Let $G = \langle A \mid R \rangle$ be a finitely presented group, given by a finite presentation. If $G$ is abelian, then we can verify this fact: simply verify the fact that $[a, b] = 1$ for all generators $a, b \in A$ (this can be done by enumerating all words equal to $1$ in the group). That is, the property "being abelian" is a semi-decidable property (however, the property of "being non-abelian" is not semi-decidable, by the [Adian–Rabin theorem](https://en.wikipedia.org/wiki/Adian%E2%80%93Rabin_theorem), cf. also [this translation](https://arxiv.org/abs/2208.08560)). I recently bumped into the following question(s), but couldn't figure out any obvious way to answer either:


Is the property of being solvable semi-decidable? Is the property of **not** being solvable semi-decidable?


That is, is it possible to verify that a group is solvable? Is it possible to verify that a group is *not* solvable? The same questions above can also be asked for solvability of bounded derived length; for example, can one verify whether $G$ is a metabelian group (i.e. derived length $\leq 2$)? For reference, nilpotency is semi-decidable, i.e. there is a procedure for verifying nilpotency, see the article ["Verifying nilpotence" by Charles Sims](https://doi.org/10.1016/S0747-7171(87)80002-0).



Is solvability semi-decidable?
The Adian–Rabin theorem, although it is not usually stated like this, says that Markov properties are not co-semi-decidable, thus « not being metabelian » or « not being solvable » are not semi-decidable.
However, whether « being solvable » or « being solvable of derived length $\le n$ » is semi-decidable is an open problem.
Giving a precise classification of these properties in terms of the arithmetical hierarchy is stated as an open problem in Miller’s survey article [Decision Problems for Groups — Survey and Reflections](https://doi.org/10.1007/978-1-4613-9730-4_1).  

It is natural to expect that « being solvable of derived length $\le n$ » should be $\Pi\_2$ complete —i.e. equivalent to an infinite conjunction of semi-decidable properties, but not semi-decidable. Also, « being solvable » should be $\Sigma\_3$ complete.


I think that these questions are both important and very basic and they definitely deserve some attention.


