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Q: What is the "simplest" finite group $G$ for which we don't know how to realise it as a Galois group over $\mathbf{Q}$ ?

So here the word simplest might be interpreted in a broad sense. If you want something precise you may take the group of smallest order but I prefer to leave the question as it is. Also since naturally one classifies finite groups into families one may also ask the following

Q: What is the "simplest" example of a family of finite groups for which the inverse Galois problem is unknown?

flag	asked Jul 12 2011 at 17:39 Hugo Chapdelaine 3,337●3●18

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Among the sporadic family, I think there is only one group for which the inverse problem is unknown-- $M_{23}$. See mathoverflow.net/questions/13851/… – Pace Nielsen Jul 12 2011 at 17:46

An almost duplicate of mathoverflow.net/questions/13851/… ? – Mark Sapir Jul 12 2011 at 19:03

Thanks Mark for the link – Hugo Chapdelaine Jul 12 2011 at 19:37

Felipe Voloch · Answer 1 · 2011-07-12 19:52:50Z UTC

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I am not an expert and I might be misrembering some talks I've attended. Anyway, $SL_2(\mathbb{F}_q)$ can be done for prime $q$ by using torsion on non-CM elliptic curves. But I don't think it's been done for general prime powers $q$. Also, what about $SL_3$?

answered Jul 12 2011 at 19:52

Felipe Voloch
16.6k●2●36●61

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The $p$-torsion on a generic elliptic curve gives not ${\rm SL}_2({\bf F}_p)$ but ${\rm GL}_2({\bf F}_p)$ with cyclotomic character. In many cases this can be twisted to get ${\rm SL}_2({\bf F}_p)$, but it's not immediate. Serre discusses this in some detail in his Topics in Galois Theory (1992). – Noam D. Elkies Jul 13 2011 at 6:13

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For general powers of q, there are some results (which do not cover all cases). In arxiv.org/abs/math/0610860 and in arxiv.org/abs/0905.1288 they prove that $PSL_2(\mathbb{F}_q)$ and $PGL_2(\mathhbb{F}_q)$ is the Galois group of an extension for almost al $q$ (in the second work they prove that fixed n, in a density one set of primes $\ell$, $PSL_2(\mathbb{F}_{\ell^n})$ is obtained). Then you can just lift the representation to get a rep. of SL_2 if you want. – A. Pacetti Jul 13 2011 at 7:59

I have an issue with all these remarks. If $\overline{\rho}: \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \rightarrow \mathrm{SL}_2(\mathbf{F}_p)$ has surjective image, then $\overline{\rho}$ is "even" in Serre's sense. So it can't possibly come (after twisting or otherwise) from a modular form. In fact, I wouldn't be surprised in the inverse Galois problem is unknown for $\mathrm{SL}_2(\mathbf{F}_p)$ for $p \ge 13$ or so. – Lavender Honey Jul 13 2011 at 18:28

@Noam, presumably Serre is twisting to $\mathrm{PSL}_2(\mathbf{F}_p)$? I could imagine doing this from some modular representation when $-1 \in \mathbf{F}^{2}_p$, for example. – Lavender Honey Jul 13 2011 at 18:29

@Pacetti, You cannot "just lift" a Galois representation with image in $\mathrm{PSL}_2(\mathbf{F}_p)$ to $\mathrm{SL}_2(\mathbf{F}_p)$; there are non-trivial obstructions. For example, suppose that $p \equiv 1 \mod 4$ and the image of complex conjugation is the diagonal matrix with entries $1$ and $-1$ (or $i$ and $-i$ if you prefer). No lift of this element to $\mathrm{SL}_2(\mathbf{F}_p)$ has order two, so one can't possibly find a lift. – Lavender Honey Jul 13 2011 at 18:35

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On the inverse Galois problem

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