Let $G_1$ and $G_2$ be the groups with the following presentations:

$$G_1=\langle a,b \;|\; (ab)^2=a^{-1}ba^{-1}, (a^{-1}ba^{-1})^2=b^{-2}a, (ba^{-1})^2=a^{-2}b^2 \rangle,$$

$$G_2=\langle a,b \;|\; ab=(a^{-1}ba^{-1})^2, (b^{-1}ab^{-1})^2=a^{-2}b, (ba^{-1})^2=a^{-2}b^2 \rangle,$$

Are these groups torsion-free?

Motivation: In both of these groups $1+a+b$ as an element of the group algebra $\mathbb{F}_2[G_i]$ over the field with two elements is a zero divisor. Thus one has a counterexample for the Kaplansky zero divisor conjecture if one of $G_i$s is torsion-free!


$$(1+a+b)(aba^{-1}ba^{-1}+a^{-1}b^2a^{-1}ba^{-1}+a^{-1}ba^{-1}+b^{-1}a^{-1}+b^{-1}a^{-2}ba^{-1}+ab+1+ba^{-1}ba^{-1}+baba^{-1}ba^{-1}+b+bab+a^{-2 }ba^{-1}+a^{-1}+b^{-1}a)=0$$

  • $\begingroup$ What is the zero divisor-mate of 1+a+b? $\endgroup$ Commented Feb 23, 2016 at 14:09
  • $\begingroup$ What is the structure of $G/G'$? Is it torsion free? $\endgroup$ Commented Feb 23, 2016 at 14:17
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    $\begingroup$ @AliTaghavi. I will shortly write the mates. Those are of length 14. $\endgroup$ Commented Feb 23, 2016 at 14:35
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    $\begingroup$ The computation of abelianization is an immediate computation (if I'm correct it's $Z\times Z/11Z$ for the first and $Z\times Z/21Z$ for the second), but I don't think it helps much. $\endgroup$
    – YCor
    Commented Feb 23, 2016 at 14:52
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    $\begingroup$ I am not aware of any algorithmic methods for attempting to verify that a group defined by a finite presentation is torsion-free (unless the group can be proved to be nilpotent). I would be interested to learn about it if there are any, even if they only work in restricted circumstances. $\endgroup$
    – Derek Holt
    Commented Feb 23, 2016 at 15:29

2 Answers 2


Denote $x=ab$, $y=a^{-1}ba^{-1}$. Then the first two relations of the first group are $x^2=y$, $y^2=(yx)^{-1}$. This implies $x^4=x^{-3}$ or $x^7=1$. So the group has torsion. I leave the second group as an exercise for the others.

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    $\begingroup$ Excellent! Of course you also need to show that $x \ne 1$, which could be difficult in general, but fortunately its image in $G/[G,G]$ is nontrivial. $\endgroup$
    – Derek Holt
    Commented Feb 23, 2016 at 18:09

Similar to the answer of Mark Sapir, for the second group let $x=ba$ and $y=b^{-1}ab^{-1}$. From the second relation we have $y^2=x^{-1}y^{-1}$. So, from the first relation we have $x=y^4$. These relations implies that $y^4=y^{-3}$ or $y^7=1$.


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