I was reading about the monster group, and how hard it was to do calculations in it, and I wondered: Is there a known presentation of the monster group? I know that it is a hurwitz group, but other than that I don't know. If we have two generators a and b such that $a^2=b^3=(ab)^7=1$, what are the possible orders of $[a,b]$? I believe that the conjugacy classes are known, so if you are given the conjugacy classes of two elements x and y, can we determine the possible conjugacy classes of $xy$?
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3 Answers
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There's a 12-generator 80-relator presentation for the Monster group. Specifically, we have 78 relators for the Coxeter group Y443:
- $12$ relators of the form $x^2 = 1$, one for each node in the Coxeter-Dynkin diagram;
- $11$ relators of the form $(xy)^3 = 1$, one for each pair of adjacent nodes;
- $55$ relators of the form $(xy)^2 = 1$ (commutators), one for each pair of non-adjacent nodes;
together with a single 'spider' relator, $(a b_1 c_1 a b_2 c_2 a b_3 c_3)^{10} = 1$, which results in the group $M \times C_2$. We can get rid of the $C_2$ by quotienting out by an eightieth relation, $x = 1$, where $x$ is the unique non-identity element in the centre of the group.
See http://www.maths.qmul.ac.uk/~jnb/web/Pres/Mnst.html for the explicit Coxeter-Dynkin diagram.
answered Sep 15, 2013 at 14:10
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$\begingroup$ So, it is x^2=1 for each node, (xy)^3 for each pair of adjacent nodes, and [x,y]^2 ([x,y] is the commutator of x and y) for each pair of non-adjacent nodes. Then the "spider" relation, and quotienting out the center. Wow, that is very interesting. $\endgroup$– ThomasCommented Sep 16, 2013 at 2:00
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$\begingroup$ Also, what is the center element? It doesn't say on the page. $\endgroup$– ThomasCommented Sep 16, 2013 at 2:08
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$\begingroup$ What would happen if you did not include the spider relation? $\endgroup$– ThomasCommented Sep 16, 2013 at 3:58
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$\begingroup$ without the spider, you get some weird infinite Coxeter group. $\endgroup$ Commented Sep 16, 2013 at 6:54
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1$\begingroup$ Interesting. What about the spider relation? How was that found? Were they intentionally trying to find the monster group, or was it an accident? $\endgroup$– ThomasCommented Sep 17, 2013 at 11:12
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A presentation for $M$ was constructed by A.A.Ivanov. This presentation is not so easy to describe; it arises from an amalgam of parabolics in certain diagram geometry. S.Norton then proved that the Y-presentation from ATLAS indeed describes the "Bimonster" (see Derek Holt's comment to the previous answer).
answered Sep 15, 2013 at 11:47
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$\begingroup$ I never would have thought it would be the wreath product of the monster by Z2. The link you gave me does not let me see where the monster group is derived, let alone the whole chapter. Could someone email me a copy? $\endgroup$– ThomasCommented Sep 15, 2013 at 14:03
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$\begingroup$ sure, but where to? your profile doesn't reveal any email... $\endgroup$ Commented Sep 15, 2013 at 17:21
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1$\begingroup$ dl.dropboxusercontent.com/u/11004520/CBO9780511629259A013.pdf $\endgroup$ Commented Sep 15, 2013 at 17:29
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Bimonster can be presented by the Coxeter group of the 26-node graph of the projective plane of order 3: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.23.2634 (26 Implies The Bimonster, by John H. Conway and Christopher S. Simons). See also http://citeseer.uark.edu:8080/citeseerx/viewdoc/summary;jsessionid=57D5F004B2D42A5AE3CF28FD1C29E8E7?doi=10.1.1.137.1363 (An Elementary Approach to the Monster, by Christopher S. Simons).
answered Sep 15, 2013 at 15:00