Deriving a relation in a group based on a presentation - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:21:37Z http://mathoverflow.net/feeds/question/15180 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15180/deriving-a-relation-in-a-group-based-on-a-presentation Deriving a relation in a group based on a presentation Steve D 2010-02-13T10:42:44Z 2010-02-14T14:57:51Z <p>Suppose I have the group presentation $G=\langle x,y\ |\ x^3=y^5=(yx)^2\rangle$. Now, $G$ is isomorphic to $SL(2,5)$ (see my proof <a href="http://www.artofproblemsolving.com/Forum/viewtopic.php?t=328702" rel="nofollow">here</a>). This means the relation $x^6=1$ should hold in $G$. I was wondering if anyone knows how to derive that simply from the group presentation (not using central extensions, etc.). Even nicer would be an example of how software (GAP, Magma, Magnus, etc.) could automate that.</p> http://mathoverflow.net/questions/15180/deriving-a-relation-in-a-group-based-on-a-presentation/15189#15189 Answer by Joel David Hamkins for Deriving a relation in a group based on a presentation Joel David Hamkins 2010-02-13T15:42:28Z 2010-02-13T15:42:28Z <p>Let me give a totally useless pure-existence answer, while we wait for someone to show up with a better answer.</p> <p>Namely, if it is true that those relations imply x<sup>6</sup> = 1, then there definitely will be an elementary proof of this, using only the group axioms and the relations. That is, if it is true, then you can be confident that there is a elementary proof, involving just playing with group elements and equations in that group.</p> <p>This is a consequence of Goedel's Completeness theorem, which says that every statement true in all models of a first order theory has a proof from that theory. In your case, if those relations imply that identity in all groups, then there will be a first order proof of this from the group axioms.</p> <p>As for automating such questions, of course the <a href="http://en.wikipedia.org/wiki/Word%5Fproblem%5Ffor%5Fgroups" rel="nofollow">word problem</a> is undecidable, so in general it is impossible to automate the general question of determining whether a given identity is a consequence of a given set of relations. </p> <p>But your question is not an instance of the word problem, since you are not asking whether the identity holds, but rather, you claim to know that it holds, and want an elementary proof of that. This problem is in principal computable. The reason is that the set of identities that hold in a given presentation is computably enumerable---one can just search through the collection of all proofs, until the desired proof is found. Again, the completeness theorem ensures that there will be such a proof, and so there is a computable procedure to find it.</p> <p>I apologize for my useless answer.</p> http://mathoverflow.net/questions/15180/deriving-a-relation-in-a-group-based-on-a-presentation/15191#15191 Answer by Victor Miller for Deriving a relation in a group based on a presentation Victor Miller 2010-02-13T16:23:16Z 2010-02-13T18:07:23Z <p>The theory (and practice) of automatic groups is the most generally useful systematic way to deal with these things. There is a nice package written by Derek Holt and his associates called kbmag (available for download here: <a href="http://www.warwick.ac.uk/~mareg/download/kbmag2/" rel="nofollow">http://www.warwick.ac.uk/~mareg/download/kbmag2/</a> ). There is a book "Word Processing in Groups" by Epstein, Cannon, Levy, Holt, Paterson and Thurston that describes the ideas behind this approach. It's not guaranteed to work (not all groups have an "automatic" presentation) but it is surprisingly effective.</p> <p>I made up a short input file for kbmag, and it immediately came back with a "confluent" system of relations (a particular system which has a technical property that when you just do a series of string substitutions you always get the same answer no matter what order you do them in). For your edification, here they are (xi and yi are x^-1 and yi^-1 respectively, idWord is 1 [edited to show the derivations from kbmag):</p> <pre><code>#Initial equation number 1: #x*xi -&gt; IdWord #Initial equation number 2: #xi*x -&gt; IdWord #Initial equation number 3: #y*yi -&gt; IdWord #Initial equation number 4: #yi*y -&gt; IdWord #Initial equation number 5: #y^4 -&gt; x^3*yi #Initial equation number 6: #y*x*y -&gt; x^2 #New equation number 7, from overlap 5, 3: #x^3*yi^2-&gt;y^3 #New equation number 8, from overlap 4, 5: #yi*x^3-&gt;y^4 #New equation number 9, from overlap 6, 3: #x^2*yi-&gt;y*x #New equation number 10, from overlap 4, 6: #yi*x^2-&gt;x*y #New equation number 11, from overlap 2, 7: #xi*y^3-&gt;y*x*yi #New equation number 12, from overlap 2, 9: #xi*y*x-&gt;x*yi #New equation number 13, from overlap 10, 1: #x*y*xi-&gt;yi*x #New equation number 14, from overlap 1, 11: #x*y*x*yi-&gt;y^3 #New equation number 15, from overlap 11, 3: #y*x*yi^2-&gt;xi*y^2 #New equation number 16, from overlap 12, 1: #x*yi*xi-&gt;xi*y #New equation number 17, from overlap 2, 13: #xi*yi*x-&gt;y*xi #New equation number 18, from overlap 4, 15: #yi*xi*y^2-&gt;x*yi^2 #New equation number 19, from overlap 2, 16: #xi^2*y-&gt;yi*xi #New equation number 20, from overlap 17, 1: #y*xi^2-&gt;xi*yi #New equation number 21, from overlap 18, 3: #x*yi^3-&gt;yi*xi*y #New equation number 22, from overlap 19, 3: #yi*xi*yi-&gt;xi^2 #New equation number 23, from overlap 2, 21: #xi*yi*xi*y-&gt;yi^3 #New equation number 24, from overlap 23, 3: #yi^4-&gt;xi*yi*xi #New equation number 25, from overlap 3, 24: #y*xi*yi*xi-&gt;yi^3 #New equation number 26, from overlap 25, 2: #yi^3*x-&gt;y*xi*yi #New equation number 27, from overlap 3, 26: #y^2*xi*yi-&gt;yi^2*x #New equation number 28, from overlap 27, 4: #yi^2*x*y-&gt;y^2*xi #New equation number 29, from overlap 3, 28: #y^3*xi-&gt;yi*x*y #New equation number 30, from overlap 29, 2: #yi*x*y*x-&gt;y^3 #New equation number 31, from overlap 5, 5: #y*x^3-&gt;x^3*y #New equation number 32, from overlap 5, 6: #y^3*x^2-&gt;x*y*x^2*y #New equation number 33, from overlap 8, 7: #yi*x*y^3-&gt;x*y^2*x*yi #New equation number 34, from overlap 7, 8: #y*x^2*y*x-&gt;x^2*y^3 #New equation number 35, from overlap 11, 6: #y*x*yi*x*y-&gt;xi*y^2*x^2 #New equation number 36, from overlap 12, 9: #xi*y^2*x-&gt;x*yi*x*yi #New equation number 37, from overlap 10, 13: #yi*x*yi*x-&gt;x*y^2*xi #New equation number 38, from overlap 11, 15: #y*x*yi*x*yi^2-&gt;xi*y^2*xi*y^2 #New equation number 39, from overlap 12, 16: #xi*y*xi*y-&gt;x*yi^2*xi #New equation number 40, from overlap 17, 13: #y*xi*y*xi-&gt;xi*yi^2*x #New equation number 41, from overlap 18, 15: #yi*x*yi^2*xi*y-&gt;x*yi^2*x*yi^2 #New equation number 42, from overlap 18, 20: #yi*xi*y*xi*yi-&gt;x*yi^2*xi^2 #New equation number 43, from overlap 19, 20: #yi*xi^3-&gt;xi^3*yi #New equation number 44, from overlap 17, 21: #y*xi*yi^3-&gt;xi*yi^2*xi*y #New equation number 45, from overlap 23, 20: #yi^3*xi^2-&gt;xi*yi*xi^2*yi #New equation number 46, from overlap 22, 24: #yi*xi^2*yi*xi-&gt;xi^2*yi^3 #New equation number 47, from overlap 25, 19: #yi^3*xi*y-&gt;y*xi*yi^2*xi #New equation number 48, from overlap 22, 26: #xi^2*yi^2*x-&gt;x*yi^2*xi^2 #New equation number 49, from overlap 27, 26: #yi^2*x*yi^2*x-&gt;y*xi*yi^2*x*yi #New equation number 50, from overlap 49, 1: #y*xi*yi^2*xi*y-&gt;yi^2*x*yi^2 #New equation number 51, from overlap 7, 28: #x^3*y^2*xi-&gt;y^2*x^2 #New equation number 52, from overlap 27, 28: #yi*x*y^2*xi*y-&gt;y^2*xi*y^2*xi #New equation number 53, from overlap 29, 12: #yi*x*y^2*x-&gt;y^3*x*yi #New equation number 54, from overlap 31, 7: #x^3*y^2*x*yi-&gt;y*x^2*y^3 #New equation number 55, from overlap 32, 13: #x*y*x^2*y^2*xi-&gt;y^3*x*yi*x #New equation number 56, from overlap 32, 14: #x*y*x^2*y^2*x*yi-&gt;y^2*x^2*y^2 #New equation number 57, from overlap 2, 56: #y*x^2*y^2*x*yi-&gt;x^2*y^2*xi*y^2 #New equation number 58, from overlap 56, 4: #y^2*x^2*y^3-&gt;x*y*x^2*y^2*x #New equation number 59, from overlap 57, 4: #x^2*y^3*x*yi-&gt;y*x^2*y^2*x #New equation number 60, from overlap 33, 34: #y^2*x^2*y^2*xi*y^2-&gt;x*y^2*x^2*y^2*x #New equation number 61, from overlap 11, 35: #x^2*y^2*xi*y^2*xi-&gt;y*x^2*y^2*xi*y #New equation number 62, from overlap 35, 25: #x*yi*x*yi^2*xi-&gt;xi*y^2*xi*y #New equation number 63, from overlap 35, 32: #x^2*y^2*x^2*y^2*xi-&gt;y*x^2*y^2*x^2*y #New equation number 64, from overlap 33, 35: #y^2*x^2*y^2*xi*y-&gt;yi*x*yi^2*x*yi^2 #New equation number 65, from overlap 64, 3: #y^2*x^2*y^2-&gt;yi^2*x*yi^2 #New equation number 66, from overlap 4, 64: #y*x^2*y^2*xi*y-&gt;xi*yi^2*x*yi^2*xi #New equation number 67, from overlap 65, 3: #y*x^2*y-&gt;xi*yi^2*xi #New equation number 68, from overlap 4, 66: #xi*y*xi*yi^2*xi-&gt;x^2*y^2*xi*y #New equation number 69, from overlap 67, 3: #xi*yi*xi^2-&gt;y*x^2 #New equation number 70, from overlap 4, 67: #xi^2*yi*xi-&gt;x^2*y #New equation number 71, from overlap 68, 2: #x^2*y^2*x-&gt;xi*y*xi*yi #New equation number 72, from overlap 1, 69: #x*y*x^2-&gt;yi*xi^2 #New equation number 73, from overlap 69, 2: #x^3*y-&gt;xi*yi*xi #New equation number 74, from overlap 70, 2: #x^2*y*x-&gt;xi^2*yi #New equation number 75, from overlap 71, 1: #xi*yi^3-&gt;x^2*y^2 #New equation number 76, from overlap 2, 71: #yi*xi^2*yi-&gt;x*y^2*x #New equation number 77, from overlap 72, 1: #xi^3*yi-&gt;x*y*x #New equation number 78, from overlap 73, 3: #xi^3-&gt;x^3 #New equation number 79, from overlap 75, 4: #x^2*y^3-&gt;xi*yi^2 #New equation number 80, from overlap 1, 78: #x^4-&gt;xi^2 #New equation number 81, from overlap 2, 79: #xi^2*yi^2-&gt;x*y^3 #New equation number 82, from overlap 29, 36: #y^3*x*yi*x*yi-&gt;x*y^2*x*yi*x #New equation number 83, from overlap 7, 37: #y^3*x*yi*x-&gt;yi^2*xi*y*xi #New equation number 84, from overlap 4, 83: #xi*yi^2*xi^2-&gt;y*x*yi*x #New equation number 85, from overlap 1, 84: #yi^2*xi^2-&gt;y^3*x #New equation number 86, from overlap 9, 37: #yi^3*xi-&gt;y^2*x^2 #New equation number 87, from overlap 38, 8: #x^2*y^2*xi*y^2-&gt;xi*yi^2*x*yi^2 #New equation number 88, from overlap 11, 38: #xi*y^2*xi*y^2*xi*y-&gt;y^2*x*yi*x*yi^2 #New equation number 89, from overlap 88, 3: #xi*y^2*xi*y^2*xi-&gt;y*xi*y^2*xi*y #New equation number 90, from overlap 38, 37: #y*xi*yi^2*x*yi^2*xi-&gt;yi*x*yi^2*x*yi^2 #New equation number 91, from overlap 39, 39: #x*y^2*x^2-&gt;yi*xi*y*xi #New equation number 92, from overlap 41, 39: #yi*x*yi^2*x*yi^2*xi-&gt;x*yi*x*yi^2*x*yi^2 #New equation number 93, from overlap 42, 47: #xi*y*xi*yi^2*x*yi^2-&gt;x*y^2*xi*y^2*xi*y #New equation number 94, from overlap 93, 4: #x*y^2*xi*y^2*xi*y^2-&gt;xi*y*xi*yi^2*x*yi #New equation number 95, from overlap 2, 94: #y^2*xi*y^2*xi*y^2-&gt;x*y^2*x*yi*x*yi </code></pre> <p>#68 eqns; total len: lhs, rhs = 299, 246; 77 states; 0 secs. max len: lhs, rhs = 8, 8.</p> <p>#System is confluent.</p> <p>#Halting with 68 equations. #Exit status is 0</p> http://mathoverflow.net/questions/15180/deriving-a-relation-in-a-group-based-on-a-presentation/15259#15259 Answer by Steve D for Deriving a relation in a group based on a presentation Steve D 2010-02-14T14:57:51Z 2010-02-14T14:57:51Z <p>OK, here is the derivation, based completely on the amazing information provided by Victor Miller (who I should also thank for letting me know about kbmag). First, some identities:</p> <p>(1) From $x^3=xyxy$ we get: (a) $x^2=yxy$; (b) $xyx^{-1}=y^{-1}x$; (c) $x^{-1}yx=xy^{-1}$.</p> <p>(2) From $y^5=xyxy$ we get: (a) $y^4=xyx$; (b) $x^{-1}y^3=yxy^{-1}$; (c) $y^3x^{-1}=y^{-1}xy$.</p> <p>(3) From (1a) and (3b) we get $(yxy)(yxy^{-1})=(x^2)(x^{-1}y^3) = xy^3$; so $xy^2xy^{-1}=y^{-1}xy^3$.</p> <p>(4) From (2b) and (1b) we get $(yxy^{-1})(xyx^{-1}) = (x^{-1}y^3)(y^{-1}x) = x^{-1}y^2x$, so that $yxy^{-1}xy=x^{-1}y^2x^2$.</p> <p>(5) From (2c) we get $y^2x^{-1}y^{-1}=y^{-2}x$; squaring that yields $y(yx^{-1}yx^{-1})y^{-1}=y^{-2}xy^{-2}x$. (1c), inverse, squared, shows this is the same as $yx^{-1}y^{-2}xy^{-1}=y^{-2}xy^{-2}x$.</p> <p>(6) Similar to (5). From (2c) we get $y^2x^{-1}=y^{-2}xy$; squaring that yields $y^2x^{-1}y^2x^{-1}=y^{-1}(y^{-1}xy^{-1}x)y$. (1b) squared shows this is the same as $y^2x^{-1}y^2x^{-1}=y^{-1}xy^2x^{-1}y$.</p> <p>OK, now consider the word $(y^{-1}xy^3)xy^{-1}xy$. From (3) this is $xy^2x(y^{-1}xy^{-1}x)y$, which from (1b) squared is $xy^2x(xy^2x^{-1})y=xy^2x^2y^2x^{-1}y$.</p> <p>This word can also be written as $y^{-1}xy^2(yxy^{-1}xy)$, which from (4) is $y^{-1}xy^2(x^{-1}y^2x^2)$. So the previous two computations show $y^2x^2y^2xy^{-1}=x^{-1}(y^{-1}xy^2x^{-1}y)yx^2$</p> <p>$=x^{-1}y^2x^{-1}y^2(x^{-1}yx)x$ ...... from (6)</p> <p>$=x^{-1}y^2(x^{-1}y^2x)y^{-1}x$ ....... from (1c)</p> <p>$=(x^{-1}y^2x)y^{-1}xy^{-2}x$ ......... from (1c) squared</p> <p>$=xy^{-1}x(y^{-2}xy^{-2}x)$ ........... from (1c) squared</p> <p>$=xy^{-1}(xyx^{-1})y^{-2}xy^{-1}$ ..... from (5)</p> <p>$=x(y^{-2}xy^{-2}x)y^{-1}$ .............from (1b)</p> <p>$=(xyx^{-1})y^{-2}xy^{-2}$ ............ from (5)</p> <p>$=y^{-1}xy^{-2}xy^{-2}$ ............... from (1b).</p> <p>So $y^2x^2y^2x^{-1}y=y^{-1}xy^{-2}xy^{-2}$, or $y^2x^2y^2=y^{-1}xy^{-2}xy^{-3}x$. But $y^{-1}xy^{-2}x(y^{-3}x)=y^{-1}xy^{-2}(xyx^{-1})y^{-1}=y^{-1}x(y^{-3}x)y^{-1}=y^{-1}(xyx^{-1})y^{-2}=y^{-2}xy^{-2}$ (through repeated application of (2b) inverse, and (1b)).</p> <p>Thus $y^2x^2y^2=y^{-2}xy^{-2}$, or $y^4x^2y^4=x$, and from (2a) we get $xyx^4yx=x$, or $1=yx^4yx=(yxy)x^4=x^6$ (the second follows from $x^3$ being central and the third from (1a)).</p> <p>Done! </p>