Combinatorial problem in $G(54, \, 5)$ - Reprise

Question

This post is a follow-up to my previous post MO479127. I am trying to concentrate on a subset of relations, hoping to find some structure on the set of solutions that explains why the whole set of relations in the aforementioned post has no solutions.

I am working on a problem in Combinatorial Group Theory related to the construction of some special fibred surfaces in Algebraic Geometry via the representations of some braid group in genus $2$. I would like to have a conceptual proof of the fact described below.

Let us consider the group $G=G(54, \, 5) = (C_3 \times C_3) \rtimes C_6$, whose presentation is $$G=\langle a, \, b, \, x \; | \; a^3=b^3=x^6=1, \, [a, \, b]=1, \, xax^{-1} =a^{-1}b^{-1}, \, xbx^{-1}=b^{-1} \rangle$$

This group is monolithic, namely, the intersection $\operatorname{Mon}(G)$ of all non-trivial, normal subgroups is different from $1$; in fact, we have $\operatorname{Mon}(G) = \Phi(G)=\langle b \rangle \simeq C_3$.

I am looking for ordered $4$-tuples $$(\mathsf{t}_{11}, \, \mathsf{r}_{12}, \, \, \mathsf{r}_{22}, \, \mathsf{t}_{22})$$ of non-trivial elements in $G$, where the following relations hold: \begin{equation} \renewcommand{\t}{\mathsf{t}} \renewcommand{\r}{\mathsf{r}} \newcommand{\z}{\mathsf{z}} (R_6) \; [\r_{12}, \, \r_{22}]=1 \\ (T_1) \; [\t_{11}, \, \r_{22}]=1 \\ (T_3) \; [\t_{11}, \, \t_{22}]=1 \\ (R_8) \; [\r_{12}, \, \t_{22}]=b^{-1} \\ \end{equation} Doing computations with GAP4, I verified the following

Proposition 1. There are precisely 10908 such $4$-tuples.

I tried to give a proof of this result by hand, but I soon got stuck in some messy and not-so-enlightening case-by-case analysis. I hope that there is some smart method based on standard group theory techniques to verify the computation, so let me ask the

Question. Is there a clever way to prove the proposition above without using the computer?

Answer 1

Set $V=\mathbb F_3^2$, $$ p=\pmatrix{0\\-1}, \; q=\pmatrix{1\\0}\in V, \quad A=\pmatrix{-1&1\\0&-1}. $$ Then the subgroup in $AGL(V)$ generated by $a\colon v\mapsto v+p$, $b\colon v\mapsto v+q$, $x\colon v\mapsto Ax$ is isomorphic to $G$, so we may assume $G$ is that subgroup.

Set $U=\langle q\rangle$. Each element in $G$ is uniquely determined as $v\mapsto A^kv+r$, where $0\leq k\leq 6$ and $r\in V$, so we denote such element by $(k,r)$. Now the following conditions for elements $c=(k,r)$, $d=(\ell,s)$ are easy to check:

$$ [c,d]=1 \iff A^k(A^\ell v+s)+r=A^\ell(A^k v+r)+s \iff (A^k-I)s=(A^\ell-I)r, $$ which is easy to descrbe for every certain pair (k,\ell)$;

$$ [c,d]=b^{-1}\iff A^k(A^\ell v+s)+r=A^\ell(A^k (v-q)+r)+s \iff (A^k-I)s=(A^\ell-I)r-A^\ell q, $$ which is also not bad.

I hope now the computation is easy, at least as the sum over all pairs $(k,\ell)$.