The relationship between SL(2,Z) and Gal(Qbar,Q) (caveat: I'm not a number-theorist or Langlands-programme-er, and I don't expect to understand all the answers to this question, but I figured they might be useful to someone besides me).
I've been making videos of symmetries of Klein's $j(\tau)$:
https://math.stackexchange.com/questions/466975/elements-of-sl2-mathbbz-which-fix-roots-of-kleins-absolute-invariant-j
https://math.stackexchange.com/questions/338332/visualizing-functions-invariant-or-almost-under-modular-transformation
in an attempt to have a better visual feel of the symmetries of $\mathbf{SL}(2,\mathbb{Z})$. 
I know that $\mathbf{SL}(2,\mathbb{Z})$ and $\mathrm{Gal}(\mathbb{\bar{Q}},\mathbb{Q})$ are related -- for instance this paper by Ribet: http://math.berkeley.edu/~ribet/Articles/motives.pdf but that's outside of my scope of knowledge, and that $\mathrm{Gal}(\mathbb{\bar{Q}},\mathbb{Q})$ is something which no one quite understands yet, so:


*

*Is $\mathbf{SL}(2,\mathbb{Z})$ a subgroup of $\mathrm{Gal}(\mathbb{\bar{Q}},\mathbb{Q})$? and if so,

*What properties does their  quotient $\mathrm{Gal}(\mathbb{\bar{Q}},\mathbb{Q})/\mathbf{SL}(2,\mathbb{Z})$  have?
 A: The answer to the question as stated is no. The reason is that $\text{SL}_2(\mathbb{Z})$ contains an element of order $4$, while $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ does not. In fact the following more general claim holds.
Proposition: Let $K$ be a field and let $g \in \text{Gal}(\overline{K}/K)$ have finite order. Then $g^2 = \text{id}$.
Proof. Let $L$ be the fixed field of $g$. Then $\overline{K}/L$ is a finite extension of degree the order of $g$, but $\overline{K}$ is algebraically closed, so by the Artin-Schreier theorem $[\overline{K} : L] = 1$ or $2$. $\Box$
A: I edit my answer due to two downvotes and the comments. It is an easy argument, that if an embedding would exist the quotient would look horrible. QY shows no embedding is possible, so to some extent my answer is negligible.
$Gal( \overline{\mathbb Q}, \mathbb Q)$ is a profinite group, in particular compact, and $SL_2(\mathbb{Z})$ is a discrete infinite group. Every embedding of an infinite discrete group in a compact group is dense, and no nice topological quotient exists. Usually, the cosets are not even measurable, look at $\mathbb{R} / \mathbb{Q}$.
You might be able to find $SL_2(\mathbb{Z})$ as a subgroup of a quotient, though. It is at least conjectured that $SL_2(\mathbb{Z}/N)$ turns up as quotients for every $N$. If they would turn up in a consistent manner, you might build the projective limit, i.e., find the profinite completion $\otimes_p SL_2(\mathbb{Z}_p)$  of $SL_2(\mathbb{Z})$ as a quotient, which has $SL_2(\mathbb{Z})$ as a discrete subgroup.
