From: (Torsten Sillke) Subject: A Word Problem Date: Aug 92 A Word-Problem: Let G = < a,b | aabb=baba, bbaa=abab > the free group of two generators a and b with the two relations aabb=baba and bbaa=abab. Show that aaabbb unequal bababa. :: This Problem was solved imediatly by David Sibley. :: Then there were contributions by Noam Elkies and Marston Conder. :: Later I learned, that the problem had been solved already by :: J H Conway (see AMM 97 (1990) 757-773). --------------------------------------------------------------------- Subject: Re: A Word Problem Date: Tue, 11 Aug 92 19:07:31 MESZ I believe I have the example you are looking for. I was led to this by the theoretical material I sent you the other day. Let H be the quaternion group of order 8. H admits an automorphism x of order 3 which permutes i, j, k cyclically. Let K be the semi-direct product of Z_3 acting on H in this way. Let G be the central product of K and H. That is, G = (KxH)/<(-1,-1)>. Let a = (-x,i) and b = (-xi,j) in G. One readily verifies that a^2*b^2 = (ba)^2 and b^2*a^2 = (ab)^2, yet a^3*b^3 is not (ba)^3. I leave it to you to independently verify my calculations. David Sibley sibley@math.psu.edu --------------------------------------------------------------------- Article: 12518 of sci.math From: elkies@ramanujan.harvard.edu (Noam Elkies) Subject: Re: A Word-Problem Date: 11 Aug 92 17:27:30 GMT Organization: Harvard Math Department In article sibley@math.psu.edu writes: :In article 26212@unibi.uni-bielefeld.de, :uphya159@unibi.uni-bielefeld.de (0118) writes: :>A Word-Problem: :> :>Let G = < a,b | aabb=baba, bbaa=abab > the free group of :>two generators a and b with the two relations aabb=baba and bbaa=abab. :> :>Show that aaabbb unequal bababa. : :Here are two permutations and a verification by GAP. : :gap> A; :( 1, 3, 6,16)( 2,20,22,27,11,13,15,23, 5, 7,12,31)( 4,10,17, 9) :( 8,30,32,18,21,26,28,29,14,19,24,25) :gap> B; :( 1,28,22,17,15,14, 6, 8, 5, 4, 2,32)( 3,19,31,10,27,25,16,18,13, 9, 7,26) :(11,21,12,24)(20,29,23,30) :gap> A^2*B^2=(B*A)^2; :true :gap> B^2*A^2=(A*B)^2; :true :gap> A^3*B^3=(B*A)^3; :false ...which suggests several questions: 1) Where does this specific word-problem come from? Perhaps trying to prove that no rectangle of odd area can be tiled with L-triominos if only two orientations (related by 180-degree rotation) are allowed? 2) Now that we have a group-theoretic proof, is there an elementary proof of the same result about tilings? Note that if all four orientations (or even three of the four orientations) are allowed then a 5x9 rectangle can be tiled with L-triominos. 3) How did David Sibley find the two premutations above? Could they be affine linear transformations on a 5-dimensional vector space over Z/2? The cycle structures (4)(4)(12)(12) of both A and B are compatible with this guess. --Noam D. Elkies (elkies@zariski.harvard.edu) Dept. of Mathematics, Harvard University --------------------------------------------------------------------- Article 12565 of sci.math: From: cgodsil@watserv1.uwaterloo.ca (C. Godsil - C and O) Subject: Re: A Word-Problem Date: 12 Aug 92 20:27:15 GMT Organization: University of Waterloo In article <1992Aug9.172411.26212@unibi.uni-bielefeld.de> uphya159@unibi.uni-bielefeld.de (0118) writes: >A Word-Problem: > >Let G = < a,b | aabb=baba, bbaa=abab > the free group of >two generators a and b with the two relations aabb=baba and bbaa=abab. > >Show that aaabbb unequal bababa. > >I've tried all homomorphisms from G to S10 without success. >(e.g. a->(123), b->(142) shows ab unequal ba but aaabbb=bababa) > >Torsten Sillke The group G = < a, b : a^2*b^2=(b*a)^2, b^2*a^2=(a*b)^2 > is soluble: Let K be the subgroup generated by p = a*b and q = b*a . This subgroup is normal, with quotient G/K cyclic (NB: p^a = q, p^b = p^-1*q^2, q^a = q^-1*p^2, & q^b = p ). Let L be the subgroup generated by u = p^3 and v = q^3 . This subgroup too is normal (with u^a = v = u^b and v^a = u = v^b ), and also Abelian (since in fact both p and q centralize both u and v). Further, the quotient K/L is a homomorphic image of the (3,3,3) triangle group < p, q : p^3 = q^3 = (p*q)^3 = 1 >, which is Abelian-by-cyclic, and is therefore soluble. Now since G/K is cyclic, K/L is soluble, and L is Abelian, it follows that G itself is soluble. Here's a finite matrix representation of G over GF(2): A = [ 0 1 0 0 0 ] B = [ 1 0 1 1 0 ] [ 0 0 1 0 0 ] [ 0 0 1 0 0 ] [ 1 0 0 0 0 ] [ 1 0 0 0 0 ] [ 0 0 0 1 0 ] [ 1 1 1 0 0 ] [ 1 0 0 0 1 ] [ 0 0 0 0 1 ] A^3*B^3 = [ 0 1 1 1 0 ] B^3*A^3 = [ 0 1 1 1 0 ] [ 1 0 1 1 0 ] [ 1 0 1 1 0 ] [ 1 1 0 1 0 ] [ 1 1 0 1 0 ] [ 1 1 1 0 0 ] [ 1 1 1 0 0 ] [ 0 0 0 1 1 ] [ 1 1 1 0 1 ] (A^3*B^3)^2 = (B^3*A^3)^2 = [ 1 0 0 0 0 ] [ 0 1 0 0 0 ] [ 0 0 1 0 0 ] [ 0 0 0 1 0 ] [ 1 1 1 1 1 ] These matrices generate a factor group of order 96, with centre of order 2, and clearly isomorphic to a subgroup of AGL(4,2). Marston Conder (marston@dibbler.uwaterloo.ca) 12 August 1992 --------------------------------------------------------------------- From: (Torsten Sillke) Date: 13 Aug 92 This is the problem the word-problem comes from: 1 1 1 2 2 3 3 2 3 3 3 3 1 1 2 1 2 2 1 2 2 1 1 2 3 3 1 1 2 3 3 1 3 2 1 3 2 2 3 1 1 2 2 3 3 o This figure shows that you can tile a triangle T9 with T2: o o QUESTION: which T(n) can be tiled with T2? SOLUTION: The number of points of T(n) is ( n+1 \choose 2 ) = t(n). 3 | t(n) iff n = 0, 2 (mod 3). CASE I: if n = 0, 2, 9, 11 (mod 12) then T(n) is tileable. From T9 you can get T11 and T12 adding: 1 2 2 1 2 2 1 2 2 1 1 1 2 1 1 2 1 1 2 1 1 or 1 1 2 2 3 3 1 1 2 2 2 1 3 2 1 3 2 1 3 2 1 2 2 3 3 1 1 2 2 3 3 1 1 With T11, T12, and k times 1 2 2 1 2 2 1 2 2 1 2 2 1 1 2 1 1 2 1 1 2 1 1 2 you can construct T(n+12) from T(n). CASE II: if n = 3, 5, 6, 8 (mod 12) then T(n) is not tileable. This is the hard case. You can look at the equivalent triomino puzzle. In this case T5 would look like: o The allowed L-triominoes are: o o o o o o o o = T5 o o and o o o o o o o o o o (only 180-degree rotation) Now let's try group theory: You can read about this technique in the article: Dmitry V. Fomin, Getting it together with "polyominoes", quantum 2:2 (Nov/Dec 1991) 20-23, 61 So you get the group: G = < a, b| aabb=baba, bbaa=abab >. If a^n b^n <> (ba)^n for n = 3, 5, 6, 8 (mod 12) then is T(n) not tileable. Is a^n b^n = (ba)^n for some n = 3, 5, 6, 8 (mod 12) then a^3 b^3 = (ba)^3. This you can see be the same technique as in CASE I. You have the additional possibility of reduction in the group. If you show a^3 b^3 <> (ba)^3 then CASE II is solved. David Sibley found a homomorphism, which shows this inequality. So CASE II is proofed. Annotation: Noam Elkies is right, that the homomorphis of David Sibley shows that it is implossible to tile an odd rectangle with the L-triomino (only 180-degree rotation allowed) too. In this case you have to show that a^3 b <> b a^3. Torsten Sillke