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The modular group is the group $G$ consisting of all linear fractional transformations $\phi$ of the form $$\phi(z)=\frac{az+b}{cz+d}$$ where $a,b,c,d$ are integers and $ad-bc=1$. I have read that $G$ is generated by the transformations $\tau(z)=z+1$ and $\sigma(z)=-1/z$. Is there an easy way to prove this? In particular, is there a proof that uses the relation between linear fractional transformations and matrices? Any good reference would be helpful.

Thank you, Malik

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up vote 11 down vote accepted

Yes; this statement is essentially equivalent to the Euclidean algorithm. I discuss these issues in this old blog post. (A very brief sketch: by applying the generators and the inverses to an arbitrary element of the modular group it is possible to perform the Euclidean algorithm on $a$ and $c$ (or maybe it's $a$ and $b$). The rest is casework.) You can think of this as a form of row reduction, which is generalized by the notion of Smith normal form.

There is also a geometric proof using the action on the upper half plane which is given, for example, in the relevant section of Serre's Course in Arithmetic.

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great, thanks! Nice blog post. – Kalim Sep 24 '10 at 0:19
@Malik Younsi: I should mention that there are a few ways to go about the geometric proof. The most sophisticated one gives you a different set of generators; it realizes the modular group as a free product of a cyclic group of order 2 and a cyclic group of order 3 based on the fact that it is (I think I'm saying this right) the orbifold fundamental group of the moduli stack of elliptic curves, but I think this does not really qualify as an "easy proof." – Qiaochu Yuan Sep 24 '10 at 5:21

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