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Modular Group

The modular group $\SL_2(\Z)$, sometimes called the full modular group, is the group of $2\times 2$ matrices with integer entries, and determinant 1. An example of such a matrix is

$$ G= \begin{bmatrix}1&1\\0&1\end{bmatrix} $$

Another commonly accepted definition of the modular group is the projective special linear group $\PSL_2({\Z})$, which is defined as

$$ \PSL_2(\Z) \cong\SL_2(\Z)/\{\pm I\}, $$

where $R$ is a commutative ring with identity. We will mostly be dealing with the first definition.

Generators of the Modular Group

We now wish to explore two different ways to generate the modular group. There are two common ones that we use, both based on the same set of matrices, namely $T$ and $S$, defined below.

First Generating Set

There are two sets of generators for $\SL_2(\Z)$ that we focus on. To begin, we define two matrices, $S$ and $T$, given as

$$ S := \begin{bmatrix}0&-1\\1&0\end{bmatrix}, \qquad T := \begin{bmatrix}1&1\\0&1\end{bmatrix} $$

Note that $S^2 = -I$, thus $\text{ord}(S)=4$. Additionally, observe that

$$ T^n = \begin{bmatrix}1&n\\ 0 & 1\end{bmatrix}. $$

It is immediate that $T$ does not have finite order.

To see that these two matrices generate $\SL_2(\Z)$, begin by fixing some matrix $A = \begin{bmatrix}a&b\ c&d\end{bmatrix}$, and consider two cases; $c\neq 0,$ and $c=0$.

Case I: $c=0$

Suppose $c=0$. Since $\det A = ad-bc$, and $c=0$, we have that either $a=d=1$, or $a=d=-1$. In either case, we can rewrite $A$ as $A = S^{2n}\begin{bmatrix}1&b'\\0&1\end{bmatrix},$ where $b'=(-1)^nb$. Note though, that this left-hand matrix is of the form $T^{b'}$, thus, $A$ can be written as $S^{2n}T^{b'}$, which is contained in the set $\langle S, T\rangle$.

Case II: $c\neq 0$

Let $A$ be as above. By the Division Algorithm, we can rewrite $a=cq+r$, for $q,r\in\Z$, and $0\leq r < c$. We now multiply $A$ by $T^{-q}$, yielding

$$ T^{-q}A = \begin{bmatrix}a-cq&b-dq\\ c & d\end{bmatrix} $$

Observe now that the first entry is equal to $r$, so we can rewrite this as

$$ T^{-q}A = \begin{bmatrix}r & *\\c & * \end{bmatrix}. $$

We put astrixes for the second column, as the value don't matter for this. We now can swap the first and second rows, whilst adding a negative sign, by left-multiplying by $S$, yielding

$$ ST^{-q}A = \begin{bmatrix}-c & *\\ r & *\end{bmatrix} $$

We now repeat this process until $r=0$, which is guaranteed in finite steps due to the Euclidean Algorithm. We have now reduced this to a matrix of the form in Case IModular Group
The modular group $\SL_2(\Z)$, sometimes called the full modular group, is the group of $2\times 2$ matrices with integer entries, and determinant 1. An example of such a matrix is

$$
G= \begin{bmatrix}1&1\0&1\end{bmatrix}
$$

Another commonly accepted definition of the modular group is the projective special linear group $\PSL_2({\Z})$, which is defined as

$$
\PSL_2(\Z) \cong\SL_2(\Z)/\{\pm I\},
$$

where $R$ is a commutative ring with identity. We will mostly be dealing with the ..., and thus, $\langle S, T\rangle$ generates $\SL_2(\Z)$.

A Second Generating Set

We now attempt to show that the set $\langle S, ST\rangle$ generates $\SL_2(\Z)$. What is important here, is that both $S$ and $ST$ are of finite order, $4$ and $6$ respectively. To show that $S$ and $ST$ generate $\SL_2(\Z)$, we show that $\langle S, ST\rangle =\langle S, T\rangle$, where it suffices to show that the generators of each are contained in the other. It is immediate that $S$ and $ST$ are contained within $\langle S, T\rangle$, and that the $S$ and $T$ must be in $\langle S, ST\rangle$, as $T=S^{-1}ST\in \langle S, ST\rangle$. Thus, $\langle S, ST\rangle$ generate $\SL_2(\Z)$.

Homomorphisms of $\SL_2(\Z)$ into $\mathbb{C}^{\times}$

Since we know that $\SL_2(\Z)$ is generated by two elements, of orders 4 and 6 respectively, we can investigate homomorphisms of $\SL_2(\Z)$ into $\mathbb{C}^{\times}$. Recall that a group homomorphism is uniquely determined by the image of its generators. Thus, we know that for any $\varphi : \SL_2(\Z)\to \mathbb{C}^{\times}$, we have that

$$ \varphi(S)^4=\varphi(S^4)=\varphi(I)=1 $$

Therefore, the image of $S$ under $\varphi$ is a fourth root of unity. Similarly,

$$ \varphi(ST)^6 = \varphi((ST)^6)=\varphi(I)=1 $$

Thus, $\varphi$ maps $ST$ to a sixth root of unity. Therefore, we can deduce that the image of $\varphi$ are in fact the 12th roots of unity of $\mathbb{C}$.