Talk:Higman–Sims group

Latest comment: 5 years ago by Scott Tillinghast, Houston TX in topic Stabilizer of edge

Presentation

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Is there any value to having the presentation for HS on this page? It has been copied from the Atlas (http://brauer.maths.qmul.ac.uk/Atlas/spor/HS/) and is hopeless (because completely unmotivated) as a definition. Unless I hear strong objections, I shall remove it.

I have removed it. The previous content was:
HS can also be defined in terms of generators a and b and the following relations:
 
 
 
 
--Huppybanny 09:48, 27 October 2005 (UTC)Reply

Representation

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The Mathieu group M22 occurs as a maximal subgroup of HS. One permutation matrix representation of M22 fixes a 2-3-3 triangle with vertices (024), (5,123), and (1,5,122). A non-monomial matrix is needed to complete generation of a representation of HS. Scott Tillinghast, Houston TX (talk) 21:28, 21 December 2015 (UTC)Reply

Wilson (2009) gives an example of a Higman-Sims graph within the Leech lattice, permuted by the representation of M22 above:

  • 22 points of shape (1,1,-3,121)
  • 77 points of shape (2,2,26,016)
  • A 100th point (4,4,022)

Differences of adjacent points are of type 3; those of non-adjacent ones are of type 2. Scott Tillinghast, Houston TX (talk) 16:08, 23 December 2015 (UTC)Reply

Problem: find an involution fixing the aforesaid 2-3-3 triangle and transposing all 100 points of the aforesaid graph. What is its trace? Co3 contains involutions of trace 0 but not of trace -8. Scott Tillinghast, Houston TX (talk) 09:07, 29 December 2015 (UTC)Reply

A 2-3-3 triangle?

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Not if the origin is one vertex. Rather a translation of a 2-3-3 triangle. It would define the same vector space over the rationals as the original triangle, but not the same Z-module. Scott Tillinghast, Houston TX (talk) 15:56, 15 January 2016 (UTC)Reply

Disregard. Scott Tillinghast, Houston TX (talk) 04:02, 25 January 2018 (UTC)Reply

Format problem

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For some reason my new table of conjugacy classes goes to the end of the page. Scott Tillinghast, Houston TX (talk) 21:25, 24 January 2018 (UTC)Reply

Corrected! Changed |-} to |} at end.

Stabilizer of edge

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An involution class 2A (trace 8) transposes 40 non-edges, fixes 20 vertices in the Higman-Sims graph.

The stabilizer of an edge includes the group M21 = PSL(3,4), fixing the 100th vertix C and one of the 22 points. It seems the full stabilizer PSL(3,4):2 of that edge is generated by an involution class 2B (trace 0) which would transpose 50 edges. Scott Tillinghast, Houston TX (talk) 05:53, 2 June 2019 (UTC)Reply

I have not found any matrix construction of a 2B involution in the literature. It would exchange the 100th vertex C with one of the 22 points, could exchange the other 21 points with hexads. There would be 56 hexads left over. Scott Tillinghast, Houston TX (talk) 18:14, 2 June 2019 (UTC)Reply

A class 2B matrix gives a bonus: with M23 a representation of Co3. Scott Tillinghast, Houston TX (talk) 22:56, 2 June 2019 (UTC)Reply

An involution class 2A can only transpose non-edges, because it is conjugate to one in the M22 that exchanges hexads only with hexads. A transposed pair of adjacenr (disjoint) hexads would imply a dodecad whose complement includes an octad. Scott Tillinghast, Houston TX (talk) 12:48, 4 June 2019 (UTC)Reply

But Wilson (2009) [p. 213] constructs a monomial class 2B matrix within a representation of the maximal subgroup 24S6, stabilizer of a non-edge. This involution transposes a non-edge. Scott Tillinghast, Houston TX (talk) 03:31, 9 June 2019 (UTC)Reply

The orbits of M21 are 1 (point 1), 1 (100th vertix), 21 (points), 21 (hexads containing point 1), 56 (hyper-ovals). The 2 fixed points form an edge. Magliveras may be wrong that the normalizer M21:2 has orbits 2, 42, 56. Scott Tillinghast, Houston TX (talk) 20:22, 10 June 2019 (UTC)Reply

Perhaps the edge is transposed by an element of order 4. In that case a stabilizer M21.2 would not be a split extension. Scott Tillinghast, Houston TX (talk) 17:10, 16 June 2019 (UTC)Reply