
Re: Group presentation: deriving a relation
Posted:
Feb 2, 2010 4:59 AM


On 1 Feb, 23:50, "John R. Martin" <sdma...@gmail.com> wrote: > > In article > > <1517466167.90120.1265053913404.JavaMail.root@gallium. > > mathforum.org>, > > "John R. Martin" <sdma...@gmail.com> wrote: > > > > I have the following group presentation: > > > <a,b  a^5 = b^3 = (ab)^2>. Note that I'm not > > requiring a^5=1, just that > > > these three elements are equal. I plugged this > > into GAP, and it told me the > > > group is SL(2,5). That means b should have finite > > order. Is there a way, > > > using just the two relations given, to show b^n=1 > > for some n? > > > GAP might be using the ToddCoxeter algorithm. You > > can use it, too, > > just not as fast. It's in many textbooks, and no > > doubt many websites. > > >  > > Gerry Myerson (ge...@maths.mq.edi.ai) (i > u for > > email) > > The problem I have with the ToddCoxeter algorithm is I would normally use it on cosets of a nontrivial subgroup, like <a> or <b>. But it's not hard to see that G/<bbb> is A_5. So I don't see what I would gain from the ToddCoxeter algorithm, unless I used the subgroup {1}. My other question about using Todd_Coxeter: once the diagram is complete, can I use it to read off manipulations? That is, can a complete ToddCoxeter diagram allow me to figure out how to manipulate the relations to show b^6=1? > JR
The answer to that last question is yes. A completed ToddCoxeter enumeration over the identity subgroup gives you the Cayley graph of the group, from which you can read off any relator in the group. But you would probably not want to do that by hand with a group of order 120.
A better way to do this by computer is to use the modifies Todd Coxeter algorithm, which computes a presentation of the subgroup used for the enumeration at the same time as doing the enumeration. So you could do this over the subgroup <b> of index 20, and get a presentation of <b> at the same time. You can do that in GAP like this:
gap> F:=FreeGroup(2);; gap> H:=Subgroup(G, [G.2] );; gap> P:=PresentationSubgroupMtc( G, H ); #I there are 3 generators and 6 relators of total length 21 #I there are 1 generator and 1 relator of total length 6 <presentation with 1 gens and 1 rels of total length 6> gap> TzPrintRelators(P); #I 1. _x1^6
So b has order 6. You could conceivably do the modified ToddCoxeter by hand (index 20), but it would still be hard work.
But with a little more theory it becomes easier still. It is easy to see that your presentation defines a perfect central extension of the group < a,b  a^5=b^3=(ab)^2=1 >, and if you know that this group is A_5 and that the Schur cover of A_5 is SL(2,5), then you know that your group must be either A_5 or SL (2,5). To verify that it is SL(2,5) you just need to find matrices a,b in SL(2,5) that satisfy the relations of your presentation, and I will leave that to you!
Derek Holt.

