Generators of Special Linear Group of Order 2 over Integers

Theorem
Let:
 * $ S = \begin{pmatrix}

0 & - 1 \\ 1 & 0 \end{pmatrix}$ and:
 * $T = \begin{pmatrix}

1 & 1 \\ 0 & 1 \end{pmatrix}$

Then $S$ and $T$ are generators for the special linear group of order $2$ over $\Z$.

Proof
Let:
 * $ g = \begin{pmatrix}

a & b \\ c & d \end{pmatrix}$

be an element of $\SL {2, \Z}$.

Observe that:
 * $T^n = \begin{pmatrix}

1 & n \\ 0 & 1 \end{pmatrix}$

so that:


 * $T^n g = \begin{pmatrix}

a + nc & b + nd \\ c & d \end{pmatrix}$

Also note that:
 * $S^2 = -I$

and:
 * $S g = \begin{pmatrix}

-c & -d \\ a & b \end{pmatrix}$

We now describe an algorithm to reduce $g$ to the identity matrix only using multiplication by $S$ and $T$.

This will show that $g^{-1}$ can be written as a word in $S$ and $T$.

We can then take the inverse of this expression in $S$ and $T$ and it too will be a word in $S$ and $T$.

Recall $g = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.

If $c = 0$ then, since $\map \det g = 1$, we must have $a d = 1$.

Thus $a = d = \pm 1$. In this case:


 * $g = \pm\begin{pmatrix}

1 & b' \\ 0 & 1 \end{pmatrix} = \pm T^{b'}$

where $b' = \pm b$.

Since $S^2 = -I$ we either have $g = T^{b'}$ or $g = S^2 T^{b'}$.

If $c \ne 0$ we proceed as follows.

Without loss of generality we may suppose $\size a \ge \size c$, for if $\size a < \size c$ we could apply $S$ to swap $a$ and $c$.

By the Division Algorithm, we can write $a = c q + r$ where $q, r \in \Z$ and $0 \le r < \size c$.

Then $T^{-q} g$ has upper left entry $a - q c = r$ which is smaller than $\size c$.

Applying $S$ switches the rows (with a sign change), which then makes our lower left entry $-r$.

We have applied a combination of $S$ and $T$ to turn $g$ into a matrix where the absolute value of the lower left entry is strictly less than the one we started with.

If $r = 0$ then we're in our original case, and we know we can multiply by a multiple of $T$ and $S$ to get to the identity matrix.

Note that this process can have at most $r$ steps, since there are only $r$ natural numbers between $r$ and $0$.

Thus the process terminates in a finite number of steps.