# Finite quotients of the absolute Galois group of $\mathbb{Q}$ via torsion of elliptic curves

Let $$E/\mathbb{Q}$$ be an elliptic curve and let $$p$$ be a prime. Then there is an action of the absolute Galois group of $$\mathbb{Q}$$ on $$E[p]$$ that factors through a finite quotient.

Does any finite quotient of the absolute Galois group of $$\mathbb{Q}$$ arise this way for some $$E$$ and $$p$$? What if we consider abelian varieties instead of elliptic curves?

• Every such action is odd, meaning the complex conjugation is mapped to multiplication by $-1$ (this is certainly true for elliptic curves and I believe also for abelian varieties). So any quotient which identifies complex conjugation and identity won't arise this way. – Wojowu Apr 25 at 16:10
• If your finite group has a 2-dimensional representation, then en.wikipedia.org/wiki/Serre%27s_modularity_conjecture implies that it comes from a modular form if it is odd etc. But only few of those come from elliptic curves. – Chris Wuthrich Apr 25 at 16:44
• Even something stronger holds: The field obtained by adjoing all $E[p]$ to $\mathbb{Q}$, for all ellliptic curves $E$ and all $p$, is still very far away from the algebraic closure $\overline{\mathbb{Q}}$ (in fact it is a Hilbertian field). I know of no similar result when running over all abelian varieties though. – Arno Fehm Apr 25 at 18:21

For elliptic curves, the answer is no, as Chris Wuthrich and Aron Fehm point out in the comments. In fact I think every extension with Galois group a finite simple group not of the form $$PSL_2(\mathbb F_p)$$ for any $$p$$ cannot arise this way.
For abelian varieties, the answer is positive. In fact it's sufficient to take just the $$2$$-torsion, and even just the $$2$$-torsion of Jacobians of hyperelliptic curves. Given an arbitrary extension, take the minimal polynomial $$f$$ of a generator $$\alpha$$ of that extension, form the hyperelliptic curve with equation $$y^2 = f(x)$$, and take its Jacobian. (If the degree of $$f$$ is at most $$4$$, you will need to add some linear factors as well to make it hyper-.) The Galois group of the $$2$$-torsion of the hyperilliptic curve will be a subgroup of $$S_n$$, generated by the Galois action on the roots of this polynomial (plus the point at $$\infty$$, if the degree is odd) and will therefore include the Galois group of your field.