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$\newcommand{\orb}{\mathrm{orb}}$Let $T$ ($K$) be the torus (Klein bottle) with one cone point of order $q\geq 2$. The presentation of their orbifold fundamental groups are easy to find. Namely,

$$\pi_1^{\orb}(T)=\{a,b\ |\ (aba^{-1}b^{-1})^q=1\}.$$

$$\pi_1^{\orb}(K)=\{a,b\ |\ (aba^{-1}b)^q=1\}.$$

Want to know the centre of these two groups. Are they trivial?

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3 Answers 3

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These groups have trivial centers. As one proof, they are both fuchsian and so embed in $\mathrm{PSL}(2, \mathbb{R})$ (well, the orientation preserving subgroups do). However, elements of $\mathrm{PSL}$ that commute must have common fixed points on the "circle at infinity" of the hyperbolic plane.

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    $\begingroup$ Thanks a lot Sam! I was expecting it to be true. For more than one cone points case, I have a direct proof of the triviality of the centre, but this case was troubling. $\endgroup$
    – RKS
    Commented Nov 21, 2022 at 17:52
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The groups have trivial center, as pointed out by Sam Nead. Another, more combinatorial, way to show this is to apply the algorithm from [1], which decides whether any given one-relator group has a non-trivial center or not. If one wishes to be very explicit, one can even compute an explicit generating set for a finite index surface group inside the groups, see [2], particularly §5, where the Reidemeister-Schreier method is used to find a one-relator presentation on $4q-2$ generators for such a finite index subgroup.

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[1] Baumslag, G.; Taylor, Tekla, The centre of groups with one defining relator, Math. Ann. 175, 315-319 (1968). ZBL0157.34901.

[2] Baumslag, Gilbert; Troeger, Douglas, Virtually free-by-cyclic one-relator groups. I., Fine, Benjamin (ed.) et al., Aspects of infinite groups. A Festschrift in honor of Anthony Gaglione. Papers of the conference, Fairfield, USA, March 2007 in honour of Anthony Gaglione’s 60th birthday. Hackensack, NJ: World Scientific (ISBN 978-981-279-340-9/hbk). Algebra and Discrete Mathematics (Hackensack) 1, 9-25 (2008). ZBL1188.20023.

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    $\begingroup$ Thanks Carl for the detailed reply. I will look at the references. $\endgroup$
    – RKS
    Commented Nov 21, 2022 at 17:53
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    $\begingroup$ This is a sledgehammer to crack a nut! Plus note that Sam Nead's argument works verbatim for all hyperbolic orbifolds. $\endgroup$
    – HJRW
    Commented Nov 22, 2022 at 8:52
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    $\begingroup$ @HJRW Agreed on the sledgehammer (for the first part!), but it is such a nice paper that it is worth reading anyway :-) For the second part, it is nice to see more explicit generating sets, I think. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Nov 22, 2022 at 9:10
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    $\begingroup$ @Carl-FredrikNybergBrodda: Indeed! My point is that the judgement of both Gromov and posterity is that Dehn was right. I persist with this point because it's still a serious issue for the field: I regularly see people working on the combinatorial side exaggerating the difficulty of results that are trivial from a geometric point of view. $\endgroup$
    – HJRW
    Commented Nov 22, 2022 at 11:56
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    $\begingroup$ @HJRW Agreed! (For the most part, with the remainder coming down to my ignorance; I confess I may at times be one of those people you speak of, as much as I try to avoid it...). $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Nov 22, 2022 at 12:07
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The centre of any non-cyclic one-relator group with torsion is trivial.

Moreover,

the centraliser of any non-identity element of a one-relator group with torsion is cyclic.

This is B. B. Newman’s theorem (1973).

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    $\begingroup$ Thanks Anton for the precise reference. I find this more general fact interesting. $\endgroup$
    – RKS
    Commented Nov 23, 2022 at 15:11

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