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Let $G(4, 7)$ be an abstract group with the presentation $$\langle a,b,c | a^2 = b^2 = c^2 = 1, (ab)^4 = (bc)^4 = (ca)^4 = 1, (acbc)^7 = (baca)^7 = (cbab)^7 = 1 \rangle $$ Richard Schwartz considered a representation of $G(4,7)$ into $PU(2,1)$ and extensively studied the image group denoted $\Gamma(4,7)$, for instance, the topology of the complex hyperbolic surface $H^{2}_{\mathbb{C}}/\Gamma(4, 7)$. But I'm more curious about $G(4,7)$ itself.

Is $G(4,7)$ a Coxeter group?

Whether $G(4,7)$ is a torsion group in which every element has finite order?

Edit: The second question has been answered in negative thanks to comments from Tom and Derek. I'd like to offer further context regarding the first question. Some topologists including me are interested in the end behavior of $G(4,7)$, for instance, is its boundary a Menger curve? Note that Jordane Granier in 2015 dissertation considered a similar group $$\langle \iota_1,\dots, \iota_6|\iota_{i}^{3} = id, \iota_{i}\iota_{i+1} = \iota_{i+1}\iota_{i}, i\in \mathbb{Z}/6\mathbb{Z} \rangle$$ It has been shown that this is a hyperbolic group with a Menger curve boundary. This was proved using the classic characterization of hyperbolic group boundaries by Kapovich-Kleiner. A pivotal aspect of Granier's proof involved geometrizing the group with the aid of Dymara-Osajda's building blocks. Given that $G(4,7)$ is hyperbolic, it seem natural to turn to Kapovich-Kleiner's characterization. Yet, an appropriate geometrization is necessary. This leads us to the first question, which, if answered affirmatively, would simplify the task. We do suspect it cannot be true, however, so far we don't know how to disprove it.

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    $\begingroup$ 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? $\endgroup$
    – Tom De Medts
    Commented Oct 5, 2023 at 10:32
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    $\begingroup$ @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. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Oct 5, 2023 at 10:43
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    $\begingroup$ @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 $2$). $\endgroup$
    – Tom De Medts
    Commented Oct 5, 2023 at 10:49
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    $\begingroup$ Interestingly, this group seems to have $\mathop{\rm PSL}(2,71)$ and $\mathop{\rm PGL}(2,7)$ as quotients. $\endgroup$
    – Dave Benson
    Commented Oct 5, 2023 at 11:46
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    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Commented Oct 5, 2023 at 13:35

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