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) ]
- $\begingroup$Can you post the list of the $7$-uples here?$\endgroup$Alex– Alex2020-12-18 21:16:42 +00:00CommentedDec 18, 2020 at 21:16
- $\begingroup$@Alex: I added the list in the question$\endgroup$Francesco Polizzi– Francesco Polizzi2020-12-18 21:26:48 +00:00CommentedDec 18, 2020 at 21:26
- 3$\begingroup$Yes, of course I know this.$\endgroup$Francesco Polizzi– Francesco Polizzi2020-12-18 21:53:14 +00:00CommentedDec 18, 2020 at 21:53
- 2$\begingroup$Is there any way you can show first that the seven elements together generate a Sylow $2$-subgroup of $S_4$ in which $z$ is central? Then relations R5 and R10 become vacuous, and the right sides of R7 and R9 become trivial. Now all of your relations tell you that some pairs of elements commute, and some do not.$\endgroup$John Shareshian– John Shareshian2020-12-19 07:01:28 +00:00CommentedDec 19, 2020 at 7:01
- 1$\begingroup$Here's a simple observation (which might go towards working out that these elements generate a $2$-Sylow subgroup): In $S_4$, the centralizer of any order $3$ element is simply the cyclic subgroup generated by it. So if any of the $r_{ij}$ or $t_{22}$ were of order $3$, they all would lie in the same $3$-Sylow subgroup, and contradict $R_8$.$\endgroup$Achim Krause– Achim Krause2020-12-19 13:06:37 +00:00CommentedDec 19, 2020 at 13:06
1 Answer1
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$C_{S_4}(z)$.)
- $\begingroup$Thanks for the nice answer. By "$2$-element" you mean an element whose order is a power of $2$, not necessarily an involution, right?$\endgroup$Francesco Polizzi– Francesco Polizzi2020-12-20 17:07:44 +00:00CommentedDec 20, 2020 at 17:07
- 2$\begingroup$That's correct. Every element of $S_4$ either has order three or is a $2$-element, in the language I am using.$\endgroup$John Shareshian– John Shareshian2020-12-20 17:46:35 +00:00CommentedDec 20, 2020 at 17:46
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