(a′′z+b′′)/(c′′z+d′′),where
a′′=a′a+b′c, b′′=a′b+b′d,
c′′=c′a+d′c, d′′=c′b+d′d,and hence
a′′d′′−b′′c′′=(a′d′−b′c′)(ad−bc)= 1.It follows that the setΓof all modular transformations is a group. Moreover, compo-
sition of modular transformations corresponds to multiplication of the corresponding
matrices:
(
a′′ b′′
c′′ d′′
)
=
(
a′ b′
c′ d′)(
ab
cd)
.
However, the same modular transformation is obtained if the signs ofa,b,c,dare all
changed (and in no other way). It follows that themodular groupΓis isomorphic to
the factor groupSL 2 (Z)/{±I}of the special linear groupSL 2 (Z)of all 2×2integer
matrices with determinant 1 by its centre{±I}.
Proposition 12The modular groupΓis generated by the transformations
T(z)=z+ 1 , S(z)=− 1 /z.Proof It is evident thatS,T ∈Γ andS^2 =Iis the identity transformation. Any
g∈Γhas the form
g(z)=(az+b)/(cz+d),wherea,b,c,d∈Zandad−bc=1. Ifc=0, thena=d=±1andg=Tm,
wherem=b/d∈Z. Similarly ifa=0, thenb=−c=±1andg=STm,where
m=d/c∈Z. Suppose now thatac=0. For anyn∈Zwe have
ST−ng(z)=(a′z+b′)/(c′z+d′),wherea′=−c,b′=−d,c′=a−ncandd′=b−nd. We can choosen=m 1 so
that forg 1 =ST−m^1 gwe have|c′|<|a|and hence|a′|+|c′|<|a|+|c|.Ifa′c′=0,
the argument can be repeated withg 1 in place ofg. After finitely many repetitions we
must obtain
ST−mk···ST−m^1 g=TmorSTm.