Let $S_4 = \left(\begin{array}{cc}0&-1 \\\ 1&0 \end{array}\right) \textrm{ and } S_6 = \left(\begin{array}{cc} 1&-1 \\\ 1&0\end{array}\right)$. Serre proves in his book on trees that $SL_2(\mathbb{Z}) \cong \mathbb{Z}/4 *_{\mathbb{Z}/2} \mathbb{Z}/6$, and $S_4$ and $S_6$ are the elements corresponding to the generators of $\mathbb Z/4$ and $\mathbb Z/6$ (I'm not sure if this is related to my question). Then let $a = S_4 S_6$ and $b = S_4 S_6^2$. I believe every element of $SL_2(\mathbb Z)$ can be written as $S_6^d w S_6^e$, where $w$ is a word in $a$ and $b$ but not $a^{-1}$ or $b^{-1}$.

I wrote a program (for other purposes) that seems to show that there aren't any relations between $a$ and $b$ that have length 15 or less and don't involve $a^{-1}$ or $b^{-1}$. I'm not certain that the program is right, but if it is, one might make a naive guess that these two elements generate a free group. This makes me suspicious.

1) Does $SL_2(\mathbb Z)$ contain a free group (of rank > 1)? If it does, is there an easy way to determine whether the subgroup generated by $a$ and $b$ is free?

2) A slightly less naive guess is that $a$ and $b$ generate a free monoid in $SL_2(\mathbb Z)$. Is there a reason why $SL_2(\mathbb Z)$ can't contain a free monoid, or an example showing that it does?

EDIT: Thanks for the quick replies. As Robin and Jack pointed out, $a$ and $b$ generate SL(2,Z), so clearly don't generate a free group. Also, there are free subgroups that are easy to write down. I'm still curious about #2, though.

`$S_4^{a_1} S_6^{b_1} \dots S_4^{a_n} S_6^{b_n}$`

where $a_i = 1$ and $b_i \in \{1,2\}$ and we can optionally omit the leading and/or trailing factors. From this it's easy to solve any word problem in your $a$ and $b$ (almost as easy as in a free group) – Victor Miller Mar 18 '10 at 22:04