# Topology/Free group and presentation of a group

## Free monoid spanned by a set

Let ${\displaystyle V}$  be a vector space and ${\displaystyle v_{1},\ldots ,v_{n}}$  be a basis of ${\displaystyle V}$ . Given any vector space ${\displaystyle W}$  and any elements ${\displaystyle w_{1},\ldots ,w_{n}\in W}$ , there is a linear transformation ${\displaystyle \varphi :V\rightarrow W}$  such that ${\displaystyle \forall i\in \{1,\ldots ,n\},\,\varphi (v_{i})=w_{i}}$ . One could say that this happens because the elements ${\displaystyle v_{1},\ldots ,v_{n}}$  of a basis are not "related" to each other (formally, they are linearly independent). Indeed, if, for example, we had the relation ${\displaystyle v_{1}=\lambda v_{2}}$  for some scalar ${\displaystyle \lambda }$  (and then ${\displaystyle v_{1},\ldots ,v_{n}}$  wasn't linearly independent), then the linear transformation ${\displaystyle \varphi }$  could not exist.

Let us consider a similar problem with groups: given a group ${\displaystyle G}$  spanned by a set ${\displaystyle X=\{x_{i}:i\in I\}\subseteq G}$  and given any group ${\displaystyle H}$  and any set ${\displaystyle Y=\{y_{i}:i\in I\}\subseteq H}$ , does there always exist a group morphism ${\displaystyle \varphi :G\rightarrow H}$  such that ${\displaystyle \forall i\in I,\,\varphi (x_{i})=y_{i}}$ ? The answer is no. For example, consider the group ${\displaystyle G=\mathbb {Z} _{n}=\mathbb {Z} /n\mathbb {Z} }$  which is spanned by the set ${\displaystyle X=\{1\}}$ , the group ${\displaystyle H=\mathbb {R} }$  (with the adition operation) and the set ${\displaystyle Y=\{2\}}$ . If there exists a group morphism ${\displaystyle \varphi :\mathbb {Z} _{n}\rightarrow \mathbb {R} }$  such that ${\displaystyle \varphi (1)=2}$ , then ${\displaystyle n2=n\varphi (1)=\varphi (n\,1)=\varphi (0)=0}$ , which is impossible. But if instead we had choose ${\displaystyle G=\mathbb {Z} }$ , then such a group morphism does exist and it would be given by ${\displaystyle \varphi (t)=2t}$ . Indeed, given any group ${\displaystyle H}$  and any ${\displaystyle y\in H}$ , we have the group morphism ${\displaystyle \varphi :\mathbb {Z} \rightarrow H}$  defined by ${\displaystyle \varphi (t)=y^{t}}$  (in multiplicative notation) that verifies ${\displaystyle \varphi (1)=y}$ . In a way, we can think that this happens because the elements of the set ${\displaystyle X=\{1\}\subseteq \mathbb {Z} }$  (that spans ${\displaystyle \mathbb {Z} }$ ) don't verify relations like ${\displaystyle nx=1}$  (like ${\displaystyle \mathbb {Z} _{n}}$ ) or ${\displaystyle xy=yx}$ . So, it seems that ${\displaystyle \mathbb {Z} }$  is a group more "free" that ${\displaystyle \mathbb {Z} _{n}}$ .

Our goal in this section will be, given a set ${\displaystyle X}$ , build a group spanned by the set ${\displaystyle X}$  such that it will be the most "free" possible, in the sense that it doesn't have to obey relations like ${\displaystyle x^{n}=1}$  or ${\displaystyle xy=yx}$ . 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 ${\displaystyle X}$ , 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 ${\displaystyle x_{1}\ldots x_{n}}$  that we will use for the element of this monoid meets this idea that the elements of this monoid are the words ${\displaystyle x_{1}\ldots x_{n}}$  where ${\displaystyle x_{1},\ldots ,x_{n}}$  are letters of the alphabet ${\displaystyle X}$ . Here is the definition of this monoid.

Definition Let ${\displaystyle X}$  be a set.

1. We denote the ${\displaystyle n}$ -tuples ${\displaystyle (x_{1},\ldots ,x_{n})}$  with ${\displaystyle x_{i}\in X}$  and ${\displaystyle n\in \mathbb {N} }$  by ${\displaystyle x_{1}\ldots x_{n}}$ .
2. We denote ${\displaystyle ()}$ , that is ${\displaystyle (x_{1},\ldots ,x_{n})}$  with ${\displaystyle n=0}$ , by ${\displaystyle 1}$ .
3. We denote by ${\displaystyle FM(X)}$  the set ${\displaystyle \{x_{1}\ldots x_{n}:n\in \mathbb {N} ,x_{i}\in X\}}$ .
4. We define in ${\displaystyle FM(X)}$  the concatenation operation ${\displaystyle *}$  by ${\displaystyle x_{1}\ldots x_{m}*y_{1}\ldots y_{n}=x_{1}\ldots x_{m}y_{1}\ldots y_{n}}$ .

Next we prove that this monoid is indeed a monoid. It's an easy to prove result, we need to show associativity of ${\displaystyle *}$  and that ${\displaystyle 1*x=x*1=x}$ .

Proposition ${\displaystyle (FM(X),*)}$  is a monoid with identity ${\displaystyle 1}$ .

Proof The operation ${\displaystyle *}$  is associative because, given any ${\displaystyle x_{1}\ldots x_{m},y_{1}\ldots y_{n},z_{1}\ldots z_{p}\in FM(X)}$ , we have

${\displaystyle (x_{1}\ldots x_{m}*y_{1}\ldots y_{n})*z_{1}\ldots z_{m}}$
${\displaystyle =x_{1}\ldots x_{m}y_{1}\ldots y_{n}*z_{1}\ldots z_{m}}$
${\displaystyle =x_{1}\ldots x_{m}y_{1}\ldots y_{n}z_{1}\ldots z_{m}}$
${\displaystyle =x_{1}\ldots x_{m}*(y_{1}\ldots y_{n}z_{1}\ldots z_{m})}$
${\displaystyle =x_{1}\ldots x_{m}(y_{1}\ldots y_{n}*z_{1}\ldots z_{m})}$ .

It's obvious that ${\displaystyle (FM(X),*)}$  has the identity ${\displaystyle 1}$  as ${\displaystyle 1*x_{1}\ldots x_{n}=x_{1}\ldots x_{n}}$  by the definition of ${\displaystyle 1}$  and ${\displaystyle *}$ . ${\displaystyle \square }$

Following the idea that the monoid ${\displaystyle (FM(X),*)}$  is the most "free" monoid spanned by ${\displaystyle X}$ , we will call it the free monoid spanned by ${\displaystyle X}$ .

Definition Let ${\displaystyle X}$  be a set. We denote the free monoid spanned by ${\displaystyle X}$  by ${\displaystyle (FM(X),*)}$ .

Examples

1. Let ${\displaystyle X=\{x\}}$ . Then ${\displaystyle FM(X)=\{1,x,xx,xxx,\ldots \}}$  and, for example, ${\displaystyle xx*xxx=xxxxx}$ .
2. Let ${\displaystyle X=\{x,y,z\}}$ . Then ${\displaystyle 1,x,y,z,xxx,yxz,xyzzz\in FM(X)}$  and, for example, ${\displaystyle xxx*yxz=xxxyxz}$ .

## Free group spanned by a set

Now let us construct the more "free" group spanned by a set ${\displaystyle X}$ . Informally, what we will do is insert in the monoid ${\displaystyle FM(X)}$  the inverse elements that are missing in it for it to be a group. In a more precise way, we will have a set ${\displaystyle {\bar {X}}}$  equipotent to ${\displaystyle X}$ , choose a bijection from ${\displaystyle X}$  to ${\displaystyle {\bar {X}}}$  and in this way achieve a "association" between the elements of ${\displaystyle X}$  and the elements of ${\displaystyle {\bar {X}}}$ . Then we face ${\displaystyle x_{1}\ldots x_{n}\in FM(X)}$  (with ${\displaystyle x_{1},\ldots ,x_{n}\in X}$ ) as having the inverse element ${\displaystyle {\overline {x_{n}}}\ldots {\overline {x_{1}}}}$  (with ${\displaystyle {\overline {x_{1}}},\ldots ,{\overline {x_{n}}}\in X}$ ), where ${\displaystyle x_{1},\ldots ,x_{n}\in X}$  is is associated with ${\displaystyle {\overline {x_{1}}}\ldots {\overline {x_{n}}}}$ , respectively. Let us note that the order of the elements in ${\displaystyle {\overline {x_{n}}}\ldots {\overline {x_{1}}}}$  is "reversed" because the inverse of the product ${\displaystyle x_{1}\ldots x_{n}=x_{1}*\cdots *x_{n}}$  must be ${\displaystyle x_{n}^{-1}*\cdots *x_{1}^{-1}}$ , and the ${\displaystyle x_{1}^{-1},\ldots ,x_{n}^{-1}}$  are, respectively, ${\displaystyle {\overline {x_{1}}},\ldots ,{\overline {x_{n}}}}$ . The way we do that ${\displaystyle {\overline {x_{n}}}\ldots {\overline {x_{1}}}}$  be the inverse of ${\displaystyle x_{1}\ldots x_{n}}$  is to take a congruence relation ${\displaystyle R}$  that identifies ${\displaystyle x_{1}\ldots x_{n}{\overline {x_{n}}}\ldots {\overline {x_{1}}}}$  with ${\displaystyle 1}$ , and pass to the quotient ${\displaystyle FM(X\cup {\bar {X}})}$  by this relation (defining then, in a natural way, the binary operation of the group, ${\displaystyle [u]_{R}\star [v]_{R}=[u*v]_{R}}$ ). By taking the quotient, we are formalizing the intuitive idea of identifying ${\displaystyle x_{1}\ldots x_{n}{\overline {x_{n}}}\ldots {\overline {x_{1}}}}$  with ${\displaystyle 1}$ , because in the quotient we have the equality ${\displaystyle [x_{1}\ldots x_{n}{\overline {x_{n}}}\ldots {\overline {x_{1}}}]_{R}=[1]_{R}}$ . Let us give the formal definition.

Definition Let ${\displaystyle X}$  be a set. Let us take another set ${\displaystyle {\overline {X}}}$  equipotent to ${\displaystyle X}$  and disjoint from ${\displaystyle X}$  and let ${\displaystyle f:X\rightarrow {\overline {X}}}$  be a bijective application.

1. For each ${\displaystyle x\in X}$  let us denote ${\displaystyle f(x)}$  by ${\displaystyle {\bar {x}}}$ , for each ${\displaystyle x\in {\overline {X}}}$  let us denote ${\displaystyle f^{-1}(x)}$  by ${\displaystyle {\bar {x}}}$  and for each ${\displaystyle x_{1}\ldots x_{n}\in FM(X\cup {\overline {X}})}$  let us denote ${\displaystyle {\overline {x_{n}}}\ldots {\overline {x_{1}}}}$  by ${\displaystyle {\overline {x_{1}\ldots x_{n}}}}$ .
2. Let ${\displaystyle R}$  be the congruence relation of ${\displaystyle FM(X\cup {\overline {X}})}$  spanned by ${\displaystyle G=\{(u*{\bar {u}},1):u\in X\cup {\overline {X}}\}}$ , this is, ${\displaystyle R}$  is the intersection of all the congruence relations in ${\displaystyle FM(X\cup {\overline {X}})}$  wich have ${\displaystyle G}$  as a subset. We denote the quotient set ${\displaystyle FM(X\cup {\overline {X}})/R}$  by ${\displaystyle FG(X)}$ .

Frequently, abusing the notation, we represent an element ${\displaystyle [u]_{R}\in FG(X)}$  simply by ${\displaystyle u}$ .

Because the operation ${\displaystyle [u]_{R}\star [v]_{R}=[u*v]_{R}}$  that we want to define in ${\displaystyle FM(X\cup {\bar {X}})/R}$  is defined using particular represententes ${\displaystyle u}$  and ${\displaystyle v}$  of the equivalence classes ${\displaystyle [u]_{R}}$  and ${\displaystyle [v]_{R}}$ , a first precaution is to verify that the definition does not depend on the chosen representatives. It's an easy verification.

Lemma Let ${\displaystyle X}$  be a set. It is well defined in ${\displaystyle FG(X)}$  the binary operation ${\displaystyle \star }$  by ${\displaystyle [u]_{R}\star [v]_{R}=[u_{r}*v_{r}]_{R}}$  (where ${\displaystyle R}$  is the congruence relation of the previous definition).

Proof Let ${\displaystyle u,u',v,v'\in FM(X)}$  be any elements such that ${\displaystyle [u]_{R}=[u']_{R}}$  and ${\displaystyle [v]_{R}=[v']_{R}}$ , this is, ${\displaystyle uRu'}$  and ${\displaystyle vRv'}$ . Because ${\displaystyle R}$  is a congruence relation in ${\displaystyle FM(X\cup {\bar {X}})}$ , we have ${\displaystyle u*vRu'*v'}$ , this is, ${\displaystyle [u*v]_{R}=[u'*v']_{R}}$ . ${\displaystyle \square }$

Because the definition is valid, we present it.

Definition Let ${\displaystyle X}$  be a set. We define in ${\displaystyle FG(X)}$  the binary operation ${\displaystyle \star }$  by ${\displaystyle [u]_{R}\star [v]_{R}=[u_{r}*v_{r}]_{R}}$ .

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

Proposition Let ${\displaystyle X}$  be a set. ${\displaystyle (FG(X),\star )}$  is a group with identity ${\displaystyle [1]_{R}}$  and where ${\displaystyle \forall [u]_{R}\in FG(X),\,{[u]_{R}}^{-1}=[{\bar {u}}]_{R}}$ .

Proof

1. ${\displaystyle (FG(X),\star )}$  is associative because ${\displaystyle \forall [u]_{R},[v]_{R},[w]_{R}\in FG(X),\,([u]_{R}\star [v]_{R})\star [w]_{R}=([u*v]_{R})\star [w]_{R}=[([u]_{R}*[v]_{R})*w]_{R}=}$  ${\displaystyle [u*(v*w)]_{R}=[u]_{R}\star [v*w]_{R}=[u]_{R}\star ([v]_{R}\star [w]_{R}).}$
2. Let us see that ${\displaystyle [1]_{R}}$  is the identity ${\displaystyle (FG(X),\star )}$ . Let ${\displaystyle [u]_{R}\in FG(X)}$  be any element. We have ${\displaystyle [u]_{R}\star [1]_{R}=[u*1]_{R}=[u]_{R}}$  and, in the same way, ${\displaystyle [1]_{R}\star [u]_{R}=[u]_{R}}$ .
3. Let ${\displaystyle [u]_{R}\in FG(X)}$  be any element and let us see that ${\displaystyle [u]_{R}\star [{\bar {u}}]_{R}=[1]_{R}}$ . We have ${\displaystyle [u]_{R}\star [{\bar {u}}]_{R}=[u*{\bar {u}}]_{R}}$  and, by definition of ${\displaystyle R}$ , ${\displaystyle u*{\bar {u}}R1}$ , this is, ${\displaystyle [u*{\bar {u}}]_{R}=[1]_{R}}$ , therefore ${\displaystyle [u]_{R}\star [{\bar {u}}]_{R}=[1]_{R}}$  and, in the same way, ${\displaystyle [{\bar {u}}]_{R}\star [u]_{R}=[1]_{R}}$ . ${\displaystyle \square }$

In the same way that we did with the free monoid, we will call free group spanned by the set ${\displaystyle X}$  to the more "free" group spanned by this set.

Definition Let ${\displaystyle X}$  be a set. We call free group spanned by ${\displaystyle X}$  to ${\displaystyle (FG(X),\star )}$ .

Example Let ${\displaystyle X=\{x\}}$ . Let us choose any set ${\displaystyle {\bar {X}}=\{y\}}$  disjoint (and equipotent) of ${\displaystyle X}$ . Let ${\displaystyle f:X\rightarrow {\bar {X}}}$  be any (in fact, the only) bijective application of ${\displaystyle X}$  in ${\displaystyle {\bar {X}}}$ . Then we denote ${\displaystyle f(x)=y}$  by ${\displaystyle {\bar {x}}}$  and we denote ${\displaystyle f^{-1}(y)=x}$  by ${\displaystyle {\bar {y}}}$ . We regard ${\displaystyle x}$  and ${\displaystyle y}$  as inverse elements. Let ${\displaystyle R}$  be the congruence relation of ${\displaystyle FM(X\cup {\bar {X}})}$  spanned by ${\displaystyle G=\{(1,1),(x{\bar {x}},1),(xx{\bar {x}}{\bar {x}},1),\ldots \}}$ . ${\displaystyle FG(X)=FM(\{x,{\bar {x}}\})/R}$  is the set of all "words" written in the alphabet ${\displaystyle \{[x]_{R},[{\bar {x}}]_{R}\}}$ . For example, ${\displaystyle [1]_{R},[x]_{R},[{\bar {x}}]_{R},[xx{\bar {x}}xx]_{R}\in FG(X)}$ .

We have ${\displaystyle G\subseteq R}$  and, for example, ${\displaystyle (xx{\bar {x}},1)\in R}$ , because ${\displaystyle (x{\bar {x}},1)\in G\subseteq R}$  (therefore ${\displaystyle x{\bar {x}}R1}$ ) and because ${\displaystyle R}$  is a congruence relation, we can "multiply" both "members" of the relation ${\displaystyle x{\bar {x}}R1}$  by ${\displaystyle x}$  and obtain ${\displaystyle xx{\bar {x}}Rx}$ . We see ${\displaystyle x{\bar {x}}R1}$  as meaning that in ${\displaystyle FG(X)}$  We have ${\displaystyle xx{\bar {x}}=x}$  (more precisely, ${\displaystyle [xx{\bar {x}}]_{R}=[x]_{R}}$ ), and we think in this equality as being the result of one ${\displaystyle x}$  "cut out" with ${\displaystyle {\bar {x}}}$  in ${\displaystyle xx{\bar {x}}}$ .

Given ${\displaystyle u\in FM(X\cup {\bar {X}})}$ , let us denote the exact number of times that the "letter" ${\displaystyle x}$  appears in ${\displaystyle u}$  by ${\displaystyle |u|_{x}}$  and let us denote the exact number of times the "letter" ${\displaystyle {\bar {x}}}$  appears in ${\displaystyle u}$  by ${\displaystyle |u|_{\bar {x}}}$ . Then "cutting" ${\displaystyle x}$ 's with ${\displaystyle {\bar {x}}}$ 's it remains a reduced word word with ${\displaystyle |u|_{x}-|u|_{\bar {x}}}$  times the letter ${\displaystyle x}$  (if ${\displaystyle |u|_{x}-|u|_{\bar {x}}<0}$ , let us us consider that there aren't any letters ${\displaystyle x}$  and remains ${\displaystyle -(|u|_{x}-|u|_{\bar {x}})}$  times the letter ${\displaystyle {\bar {x}}}$ ). Let us denote ${\displaystyle |u|_{x}-|u|_{\bar {x}}}$  by ${\displaystyle |u|_{x-{\bar {x}}}}$ . We have

1. ${\displaystyle [u]_{R}=[v]_{R}}$  if and only if ${\displaystyle |u|_{x-{\bar {x}}}=|v|_{x-{\bar {x}}}}$  and
2. ${\displaystyle \forall [u]_{R},[v]_{R}\in FG(X),\,|uv|_{x-{\bar {x}}}=|u|_{x-{\bar {x}}}+|v|_{x-{\bar {x}}}}$ .

In this way, each element ${\displaystyle [u]_{R}\in FG(X)}$  is determined by the integer number ${\displaystyle |u|_{x-{\bar {x}}}}$  and the product ${\displaystyle \star }$  of two elements ${\displaystyle [u]_{R},[v]_{R}\in FG(X)}$  correspondent to the sum of they associated integers numbers ${\displaystyle |u|_{x-{\bar {x}}}}$  and ${\displaystyle |v|_{x-{\bar {x}}}}$ . Therefore, it seems that the group ${\displaystyle (FG(X),\star )}$  is "similar" to ${\displaystyle (\mathbb {Z} ,+)}$ . indeed ${\displaystyle (FG(X),\star )}$  is isomorph to ${\displaystyle (\mathbb {Z} ,+)}$  and the application ${\displaystyle |\cdot |_{x-{\bar {x}}}:FG(X)\rightarrow \mathbb {Z} }$  is a group isomorphism.

## Presentation of a group

Informally, it seems that ${\displaystyle \mathbb {Z} _{n}}$  is obtained from the "free" group ${\displaystyle \mathbb {Z} }$  imposing the relation ${\displaystyle nx=1}$ . Let us try formalize this idea. We start with a set ${\displaystyle X}$  that spans a group ${\displaystyle G}$  that que want to create and a set of relations ${\displaystyle R}$  (such as ${\displaystyle x^{n}=1}$  or ${\displaystyle xy=yz}$ ) that the elements of ${\displaystyle G}$  must verify and we obtain a group ${\displaystyle G/R}$  spanned by ${\displaystyle G}$  and that verify the relations of ${\displaystyle R}$ . More precisely, we write each relation ${\displaystyle u=v}$  in the form ${\displaystyle uv^{-1}=1}$  (for example, ${\displaystyle xy=yx}$  is written in the form ${\displaystyle xyx^{-1}y^{-1}=1}$ ) and we see each ${\displaystyle uv^{-1}}$  as a "word" of ${\displaystyle FG(X)}$ . Because ${\displaystyle R}$  doesn't have to be a normal subgroup of ${\displaystyle G}$ , we can not consider the quotient ${\displaystyle FG(X)/R}$ , so we consider the quotient ${\displaystyle FG(X)/N}$  where ${\displaystyle N}$  is the normal subgroup of ${\displaystyle FG(X)}$  spanned by ${\displaystyle R}$ . In ${\displaystyle G/N}$ , we will have ${\displaystyle uv^{-1}N=1N}$ , which we see as meaning that in ${\displaystyle G/N}$  the elements ${\displaystyle uv^{-1}}$  and ${\displaystyle 1}$  are the same. In this way, ${\displaystyle FG(X)/N}$  will verify all the relations that we want and will be spanned by ${\displaystyle X}$  (more precisely, by ${\displaystyle \{xN:x\in X\}}$ ). Let us formalize this idea.

Definition Let ${\displaystyle G}$  be a group. We call presentation of ${\displaystyle G}$ , and denote by ${\displaystyle }$ , to a ordered pair ${\displaystyle (X,R)}$  where ${\displaystyle X}$  is a set, ${\displaystyle R\subseteq FG(X)}$  and ${\displaystyle G\simeq FG(X)/N}$ , where ${\displaystyle N}$  is the normal subgroup of ${\displaystyle FG(X)}$  spanned by ${\displaystyle R}$ . Given a presentation ${\displaystyle }$ , we call spanning set to ${\displaystyle X}$  and set of relations to ${\displaystyle R}$ .

Let us see some examples of presentations of the free group ${\displaystyle FG(X)}$  and the groups ${\displaystyle \mathbb {Z} _{n}}$ , ${\displaystyle \mathbb {Z} \oplus \mathbb {Z} }$ , ${\displaystyle \mathbb {Z} _{m}\oplus \mathbb {Z} _{n}}$  and ${\displaystyle S_{3}}$ . 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.

Examples

1. Let ${\displaystyle X}$  be a set. ${\displaystyle }$  is a presentation of ${\displaystyle FG(X)}$  because ${\displaystyle FG(X)\simeq FG(X)/\{1\}}$ , where ${\displaystyle \{1\}}$  is the normal subgroup of ${\displaystyle FG(X)}$  spanned by ${\displaystyle \emptyset }$ . In particular, ${\displaystyle <\{x\}:\emptyset >}$  is a presentation of ${\displaystyle (\mathbb {Z} ,+)\simeq FG(\{x\})}$ , more commonly denoted by ${\displaystyle }$ . Another presentation of ${\displaystyle (\mathbb {Z} ,+)}$  is ${\displaystyle <\{x,y\}:\{xy^{-1}\}>}$ , more commonly denoted by ${\displaystyle }$ . Informally, in the presentation ${\displaystyle }$  we insert a new element ${\displaystyle y}$  in the spanning set, but we impose the relation ${\displaystyle xy^{-1}=1}$ , this is, ${\displaystyle x=y}$ , which is the same as having not introduced the element ${\displaystyle y}$  and have stayed by the presentation ${\displaystyle }$ .
2. Let ${\displaystyle X=\{x\}}$ . ${\displaystyle }$  (where ${\displaystyle x^{n}=x\star \cdots \star x\in FG(X)}$  ${\displaystyle n}$  times) is a presentation of ${\displaystyle \mathbb {Z} _{n}}$ . Indeed, the subgroup of ${\displaystyle FG(X)}$  spanned by ${\displaystyle \{x^{n}\}}$  is ${\displaystyle N=\{\ldots ,{\bar {x}}^{2n},{\bar {x}}^{n},1,x^{n},x^{2n},\ldots \}\simeq n\mathbb {Z} }$  and ${\displaystyle FG(X)\simeq \mathbb {Z} }$ , therefore${\displaystyle \mathbb {Z} _{n}=\mathbb {Z} /n\mathbb {Z} \simeq FG(X)/N}$ . Is more common to denote ${\displaystyle <\{x\}:\{x^{n}\}>}$  by ${\displaystyle }$ .
3. Let ${\displaystyle X=\{x,y\}}$  (with ${\displaystyle x}$  and ${\displaystyle y}$  distinct) and ${\displaystyle R=\{xyx^{-1}y^{-1}\}}$ . ${\displaystyle }$  is a presentation of ${\displaystyle \mathbb {Z} \oplus \mathbb {Z} }$ . Informally, what we do is impose comutatibility in ${\displaystyle FG(X)}$ , this is, ${\displaystyle xy=yx}$ , this is, ${\displaystyle xyx^{-1}y^{-1}=1}$ , obtaining a group isomorph to ${\displaystyle \mathbb {Z} \oplus \mathbb {Z} }$ . It's more usually denote ${\displaystyle <\{x,y\}:\{xyx^{-1}y^{-1}\}>}$  by ${\displaystyle }$ .
4. Let ${\displaystyle X=\{x,y\}}$  and ${\displaystyle R=\{xyx^{-1}y^{-1},x^{m},y^{n}\}}$ . ${\displaystyle }$  is a presentation of ${\displaystyle \mathbb {Z} _{m}\times \mathbb {Z} _{n}}$ . Informally, what we do is impose comutability in the same way as in the previous example, and we impose ${\displaystyle x^{m}=1}$  and ${\displaystyle x^{n}=1}$  to obtain ${\displaystyle \mathbb {Z} _{m}\times \mathbb {Z} _{n}}$  instead ${\displaystyle \mathbb {Z} \oplus \mathbb {Z} }$ . It's more common denote${\displaystyle <\{x,y\}:\{xyx^{-1}y^{-1},x^{m},y^{n}\}>}$  by ${\displaystyle }$ .
5. ${\displaystyle <\{a,b,c\}:\{aa,bb,cc,abac,cbab\}>}$ , more commonly written ${\displaystyle }$ , is a presentation of ${\displaystyle S_{3}}$ , the group of the permutations of ${\displaystyle \{1,2,3\}}$  with the composition of applications. To verify this, one can verify that any group with presentation ${\displaystyle }$  as exactly six elements ${\displaystyle id}$ , ${\displaystyle a}$ , ${\displaystyle b}$ , ${\displaystyle c}$ , ${\displaystyle a}$ , ${\displaystyle ab}$  and ${\displaystyle ac}$ , and that the multiplication of this elements results in the following Cayley table that is equal to the Cayley table of ${\displaystyle S_{3}}$ . Just to give an idea how this can be achieved, a group with presentation ${\displaystyle }$  as exactly the elements ${\displaystyle id}$ , ${\displaystyle a}$ , ${\displaystyle b}$ , ${\displaystyle c}$ , ${\displaystyle a}$ , ${\displaystyle ab}$  and ${\displaystyle ac}$  because none of this elements are the same (the relations ${\displaystyle a^{2}=b^{2}=c^{2}=abac=cbab=1}$  don't allow us to conclude that two of the elements are equal) and because "another" elements like ${\displaystyle bc}$  are actually one of the previous elements (for example, from ${\displaystyle cbab=id}$  we have ${\displaystyle cb=ab}$ , and taking inverses of both members, we have ${\displaystyle b^{-1}c^{-1}=b^{-1}a^{-1}}$ , which, using ${\displaystyle a^{2}=b^{2}=c^{2}=id}$ , this is, ${\displaystyle a=a^{-1}}$ , ${\displaystyle b=b^{-1}}$  and ${\displaystyle c=c^{-1}}$ , results in ${\displaystyle bc=ba}$ ). Then, using the relations of the presentation, one can compute the Cayley table. For example, ${\displaystyle a(ab)=b}$  because we have the relation ${\displaystyle a^{2}=1}$ . Another example: we have ${\displaystyle b(ac)=a}$  because we can multiply both members of the relation ${\displaystyle abac=id}$  by ${\displaystyle a}$  and then use ${\displaystyle a^{2}=id}$ . One could have suspected of this presentation by taking ${\displaystyle a=(1\ 2)}$ , ${\displaystyle b=(1\ 3)}$  and ${\displaystyle c=(2\ 3)}$  and then, trying to construct the Cayley table of ${\displaystyle S_{3}}$ , found out that it was possible if one know that ${\displaystyle a^{2}=b^{2}=c^{2}=abac=cbab=1}$ .
${\displaystyle \times }$  ${\displaystyle id}$  ${\displaystyle a}$  ${\displaystyle b}$  ${\displaystyle c}$  ${\displaystyle ab}$  ${\displaystyle ac}$
${\displaystyle id}$  ${\displaystyle id}$  ${\displaystyle a}$  ${\displaystyle b}$  ${\displaystyle c}$  ${\displaystyle ab}$  ${\displaystyle ac}$
${\displaystyle a}$  ${\displaystyle a}$  ${\displaystyle id}$  ${\displaystyle ab}$  ${\displaystyle ac}$  ${\displaystyle b}$  ${\displaystyle c}$
${\displaystyle b}$  ${\displaystyle b}$  ${\displaystyle ac}$  ${\displaystyle id}$  ${\displaystyle ab}$  ${\displaystyle c}$  ${\displaystyle a}$
${\displaystyle c}$  ${\displaystyle c}$  ${\displaystyle ab}$  ${\displaystyle ac}$  ${\displaystyle id}$  ${\displaystyle a}$  ${\displaystyle b}$
${\displaystyle ab}$  ${\displaystyle ab}$  ${\displaystyle c}$  ${\displaystyle a}$  ${\displaystyle b}$  ${\displaystyle ac}$  ${\displaystyle id}$
${\displaystyle ac}$  ${\displaystyle ac}$  ${\displaystyle b}$  ${\displaystyle c}$  ${\displaystyle a}$  ${\displaystyle id}$  ${\displaystyle ab}$

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 ${\displaystyle (G,\times )}$  be a group.

1. The application ${\displaystyle \varphi :FG(G)\rightarrow G}$  defined by ${\displaystyle \varphi ([x_{1}]_{R}\star \cdots \star [x_{n}]_{R})=x_{1}\times \cdots \times x_{n}}$  (where ${\displaystyle x_{1},\ldots ,x_{n}\in G}$ ) is an epimorphism of groups.
2. ${\displaystyle }$  is a presentation of ${\displaystyle (G,\times )}$ .

Proof

1. ${\displaystyle \varphi }$  is well defined because every element of ${\displaystyle FG(X)}$  as a unique representations in the form ${\displaystyle [x_{n}]\star \cdots [x_{n}]_{R}}$  with ${\displaystyle x_{1},\ldots ,x_{n}\in G}$ , with the exception of ${\displaystyle [1]_{R}}$  appears several times in the representation, but that doesn't affect the value of ${\displaystyle x_{1}\times \cdots \times x_{n}}$  Let ${\displaystyle [x_{1}]_{R}\star \cdots \star [x_{m}]_{R},[y_{1}]_{R}\star \cdots \star [y_{n}]_{R}\in FG(X)}$  be any elements, where ${\displaystyle x_{1},\ldots ,x_{m},y_{1},\ldots ,y_{n}\in G}$ . We have ${\displaystyle \varphi (([x_{1}]_{R}\star \cdots \star [x_{m}]_{R})\star ([y_{1}]_{R}\star \cdots \star [y_{n}]_{R}))=(x_{1}\times \cdots \times x_{m})\times (y_{1}\times \cdots \times y_{n})=}$  ${\displaystyle \varphi ([x_{1}]_{R}\star \cdots \star [x_{m}]_{R})\times \varphi ([y_{1}]_{R}\star \cdots \star [y_{n}]_{R})}$ , therefore ${\displaystyle \varphi }$  is a morphism of groups. Because ${\displaystyle \forall x\in G,\,\varphi ([x]_{R})=x}$ , then ${\displaystyle G}$  is a epimorphism of groups.
2. Using the first isomorphism theorem (for groups), we have ${\displaystyle FG(G)/{\textrm {ker}}\,\varphi \simeq \varphi (G)=G}$ , therefore ${\displaystyle }$  is a presentation of ${\displaystyle (G,\times )}$ . ${\displaystyle \square }$

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