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From: Derek Holt on 2 Feb 2010 04:59
On 1 Feb, 23:50, "John R. Martin" <sdma...(a)gmail.com> wrote:
> > In article
> > <1517466167.90120.1265053913404.JavaMail.root(a)gallium.
> > mathforum.org>,
> >  "John R. Martin" <sdma...(a)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 Todd-Coxeter algorithm. You
> > can use it, too,
> > just not as fast. It's in many textbooks, and no
> > doubt many websites.
>
> > --
> > Gerry Myerson (ge...(a)maths.mq.edi.ai) (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
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 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
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.

From: Arturo Magidin on 2 Feb 2010 11:25
On Feb 2, 1:19 am, William Elliot <ma...(a)rdrop.remove.com> wrote:
> On Mon, 1 Feb 2010, 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?
> > Some things I've derived so far:
> > (ab)^2 = (ba)^2
>
> Then, a^5 = a^3;  a^2 = e = b^2.

How so?

--
Arturo Magidin
From: William Elliot on 3 Feb 2010 00:50

On Tue, 2 Feb 2010, Arturo Magidin wrote:

> On Feb 2, 1:19�am, William Elliot <ma...(a)rdrop.remove.com> wrote:
>> On Mon, 1 Feb 2010, 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?
>>> Some things I've derived so far:
>>> (ab)^2 = (ba)^2
>>
>> Then, a^5 = a^3; �a^2 = e = b^2.
>
> How so?
>
Whoops, a and b aren't variables.
From: Derek Holt on 3 Feb 2010 04:30
On 3 Feb, 05:50, William Elliot <ma...(a)rdrop.remove.com> wrote:
> On Tue, 2 Feb 2010, Arturo Magidin wrote:
> > On Feb 2, 1:19 am, William Elliot <ma...(a)rdrop.remove.com> wrote:
> >> On Mon, 1 Feb 2010, 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?
> >>> Some things I've derived so far:
> >>> (ab)^2 = (ba)^2
>
> >> Then, a^5 = a^3;  a^2 = e = b^2.
>
> > How so?
>
> Whoops, a and b aren't variables.

You were presumably assuming that ab=ba. If you do that, then b^3=(ab)
^2 gives b=a^2, then a^5=b^3 gives a=1, so b=1. This proves that the
group is perfect - equal to its own derived group.

Derek Holt.
From: Mark on 3 Feb 2010 13:03
Lern hau two uze a keyboard, you spamming
vat of partially digested
horse
testile
pus.

"M. M i c h a e l M u s a t o v" <marty.musatov(a)gmail.com> wrote in message
news:18be769a-ff77-4af0-890d-2ebed15a1083(a)h2g2000yqj.googlegroups.com...

"My name is Marty Mustav Error and I am a third-degree google-posting
fuckwit"



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