Please forgive me if this question is ill-formed. I don't know much about decidability.
Some Background:
There are problems in combinatorial group theory that are undecidable, such as the word problem.
I don't know of any results either way in terms of the decidability of properties of sets of groups.
One property I'm working on for some finitely presented groups is the Tits Alternative (although I'm not looking at decidability).
Definition 1: A class $\mathcal{G}$ of groups satisfies the Tits alternative if for any $G$ in $\mathcal{G}$ either $G$ has a free, non-abelian subgroup or $G$ has a solvable subgroup of finite index.
Example 1: The class of finitely generated linear groups satisfies the Tits alternative.
The Question:
Is the Tits Alternative decidable for any given (suitably defined$^\dagger$) class $\mathcal{G}$ of groups?
Thoughts:
I'm not sure if the question even makes sense. There might be some technicalities I am unaware of that render it nonsense.
One thing I suspect is that
A hunch: If the problem of whether a given group is solvable is undecidable, then the Tits Alternative is undecidable.
This appears to be obvious but I'm not sure how to prove it.
(Edit: The hunch is wrong. See YCor's comment below.)
I doubt that this question has an answer yet. After a brief search through the literature, I couldn't find anything.
Perhaps I'm asking for too much then . . .
Nevertheless, I'm curious.
It's certainly not a question I think I could answer myself. It's outside the scope of my research.
Let me know what you think :)
$\dagger$ See the first comment for why I added this.
It is proved here that it is decidable whether a finitely generated linear group is solvable by finite, so the Tits alternative is decidable for this class of groups.
I think there is some underlying assumption here that the linear group is defined over a field in which exact computation is possible, such as a number field or a function field.