# Generators and Relations

In this post we are going to face the very last universal problem, it won’t be so hard though it won’t be as easy as the previous problem. However, the solution for the problem is great! we will use it a lot, since it can be really comfortable in some cases. Ok, let’s do it:

## The problem

In the previous problem we started with only one set $X$ and a function $f$ from it to a group. $f$ had no restrictions – it was free.

Now the case is a bit different – we do want to set some restrections, then the second thing we are given is a set $R$ of restrections, which are just words made of elements of the set $X$ and their ‘inverses’. For example:

X=\{x_1,x_2,x_3\},  R=\{x_1x_2\ ,\ x_2x_1^{-1}\ ,\ x_3^{-1}x_3^{-1}x_1\}

And now we won’t ‘allow’ any function $f:X\to K$ (where $K$ is a group), we want $f$ to respect the restrections, that is:

\hat f(r)=1_K

where $r=x_{i_1}^{j_1}\cdots x_{i_k}^{j_k}\in R$ and $\hat{f}$ is defined as:

\hat f(r)=\hat f(x_{i_1}^{j_1}\cdots x_{i_k}^{j_k}) = f(x_{i_1})^{j_1}\cdots f(x_{i_k})^{j_k}=1_K


As always, we are looking for a universal group $U$ with a function $i:X\to U$ that respects the realtion:

That is, for any given group $K$ with a function $f$ that respects the realtion, $f$ must ‘go through’ $U$. In other words, there exists a unique homomorphism $L:U\to K$ such that:

f=L\circ i

### Solving the problem

Since the diagram is the exact same thing as the previous problem, the first thing that comes to your head is probably: “maybe the free group is the solution here too?”

Well, no… The free group’s homomorphism may not / will not respect the restrections. However, we can use a trick we’ve already used twice before – we divide it by a normal subgroup.

So we’ll take the (minimal) normal subgroup $N$ which is genarated by the set:

A=\{\ \hat{i}(r):r\in R\ \}

So now the diagram looks like:

We still need to find the map from $X$ to $F(X)/N$ and the homomorphism $L:F(x)/N\to K$.

Let’s present the diagram a bit different (recall that $\rho:F(X)\to F(X)/N$ is the natural projection, $\vec{x}\mapsto \vec{x} N$):

Now it’s pretty cleat what the map from $X$ to $F(X)/N$ should be, we’ll define it as:

i_N=\rho\circ i \\
i_N(x) =\rho(i({x}))=\rho(x)=\overline{x}\in F(X)/N

Now, for every $r\in R$:

\hat{i}_N(r)\in N\Rightarrow \hat{i}_N(r)=1_{F(X)/N}

So $\hat{i}_N$ indeed respects the restrections!

The only thing left to find is the homomorphism $L:F(X)/N\to K$. For that, we’ll use only part of the diagram:

This is something we are already familiar with, As we’ve proved in this post, we know that $L^\prime$ induces a well defined homomrphism $L$ if and only if $N$ is contained in the kernel of $L^\prime$.

Moreover, $L$ will be unique since $\rho$ is onto, and every element of the quotient group can be presented as $\rho(\vec{x})$ for some $\vec{x}\in F(X)$, and $L(\rho(\vec{x}))=L^\prime$.

Ok, so let’s find out if $N\sube \ker L^\prime$:

Recall that it is enough to verify that the genarators of normal subgroup are in the kernel, since the other elements in the subgroup are ‘made of’ them. Therefore, we need to verify that:

L^\prime(\hat{i}(r))=1_K \Rightarrow L^\prime\circ \hat{i}(r)=1_K

But wait… this composition is just $\hat{f}$!

And by the definition of $f$, we know that it respects the restrections! Thus, $N$ is indeed contained the kernel of $L^\prime$, as we wanted.

#### The final solution

Perfect – we’ve solved the problem, as it turns out, the universal group that solves it is the group:

F(X)/N

And, as all the other solutions, it has a special name:

It is called a representation of a group with generators and relations (pretty long name…) and we denote it as:

\langle X| R\rangle

$X$ is called the set of genarators and $R$ is called the set of the relations.

Note how comfortable this notation is – it tells us the whole information about the group in a ‘compact’ way.

Let’s see an example: Supopse that the set of genarators is $X=\{\sigma,\tau\}$ And the set of relations is: $R=\{\sigma^n, \tau^2, \tau^{-1}\sigma\tau^{-1}\sigma\}$. Then the group is:

\langle \sigma,\tau | \sigma^n\ , \ \tau^2\ , \ \tau^{-1}\sigma\tau^{-1}\sigma\rangle

Are you familiar with this group? If you’ve studied group theory you probably do – this is the Dihedral group$D_n$.

Pretty cool right? note that we can present any group we want in that way – for example, if we want to present the cyclic group $\mathbb{Z}_n$ we can write:

\langle a|a^n\rangle

(Convince yourself that it works!). And what about the infinite cyclic group – $\mathbb{Z}$? Easy, this is just the set genarated by one element with no restrections – and that is $F_1$:

\lang a|\rangle=F_1\cong \mathbb{Z}

Even without showing any special property of this solution, I think that you can already understand how it can be very comfotable in some cases.

## Summary

That’s it! We are done with the universal problem, and the next post is going to be about topology again – As I said in the previous post, we are going to prove Seifert–van Kampen theorem.