# Showing that a group is a free group

1. Nov 14, 2009

### luzerne

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

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

2. Relevant equations

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.

3. The attempt at a solution

Here's what i have so far:
Let G = <x,y,z | yxyz-2 = 1>
We can show that x = y-1zy-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 m= yi zj for some i,j. I can show that if g(m) = 1, then m=1 $$\Rightarrow$$ yi = 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 g(m) = 1 $$\Leftrightarrow$$ 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. Nov 14, 2009

### rasmhop

Why would every element be of the form $y^i z^j$? This only holds for free abelian groups. Could you express yzy in that form?

These are some pretty good observations though isolating x you should have gotten $x=y^{-1}z^2y^{-1}$, and you're close to cracking it. You have used the universal property of F to define a homomorphism $g:F\to G$. If we could also find a homomorphism $g' :G \to F$ such that $g \circ g'$ and $g' \circ g$ are both identity maps, then we have an isomorphism from G to F. To see this we want to exploit the universal property of the free group $H=\langle x,y,z\rangle$, but we need to connect it to G somehow. This connection comes from the following observation:
Consider a presentation $G=\langle R|S\rangle$, then letting N be the normal closure of S we have G=F(R)/N and we can define the surjective natural projection $\pi : F(R) \to G$ by $\pi(x) = xN$. Now suppose H is a group and $\varphi : G \to H$ is a group homomorphism with the property that if $s \in S$, then $\varphi(s)$ is 1. Intuitively it seems obvious that there should exist a unique homomorphism $\psi : G \to H$ such that $\varphi = \psi \circ \pi$ (i.e. $\varphi : F \rightarrow H$ factors uniquely through $\pi : F \rightarrow G$). This is also true because for every $x \in G$ we have $\pi^{-1}(x) = xN$ and $\varphi$ maps any element of S to 1 so S is contained in $\ker\varphi$ and since kernels are normal and the normal closure is the smallest possible normal subgroup containing S we get $N \subseteq \ker \varphi$. Thus if $xn \in xN$ we have $\varphi(xn) = \varphi(x)\varphi(n) = \varphi(x)$ which shows that $\varphi$ must map any pre-image xN to the singleton $\{\varphi(x)\}$, so we can map any element $xN \in G$ to $\varphi(x)$.

Note that this result may very well be something which you can refer to in your book, or easily derive from some result in your book.

Now using this result let us define $\varphi : F(\{x,y,z\}) \to F$ by,
$$\varphi(x) = y^{-1}z^2y^{-1} \quad \varphi(y) = y \quad \varphi(z) = z$$
which by the universal property is enough to specify a unique homomorphism. It's easy to see $\varphi(yxyz^{-2}) =1$ so by the previous result we can find a homomorphism $\psi : G \to F$ such that $\psi \circ \pi = \varphi$. We have
$$(\psi \circ g)(a) = \psi(g(a)) = \psi(a) = a$$
for $a \in \{y,z\}$ so $\psi \circ g = 1_F$. For every $a \in G$ there exists at least one $b \in F(\{x,y,z\})$ such that $\pi(b) = a$ and then we have,
$$(g\circ\psi)(a) = g(\psi(\pi(b)) = g(\varphi(b))$$
$g \circ \varphi$ maps y,z to y,z since both $\varphi$ and $g$ maps them to themselves. It maps x to itself since:
$$g(\varphi(x)) = g(y^{-1}z^2y^{-1}) = y^{-1}z^2y^{-1} = x$$
Thus $g \circ \varphi = 1_G$. This shows that $F \cong G$.