Prev: Functional-Equation
Next: A call for help (was: Re: Welcome to <sci.math>. These suggestions may help you.)
From: John R. Martin on 1 Feb 2010 04:51
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
b^2 = aba
a^4 = bab
a^b = ba'
b^a = aaab'

Thanks,
JR
From: Gerry Myerson on 1 Feb 2010 16:38
In article
<1517466167.90120.1265053913404.JavaMail.root(a)gallium.mathforum.org>,
"John R. Martin" <sdmacro(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 (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: John R. Martin on 1 Feb 2010 08:50
> In article
> <1517466167.90120.1265053913404.JavaMail.root(a)gallium.
> mathforum.org>,
> "John R. Martin" <sdmacro(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 (gerry(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
From: M. M i c h a e l M u s a t o v on 2 Feb 2010 00:17
On
Feb
1,
5:50 
pm,
"John
R
.
Martin"
<sdma
.
.
.
@gmail
.
com>
wrote:
>
>
In
article
>
>
<1517466167
.
90120
.
1265053913404
.
JavaMail
.
root(a)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
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
.
.
.
@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-
Hide
quoted
text
-
>
>
-
Show
quoted
text
-
Results
1
-
10
for
<a,b
|
a^5
=
b^3
=
(ab)^2>
.
.
(0
.
14
seconds)
How
do
you
exact
Search
"A
.
B"
without
getting
results
for
"A
B
.
.
.
8
posts
-
5
authors
-
Last
post:
Oct
17,
2009
I
tried
to
search
on
exact
"A
.
B"
and
google
searches
on
"A
B"
Why
is
the
exact
search
not
exact
?
.
.
.
2
of
5
people
found
this
answer
helpful
.
Did
you
?
Sign
in
to
vote
.
.
.
.
3
of
3
people
found
this
answer
helpful
.
Did
you
?
.
.
.
2)
Imagine
you're
looking
for
the
Michael
Jackson
song
"Black
or
White
.
.
.
.
www
.
google
.
com/support/forum/p/Web+Search/thread
?
tid
.
.
.
hl=en
Formula
For
(A+B)^2,
(A-B)^2,
A^2
-
B^2
,
A^3
+
B^3
,
A^3
-B^3
.
.
.
4
.
a3+b3=
(a+b)(a2-ab+b2)
.
=
(a+b)3-3ab(a+b)
.
5
.
a3-b3=
(a-b)(a2+ab+b2)
.
=
(a-b)3+3ab(a-b)
.
.
.
.
x3+43
.
=
x3+64
.
Problem
2:
If
a+b=5
and
ab=1,find
a3+b3
.
.
.
www
.
tutor next
.
com/grade-8/algebra/
.
.
.
/formula-ab2-b2-a2-b2-a3-b3-a3-b3
Expansions
explained
with
examples
|
Tutor
Vista
5
.
(a
+
b)2
=
(a
-
b)2
+
4ab
.
7
.
(a
+
b
+
c)2
=
a2
+
b2
+
c2
+
2
(ab
+
bc
+
ca)
.
8
.
(a
+
b)3
=
a3
+
3a2b
+
3ab2
+
b3
.
=
a3
+
b3
+
3ab
(a
+
b)
.
.
.
www
.
tut
or
vista
.
com/content/math/algebra/expansions/expansion
.
php
Or
kut
getting
Worse
Day
by
day
and
the
worst
Part
is
they
cant
fix
.
.
.
27
posts
-
8
authors
-
Last
post:
Jan
7
I
reported
my
B`day
is
not
appearing
in
my
friend`s
homepage
.
.
.
.
all
.
.
.
.
Level
3
1/5/10
.
.
.
Playboy
aka
Rockstar
.
Level
2
1/5/10
.
.
.
www
.
google
.
com/support/forum/p/orkut/thread
?
tid
.
.
.
hl=en
Simplify
A
2
B
2
Ab
|
Tut
or
Vista
.
com
AB
=
5
units
.
Point
A
has
coordinates
(2,
5)
.
Find
the
possible
number
of
locations
for
B
if
the
coordinates
has
to
lie
in
positive
quadrant
.
=>
4
or
2
or
3
.
.
.
www
.
tut or vista
..
com/search/simplify-a-2-b-2-ab
Hammond
organ
-
Wikipedia,
the
free
encyclopedia
The
B-2
/
C-2
organs
were
produced
from
December
1949–December
1954
.
The
B-3/C-3
were
produced
from
January
1955–1974
.
The
A-100
series
was
produced
from
.
.
.
en
.
wikipedia
.
or
g/wiki/Hammond_organ
Simplify
A
2
A
3
B
|
Tut
or
Vista
.
com
Simplify:
2
a
3
×
5
b
2
=>
a
3
b
3
or
2
a
3
b
3
or
20
a
4
b
3
or
10
a
3
b
2
.
.
.
.
Simplify:
2
a
(4
b
+
3
c
)
=>
2
bc
+
3
ac
or
8
ab
+
6
ac
or
2
ab
+
3
ac
or
.
.
.
www
.
tut
or
vista
.
com/search/simplify-a-2-a-3-b
Express
in
simplest
form
5
over
a+b
-
5
over
a-b
over
10
over
a^2
.
.
.
Jan
23,
2010
.
.
.
-(5
(a^2
b-10
a-b^3))/(-a^3
b-a^2
b^2+10
a^2+a
b^3+10
a
b+b^4-50)
.
.
.
integral
5/(a+b-5/(a-b/(10/(a^2-b^2))))
da
=
5
RootSum[-#1^3 b-#1^2
.
.
.
answers
.
yahoo
.
com/question/index
?
qid=20100123123051AAL5uL2
\
Simplify
A
2
2ab
B
2
Ab
|
Tut
or
Vista
.
com
If
M
is
the
midpoint
of
AB,
where
the
coordinates
of
A
and
M
are
(4,
-3)
and
(-2,
5)
,
then
find
the
coordinates
of
B
..
=>
(1,
1)
or
(2,
2)
or
(0,
7)
or
(-8,
.
.
.
www
.
tut
or
vista
.
com/search/simplify-a-2-2ab-b-2-ab
Formula
For
$(A+B)^2$,
$(A-B)^2$,
$A^2
-
B^2$
,
$A^3
+
B^3$
,
$A^3
.
.
.
(a
+
b)3
-
3ab(a
+
b)
.
5
.
a3
-
b3=
(a
-
b)(a2
+
ab
+
b2)
.
=
(a
-
b)3
+
3ab(a
-
b)
.
.
.
.
x3 + 43
.
=
x3
+
64
.
Problem
2:
If
a
+
b=5
and
ab=1,find
a3
+
b3
.
.
.
www
.
tut
or
next
.
com/formula-ab2-b2-a2-b2-a3-b3-a3-b3/738
1
2
3
4
5
6
7
8
9
10
Next
http://www
.
meami
.
org/
?cx=000961116824240632825%3A5n3yth9xwbo&cof=FORID%3A9%3B+NB
%3A1&ie=UTF-16&q=%3Ca%2Cb+%7C+a%5E5+%3D+b%5E3+%3D+(ab)%5E2%3E
.
#977
From: William Elliot on 2 Feb 2010 02:19
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.

> b^2 = aba
> a^4 = bab

> a^b = ba'
> b^a = aaab'

a' = a^-1 ?
a^b = bab' ?

----
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