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.



Relations between two particular elements of SL_2(Z)?
Certainly $\mathrm{SL}(2,\mathbb{Z})$ contains a free group.
For instance $\Gamma(2)$, the subgroup of all matrices congruent
to the identity modulo $2$, is free of rank $2$. The matrices
$\left(\begin{array}{cc}1&2\\\ 0&1\end{array}\right)$
and
$\left(\begin{array}{cc}1&0\\\ 2&1\end{array}\right)$
freely generate $\Gamma(2)$. 
This can be proved by considering the action on the upper half-plane
or by careful examination of reduced words.
There's a nice proof in chapter 18 of David Ullrich's
book *Complex Made Simple*.


Your $a$ and $b$ don't generate a free group alas, since they
generate all of $\mathrm{SL}(2,\mathbb{Z})$.


Re the edited question. Let's write
$$T=\left(\begin{array}{cc}1&1\\\ 0&1\end{array}\right)\qquad
\textrm{and}\qquad 
U=\left(\begin{array}{cc}1&0\\\ 1&1\end{array}\right).$$
As both Jack and I pointed out, $T^2$ and $U^2$ generate
a free subgroup of rank $2$. Now it's an easy exercise to prove that
$T$ and $U$ freely generate a free monoid of rank $2$ (because
their entries are non-negative). On the other hand, they generate
thw whole group $\mathrm{SL}(2,\mathbb{Z})$ which is certainly
not free. Your matrices $a$ and $b$ are, if my calculations
are right, $-U^{-1}$ and $-T^{-1}$. The matrix $S\_4$ conjugates $T$
and $U$ into $U^{-1}$ and $T^{-1}$ so $U^{-1}$ and $T^{-1}$ freely
generate a free monoid of rank $2$. The same must be true of
$U^{-1}$ and $T^{-1}$, that is, of $a$ and $b$.


