Free product with amalgamation

We are now facing the fourth universal problem – the most challenging one yet! Don’t worry though, I’ll try to explain it as simple and organized as possible. Moreover, despite it being challenging, it is also the most rewarding problem, as we’ll see in the future. Let’s do it:

The problem

Now there are three given groups: G,\color{blue}G\color{black},\color{red}H. However, Those group also come with two homomorphisms:

\color{blue}i\color{black}:P\to \color{blue}G\\
\color{red}j\color{black}:P\to \color{red}H

Now, for any given group K, we are interested in two homomorphisms:

\color{blue}\varphi\color{black}:\color{blue}G\color{black}\to K \\
\color{red}\psi\color{black}:\color{red}H\color{black}\to K

Such that, intuitively, no matter which group we choose to “go through” we will get the same homomorphism.

Formally, we want that:


In other words, we want the diagram to ‘commute’. But, as we already know from the previous cases, this is not so hard to find such two (find them!). The interesting question is:

“Is there some unique universal group U such that all the homomorphisms like \varphi and \psi must ‘go through’?”

So we are looking for a group U equipped with two homomorphisms:

\color{blue}\ell_G\color{black}:\color{blue}G\color{black}\to U \\
\color{red}\ell_H\color{black}:\color{red}H\color{black}\to U

And a unique homomorphism from L:U\to K, (where K is some group with two homomorphims like in the above) such that the diagram:

Satisfies two things:

First, we want U and it’s homomorphisms to satisfy the desired property, that is:


Secondly, we also want the ‘usual thing’, which is that \varphi,\psi are both ‘go through’ U. That is:

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

Same as last post, I won’t prove the uniqueness of U, since, again, this is the exact same proof for all the universal problems, if you want to see one – you can find it here.

First attempt

Let’s try to use the following insight:

If we ignore P and the homomorphisms that comes out of it, the diagram that we will get is:

Looks familiar? Of course it is – this is just the diagram that the free product solves! So maybe the free product is the solution? Let’s try it:

Now U=G* H and the homomorphisms \color{blue}\ell_G\color{black},\color{red}\ell_H are the homomorphisms \color{blue}i_G\color{black},\color{red}i_H respectivly.

(recall that i_G(g)=[(g)],i_H(h)=[(h)])

So the diagram is:

And in order for it to be the solution, we need that:


That is, for every p\in P:


Which is equivalent to:


And that is… not true at all! Why should those terms should be equals? suppose that i(p)=g,j(p)=h for some g\in G, h\in H that are not the identity. Thus:


So this attempt fails…

Wait, not everything is lost!

Maybe we can fix it, maybe we can force elements of the form:


To be the identity. Does that remind you something? That’s a really similar problem to what we’ve faced against in the problem of the Abelianization. Let’s just take a normal subgroup that contains the set:

A=\{(\color{blue}i_G\color{black}\circ\color{blue}i\color{black}(p))(\color{red}i_H\color{black}\circ\color{red}j\color{black}(p))^{-1}:p\in P\}

and divide the group by it.

So in the case of the abelianization, we were lucky and the sub-group that was genarated by the set we wanted to be equivalent to the identity was indeed normal.

However, that’s not the case here, the subgroup \langle A\rangle may-not be normal!

So how can we solve it? We can pick N to be the minimal normal sub-group that contains A. But what is it exactly? Well, there are two ways to construct it:

  1. From the outside – take the intersection of all the normal subgroups in G*H that contains A.
  2. From the inside – The normal subgroup will be the set A with all of the products of elements of A, along with all the elements of the form g^{-1}ag where a is an element of the subgroup that generated by A.

So now the candidate for the solution is the group (G*H) / N.

Before I’ll draw the new diagram, I want to denote

\rho:G*H\to(G*H)/H \\

As the natural projection to the quotient group.

Ok, I’ll present now the new diagram, but I’ll add some extra stuff there that will help us to see things clearer.

Yes, I know, This diagram is massive!

So now we have:

\color{blue}\ell_G\color{black}=\rho\circ\color{blue}i_G\color{black} \\

And we want two things:

(G*H)/N and it’s homomorphisms to satisfy the desired property

That is, for every p\in P:


We can use the relations in the above to present the expression as:


Which is equivalent to:


And that’s exactly the same as showing that:


We can now use the properties of a homomorphism to simplify the expression:

\rho(\ (\color{blue}i_G\color{black}\circ\color{blue}i\color{black}(p))(\color{red}i_H\color{black}\circ\color{red}j\color{black}(p))^{-1}\ )=1_{(G*H)/N}

But by definition, we know that (\color{blue}i_G\color{black}\circ\color{blue}i\color{black}(p))(\color{red}i_H\color{black}\circ\color{red}j\color{black}(p))^{-1}\in N, that’s exactly the reason why we’ve even defined this normal subgroup!

Therefore, the last expression is indeed true, hence, all the previous expressions are also true, and in particular, we conclude that:


As we wanted!

\varphi,\psi are both ‘go through’ U

That is, we need to find a unique homomorphism L:(G*H)/N\to K such that:

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

First, I’ll consider only part of the digram, in fact, we’ve already discussed:

I’ve added this green arrow with a function M. But why? If we ignore the quotient group in the middle, we have the diagram that G*H solves. In this problem, we’ve seen that there exists a unique M such that:

\color{blue}\varphi\color{black}=M\circ\color{blue}i_G\color{black} \\

Now if we’ll only consider at the groups G*H, (G*H)/N, K, we have the following diagram:

And we’ve already proved here, that M induces a well defined homomrphism L if and only if N\sube \ker M. Moreover, since \rho is onto, we can present each element of (G*H)/N as \rho(x) for some x\in G*H. Therefore, L is unique and defined as:


Therefore, all we have to show is that N is indeed contained in the kernel of M, if it is, then we are done!

First of all, note that it’s enough to show that:


For some p\in P. This follows from the fact that every element of N is a product of such elements or a conjugate of such products. We can use the fact that M is a homomorphism and write the expression as:


Which is equivalent to:


And we can write this expression as:


But that’s exactly the same as:


We know what those compositions are:


And this expression is true! why? recall that this was just ‘part of the ruels’. The assumption was that those homomorphisms satisfiy the proeprty and we were looking for some universal group, where those homomorphism ‘go thorugh’.

Great, so we can now conclude that N is indeed in the kernel of M, hence, L is unique and well defined, as we wanted!

The free product with amalgamation

Great, so we’ve finally solved the problem! We found out that the group (G*H)/N is the solution for the problem.

This group is very important, it even has a name. It is called The free product with amalgamation of G and H, and we denote it as:


That’s a pretty detailed symbol – it tells us who are exactly the groups and the homomorphisms. Sometimes I’ll denote it as G*_{p}H

The diagram now is:

And that’s it, we’ve solved this massive problem.

We now have only two more universal problem to solve, which are pretty easy, and then – we’ll finally go back to topology and use all those universal solutions to deduce some outstanding results!

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