Conway's "Murder Weapon"

Question

The following quote is an excerpt from an interview with John Conway:

Coxeter came to Cambridge and he gave a lecture, then he had this problem for which he gave proofs for selected examples, and he asked for a unified proof. I left the lecture room thinking. As I was walking through Cambridge, suddenly the idea hit me, but it hit me while I was in the middle of the road. When the idea hit me I stopped and a large truck ran into me and bruised me considerably, and the man considerably swore at me. So I pretended that Coxeter had calculated the difficulty of this problem so precisely that he knew that I would get the solution just in the middle of the road. In fact. I limped back after the accident to the meeting. Coxeter was still there, and I said, "You nearly killed me." Then I told him the solution. It eventually became a joint paper. Ever since, I've called that theorem "the murder weapon." One consequence of it is that in a group if $a^2 = b^3 = c^5 =(abc)^{-1}$, then $c ^{610} =1$.

Does anyone know where one can find a statement of this theorem or the publication information on the paper Conway and Coxeter wrote?

This quote is from Math. Intelligencer 23 (2001), no. 2,pp.8-9.

obligatory xkcd comic – Tasos Papastylianou Sep 29 '17 at 06:35 — Tasos Papastylianou, Sep 29 '17 at 06:35
No offense, but mathematicians should be worth less points. – Little Endian Sep 29 '17 at 15:50 — Little Endian, Sep 29 '17 at 15:50

score 31 · Accepted Answer · edited Jun 12 '20 at 10:38

According to Siobhan Roberts (2006) King of Infinite Space: Donald Coxeter, the Man Who Saved Geometry p.359 (Google Books):

This is a simple statement of Coxeter's “Murder Weapon”: If “$A^p=B^q=C^r=ABC=1$” defines a finite group, then “$A^p=B^q=C^r=ABC=Z$” implies $Z^2=1$. Conway; and Conway, interview, November 26, 2005; Conway with Coxeter and G. C. Shepherd, “The Centre of a Finitely Generated Group,” in Tensor, 1972 (contains the proof for “The Murder Weapon”).

The latter reference apparently means this citation: http://www.ams.org/mathscinet-getitem?mr=333001

MR333001 20F05

Conway, J. H.; Coxeter, H. S. M.; Shephard, G. C. The centre of a finitely generated group. Commemoration volumes for Prof. Dr. Akitsugu Kawaguchi's seventieth birthday, Vol. II. Tensor (N.S.) 25 (1972), 405–418; erratum, ibid. (N.S.) 26 (1972), 477.

Review: https://zbmath.org/0236.20029, written by Coxeter:

The polyhedral group $(l,m,n)$, defined by $a^l=b^m=c^n=abc=1$, is finite if and only if $g>0$, where $g=2(l^{-1}+m^{-1}+n^{-1}-1)^{-1}$, and then its order is $g$. The same condition for finiteness applies also to the binary polyhedral group $\langle l,m,n\rangle$, defined by $a^l=b^m=c^n=abc$, whose order is $2g$. In fact, the central element $z=a^l=b^m=c^n=abc$ is of period $2$. This last statement is not at all obvious. The first step is to observe that the Cayley diagram for $(l,m,n)$ is identical with the coset diagram for $\langle l,m,n\rangle$ relative to the subgroup generated by $z$. This diagram covers the inversive plane, or the $2$-sphere, with a "map" which consists of $g/l$ $l$-gons representing $a^l$, $g/m$ $m$-gons representing $b^m$, $g/n$ $n$-gons representing $c^n$, and $g$ negatively oriented triangles representing $abc$. The procedure is to compare the results of shrinking the peripheral circuit (or any other circuit) to a single point (or "point-circuit") in two distinct ways, corresponding to the two discs into which the circuit decomposes the $2$-sphere. More symmetrically, any point-circuit (such as the point at infinity of the inversive plane) may be continuously deformed into another point-circuit by contracting over all the oriented polygons in the Cayley diagram. The same procedure is applied in more complicated cases. For instance, in the group for which $z$ is central while $a^l=z^p$, $b^m=z^q$, $c^n=z^r$, $abc=z^s$ and $g$ (as defined above) is a positive integer, the period of $z$ is $g|pl^{-1}+qm^{-1}+rn^{-1}-s|$.

score 13 · Answer 2 · answered Sep 29 '17 at 03:53

13

A Google search for "Conway and Coxeter murder" came up with the book "King of Infinite Space: Donald Coxeter, the Man Who Saved Geometry".

A search via Google inside this book came up with this on page 359, note 14:

This is a simple consequence of Coxeter's "murder weapon": "If $a^p = b^q = c^r = abc = 1$ defines a finite group, then $a^p=b^q=c^r = abc = z$ implies $z^2 = 1$."

The reference is Conway, Coxeter, and Shepherd, Tensor, 1972.

answered Sep 29 '17 at 03:53

marty cohen

110,450

2

Correct me if I'm wrong, but wouldn't $a^p = b^q = c^r = abc = 1$ and $a^p=b^q=c^r = abc = z$ together imply $z = 1$ trivially? – Arthur Sep 29 '17 at 07:07
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@Arthur The statement means if the group with presentation $\langle a,b,c,|a^p=b^q=c^r=abc=1\rangle$ is finite, then for any group $G$ and for $w,x,y,z \in G$ if $w^p=x^q=wxy=z$, then $z^2=1$. – Nex Sep 29 '17 at 07:15
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@Nex Aah, so the two equalities are not in the same group. A bit confusing that they're using the same letters, but sure. At least that explains why it's not trivial. – Arthur Sep 29 '17 at 07:22
When you wrote "This is a simple consequence", did you mean to write "This is a simple statement"? – Sophie Swett Sep 29 '17 at 16:24
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I was quoting the book. – marty cohen Sep 29 '17 at 16:33

Conway's "Murder Weapon"

2 Answers2