Showing that a group is a free group

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In summary, the conversation discusses how to prove that a group generated by x, y, and z with a single relation is a free group. The conversation includes the definition of a free group and the mapping property of free groups. The conversation also mentions using the first isomorphism theorem and the universal property of free groups to show that the group in question is isomorphic to the free group. The conversation ends with a summary of the proposed solution, including the use of a homomorphism \varphi and the existence of a unique homomorphism \psi to prove that the group is isomorphic to the free group.
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luzerne
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Homework Statement


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.



Homework 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.


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 [tex]\Rightarrow[/tex] 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 [tex]\Leftrightarrow[/tex] 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!
 
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  • #2
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 [tex]\Rightarrow[/tex] 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 [tex]\Leftrightarrow[/tex] 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...)
Why would every element be of the form [itex]y^i z^j[/itex]? This only holds for free abelian groups. Could you express yzy in that form?

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

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

1. What is a free group?

A free group is a type of mathematical group that is defined by a set of elements and a set of operations, without any additional constraints or relations. This means that the elements can be combined in any way using the specified operations, without any restrictions.

2. How do you show that a group is a free group?

To show that a group is a free group, you must demonstrate that it satisfies the defining properties of a free group. This includes showing that the group has a set of elements, a set of operations, and that the elements can be combined in any way using the operations, without any additional constraints or relations.

3. What are the defining properties of a free group?

The defining properties of a free group include a set of elements, a set of operations, and the ability for the elements to be combined in any way using the operations, without any additional constraints or relations. Additionally, any element in a free group can be uniquely expressed as a combination of the other elements using the specified operations.

4. What are some examples of free groups?

The most well-known example of a free group is the free group on two generators, which is commonly denoted as F2. This group has two elements, a and b, and the operations of multiplication and inverse, where any expression of a and b using these operations is a valid element of the group. Other examples include free abelian groups and free product groups.

5. How are free groups used in mathematics?

Free groups are used in a variety of mathematical fields, including abstract algebra, topology, and geometric group theory. They are also used in computer science and cryptography for their properties of being easily generated and manipulated. Additionally, free groups have important applications in the study of group presentations and group actions.

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