Suppose we have a finitely presented group $G$ with solvable word problem. (For instance, the command RWSGroup in Magma terminates giving us a finite [but possibly gigantic] rewrite system.) Is there then an algorithm to determine whether $G$ has an element of order 2?

If so, where is the best place to find such an implementation?

If not, what more needs to assumed? What standard methods exist to disprove the existence of 2-torsion?

Edited to add: Here is a specific example with three generators and four relations. Take $$ G=\langle x,y,z\ :\ xy^{-1}x^{-1}y^{-1}z^{-2}=1,\ x^{-1}z^{-2}xz^{-2}=1,\\ xyx^{-1}yz^{-2}=1,\ y^{2}xzx^{-1}z^{-1}=1\rangle. $$ I can prove that this group is not a unique product group, but I wish to show it is torsion-free. I can show it has no odd torsion. Any advice?

  • 3
    $\begingroup$ This sounds to me like the sort of problem that could be unsolvable. It is solvable for special classes of groups, such as hyperbolic groups and 1-relator groups. en.wikipedia.org/wiki/Hyperbolic_group $\endgroup$
    – Ian Agol
    May 25, 2017 at 4:21
  • $\begingroup$ @Ian I was afraid of that. Perhaps sometime next week I'll post the group presentation I'm specifically trying to work on. (I was hoping just for general guidelines of what techniques I might try, but perhaps people need a specific example.) $\endgroup$ May 27, 2017 at 1:10
  • 2
    $\begingroup$ @IanAgol It is indeed undecidable in general, as not having $2$-torsion is a Markov property. $\endgroup$
    – ADL
    Jun 13, 2017 at 9:57

2 Answers 2


Using a mixture of computation and thought I believe that I have established that this group is indeed torsion-free. I don't know of any general approach to solving that particular problem. Even if the group is hyperbolic (which this example is not, because it has free abelian subgroups of rank $2$), I am not aware of any practical implementable algorithm for deciding torsion-freeness.

Here is some Magma code for this problem. Following Pace Nielsen's suggestion that there might be an easily identifiable subgroup of index $32$, I found such a subgroup $K$ of index $4$, and we can compute a presentation of it on its four defining generators.

> G<x,y,z>:=Group<x,y,z|x*y^-1*x^-1=z^2*y, x*z^-2=z^2*x, x*y*x^-1*y=z^2,
>                       y^2*x*z=z*x >;
> K<a,b,c,d> := sub<G | x^2, z^2, x*z*y^-1, y^2>;
> Index(G,K);
> Rewrite(G,~K);
> K;
Finitely presented group K on 4 generators
Index in group G is 4 = 2^2
Generators as words in group G
    a = x^2
    b = z^2
    c = x * z * y^-1
    d = y^2
    (c^-1, a) = Id(K)
    (a^-1, b) = Id(K)
    (a^-1, d^-1) = Id(K)
    (d^-1, b^-1) = Id(K)
    (b, c) = Id(K)
    d * c * b^-1 * d^-1 * c^-1 * b^-1 = Id(K)
    b^-1 * a * c^-1 * a^-1 * b * c = Id(K)

We see that all pairs of generators of $K$ commute except for $c$ and $d$. The final relation collapses completely, and the preceding one is equivalent to $dcd^{-1}c^{-1} =b^2$, so $K$ is a direct product of $\langle a \rangle$ and a torsion-free nilpotent group of class $2$ generated by $b,c,d$. So $K$ is torsion-free.

> Transversal(G,K);
{@ Id(G), x, y, z @}

So a nontrivial torsion-element would have to have order $2$ and be of the form $xk$, $yk$ or $zk$ for some $k \in K$. I spent some time trying to prove that this does not happen, but eventually found an easy way to do it by calculating in a finite quotient. We let $K_2$ be the kernel of the map of $K$ on the to the largest elementary abelian quotient of $K$, which has order $16$, and then check for corresponding elements of order $2$ in $G/K_2$. We see that there are none, so $G$ is torsion-free.

> PK, phi := ElementaryAbelianQuotient(K,2);
> Order(PK);
> K2 := Kernel(phi);
> Index(K,K2);
> T2 := Transversal(K,K2);
> exists{k : k in T2 | (x*k)^2 in K2 };
> exists{k : k in T2 | (y*k)^2 in K2 };
> exists{k : k in T2 | (z*k)^2 in K2 };

For your given group presentation, I think one may reduce the problem of finding torsion elements a bit.

We may rewrite the relators as:

$xyx^{-1}= y^{-1}z^{-2}, xyx^{-1} = z^2 y^{-1}, xz^2x^{-1}=z^{-2}, xzx^{-1}=y^{-2}z$.

We get the resulting relations $xyx^{-1}=y^{-1}z^{-2}=z^2y^{-1}, xz^2x^{-1}=z^{-2}=(y^{-2}z)^2=(xzx^{-1})^2$.

Thus, the subgroup generated by $y,z$ has the relations $y=z^2yz^2, z^3=y^2z^{-1}y^2$. Moreover, the generator $x$ conjugates this subgroup into itself, by the homomorphism $\varphi(y)=z^2y^{-1}, \varphi(z)=y^{-2}z$. After a laborious computation, I think I can show that $\varphi$ is an automorphism of the two-generator group $\langle y,z | y=z^2yz^2, z^3=y^2z^{-1}y^2\rangle$, with inverse $\varphi^{-1}(y)=y^{-1}z^{-2}, \varphi^{-1}(z)=y^{-2}z$. Hence your group is an HNN extension of this group by the automorphism $\varphi$. Thus, any torsion must lie in the subgroup $\langle y,z | y=z^2yz^2, z^3=y^2z^{-1}y^2\rangle$.

I'm not sure how to show this subgroup is torsion-free (how do you show there is no odd torsion?). One may show that $y^2$ commutes with $z^2$ as a consequence of the first relator. The abelianization of this group is $(\mathbb{Z}/4)^2$. It might be worth looking at a presentation of the kernel to the abelianization.

  • $\begingroup$ One way to prove there is no odd torsion, shown to me by my friend Steve Humphries, is to show that there exists a normal subgroup of $G$ of index 32 (probably the kernel of the abelianization, but I might be wrong), which has a natural representation as a subgroup of the upper triangular matrices over $\mathbb{Q}$ with identity down the main diagonal (which are torsion-free). $\endgroup$ Jun 1, 2017 at 17:12
  • $\begingroup$ Ah, so it's virtually nil potent. Then you just have to see if the sequence splits. $\endgroup$
    – Ian Agol
    Jun 1, 2017 at 17:34

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