# The Free Group

So, in this post we will face the fifth universal problem. It’s really not difficult, though it will be extremely useful, and not just for our purposes! We will define the free group (over a given set). Since this one is really important, After solving the universal problem, I want to show some properties of it which will help us a lot in the future.

## The problem

As I said, the problem is really not that complicated, it’s not like the previous problem where we started with three groups and two homomorphisms.

Now the case is pretty simple, all we have is a set. We would like to map this set to some group $K$ using some function $f$. Note that this map crosses categories, it is from the category of sets to the category of groups.

The catch here is that $f$ is just a function, it is not a homomorphism, it doesn’t have to be one-to-one, onto… It can be literally everything!

As always, the question is: “does a unique universal group $U$ with homomrphism $i:X\to U$ such that all the homomorphisms go ‘through it’ exists?”

Formally, we are asking the existence of a unique $L:U\to K$ such that $f=L\circ i$.

So I won’t prove the uniqueness of $U$ since the proof is the same for all the universal problems (you can find an example here).

Now the only thing left to figure out is the existence. And to do so, we have to ask ourselves first – how do we want to map a set to a group?

### Mapping a set to a group

Here is the basic idea – we can consider the disjoint union $A$ of the set $X$ with a copy of itself – $X^\prime$:

A=X\cup X^\prime

We think of the sets $X$ and $X^\prime$ as different sets, and in order to tell where an element of $A$ is from, we add to all the elements in $X^\prime$ a little tag:

a\in A \longleftrightarrow a^\prime \in A^\prime

And now the process is similar to the one of the free product, we define $M$ to be the set of all the words made from elements of $A$ and we define an equivalence realtion on it:

(---,x,x^\prime,---)\sim(------) \\
(---,x^\prime,x,---)\sim(------)

If you think about it, maybe a better symbol for elements in $X^\prime$ is:

a\in A \longleftrightarrow a^{-1} \in A^\prime

Let’s denote by $F(X)$ the set of the equivalence classes from now on. Great, note that we can define an action on the words in the most natural way you can think of:

(---,x_1)\circ (x_2,---)=(---,x_1,x_2,---)

This action induces an action on the equivalence classes:

[(---,x_1)]\circ [(x_2,---)]=[(---,x_1,x_2,---)]

From similar arguments to those that I’ve brought about the action in the free product – this action is well defined and associative.

Obviously, the identity in under this action is the class of the empty word, and inverse elements are exactly what you think they should be. For example:

[(x_1,x_2^{-1},x_3,x_4)]^{-1}=[(x_4^{-1},x_3^{-1},x_2,x_1^{-1})]

Since:

[(x_1,x_2^{-1},x_3,x_4)]\circ[(x_4^{-1},x_3^{-1},x_2,x_1^{-1})]
=[(x_1,x_2^{-1},x_3,x_4,x_4^{-1},x_3^{-1},x_2,x_1^{-1})]=\cdots=[()]

Great, so we’ve just proved that $F(X)$ is a group! And this one has a special name – it is called the free group over a set $X$.

Now we need to verify that $F(X)$ is indeed a soultion for the problem:

### Solving the problem

First things first, we need to define the map:

i:X\to F(X)

Well, the most natural way to do so is just map $x$ to the class $[(x)]$:

i(x)=[(x)]

Let’s take another look at the diagram:

We need to find a unique $L$ such that:

f=L\circ i

That is, for every $x\in X$:

f(x)=L\circ i(x)=L(i(x))=L([(x)])

And that’s exactly how $L$ has to be defined on elements of the form $[(x)]$. But what about elements such as $[(x^{-1})]$. That’s not really a problem, since $L$ need to be a homomorphism:

L([(x^{-1})])=L([(x)]^{-1})=L([(x)])^{-1}=f(x)^{-1}

Great, so we just proved the existence and uniqueness (since we were ‘forced’ to define $L$ in that way) of $L$ (I didn’t show how $L$ acts on classes such as $[(x_1,x_2,x_3,x_5^{-1})]$, but this is really not a problem. If you want to figure this out, just decomposite the class to products of classes that we already know where they are mapped from and use the fact that $L$ is a homomorphism).

Therefore, $F(X)$ is indeed the solution of the problem.

## Properties of the free group

#### Reduced form

For any class of words will define the reduced form of it as $[x]$ such that:

• There are no two adjacent elements $a\in A, b\in A^\prime$ in $x$ such that $b=a^{-1}$. For example, the word $(x_1,x_2,x_2^{-1})$ is not reduced.

(We will also call a word ‘reduced’ if it satisfies this property)

It’s not hard to see that every class has such a word. For example, if you consider the class of the empty word, then the empty word itself is it’s reduced form.

Moreover, it turns out that the reduced form is unique. That is, if $x,y$ are two reduced words such that $[x]=[y]$, then $x=y$.

How do we prove it? By an easy induction:

Suppose that there were two reduced words in the same class. Thus, we can change one to the other using finite number of actions. For example, if we want to transform the word $(x_1,x_2)$ to the word $(x_1,x_2,x_5,x_5^{-1},x_3,x_4^{-1},x_4)$ we do the following ‘moves’:

(x_1,x_2)\sim (x_1,x_2,x_3,\color{blue}x_4^{-1},x_4\color{black})\sim (x_1,x_2,\color{blue}x_5,x_5^{-1}\color{black},x_3,x_4^{-1},x_4)

And it took us only one ‘mid-word’. So the induction is going to be on the number of ‘mid-words’ – $n$.

##### $n=1$$n=1$

So there is one ‘mid-word’ which we will call $M$:

x\sim M\sim y

Since $x$ and $y$ are both reduced, then we must extend $x$ to $M$ and shrink $M$ to $y$. But there are only two ways to extend $x$:

The first is:

\overbrace{(---)}^{x}\sim\overbrace{(--,a,a^{-1},--)}^{M}

The only way to shrink $M$ now is to cancel out $a,a^{-1}$. And this action brings us back to $x$, so $y$ must be $x$.

The second way is:

\overbrace{(---)}^{x}\sim\overbrace{(--,a^{-1},a,--)}^{M}

And from the exact same argument, we conclude that $x=y$.

##### $n>1$$n>1$

Let’s denote by $M$ the longest ‘mid-word’, and it’s ‘neighbors’ in the sequence as $m_1,m_2$:

x\sim\cdots\sim m_1\sim M\sim m_2\sim\dots\sim y

So we must extened $m_1$ to $M$ and shrink $M$ to $m_2$. From similar arguments as those in the above we conclude that $m_1=m_2$. Therefore, $M$ is ‘unnecessary’, and we can remove it from the sequnce to get a shorter one:

x\sim\cdots\sim m_1= m_2\sim\dots\sim y

We can use the induction to conclude that $x=y$, and we are done.

## Finite sets and notations

Proving that the reduced form is unique will help us a lot in the future and make our lives much easier! Can you think about how exactly?

In fact, the same exact thing applies to Free products, however, I didn’t mentioned it there and I made some ‘lies’ there when I stated, for example, that:

[(g)]\neq[(h)]\ \ \ g\in G,h\in H

Though it’s pretty clear that those classes are not the same, we still need to prove it… However, the proof has the exact same idea as the one in the above so I won’t bother presenting it here…

Before I move on to more results I want to present some new notations that will make things look more aesthetic and readable.

First, instead of writing a word in the form:

[(x_1,x_2,\dots,x_n)]

We’ll just write it as $x_1x_2\cdots x_n$ (the same thing applies to the free product as well).

In addition, If the same element appears on a word next to itself mutiple times we will write it as (for example):

[(\overbrace{x_1,x_1,\dots,x_1}^{n\text{ times}},x_2,\dots,x_n)]\to x_1^nx_2\cdots x_n \\

[(x_1^{-1},x_1^{-1},x_1^{-1})]\to x_1^{-3}

Moreover, if the set $X$ is finite, with $n$ elements, then we’ll denote the free group over $X$ as:

F_n

Great, after we’ve met those notations, let’s start presenting some results:

## What is $F_1$$F_1$?

Suppose that $X=\{a\}$. Now, it’s not hard to see that all of the reduced words are of the form:

aa\cdots a =a^n

or:

 a^{-1}a^{-1}\cdots a^{-1} = a^{-n}

And we’ll denote the empty word as:

[()]=a^0

And it’s pretty clear that (verify it):

a^na^m=a^{n+m}

For any $n,m\in \mathbb{Z}$. Hence, we can easily define an isomorphism $\varphi:F_1\to \mathbb{Z}$ as:

\varphi(a^n)=n

Therefore:

F_1\cong \mathbb{Z}

This fact will be quite useful in the future – sometimes we would want to write $F_1$ instead of $\mathbb{Z}$, why? It follows from the simple fact that the action on $\mathbb{Z}$ is ‘+’, and sometimes, we would prefer the dot instead.

## Connection with the free product

Since the free product’s and the free group’s constructions are so similar, they’ve got to be related to each other in some way, right? so as it turns out, they do have a connection, and a pretty great one:

For every two natural numbers $n,m$:

F_n*F_m=F_{n+m}

That’s great! But how do we prove it? The idea is to construct two homomorphisms:

L:F_n*F_m\to F_{n+m}

F_n*F_m\leftarrow F_{n+m} : \tilde{L}

And show that they are inverting each other. This will allow us to conclude that those are isomorphisms and that’s what we want.

For the sake of comfort, let’s denote:

A_n=\{x_1,\dots,x_n\} \\
A_m=\{y_1,\dots,y_m\}

And we need to prove that:

\overbrace{F(A_n)}^{F_n}*\overbrace{F(A_m)}^{F_m}=\overbrace{F(A_n\cup A_m)}^{F_{n+m}}

#### Proving with universal porperties

Note that every word in $F_n$ and $F_m$ is also a word in $F_{m+n}$, so that gives us two homomorphisms:

i_n:F_n\to F_{n+m} \\
i_m:F_m\to F_{n+m}

Which are just the inclusion homomorphisms:

F_n(\vec x)= \vec x \\
F_m(\vec x)= \vec x

($\vec x$ is just a short way to denote a word, which I just made up…)

And now we can use the universal property of the free product :

(recall that $\ell_n(x_i)=[(x_i)] = x_i$ and $\ell_m(y_i)=[(y_i)]=y_i$)

From the property we can conclude that there exists some homomorphism $L:F_n*F_m\to F_{n+m}$ such that:

L\circ \ell_n=i_n \\
L\circ \ell_m=i_m \\ \Downarrow \\
L(x_i)=x_i \\
L(y_i)=y_i


(Note that I am ommiting so many different brackets such as $[()]$ – what’s really happennig here is that $L$ gets a word in $F_{n}*F_m$ and translates it to a word in $F_{n+m}$, they may be writtent the same, but we need to think of them differently – $L$ is not the identity)

The universal property also assures us that $L$ is a homomorphism and that allows us to conclude it’s full behavior, for example:

L(x_1y_2x_1^{-1}y_3)=L(x_1)L(y_2)L(x_1^{-1})L(y_3)
=L(x_1)L(y_2)L(x_1)^{-1}L(y_3)=x_1y_2x_1^{-1}y_3

Perrfect! So we now have one homomorphism, what about the other one? Well, the idea is pretty much the same, we are going to use the universal property of the free group – the one we’ve just met! All we need is a function:

f:\{x_1,\dots,x_n,y_1,\dots,y_m\}\to F_{n}*F_m

But this is really simple, we can just map $x_i,y_j$ to their corresponding words:

f(x_i)=x_i \\
f(y_i)=y_i

(Again – this is not the identity). So now the diagram looks like:

Then, there exists some homomorphism $\tilde{L}:F_{n+m}\to F_n*F_m$ such that:

\tilde{L}\circ i=f \\ \Downarrow \\
\tilde{L}(x_i)=f(x_i)=x_i\\
\tilde{L}(y_i)=f(y_i)=y_i

And, same as before, we get that, for example:

\tilde{L}(x_1x_2y_3y_4^{-1})=\tilde{L}(x_1)\tilde{L}(x_2)\tilde{L}(y_3)\tilde{L}(y_4^{-1})=x_1x_2y_3y_4^{-1}

Well, it is pretty clear now that:

L\circ\tilde{L}=id_{F_{n+m}} \\


and:

\tilde{L}\circ L=id_{F_n*F_m}

Therefore, $L,\tilde{L}$ are indeed isomorphisms, hence:

F_{n}*F_m\cong F_{n+m}

as we wanted!

## Summary

After we’ve met the free group and saw some pretty interesting results, in the next post, we will present the last universal problem, after we do so, we’ll prove “Seifert–van Kampen theorem” (I recommend reading this article and paying attention to the diagram there, does it look familiar?).

After proving this theorem, we will be able to compute a lot, and I mean it, a lot of non-trivial fundamental groups!