# 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 achived, 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).