Timeline for Is G(4,7) a Coxeter group
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Nov 20, 2023 at 4:57 | comment | added | Shijie Gu | @ThomasGobet The center is finite since it's a hyperbolic group. But we don't know if it's actually centerless. | |
| Oct 9, 2023 at 11:34 | comment | added | Thomas Gobet | Do you have any information about the center of your group ? Since the abelianization is $\mathbb{Z}/2$, if it is a Coxeter group it must be irreducible, hence have trivial center since it is infinite. So if you find a nontrivial central element, you're done... | |
| Oct 7, 2023 at 12:47 | comment | added | Shijie Gu | @MoisheKohan It's not trivial. One method employed by my friend Jiming Ma involved setting $s_1 = ab$, $s_2 = bc$, and $s_3 = ca$. The resulting presentation denoted $S(4,7)$ is a subgroup of $G(4,7)$ with index 2. Upon computation, either done manually (as Jiming initially did) or through Magma $S(4,7)$ is hyperbolic. Alternatively, Derek Holt wrote a program in GAP. See mathoverflow.net/questions/238740/…. Derek's program can be used directly to confirm that $G(4,7)$ is hyperbolic. | |
| Oct 6, 2023 at 18:54 | comment | added | Moishe Kohan | How do you know that your group is hyperbolic? | |
| Oct 6, 2023 at 3:36 | history | edited | Shijie Gu | CC BY-SA 4.0 |
Add some motivations and backgrounds of the question
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| Oct 6, 2023 at 3:04 | comment | added | Shijie Gu | @NathanReading I have the same feeling that this might not be a Coxeter group. Just lacking some tools to disprove it... | |
| Oct 5, 2023 at 20:06 | comment | added | Nathan Reading | @ShijieGu I agree with Derek Holt that it seems unlikely that this is a Coxeter group. Do you have some reason to hope/suspect that it is? Or some idea what the simple reflections would be? | |
| Oct 5, 2023 at 13:51 | comment | added | Derek Holt | @DaveBenson Yes that's right. It's the only subgroup of index $8$ up to conjugacy. | |
| Oct 5, 2023 at 13:45 | comment | added | Dave Benson | @DerekHolt Presumably this is the point stabiliser for the homomorphism to $\mathrm{PGL}(2,7)$. | |
| Oct 5, 2023 at 13:35 | comment | added | Derek Holt | An easy computation shows that there is a subgroup of index $8$ with infinite abelianization, so it is certainly not a torsion group. It seems unlikely that it is a Coxeter group. | |
| Oct 5, 2023 at 13:29 | comment | added | Dave Benson | More finite quotients include $\mathop{\rm PSL}(3,8)$, $\mathop{\rm PSL}(3,13)$, $\mathop{\rm PSL}(3,29)$, $\mathop{\rm PSU}(3,27)$. | |
| Oct 5, 2023 at 13:17 | comment | added | Tom De Medts | The paper you are referring to contains the statement that $\Gamma(4,7)$ contains a (non-trivial) finite index torsion-free subgroup, so this provides a negative answer to your second question. I strongly suspect that the answer to the first question is also negative. | |
| Oct 5, 2023 at 12:57 | history | edited | Shijie Gu | CC BY-SA 4.0 |
added 49 characters in body
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| Oct 5, 2023 at 12:55 | comment | added | Shijie Gu | @TomDeMedts Yes, thanks. I'll make the change. | |
| Oct 5, 2023 at 11:55 | comment | added | Dave Benson | The matrices in $\mathop{\rm PSL}(2,71)$ for $a$, $b$, $c$ are $(0 & 1 \\ 70 & 0)$, $(13 & 18 \\ 30 & 58)$ and $(61 & 13 \\ 25 & 10)$. | |
| Oct 5, 2023 at 11:46 | comment | added | Dave Benson | Interestingly, this group seems to have $\mathop{\rm PSL}(2,71)$ and $\mathop{\rm PGL}(2,7)$ as quotients. | |
| Oct 5, 2023 at 10:49 | comment | added | Tom De Medts | @Carl-FredrikNybergBrodda Sure, I did the justification in my head before posting my comment ;-) It's easy enough: if $a=1$, then $(cbc)^7=1$ implies $b^7=1$ so also $b=1$, and then also $c=1$, so everything would collapse. But it can't, because the abelianization of the group is non-trivial (it has order $2$). | |
| Oct 5, 2023 at 10:43 | comment | added | Carl-Fredrik Nyberg Brodda | @TomDeMedts Note, however, that the fact that $a$, $b$, and $c$ have order $2$ requires some (easy) justification beyond reading the relators (they could, in principle, all have order $1$!). But the torsion point still stands of course as the trivial group is and has torsion. Also, the existence of finitely presented torsion groups is a famous open problem, and this is definitely not such a group. | |
| Oct 5, 2023 at 10:32 | comment | added | Tom De Medts | About the second question: the elements $a$, $b$ and $c$ are involutions, so surely the group has torsion. Did you perhaps intend to ask whether the group is torsion, in the sense that every element has finite order? | |
| Oct 5, 2023 at 10:16 | history | edited | Martin Sleziak |
edited tags
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| Oct 5, 2023 at 10:14 | history | asked | Shijie Gu | CC BY-SA 4.0 |