Combinatorial problem in $G(32, \, 6)$

Question

The following problem arose when studying the same type of questions in Algebraic Geometry that led me to my previous question MO379272.

Let us consider the group $G$ of order $32$ whose label in GAP4 database is $G(32, \, 6)$. It is a semi-direct product of the form $(\mathbb{Z}_2)^3 \rtimes \mathbb{Z}_4$, whose presentation is

\begin{equation*} \begin{split} G=\langle x, \, y, \, z, \, w \; \mid \; & x^2 = y^2 = z^2 = w^4 = 1, \\ &[x, \, y]=1, \, [x, \, z]=1, \, [y,\, z]=1, \\ &wxw^{-1} = x, \, wyw^{-1} = xy, \, wzw^{-1} = yz \rangle \end{split} \end{equation*} It has nilpotency class $3$; its center is $Z(G)=\langle x \rangle \simeq \mathbb{Z}_2$ and its derived subgroup is $G'=\langle x, \, y \rangle \simeq \mathbb{Z}_2 \times \mathbb{Z}_2$.

I am looking for ordered strings of eight elements in $G$ $$\mathfrak{S} = (\mathrm{r}_{11}, \, \mathrm{t}_{11}, \, \mathrm{r}_{12}, \, \mathrm{t}_{12}, \, \mathrm{r}_{21}, \, \mathrm{t}_{21}, \, \mathrm{r}_{22}, \, \mathrm{t}_{22} )$$ such that the following relations are verified: $$[\mathrm{r}_{1j}, \, \mathrm{t}_{2k}]=[\mathrm{t}_{1j}, \, \mathrm{r}_{2k}]=x^{\delta_{jk}}, \quad [\mathrm{r}_{1j}, \, \mathrm{r}_{2k}]=[\mathrm{t}_{1j}, \, \mathrm{t}_{2k}]=1 \quad (\ast)$$ for all $j, \, k \in \{1, \, 2\}$. Using GAP4, I can prove the following

Proposition. The group $G$ admits no string $\mathfrak{S}$ satisfying $(\ast)$.

So far, I was unable to reprove this result by hand; as in my previous question, I am stuck in a jungle of case-by-case considerations, possibly because I am not really a specialist in Group Theory and something easy escapes me. So let me ask the

Question. What is a way to prove the Proposition above without using the computer?

Answer 1

Here is a proof by contradiction. Assume an $8$-tuple $E$ from $G$ satisfies all of your relations.

Let $\Phi=\langle w^2,x,y \rangle \leq G$. It is straightforward to observe that

(A) $\Phi$ is abelian.

With a little more work, one can see that, for each $g \in G$,

(B) the centralizer $C_G(g)$ is nonabelian if and only if $g \in \Phi$.

As some pairs from the $8$-tuple $E$ do not commute, at least one element of $E$ does not lie in $\Phi$. WLOG, $r_{11} \not\in \Phi$. So, $C_G(r_{11})$ is abelian. Therefore, $[t_{22},r_{21}]=1$. As $[t_{22},t_{11}]=1$ and $[t_{11},r_{21}]=x$, we see that $C_G(t_{22})$ is not abelian.

Similar arguments show that $C_G(r_{21})$ and $C_G(r_{22})$ are not abelian. So, all of $t_{22}$, $r_{21}$, and $r_{22}$ lie in $\Phi$. It follows from (A) that none of $r_{12}$, $t_{11}$ or $t_{12}$ lies in $\Phi$.

As $C_G(r_{12})$ is abelian, we see that $[t_{21},r_{21}]=1$. As $[t_{21},t_{11}]=1$ and $[r_{21},t_{11}]=x$, we see that $C_G(t_{21})$ is nonabelian and so $t_{21} \in \Phi$.

We know now that all of $r_{21}$, $r_{22}$, $t_{21}$ and $t_{22}$ lie in $\Phi$, and none of these elements lies in $Z(G)=\langle x \rangle$. It follows that there is some $g \in \Phi \setminus Z(G)$ such that two of the four elements under consideration both lie in $\{g,gx\}$ and therefore have the same centralizer. The relations you gave show that this is impossible - each of the four elements from your $8$-tuple that are in $\Phi$ fails to commute with exactly one of the four elements in $G \setminus \Phi$, and no two in $\Phi$ commute with the same element in $G \setminus \Phi$.

Answer 2

For any commutative ring $R$ and $u,a\in R$ with $u$ invertible we have a permutation $\theta_{ua}(x)=ux+a$. These form a group $\text{Aff}(R)$ with $\theta_{ua}\theta_{vb}=\theta_{uv,a+ub}$. Your group is isomorphic to $\text{Aff}(R)$ with $R=(\mathbb{Z}/2)[t]/t^3$ (by taking $x=\theta_{1,t^2}$ and $y=\theta_{1,t}$ and $z=\theta_{1,1}$ and $w=\theta_{1+t,1}$). Elements $\theta_{ua}$ and $\theta_{vb}$ commute iff $(1+v)a=(1+u)b$. This point of view should make the analysis easier.