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Consider the 2-generated metabelian group $G$ with finite $\mathbf{A}^2$-presentation

$$ \langle\langle a,b \,\vert\, [a,b^{-1}][a^{-1},b][a,b]^2 \rangle\rangle, $$

where $[a,b] = a^{-1}b^{-1}ab$, i.e. all relators of the forms $[[w,x],[y,z]]$ are included with the relators above. It can be shown that all finitely-generated metabelian groups have a so-called preferred presentation: a finite $\mathbf{A}^2$-presentation of the form

$$ \langle\langle x_1,\dots,x_n \,\vert\, R_1 \cup R_2 \rangle \rangle, $$

where $R_1$ and $R_2$ are finite sets of words of certain forms. If we call words Type A if they are of the form

$$ u = x_1^{m_1}\dots x_n^{m_n} $$

with $m_i \in \mathbb{Z}$, and Type B if they are of the form

$$ w = \prod_{1 \leq i < j \leq n} u_{ij}^{-1} [x_i,x_j] u_{ij} $$

with $u_{ij}$ of Type A, then $R_1$ consists of words $w$ of Type B, and $R_2$ consists of words $uw$ with $u$ of Type A and $w$ of Type B. The idea is that $R_2$ gives the relators of $G^{\text{ab}}$ (which is finitely-generated abelian, hence finitely-presented), while $R_1$ determines the relators of $[G,G]$ as a $\mathbb{Z}[G^{\text{ab}}]$-module (which is not necessarily finitely-generated as a group, but is as a module, hence finitely-presented as a module since $\mathbb{Z}[G^{\text{ab}}]$ is Noetherian). This blog post has a nice discussion of this.

The Theory of Infinite Soluble Groups by Lennox and Robinson discusses how to develop algorithms for finitely-generated metabelian groups using preferred presentations, building off the linked paper of Baumslag, Cannonito, and Robinson. In particular, Prop 9.5.1 states that any finite $\mathbf{A}^2$-presentation can be made into a preferred one by "collecting" the generators in the right order, moving them around using the fact that commutators commute.

However, I'm a bit lost on how to apply this fact to my group $G$, and indeed any finite $\mathbf{A}^2$-presentation that involves commutators of inverses of the generators. Using commutator identities lets me rewrite everything using just $[a,b]$:

$$ \langle \langle a,b \,\vert\, b[a,b]^{-1}b^{-1}a[a,b]^{-1}a^{-1}[a,b]^2 \rangle\rangle. $$

The problem here is that I cannot simplify this relator to be of Type B, i.e. in $R_1$, because it's not a single conjugate of $[a,b]$. The "proof" of Prop 9.5.1 does not say much beyond the account I gave above, and attempting to apply it to $G$ makes me doubt its validity. I must be missing something in the assumptions and/or commutator algebra. Any assistance would greatly appreciated.

kg583
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  • I don't know whether I can help with this, but you make it clear which of the two possible definitions of a commutator you are using: $[a,b] = a^{-1}b^{-1}ab$ or $aba^{-1}b^{-1}$? – Derek Holt May 24 '24 at 08:19
  • I think something is missing here (I looked at the paper/book, and I think both are using confusing notation). For example, if $G$ is the free abelian group of rank 3, what would a preferred presentation look like? The relators of Type B would need to involve $[a,b]$, $[a,c]$, and $[b,c]$ if I'm reading that right, and that makes no sense. – Steve D May 24 '24 at 08:19

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