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** using some function . Note that this map **crosses categories**, it is from the category of sets to the category of groups.

The catch here is that 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 with homomrphism such that all the homomorphisms go ‘through it’ exists?”

Formally, we are asking the **existence **of a **unique** such that .

So I won’t prove the uniqueness of 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** of the set with a **copy** of itself – :

A=X\cup X^\prime

We think of the sets and as **different** sets, and in order to tell where an element of is from, we add to all the elements in 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 to be the set of all the **words** made from elements of 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 is:

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

Let’s denote by 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 is a **group**! And this one has a special name – it is called **the free group over a set **.

Now we need to verify that 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 to the class :

i(x)=[(x)]

Let’s take another look at the diagram:

We need to find a unique such that:

f=L\circ i

That is, for every :

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

And that’s exactly how has to be defined on elements of the form . But what about elements such as . That’s not really a problem, since 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 in that way) of (I didn’t show how acts on classes such as , 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 is a homomorphism).

Therefore, 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 such that:

- There are no two adjacent elements in such that . For example, the word 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 are two reduced words such that , then .

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 to the word 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’ – .

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

x\sim M\sim y

Since and are both reduced, then we must **extend ** to and **shrink** to . But there are only two ways to extend :

The first is:

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

The only way to shrink now is to cancel out . And this action brings us back to , so must be .

The second way is:

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

And from the exact same argument, we conclude that .

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

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

So we must **extened** to and shrink to . From similar arguments as those in the above we conclude that . Therefore, 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 , 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 (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 is finite, with elements, then we’ll denote the free group over as:

F_n

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

## What is ?

Suppose that . 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 . Hence, we can easily define an **isomorphism** 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 instead of , why? It follows from the simple fact that the action on 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 :

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 and is also a word in , 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

( 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 and )

From the property we can conclude that there exists some homomorphism 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 gets a word in and translates it to a word in , they may be writtent the same, but we need to think of them differently – is **not** the identity)

The universal property also assures us that 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 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 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, 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!