Proving Free Group Generated by A and B

[happyg1]

Homework Statement

Hello, It's me again with group theory.

The question:
Let a be a complex number such that $$|a|\geq 2$$. Prove that

$$A= \bmatrix 1 && 0\\a && 1 \endbmatrix,B=$$ $$\bmatrix 1 && a\\0 && 1 \endbmatrix$$
generate a free group.

Homework Equations

Well, I know that in a free group that there is no nontrivial reduced word that equals the identity (I think I've got that straight)...If not, correct me please.

The Attempt at a Solution

I'm still trying to get my mind around this. As I mentioned before, this is my first course in Group Theory, so here I go...

I need to show that there is no nontrivial reduced word that equals the identity, so I just decided to start manipulating the 2 matrices around to see if I could get a clearer understanding about exactly how to go about this proof. I let $$w=A*B*A^{-1}*B*A*B^{-1}$$
which gives the identity. Now, I *think* that's a reduced word, but it might not be.

Now, I know I'm wrong, but I don't know why.Any clues?
CC

 

[StatusX]

Your reasoning is correct, and if that was the identity, the group wouldn't be free. But I tried it in MATLAB and it doesn't seem to be the identity for general values of a. Sorry, I don't have any other ideas right now.

 

[happyg1]

Matlab...brilliant! I did it by hand and I probably messed up my math somewhere. I'll crank up MATLAB and have another go.
CC

 

[happyg1]

OK,
Here's what I think I need to do.

Show that AB,BA,A(invB),B(invA) do not produce the identity.

Then do I have to show that each pair of pairs can't be the identity as well?

I used MATLAB to multiply out the different combinations and I have the equation that each multiplication produces in each matrix position...I can see that if $$|a|\geq 2$$ that the entries in the matrix will not produce the identity.

I'm just not sure if this is the correct way to do this. Do I need to use induction here?...I've got the sledge hammer out on this one. Is there a more elegant way?

Thanks,
CC

 

[StatusX]

The problem is that you need to show that there's no expression of the form:

$$A^{n_1} B^{m_1} A^{n_2} B^{m_2} ...A^{n_k} B^{m_k}$$

(specifically, with all the n_i, m_i non-zero integers, expcept possibly n_1 and m_k which may be zero - this just assures the above is in reduced form), that equals the identity. No matter how many expressions of any finite length you show are not the identity, it's not enough to guarantee that multiplying these together in some order won't give the identity.

I don't see a really clever (ie, non-computational) way to do this. The only idea I have is to notice that:

$$A^n= \bmatrix 1 && 0\\na && 1 \endbmatrix$$

$$B^n= \bmatrix 1 && na\\0 && 1 \endbmatrix$$

which is valid for any integer n, so an expression of the above form reduces to:

$$\bmatrix 1 && n_1 a\\0 && 1 \endbmatrix \bmatrix 1 && 0\\m_1 a && 1 \endbmatrix ... \bmatrix 1 && n_k a\\0 && 1 \endbmatrix \bmatrix 1 && 0\\m_k a && 1 \endbmatrix$$

So try to show this can only be the identity if n_i=m_i=0. I'd suggest looking at products like

$$\bmatrix 1 && a\\0 && 1 \endbmatrix \bmatrix 1 && 0\\b && 1 \endbmatrix ... \bmatrix 1 && c\\0 && 1 \endbmatrix \bmatrix 1 && 0\\d && 1 \endbmatrix$$

and see if you can find a pattern. Hopefully someone else has a better idea.

 

[Dick]

I've already tried the brute force approach you've outlined. The expression up to m2,n2 is easily shown to be 1 only in trivial cases. Include m3,n3 and things start looking really nasty. Of course, I've made no use of |a|>=2 yet. A 'better idea' is certainly called for, but it's eluding me. Good exercise though.

 

[AKG]

Try induction.

 

[happyg1]

Hi,
I asked my professor for advice on this one and he pointed out an example that proves that A and B generate a free group when a=2. I'm not quite sure I understand the example and I'm hoping that you guys can help me out:

Consider the mappings a and b of the complex plane given by:
$$z(a)=z+2$$

$$z(b)=\frac{z}{2z+1}; z\neq -\frac{1}{2}::::\frac{1}{2}, z=-\frac{1}{2}$$

EDIT:I could not get my latex to correctly code the left big bracket for the cases there. i hope you know what it means. Please show me how to code that

These are clearly bijections, so they generate a group F of permutations of the complex plane.: We claim that F is free on {a,b}.

To see this, observe that a nonzero power of a maps the interior of the unit circle |z|=1 to the exterior, and a nonzero power of b maps the exterior of the unit circle to the interior with 0 removed: The second statement is most easily understood from the equation
$$(\frac{1}{z})b=\frac{1}{z+2}$$.*
From this it is easy to see that no nontrivial reduced word in {a,b} can equal 1.

I don't understand this. I don't know where * came from and it's not clear to me that this is true.

Then it continues:
The functions a and b are instances of the mapping of the complex plane
$$\lambda(a,b,c,d):\longmapsto \frac{az+b}{cz+d}$$
where$$ad-bc\neq 0:a,b,c,d\in \mathbb{C}$$

I don't understand that at all. Where did it come from?

Such a mapping is known as a linear fractional transformation. Now it is easy to show that the function
$$\bmatrix a && b\\c && d \endbmatrix \longmapsto \lambda(a,b,c,d)$$
is a homomorphism from $$GL(2,\mathbb{C})$$ to the group of all linear fractional transformations of $$\mathbb{C}$$ in which
$$A= \bmatrix 1 && 0\\2 && 1 \endbmatrix ,B=\bmatrix 1 && 2\\0 && 1 \endbmatrix$$
map to a and b respectively. Since no nontrivial reduced word in {a,b} can equal 1, the same is true of reduced words in {A,B}. Consequently the group <A,B> is free on {A,B}.

This looks like they just defined two mapping that work and then make up a homomorphism that works. I just don't QUITE get it.

But it does prove (i guess) that for a=2 this is a free group.

Can I use this same mapping for general values of a?

Please help my clarify this.

Thanks,
CC

 

[AKG]

$$b(1/z) = \frac{1/z}{2/z + 1} = \frac{1}{z+2}$$

Next

$$\lambda _{\alpha ,\beta ,\gamma ,\delta}\ :\ z \mapsto \frac{\alpha z + \beta}{\gamma z + \delta}$$

$$b(z) = \frac{z}{2z+1} = \frac{1z+0}{2z+1}$$

so for b, $\alpha = 1,\ \beta = 0,\ \gamma = 2,\ \delta = 1$.

$$a(z) = z+2 = \frac{1z+2}{0z+1}$$

so for a, $\alpha = \delta = 1,\ \beta = 2,\ \gamma = 0$.

 

[happyg1]

OK,
NOW it makes more sense. I just couldn't see it.

Such a mapping is known as a linear fractional transformation. Now it is easy to show that the function
$$\bmatrix a && b\\c && d \endbmatrix \longmapsto \lambda(a,b,c,d)$$
is a homomorphism from $$GL(2,\mathbb{C})$$ to the group of all linear fractional transformations of $$\mathbb{C}$$ in which
$$A= \bmatrix 1 && 0\\2 && 1 \endbmatrix ,B=\bmatrix 1 && 2\\0 && 1 \endbmatrix$$
map to a and b respectively.

Do you think I need to prove that it's a homomorphism?

CC

 

[StatusX]

From this it is easy to see that no nontrivial reduced word in {a,b} can equal 1.

Have you proved this? It's probably the hardest part, unless I'm missing something, and it's just been glossed over.

 

[Dick]

Have you proved this? It's probably the hardest part, unless I'm missing something, and it's just been glossed over.

I think the point is that mapping the matrices into the (nonlinear) linear fractional transformations with |a|>=2 gives that powers of one transformation map the the interior of the unit circle to the exterior and the other maps the exterior to the interior/{0}. The result being that no power of a's an b's can map z=0 back to 0. Given this, the linear representation of the group can't ever return to identity either. I haven't checked all of the details. Does this really work? Seemed ok.

 

[StatusX]

What if you have a non-reduced word that does turn out to be trivial? Also the maps must be on the entire riemann sphere, otherwise they won't be bijections.

 

[Dick]

What if you have a non-reduced word that does turn out to be trivial? Also the maps must be on the entire riemann sphere, otherwise they won't be bijections.

If a word turns out to be trivial, then the gig is up and we've been lied to. Do you see a flaw in the argument?

 

[StatusX]

I don't know if you're serious. I'm saying your argument seems to claim that no product of powers of a and b can give the identity. But what about a product like a⁵b^-1ba^-5? Where in your argument do you use the fact that the words are reduced, or argue that if a word is a non-reduced equivalent of the empty word, it can in fact give the identity? Again, maybe I'm missing something.

 

[Dick]

I don't know if you're serious. I'm saying your argument seems to claim that no product of powers of a and b can give the identity. But what about a product like a⁵b^-1ba^-5? Where in your argument do you use the fact that the words are reduced, or argue that if a word is a non-reduced equivalent of the empty word, it can in fact give the identity? Again, maybe I'm missing something.

The argument doesn't apply to your word, since we assume the word is an alternating (between a and b) product of non-zero powers of a and b. (BTW AKG reversed the a and b from what they previously were). Taking the constant to be 2. Then a^n as a mobius transformation is z/(2nz+1). So a^n(0)=0 and as we've been shown a^n takes the exterior of the unit circle to the interior (I'm ignoring the boundary case). b^n is z+2n, so takes the interior to the exterior. So consider say, a^2*b^(-7)*a^3*b^2. Apply that to z=0. b^2(0) is exterior. a^3 of that point is interior (but not 0). b^(-7) of that result is an exterior point. And a^2 of that is interior again, but not 0. So the product is not zero and the word can't represent the identity. This is how I understand the argument. Starting with the point z=0 the successive powers of a and b can only alternate exterior and interior/{0} so can't ever give 0. Maybe I'm missing something?

 

[StatusX]

I was missing something, that the words had to contain alternating non-zero powers of a and b. Your argument seems good.

Wow, that is a nice proof of the original question, but I can't imagine how they'd expect you to come up with that.

 

[Dick]

I doubt I would have thought of it either. I wasn't making much progress with looking at the linear representation.

 

[happyg1]

Hi,
I had to revisit my professor on this one. He explained it to me exactly the same way as Dick did above. I now see how this works, and I can apply it to the case where 2 is replaced by a. I asked him how he came up with this solution and he said it's just something that he came across one time. I NEVER would have invented this on my own. It's pretty cool.
CC