(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?



The relationship between SL(2,Z) and Gal(Qbar,Q)
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](https://mathoverflow.net/questions/10449/artin-schreier-theorem-for-rings) $[\overline{K} : L] = 1$ or $2$. $\Box$


