# Decidability in groups

This is not my area of research, but I am curious. Let $G=\left< X|R \right>$ be a finitely presented group, where $X$ and $R$ are finite. There are many questions which are undecidable for all such $G$, for example whether $G$ is trivial or whether a particular word is trivial in $G$. Is there any non-trivial question (by trivial I mean that the answer is always yes or always no) which is decidable? For instance, is there a class $S$ (non-empty and not equal to all the finitely presented groups) such you can always decide whether $G$ is in this class?

• You might like to look at the answers to this question, and the question it refers to: mathoverflow.net/questions/16532/… . – HJRW Mar 31 '11 at 9:06
• Thanks. If a question is interesting, then it will come back. – Yiftach Barnea Mar 31 '11 at 9:44
• There is active research going on about whether such problems are decidable if we assume the word problem has a solution. For instance, determining whether a group is a 3-manifold group is decidable, given a presentation and a solution to the word problem. – JeremyKun Oct 18 '11 at 2:20
• en.wikipedia.org/wiki/Adian%E2%80%93Rabin_theorem – TT_ Apr 24 at 3:42

The problem whether $$G$$ is perfect, that is $$G=[G,G]$$ is decidable because you need to abelianize all relations (replace the operation by "+" in every relation) and solve a system of linear equations over $$\mathbb Z$$. For example, if the defining relations are $$xy^{-1}xxy^5x^{-8}=1, x^{-3}y^{-2}xyx^5 = 1$$, then the Abelianization gives $$-5x+4y=0, 3x-y=0$$. Now you need to check if these two relations "kill" $$\mathbb{Z^2}$$. That means for some integers $$a,b,c,d$$ we should have $$x = a(-5x+4y)+b(3x-y), y = c(-5x+4y)+d(3x-y)$$. This gives 4 integer equations with four unknowns: $$1=-5a+3b, 0=4a-b, 1=4c-d, 0=-5c+3d$$. This systems does not have an integer solution (it implies $$7a=1$$), so $$G\ne [G,G]$$.
• Here the abelianization has the abelian group presentation $(x,y:5x=4y,3x=y)$ and this can be rewritten as $(x,y:7x=0,3x=y)$ which is a presentation of $C_7$, so the abelianization is $C_7$, so the abelianization is nonzero. – YCor Oct 20 '18 at 6:04
• In general, there's a much more efficient procedure than making an integral non-homogeneous system on $m^2$ variables. For instance, one has a matrix $n\times m$ matrix ($n$ number of generators, $m$ relators): then abelianization is zero iff this matrix has rank $n$ over every field. One checks over $\mathbf{Q}$, using standard elementary operations, and keeping the list of denominators of these. If it doesn't have rank $n$, we are done. Otherwise, we then check whether the rank is $n$ modulo every prime appearing among divisors of the listed denominators. – YCor Oct 20 '18 at 6:08