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Topic: Group presentation: deriving a relation
Replies: 9   Last Post: Feb 3, 2010 1:03 PM

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Derek Holt

Posts: 498
Registered: 12/13/04
Re: Group presentation: deriving a relation
Posted: Feb 2, 2010 4:59 AM
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On 1 Feb, 23:50, "John R. Martin" <> wrote:
> > In article
> > <1517466167.90120.1265053913404.JavaMail.root@gallium.
> >>,
> >  "John R. Martin" <> 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 Todd-Coxeter algorithm. You
> > can use it, too,
> > just not as fast. It's in many textbooks, and no
> > doubt many websites.

> > --
> > Gerry Myerson ( (i -> u for
> > email)

> The problem I have with the Todd-Coxeter algorithm is I would normally use it on cosets of a non-trivial 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 Todd-Coxeter 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 Todd-Coxeter 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 Todd-Coxeter
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

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

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 Todd-Coxeter
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
< 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.

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