# Group presentation problem

1. Jun 27, 2007

### happyg1

1. The problem statement, all variables and given/known data

Let p be a prime. Prove that the group $$<x,y|x^p=y^p=(xy)^p=1>$$ is infinite if p>2, but that if p=2, it is a Klein 4-group.

2. Relevant equations

3. The attempt at a solution

OK
I think I got the secong part. We have a theorem that says that if $$G=<x,y|x^2=y^2=(xy)^n=1>$$
where
$$n \geq 2$$
This is the dihedral group $$D_{2n}$$ of order 2n.
So if n=2 in my problem, then I have the dihedral group of order 4, which is the Klein 4-group. I showed this by writing out the multiplication tables for each one. I think I got that part.

My trouble with the first part is that I really don't know HOW to show that a group is infinite. I understand WHY the infinite dihedral group is infinite, but I can't see how to apply that here. I'm not really seeing why the p being prime has any meaning here either....

I just don't quite know where to start that one off.

Please give me a push if you can

CC

2. Jun 27, 2007

### matt grime

Find infinitely many non-isomorphic reduced words. Dunno why you bothered to cite the result you do for n=2 - if you showed it directly, for n=2, you'd find why for n>2 you might not get a finite group. Moral of the story: don't try to invoke pointless results when just playing with things will make you realise what is going on.

3. Jun 27, 2007

### happyg1

Hello,
I cited the result I did because I understand it and it's right there next the the problem. Is my method incorrect?

I don't know why the author of my textbook would put in a "pointless result"....

I will try to "play with things" and see what I get.

Matt Grime, you always make me feel stupid....Maybe I am.

CC

4. Jun 27, 2007

### matt grime

I didn't say the result in and of itself was pointless. I said citing it in this case was pointless - complete overkill. If you just played around with the data given and tried to figure out why 2 was different from generic n, then you'd be much better off. Some times the 'dumb' thing of just seeing what happens is the correct thing to do. You aren't dumb, you just don't want to seem to try things for yourself. This is how I do mathematics - just seeing what I can show from the definitions, and trying to avoid all citations of 'big' results that might obfuscate what is really going on. There are no deep theorems here, no big results to look up - it can 'just be done':

http://www.dpmms.cam.ac.uk/~wtg10/justdoit.html

99.9% of problems posted here can be solved without _any_ clever ideas, just writing out definitions and getting one's hands dirty, but seemingly a lot of people think it will require something clever and therefore don't try something obvious or tedious.

Last edited: Jun 27, 2007
5. Jun 29, 2007

### happyg1

Ok
I did as you suggested and I found that whenever p=2, the group is abelian. So that means that x and y are normal in G, and G has order 4. As an abelian group of order 4 it is isomorphic to the klein 4 group.

I played with this thing for a long time and I'm just not getting any further with the infinitely many reduced words for p a prime. I tried to list out the words for p=4 (I know, not a prime) just to see what happens. I never did get down to a finite number for that one. I could just keep adding on x^3 or y^3 or x^2 or y^2. I know I'm missing something here. Give me a hint.
I even tried to find all of the cosets for p=4 to see if I could show that it was finite...and got bogged down in all of the possibillities and trying to find out which ones could be repeats.

My professor told me to try to find something specific that works, like matricies with entries of 0 and 1 that satisfy the relations. I can't seems to come up with anything and I've been messing with this for DAYS.
I am STUCK.

Thanks
CC

6. Jul 2, 2007

### happyg1

Any thoughts?
I still haven't been able to figure out how to prove this thing is infinite for primes. I have tried to show that it's NOT infinite for non-primes, but that hasn't worked.
I know that it's abelian.
CC

Last edited: Jul 2, 2007
7. Jul 2, 2007

### NateTG

If a group is abelian, and finitely generated by elements of finite order, the it must be finite. Perhaps the group isn't abelian?

8. Jul 2, 2007

### happyg1

OK
It's not abelian. Here's what I tried for p=3:
let $$H=(Z\times Z)\ltimes Z_3=<a,b,t|[ab],a^t=b,b^t=a^{-1}b^{-1}>$$
so
$$a^t=b,b^t=a^{-1}b^{-1}$$
oh, and$$t^3=1$$

let $$c=at,d=t$$
$$H=<c,d>$$

$$c^3=at^2at^2at^2=t(t^{-1}at)atat=tbatat=t^2a^{-1}b^{-1}bat=1$$
and
$$d^3=ttt=t^3=1$$
and
$$(cd)^3=at^2at^2at^2=a(t^{-1}at)tat^2=abtat^2=abtat^2=abt^2(t^{-1}at)t=abt^{-1}bt=aba^{-1}b^{-1}=1$$
so there's a homomorphism from G onto H and H is infinite, so G must be also.
I dunno if that is right or not. I don't know how to expand that to the general case, eother. It took me FOREVER to invent this....and it's probably wrong.....
Any input will be appreciated.
CC

Last edited: Jul 2, 2007
9. Jul 3, 2007

### NateTG

It seems like the easiest way to prove that the group is infinite is to show that, say $xxy$, has inifinite order.