Topology/Free group and presentation of a group

Free monoid spanned by a set edit

Let   be a vector space and   be a basis of  . Given any vector space   and any elements  , there is a linear transformation   such that  . One could say that this happens because the elements   of a basis are not "related" to each other (formally, they are linearly independent). Indeed, if, for example, we had the relation   for some scalar   (and then   wasn't linearly independent), then the linear transformation   could not exist.

Let us consider a similar problem with groups: given a group   spanned by a set   and given any group   and any set  , does there always exist a group morphism   such that  ? The answer is no. For example, consider the group   which is spanned by the set  , the group   (with the adition operation) and the set  . If there exists a group morphism   such that  , then  , which is impossible. But if instead we had choose  , then such a group morphism does exist and it would be given by  . Indeed, given any group   and any  , we have the group morphism   defined by   (in multiplicative notation) that verifies  . In a way, we can think that this happens because the elements of the set   (that spans  ) don't verify relations like   (like  ) or  . So, it seems that   is a group more "free" that  .

Our goal in this section will be, given a set  , build a group spanned by the set   such that it will be the most "free" possible, in the sense that it doesn't have to obey relations like   or  . To do so, we begin by constructing a "free" monoid (in the same sense). Informally, this monoid will be the monoid of the words written with the letters of the alphabet  , where the identity will be the word with no letters (the "empty word"), and the binary operation of the monoid will be concatenation of words. The notation   that we will use for the element of this monoid meets this idea that the elements of this monoid are the words   where   are letters of the alphabet  . Here is the definition of this monoid.

Definition Let   be a set.

  1. We denote the  -tuples   with   and   by  .
  2. We denote  , that is   with  , by  .
  3. We denote by   the set  .
  4. We define in   the concatenation operation   by  .

Next we prove that this monoid is indeed a monoid. It's an easy to prove result, we need to show associativity of   and that  .

Proposition   is a monoid with identity  .

Proof The operation   is associative because, given any  , we have


It's obvious that   has the identity   as   by the definition of   and  .  

Following the idea that the monoid   is the most "free" monoid spanned by  , we will call it the free monoid spanned by  .

Definition Let   be a set. We denote the free monoid spanned by   by  .


  1. Let  . Then   and, for example,  .
  2. Let  . Then   and, for example,  .

Free group spanned by a set edit

Now let us construct the more "free" group spanned by a set  . Informally, what we will do is insert in the monoid   the inverse elements that are missing in it for it to be a group. In a more precise way, we will have a set   equipotent to  , choose a bijection from   to   and in this way achieve a "association" between the elements of   and the elements of  . Then we face   (with  ) as having the inverse element   (with  ), where   is is associated with  , respectively. Let us note that the order of the elements in   is "reversed" because the inverse of the product   must be  , and the   are, respectively,  . The way we do that   be the inverse of   is to take a congruence relation   that identifies   with  , and pass to the quotient   by this relation (defining then, in a natural way, the binary operation of the group,  ). By taking the quotient, we are formalizing the intuitive idea of identifying   with  , because in the quotient we have the equality  . Let us give the formal definition.

Definition Let   be a set. Let us take another set   equipotent to   and disjoint from   and let   be a bijective application.

  1. For each   let us denote   by  , for each   let us denote   by   and for each   let us denote   by  .
  2. Let   be the congruence relation of   spanned by  , this is,   is the intersection of all the congruence relations in   wich have   as a subset. We denote the quotient set   by  .

Frequently, abusing the notation, we represent an element   simply by  .

Because the operation   that we want to define in   is defined using particular represententes   and   of the equivalence classes   and  , a first precaution is to verify that the definition does not depend on the chosen representatives. It's an easy verification.

Lemma Let   be a set. It is well defined in   the binary operation   by   (where   is the congruence relation of the previous definition).

Proof Let   be any elements such that   and  , this is,   and  . Because   is a congruence relation in  , we have  , this is,  .  

Because the definition is valid, we present it.

Definition Let   be a set. We define in   the binary operation   by  .

Finally, we verify that the group that we constructed is indeed a group.

Proposition Let   be a set.   is a group with identity   and where  .


  1.   is associative because    
  2. Let us see that   is the identity  . Let   be any element. We have   and, in the same way,  .
  3. Let   be any element and let us see that  . We have   and, by definition of  ,  , this is,  , therefore   and, in the same way,  .  

In the same way that we did with the free monoid, we will call free group spanned by the set   to the more "free" group spanned by this set.

Definition Let   be a set. We call free group spanned by   to  .

Example Let  . Let us choose any set   disjoint (and equipotent) of  . Let   be any (in fact, the only) bijective application of   in  . Then we denote   by   and we denote   by  . We regard   and   as inverse elements. Let   be the congruence relation of   spanned by  .   is the set of all "words" written in the alphabet  . For example,  .

We have   and, for example,  , because   (therefore  ) and because   is a congruence relation, we can "multiply" both "members" of the relation   by   and obtain  . We see   as meaning that in   We have   (more precisely,  ), and we think in this equality as being the result of one   "cut out" with   in  .

Given  , let us denote the exact number of times that the "letter"   appears in   by   and let us denote the exact number of times the "letter"   appears in   by  . Then "cutting"  's with  's it remains a reduced word word with   times the letter   (if  , let us us consider that there aren't any letters   and remains   times the letter  ). Let us denote   by  . We have

  1.   if and only if   and
  2.  .

In this way, each element   is determined by the integer number   and the product   of two elements   correspondent to the sum of they associated integers numbers   and  . Therefore, it seems that the group   is "similar" to  . indeed   is isomorph to   and the application   is a group isomorphism.

Presentation of a group edit

Informally, it seems that   is obtained from the "free" group   imposing the relation  . Let us try formalize this idea. We start with a set   that spans a group   that que want to create and a set of relations   (such as   or  ) that the elements of   must verify and we obtain a group   spanned by   and that verify the relations of  . More precisely, we write each relation   in the form   (for example,   is written in the form  ) and we see each   as a "word" of  . Because   doesn't have to be a normal subgroup of  , we can not consider the quotient  , so we consider the quotient   where   is the normal subgroup of   spanned by  . In  , we will have  , which we see as meaning that in   the elements   and   are the same. In this way,   will verify all the relations that we want and will be spanned by   (more precisely, by  ). Let us formalize this idea.

Definition Let   be a group. We call presentation of  , and denote by  , to a ordered pair   where   is a set,   and  , where   is the normal subgroup of   spanned by  . Given a presentation  , we call spanning set to   and set of relations to  .

Let us see some examples of presentations of the free group   and the groups  ,  ,   and  . We also use the examples to present some common notation and to show that a presentation of a group does not need to be unique.


  1. Let   be a set.   is a presentation of   because  , where   is the normal subgroup of   spanned by  . In particular,   is a presentation of  , more commonly denoted by  . Another presentation of   is  , more commonly denoted by  . Informally, in the presentation   we insert a new element   in the spanning set, but we impose the relation  , this is,  , which is the same as having not introduced the element   and have stayed by the presentation  .
  2. Let  .   (where     times) is a presentation of  . Indeed, the subgroup of   spanned by   is   and  , therefore . Is more common to denote   by  .
  3. Let   (with   and   distinct) and  .   is a presentation of  . Informally, what we do is impose comutatibility in  , this is,  , this is,  , obtaining a group isomorph to  . It's more usually denote   by  .
  4. Let   and  .   is a presentation of  . Informally, what we do is impose commutability in the same way as in the previous example, and we impose   and   to obtain   instead  . It's more common denote  by  .
  5.  , more commonly written  , is a presentation of  , the group of the permutations of   with the composition of applications. To verify this, one can verify that any group with presentation   as exactly six elements  ,  ,  ,  ,  ,   and  , and that the multiplication of this elements results in the following Cayley table that is equal to the Cayley table of  . Just to give an idea how this can be achieved, a group with presentation   as exactly the elements  ,  ,  ,  ,  ,   and   because none of this elements are the same (the relations   don't allow us to conclude that two of the elements are equal) and because "another" elements like   are actually one of the previous elements (for example, from   we have  , and taking inverses of both members, we have  , which, using  , this is,  ,   and  , results in  ). Then, using the relations of the presentation, one can compute the Cayley table. For example,   because we have the relation  . Another example: we have   because we can multiply both members of the relation   by   and then use  . One could have suspected of this presentation by taking  ,   and   and then, trying to construct the Cayley table of  , found out that it was possible if one know that  .

It's natural to ask if all groups have a presentation. The following theorem tell us that the answer is yes, and it give us a presentation.

Theorem Let   be a group.

  1. The application   defined by   (where  ) is an epimorphism of groups.
  2.   is a presentation of  .


  1.   is well defined because every element of   as a unique representations in the form   with  , with the exception of   appears several times in the representation, but that doesn't affect the value of   Let   be any elements, where  . We have    , therefore   is a morphism of groups. Because  , then   is a epimorphism of groups.
  2. Using the first isomorphism theorem (for groups), we have  , therefore   is a presentation of  .  

The previous theorem, despite giving us a presentations of the group  , doesn't give us a "good" presentation, because the spanning set   is usually much larger that other spanning sets, and the set of relations   is also usually much larger then other sufficient sets of relations (it is even a normal subgroup of  , when it would be enough that it span an appropriated normal subgroup).