# Thread: Showing that a group is a free group

1. ## Showing that a group is a free group

Hi,

Here's the question: Prove that the group generated by x,y,z with the single relation $\displaystyle yxyz^{-2} = 1$ is actually a free group.

I only know the basic definition that "a free group is a group whose generators don't have any relation", i don't really know anything else about them. There is the mapping property of free groups, that says that for any funtion from a set S to a group there exists a unique homomorphism from the free group on S to the group such that its restriction to S agrees with the function, so my idea was to somehow show that the kernel of this homomorphism is trivial and use the first isomorphism theorem to show that the homomorphism is actually an isomorphism from the free group to the given group.

Here's what i have so far:
Let $\displaystyle G = <x,y,z | yxyz^{-2} = 1>$
We can show that $\displaystyle x = y^{-1}zy^{-1}$ and so y and z alone generate G. Let S = {y,z} and f:S -> G a function defined
by f(y) = y, f(z) = z. Let F = F(S) be the free group on S.
Then by the mapping property of free groups there exists a unique homomorphism g: F -> G such that g agrees with f on S. g is surjective since S is a generating set.

I am not too sure how to proceed from there. Any element of G is of the form $\displaystyle m= y^{i} z^{j}$ for some i,j. I can show that if g(m) = 1, then $\displaystyle m=1 \implies y^{i} = z^{-j}$. Now I am tempted to just say that the only way this can happen is that both i and j are 0, and so that $\displaystyle g(m) = 1 \iff m = 1$, i.e ker(g) is trivial and conclude by the first isomorphism theorem that the quotient group F/{1} is isomorphic to the image of g, i.e F is isomorphic to G (g is surjective), but I am not sure how i could justify that (or not sure if it's even true...)

Any help would be apreciated, especially any information that clarifies my ideas about free groups and generally how to show that a group is free.

Thank you!

2. Originally Posted by luzerne
Hi,

Here's the question: Prove that the group generated by x,y,z with the single relation $\displaystyle yxyz^{-2} = 1$ is actually a free group.

I only know the basic definition that "a free group is a group whose generators don't have any relation", i don't really know anything else about them. There is the mapping property of free groups, that says that for any funtion from a set S to a group there exists a unique homomorphism from the free group on S to the group such that its restriction to S agrees with the function, so my idea was to somehow show that the kernel of this homomorphism is trivial and use the first isomorphism theorem to show that the homomorphism is actually an isomorphism from the free group to the given group.

Here's what i have so far:
Let $\displaystyle G = <x,y,z | yxyz^{-2} = 1>$
We can show that $\displaystyle x = y^{-1}z^2y^{-1}$ and so y and z alone generate G. Let S = {y,z} and f:S -> G a function defined
by f(y) = y, f(z) = z. Let F = F(S) be the free group on S.
Then by the mapping property of free groups there exists a unique homomorphism g: F -> G such that g agrees with f on S. g is surjective since S is a generating set.

I am not too sure how to proceed from there. Any element of G is of the form $\displaystyle m= y^{i} z^{j}$ for some i,j. I can show that if g(m) = 1, then $\displaystyle m=1 \implies y^{i} = z^{-j}$. Now I am tempted to just say that the only way this can happen is that both i and j are 0, and so that $\displaystyle g(m) = 1 \iff m = 1$, i.e ker(g) is trivial and conclude by the first isomorphism theorem that the quotient group F/{1} is isomorphic to the image of g, i.e F is isomorphic to G (g is surjective), but I am not sure how i could justify that (or not sure if it's even true...)

Any help would be apreciated, especially any information that clarifies my ideas about free groups and generally how to show that a group is free.

Thank you!
Any subgroup of a free group is free. This is quite hard to prove, but it is really the only way I know how to show a group is free.

Can you show that then exists a homomorphism from $\displaystyle G$ to $\displaystyle F(y, z)$ that is injective? Applying the above result, and you are done.

"Any element of G is of the form $\displaystyle m= y^{i} z^{j}$ for some i,j"

This is incorrect - especially if you are trying to show your group is free! For instance, $\displaystyle zy$ cannot be written in this form. I believe this result is only generally true for finite groups $\displaystyle G=<a,b>$ then $\displaystyle g=a^ib^j$ for all $\displaystyle g \in G$.

3. Hi!

Thank you for your reply. i haven't tried to find the homomorphism yet, but I came across Tietze transformation (Tietze transformations - Wikipedia, the free encyclopedia), that let you change a presentation of a group to another equivalent presentation. I kind of considered doing something similar when I found out that x can be expressed as a product of y,z and their inverses, i wanted to replace the instance of x in the relation by $\displaystyle y^{-1}z^{2}y^{-1}$ to get 1 = 1, but I really feel like this is a circular argument...

But the elementary Tietze transformation "removing a generator" seems to justify it, so I get that my set of relator is {1}, and letting N be its normal closure, I get N = {1} and so F/N is isomorphic to G. But as i said above, this looks like a circular argument.

I have two more questions about free groups that are a bit different from this one, and I think I will need to use the result you mentioned that "subgroups of free groups are free" to solve them.

4. ## Forum for advanced group theory

I like this type of question. I am sorry I am not posting the "answer". I would like to collaborate on these type of questions. I have posted on this thread (to date unanswered). Any interest will be met with enthusiasm. Perhaps we could start another thread at the "advanced level"... (then I will be low on the totem pole, admittedly, but there are some talented folks on this site whom I would love to engage with.) Just some thoughts... Any takers?