I am working on a problem in Combinatorial Group Theory related to a construction in Algebraic Geometry, and I would like to have a conceptual proof of the fact described below.


I am looking for ordered $7$-uples $$(\mathsf{r}\_{11}, \, \mathsf{r}\_{12}, \, \mathsf{r}\_{21}, \, \mathsf{r}\_{22}, \, \mathsf{t}\_{21}, \, \mathsf{t}\_{22}, \, \mathsf{z})$$ of non-trivial elements in the symmetric group $\mathsf{S}\_4$, where $\mathsf{z}$ belongs to the Klein subgroup $\mathsf{V}\_4$ and the following relations hold:
\begin{equation} 
\renewcommand{\t}{\mathsf{t}}
\renewcommand{\r}{\mathsf{r}}
\newcommand{\z}{\mathsf{z}}
\begin{aligned}
(R\_1) & \, \, [\r\_{11}, \, \r\_{22}]=1 & (R\_6) & \, \,
[\r\_{12}, \, \r\_{22}]=1 \\
(R\_2) & \, \, [\r\_{11}, \, \r\_{21}]=1 &
(R\_7) & \, \, [\r\_{12}, \, \r\_{21}]=
\z^{-1}\,\r\_{21}\,\r\_{22}^{-1}\,\z\,\r\_{22}\,\r\_{21}^{-1} \\
(R\_3) & \, \, [\r\_{11}, \, \t\_{22}]=1 & (R\_8) & \, \,
[\r\_{12}, \, \t\_{22}]=\z^{-1} \\
(R\_4) & \, \, [\r\_{11}, \, \t\_{21}]=\z^{-1} & (R\_9) & \, \,
[\r\_{12}, \, \t\_{21}]=[\z^{-1}, \, \t\_{21}] \\
(R\_5) & \, \, [\r\_{11}, \, \z]=[\r\_{21}^{-1}, \, \z] & (R\_{10})
& \, \, [\r\_{12}, \, \z]=[\r\_{22}^{-1}, \, \z] \\
\end{aligned}
\end{equation}
Doing computations with GAP4, I obtained the



> 
> **Proposition.** There are precisely $24$ such $7$-uples, up to the natural action of $\mathrm{Aut}(\mathsf{S}\_4)$.
> 
> 
> 


I tried to give a proof of this result by hand, but I soon got stuck in some endless case-by-case analysis. I guess that there is some clever method to do the computation, so let me ask the



> 
> **Question 1.** Is there a short way to prove the proposition above without using the computer?
> 
> 
> 


**Edit 1.** As remarked by John Shareshian in the comments, in all cases the elements of the $7$-uple generate a subgroup of $\mathsf{S}\_4$ isomorphic to $\mathsf{D}\_8$ (namely, a $2$-Sylow) in which $\z$ is central.



> 
> **Question 2.** Is there a short way to prove this without using the computer?
> 
> 
> 


**Edit 2.** Here is a list of representatives for each of the 24 classes, generated by GAP4.


[ (3,4), (1,2)(3,4), (1,2)(3,4), (3,4), (1,4,2,3), (1,4,2,3), (1,2)(3,4) ]


[ (3,4), (1,2)(3,4), (1,2)(3,4), (3,4), (1,3)(2,4), (1,3)(2,4), (1,2)(3,4) ]


[ (3,4), (1,2)(3,4), (1,2)(3,4), (3,4), (1,3)(2,4), (1,4)(2,3), (1,2)(3,4) ]


[ (3,4), (1,2)(3,4), (1,2)(3,4), (3,4), (1,3,2,4), (1,4,2,3), (1,2)(3,4) ]


[ (3,4), (1,2)(3,4), (1,2)(3,4), (1,2), (1,3)(2,4), (1,3)(2,4), (1,2)(3,4) ]


[ (3,4), (1,2)(3,4), (1,2)(3,4), (1,2), (1,3)(2,4), (1,4)(2,3), (1,2)(3,4) ]


[ (3,4), (1,2)(3,4), (1,2)(3,4), (1,2), (1,3,2,4), (1,3,2,4), (1,2)(3,4) ]


[ (3,4), (1,2)(3,4), (1,2)(3,4), (1,2), (1,3,2,4), (1,4,2,3), (1,2)(3,4) ]


[ (1,2)(3,4), (1,3)(2,4), (1,3)(2,4), (1,2)(3,4), (2,4), (2,4), (1,3)(2,4) ]


[ (1,2)(3,4), (1,3)(2,4), (1,3)(2,4), (1,2)(3,4), (2,4), (1,3), (1,3)(2,4) ]


[ (1,2)(3,4), (1,3)(2,4), (1,3)(2,4), (1,2)(3,4), (1,2,3,4), (1,2,3,4), (1,3)(2,4) ]


[ (1,2)(3,4), (1,3)(2,4), (1,3)(2,4), (1,2)(3,4), (1,2,3,4), (1,4,3,2), (1,3)(2,4) ]


[ (1,2)(3,4), (1,3)(2,4), (1,3)(2,4), (1,4)(2,3), (2,4), (2,4), (1,3)(2,4) ]


[ (1,2)(3,4), (1,3)(2,4), (1,3)(2,4), (1,4)(2,3), (2,4), (1,3), (1,3)(2,4) ]


[ (1,2)(3,4), (1,3)(2,4), (1,3)(2,4), (1,4)(2,3), (1,2,3,4), (1,2,3,4), (1,3)(2,4) ]


[ (1,2)(3,4), (1,3)(2,4), (1,3)(2,4), (1,4)(2,3), (1,2,3,4), (1,4,3,2), (1,3)(2,4) ]


[ (1,2,3,4), (1,3)(2,4), (1,3)(2,4), (1,2,3,4), (2,4), (2,4), (1,3)(2,4) ]


[ (1,2,3,4), (1,3)(2,4), (1,3)(2,4), (1,2,3,4), (2,4), (1,3), (1,3)(2,4) ]


[ (1,2,3,4), (1,3)(2,4), (1,3)(2,4), (1,2,3,4), (1,2)(3,4), (1,2)(3,4), (1,3)(2,4) ]


[ (1,2,3,4), (1,3)(2,4), (1,3)(2,4), (1,2,3,4), (1,2)(3,4), (1,4)(2,3), (1,3)(2,4) ]


[ (1,2,3,4), (1,3)(2,4), (1,3)(2,4), (1,4,3,2), (2,4), (2,4), (1,3)(2,4) ]


[ (1,2,3,4), (1,3)(2,4), (1,3)(2,4), (1,4,3,2), (2,4), (1,3), (1,3)(2,4) ]


[ (1,2,3,4), (1,3)(2,4), (1,3)(2,4), (1,4,3,2), (1,2)(3,4), (1,2)(3,4), (1,3)(2,4) ]


[ (1,2,3,4), (1,3)(2,4), (1,3)(2,4), (1,4,3,2), (1,2)(3,4), (1,4)(2,3), (1,3)(2,4) ]



Combinatorial problem in $\mathsf{S}_4$
Please allow me to substitute as follows in order to avoid typos: $r\_{11}=a$, $r\_{12}=b$, $r\_{21}=c$, $r\_{22}=d$, $t\_{21}=e$, $t\_{22}=f$.


Upon examining the character tables of $S\_4$, $A\_4$, and $D\_8$, we see that there are $56$ pairs $(x,y)$ from $S\_4$ such that $[x,y]=z$, $24$ such pairs from $C\_{S\_4}(z)$ and $32$ such pairs in $A\_4$. Every such pair in $A\_4$ includes at least one $3$-cycle, while $C\_{S\_4}(z) \cong D\_8$ is a $2$-group. We conclude that if $[x,y]=z$ and neither $x$ nor $y$ has order three, then both $x$ and $y$ centralize $z$.


Since a $3$-cycle generates its own centralizer in $S\_4$, it follows from R1,R2,R3 and R6 that if any of $a,b,c,d,f$ is a $3$-cycle, then all of these elements generate the same group. This is impossible by R8. So, all of $a,b,c,d,f$ are $2$-elements.


It follows now from R8 that $b,f$ lie in $C\_{S\_4}(z)$. Also $a \in C\_{S\_4}(z)$. Indeed, if $a \in A\_4$ then $a$ is conjugate to $z$, and otherwise, by R4, both $a$ and $e$ lie in $C\_{S\_4}(z)$.


Using R5, we see that $ac \in C\_{S\_4}(z)$, from which it follows that $c \in C\_{S\_4}(z)$. Similarly, by R10, $d \in C\_{S\_4}(z)$.


By R9, the $2$-element $bz$ centralizes $e$. So, if $|e|=3$ then $b=z$, which contradicts R8. Therefore, $e$ is a $2$-element and it follows from R4 that $e \in C\_{S\_4}(z)$.


We see now that every $7$-tuple from $S\_4$ satisfying the given conditions generates a Sylow $2$-subgroup $C\_{S\_4}(z)$. As mentioned in my comment above, this allows one to simplify the relations, and counting should be not so hard thereafter.


Added later: Indeed, one gets a set of orbit representatives as follows. First fix $z$. Let $P,Q,R$ be the noncentral conjugacy classes in $C\_{S\_4}(z)$. Choose representatives $p,q,r$ for the respective classes. Now choose any one of six ordered pairs $(a,b)$ of distinct elements of $\{p,q,r\}$. For any such choice, one gets four orbit representatives $(a,b,z,z,b,a,z)$, $(a,b,z,z,bz,a,z)$, $(a,b,z,z,b,az,z)$, and $(a,b,z,z,bz,az,z)$. (Indeed, in the previous notation, $a$ and $b$ cannot be central in $C\_{S\_4}(z)$, as $[a,e]=[b,f]=z$. Moreover, $[a,b] \neq 1$ since otherwise $b \in \{a,az\}$ and $[a,e]=[b,e]$, contradicting R4 and R9. The rest follows from the adjusted relations mentioned in the comment above and the fact that the two non-commuting elelemnts $a,b$ together generate $C\_{S\_4}(z)$.)


