Abelianization

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 – $G$ . Our goal is to map it into an abelian group $K$ using an homomorphism:

\varphi:G\to K

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

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

The diagram that fits for the situation is:

So we can now ask the question formally – we are looking for a group $U$ , with a homomorphism $\color{blue}g\color{black}:G\to \color{blue}U$ such that for every abelian group $K$ , with a homomorphism $\varphi: G\to K$ , there exists a unique homomorphism $\color{blue}L\color{black}:\color{blue}U\color{black}\to K$ where:

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

Note that this those maps from $G$ are not just regular maps – they cross categories. $G$ is an object of the category of groups while $U$ and $K$ 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 $\color{blue}U\color{black},\color{red}U^\prime$ that satisfy the above, thus we have two diagrams:

And again, since the $K$ is arbitrary, we can pick $K=\color{red}U^\prime$ in the left diagram and $K=\color{blue}U$ in the right diagram.

Both diagrams yield an equation. From them we can get two equations which allow us to conclude that $L$ is an isomorphism from $\color{blue}U$ to $\color{red}U^\prime$ .

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 $G$ to an abelian group? We know nothing about $G$ , 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 $G$ (this is not the same $G$ from the problem) is a group with a commutative action, that is, for every $x,y$ 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 $G$ is a group and $N$ is a normal subgroup of $G$ , 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 $\{xyx^{-1}y^{-1}\}$ 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 $G$ , and the elements of the form $xyx^{-1}y^{-1}$ are denoted as $[x,y]$ .

In order to verify that this is indeed a normal subgroup, we have to show that for every $g\in G$ and $h\in N$ :

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 $[x,y]\in[G,G]$ and let $g\in G$ 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 $[G,G]$ is indeed a normal subgroup, therefore, we can now finally define the Abelianization of $G$ 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 $U$ to be the abelianization – $\text{Ab}(G)$ . Moreover, there is a natural homomorphism from $G$ 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 $\color{blue} L$ is. We know that it must satisfy:

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

That is, for every $g\in G$ :

\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 $\color{blue}L$ is! As I mentioned before, $\rho$ is onto thus every element in $\text{Ab}(G)$ is of the form $\rho(g)$ for some $g\in G$ . But, and this is important – $L$ may not be even well defined! Who said that if $g_1[G, G]=g_2[G,G]$ then $\varphi(g_1)=\varphi(g_2)$ . 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 $G$ be a group and $N\vartriangleleft G$ a normal subgroup. Moreover, suppose that $H$ is another group such that $\varphi:G\to H$ is an homomorphism. Finally, suppose that $\rho:G\to G/N$ 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 $L$ is well-defined. Let $g\in N$ be some element. Thus, $gN=1_GN$ , therefore:

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

As we wanted.

Conversly, suppose that $N\sube \ker\varphi$ . Let $g_1,g_2\in G$ be two elements such that $g_1N=g_2N$ . We want to show that $L(g_1N)=L(g_2N)$ :

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 $[G,G]\sube \ker\varphi$ , and it’s enough to check on an element of the form $[x,y]$ (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 $K$ ! And $K$ 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 $L$ is indeed well-defined. Since we were ‘forced’ to define $L$ 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 $G$ , 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.