Newsgroups: sci.math
From: uphya159@unibi.uni-bielefeld.de (0118)
Subject: A Word-Problem
Date: Sun, 9 Aug 92 17:24:11 GMT
Organization: Universitaet Bielefeld
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
(posted by Udo Sprute, uphya159@unibi.HRZ.Uni-Bielefeld.DE)
From: sibley@math.psu.edu (David Sibley)
Newsgroups: sci.math
Subject: Re: A Word-Problem
Date: 11 Aug 92 15:15:45 GMT
Reply-To: sibley@math.psu.edu
Organization: Dept. of Mathematics, Penn State University
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
David Sibley
sibley@math.psu.edu
From: elkies@ramanujan.harvard.edu (Noam Elkies)
Newsgroups: sci.math
Subject: Re: A Word-Problem
Date: 11 Aug 92 17:27:30 GMT
Organization: Harvard Math Department
In article <Bstrq9.MKr@cs.psu.edu> 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
Newsgroups: sci.math
From: cgodsil@watserv1.uwaterloo.ca (C. Godsil - C and O)
Subject: Re: A Word-Problem
Organization: University of Waterloo
Date: Wed, 12 Aug 1992 20:27:15 GMT
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
>(posted by Udo Sprute, uphya159@unibi.HRZ.Uni-Bielefeld.DE)
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 ]
[ 0 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
Newsgroups: sci.math
From: umatf071@unibi.uni-bielefeld.de (0105)
Subject: Re: A Word-Problem
Date: Thu, 13 Aug 92 13:57:04 GMT
Organization: Universitaet Bielefeld
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
From: elkies@ramanujan.harvard.edu (Noam Elkies)
Newsgroups: sci.math
Subject: Re: A Word-Problem
Date: 13 Aug 92 15:36:15 GMT
Organization: Harvard Math Department
In article <1992Aug13.135704.2280@unibi.uni-bielefeld.de>
umatf071@unibi.uni-bielefeld.de (0105) writes:
"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?
[...]
So it was a tiling problem, but not the one I had imagined...
" Noam Elkies is right, that the homomorphism of David Sibley shows that
" it is impossible 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.
I omitted a step here: the commutators [a^3,b^2] and [b^2,a^3] vanish
(aaabb=a(aabb)=a(baba)=(abab)a=(bbaa)a=bbaaa, and likewise aabbb=bbbaa;
in ffect I'm tiling 2x3 and 3x2 rectangles with the allowed L-triominos),
and aaabbb=bababa iff a^3 and b^3 commute since b(abab)a=b(bbaa)a;
so the group-theory identity [a^m,b^n]=1 that would result from a tiling
of any odd mxn rectangle by allowed L-triominos is equivalent to any of
[aaa,b]=1, [aaa,bbb]=1, or aaabbb=bababa.
--Noam D. Elkies (elkies@zariski.harvard.edu)
Dept. of Mathematics, Harvard University
Date: Wed, 31 Jan 2001 09:19:39 -0500
From: Aaron Meyerowitz <meyerowi@fau.edu>
Reply-To: meyerowi@fau.edu
Organization: F.A.U. Math
To: eppstein@ics.uci.edu
Subject: junkyard correction
I love and use your junkyard site frequently!
Looking for a remembered result I found it at
http://www.ics.uci.edu/~eppstein/junkyard/wordprob.html
but it didn't work!
Halfway down one finds:
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 ]
[ 0 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 ]
which MAPLE shows is wrong. Actually, for A and B
as given it is clear that any product will give
[ 0 ]
[ 0 ]
[ X 0 ]
[ 0 ]
[ 0 0 0 0 1 ]
I tried to guess the incorrect entry or entries After i gave up, I
searched and found:
http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/word_problem_L3
which has it correctly. For matrix A the entry in row 3 column 1 is "1"
and not "0". Note that the matrices are 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 ]
regards,
Aaron Meyerowitz