After solving our first universal problem, It’s time for the second one. And as you may guess from the title, we are going to meet a new term which is called “**Abelianization**“. We’ll see exactly what does that mean:

## The universal problem

Now the rules of the game are:

We are given only one group – . Our goal is to map it into an **abelian group** using an homomorphism:

\varphi:G\to K

But this is not hard… We are really asking the question: “Does an **universal ableian group** exists?” That is:

Intuitively, we are looking for some abelian group such that all of the homomorphism from to an arbitrary abelian group are going through .

The diagram that fits for the situation is:

So we can now ask the question formally – we are looking for a group , with a homomorphism such that for every **abelian group** , with a homomorphism , there exists a **unique** homomorphism where:

\varphi=\color{blue}L\color{black}\circ\color{blue} g

Note that this those maps from are not just regular maps – they **cross** **categories**. is an object of the **category of groups** while and are objects of the category of **abelian groups**!

So now that we understand the problem, we shall begin with proving uniqueness

### uniqueness

I am not going to go over the whole thing, since it is **exactly** the same as the previous problem:

We assume that there are **two** groups that satisfy the above, thus we have two diagrams:

And again, since the is arbitrary, we can pick in the **left** diagram and in the **right** diagram.

Both diagrams yield an equation. From them we can get two equations which allow us to conclude that is an isomorphism from to .

Try to do it yourself – this is the exact same procedure that I have done in the last post.

### Finding the universal group

Ok, after we know the universal group unique, we need to find one!

How can we map to an abelian group? We know nothing about , it can be **any** group! This leads me to the question – is there a way to create an abelian group from any group? If so, how can I do it? Let’s try to figure it out:

The first question I want to ask is: What is an abelian group? (Of course I am assuming that you know the answer…) An abelian group (this is **not** the same from the problem) is a group with a **commutative action**, that is, for every in this group:

xy=yx

Note that we can write this equation a little different:

xyx^{-1}y^{-1}=1

(1 is the identity element) If the group is **not** abelian, this eqaution may not be true for some elements in the group.

However, maybe we can **‘force’** the group to satisfy this equation for any two elements of it?

How can we even do such thing? How do “force” elements to be the identity? The answer is simple – we **divide** the group by a **normal subgroup**. That is, if is a group and is a normal subgroup of , we will consider the **quotient group**:

G/N

Great, so we now have a clue to a solution: Let’s just take subgroup that is **generated** by all of the elements of the form and divide the group by it! However, there is one problem though – who said that this subgroup is **normal** – we don’t know yet if we can divide the group by it or not.

### Is it normal?

Let’s check if we are lucky enough and this group is indeed a **normal** subgroup. I want to denote this subgroup as:

[G,G]=\langle\{xyx^{-1}y^{-1}:x,y\in G\}\rangle

This group has a special name – it is called the **Commutator subgroup** of , and the elements of the form are denoted as .

In order to verify that this is indeed a normal subgroup, we have to show that for every and :

ghg^{-1}\in N

And it’s enough to check that this property is valid for the elements in the generator set of the subgroup, since the rest of the elements are just products of those.

Ok, so let’s pick some and let be some arbitrary element. Now:

\color{blue}g\color{black}[x,y]\color{blue}g^{-1}\color{black}=\color{blue}g\color{black}\cdot xyx^{-1}y^{-1}\cdot \color{blue}\color{blue}g^{-1}\color{black}\color{black}=\color{blue}\color{blue}g\color{black}\color{black}\cdot x\cdot \color{blue}g^{-1}\color{black}\color{blue}g\color{black}\cdot y\cdot \color{blue}g^{-1}\color{black}\color{blue}g\color{black}\cdot x^{-1}\cdot {\color{blue}g^{-1}\color{black}\color{blue}g\color{black}\cdot }y^{-1}\cdot \color{blue}g^{-1}\color{black}

=(\color{blue}\color{blue}g\color{black}\color{black} x \color{blue}g^{-1}\color{black})(\color{blue}g\color{black}y \color{blue}g^{-1}\color{black})(\color{blue}g\color{black} x^{-1} {\color{blue}g^{-1}\color{black})(\color{blue}g\color{black} }y^{-1} \color{blue}g^{-1}\color{black})

=(\color{blue}\color{blue}g\color{black}\color{black} x \color{blue}g^{-1}\color{black}) (\color{blue}g\color{black}y \color{blue}g^{-1}\color{black}) (\color{blue}\color{blue}g\color{black}\color{black} x \color{blue}g^{-1}\color{black})^{-1} (\color{blue}g\color{black}y \color{blue}g^{-1}\color{black})^{-1}=[\color{blue}\color{blue}g\color{black}\color{black} x \color{blue}g^{-1}\color{black},\color{blue}g\color{black}y \color{blue}g^{-1}\color{black}]\in[G, G]

So a little compuatation showed that is indeed a normal subgroup, therefore, we can now finally define the **Abelianization** of as:

\text{Ab}(G)=G/[G,G]

Of course, by it’s defintion, this group is **abelian**.

### Back to the diagram

Now, it is the most natural thing to do is to define the universal group to be the abelianization – . Moreover, there is a natural homomorphism from to it:

\rho:G\to G/[G, G] \\ g\mapsto g[G,G]

And this map is **onto**.

Great, let’s see how the diagram looks like now:

The only thing left to do is to find what is. We know that it must satisfy:

\varphi=\color{blue}L\color{black}\circ\color{blue}\rho

That is, for every :

\varphi(g)=\color{blue}L\color{black}\circ\color{blue}\rho\color{black}(g)=\color{blue}L\color{black}(\color{blue}\rho\color{black}(g))=\color{blue}L\color{black}(g[G, G])

And we’ve found exactly what is! As I mentioned before, is **onto** thus every element in is of the form for some . **But**, and this is important – may not be even **well defined**! Who said that if then . It may not be true at all! We need to verify it, in order to do so, I want to prove a general theorem in group theory:

#### Quick theorem

Let be a group and a normal subgroup. Moreover, suppose that is another group such that is an homomorphism. Finally, suppose that is the natural projection.

Consider the following diagram:

Then:

L\text{ is well defined} \iff N\sube \ker\varphi

This theorem shows us exactly when the map from the qutient group is well-defined.

The proof is really easy:

Supopse that is well-defined. Let be some element. Thus, , therefore:

\varphi(g)=L(gN)=L(1_GN)=\varphi(1_G)=1_H \\ \Downarrow \\ g\in\ker\varphi

As we wanted.

Conversly, suppose that . Let be two elements such that . We want to show that :

g_1N=g_2N\Rightarrow g_2^{-1}g_1N=N\Rightarrow g_2^{-1}g_1\in N\sube\ker\varphi

\Rightarrow\varphi(g_2^{-1}g_1)=1_H\Rightarrow\varphi(g_2)^{-1}\varphi(g_1)=1_H\Rightarrow\varphi(g_1)=\varphi(g_2)\Rightarrow L(g_1N)=L(g_2N)

And that’s it.

### Back to our business

So now all we have to verify is that , and it’s enough to check on an element of the form (why?). Let’s do it:

\varphi([x,y])=\varphi(xyx^{-1}y^{-1})=\varphi(x)\varphi(y)\varphi(x)^{-1}\varphi(y)^{-1}

But this is a product in ! And is **abelian** thus:

\varphi(x)\varphi(y)\varphi(x)^{-1}\varphi(y)^{-1}=\overbrace{\varphi(x)\varphi(x)^{-1}}^{1_K}\overbrace{\varphi(y)\varphi(y)^{-1}}^{1_k}=1_k \\ \Downarrow \\ [x,y]\in\ker\varphi

So we can apply the theorem and conclude that is indeed well-defined. Since we were ‘forced’ to define in that way, then we know that it is unique as well.

## Summary

That’s it for this problem – we have defined a new group – the **Abelianization** of , and by that, we managed to find a solution for the universal problem!

In the next post, we will meet **another** new group that solves another universal problem – that is the **free product** of two groups.