Torsion-freeness of two groups with 2 generators and 3 relators and Kaplansky Zero Divisor Conjecture 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! 
$$(1+a+b)(b^{-1}a^{-2}ba^{-1}+a^{-1}ba^{-2}ba^{-1}+a^{-1}ba^{-1}+b^{-1}a^{-1}+a^{-1}b^2a^{-1}ba^{-1}+aba^{-1}ba^{-1}+1+a^{-2}ba^{-1}+a^{-1}b^{-1}a^{-1}+b+baba^{-1}ba^{-1}+ba^{-1}ba^{-1}+a^{-1}+b^{-1}a)=0$$
$$(1+a+b)(aba^{-1}ba^{-1}+a^{-1}b^2a^{-1}ba^{-1}+a^{-1}ba^{-1}+b^{-1}a^{-1}+b^{-1}a^{-2}ba^{-1}+ab+1+ba^{-1}ba^{-1}+baba^{-1}ba^{-1}+b+bab+a^{-2
}ba^{-1}+a^{-1}+b^{-1}a)=0$$

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.   
