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!