gr.group theory - Relations between two particular elements of SL_2(Z)?

Let $S_4 = left(begin{array}{cc}0&-1 \ 1&0 end{array}right) textrm{ and } S_6 = left(begin{array}{cc} 1&-1 \ 1&0end{array}right)$. 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.