The following problem arose when studying the same type of questions in Algebraic Geometry that led me to my previous question [MO379272](https://mathoverflow.net/questions/379272/combinatorial-problem-in-mathsfs-4).


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?
> 
> 
> 



Combinatorial problem in $G(32, \, 6)$
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 $G \setminus \Phi$.


