Topology/Free group and presentation of a group

Free monoid spanned by a set
Let $$V$$ be a vector space and $$v_1,\ldots,v_n$$ be a basis of $$V$$. Given any vector space $$W$$ and any elements $$w_1,\ldots,w_n \in W$$, there is a linear transformation $$\varphi : V \rightarrow W$$ such that $$\forall i \in \{1,\ldots,n\}, \, \varphi(v_i) = w_i$$. One could say that this happens because the elements $$v_1,\ldots,v_n$$ of a basis are not "related" to each other (formally, they are linearly independent). Indeed, if, for example, we had the relation $$v_1 = \lambda v_2$$ for some scalar $$\lambda$$ (and then $$v_1,\ldots,v_n$$ wasn't linearly independent), then the linear transformation $$\varphi$$ could not exist.

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

Our goal in this section will be, given a set $$X$$, build a group spanned by the set $$X$$ such that it will be the most "free" possible, in the sense that it doesn't have to obey relations like $$x^n = 1$$ or $$xy = yx$$. 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 $$X$$, 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 $$x_1 \ldots x_n$$ that we will use for the element of this monoid meets this idea that the elements of this monoid are the words $$x_1 \ldots x_n$$ where $$x_1,\ldots,x_n$$ are letters of the alphabet $$X$$. Here is the definition of this monoid.

Definition Let $$X$$ be a set.
 * 1) We denote the $$n$$-tuples $$(x_1,\ldots,x_n)$$ with $$x_i \in X$$ and $$n \in \mathbb{N}$$ by $$x_1 \ldots x_n$$.
 * 2) We denote $$$$, that is $$(x_1,\ldots,x_n)$$ with $$n = 0$$, by $$1$$.
 * 3) We denote by $$FM(X)$$ the set $$\{x_1 \ldots x_n : n \in \mathbb{N}, x_i \in X\}$$.
 * 4) We define in $$FM(X)$$ the concatenation operation $$*$$ by $$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 $$*$$ and that $$1*x=x*1=x$$.

Proposition $$(FM(X),*)$$ is a monoid with identity $$1$$.

Proof The operation $$*$$ is associative because, given any $$x_1 \ldots x_m,y_1 \ldots y_n,z_1 \ldots z_p \in FM(X)$$, we have


 * $$(x_1 \ldots x_m * y_1 \ldots y_n) * z_1 \ldots z_m$$
 * $$= x_1 \ldots x_m y_1 \ldots y_n * z_1 \ldots z_m$$
 * $$= x_1 \ldots x_m y_1 \ldots y_n z_1 \ldots z_m$$
 * $$= x_1 \ldots x_m * (y_1 \ldots y_n z_1 \ldots z_m)$$
 * $$= x_1 \ldots x_m ( y_1 \ldots y_n * z_1 \ldots z_m)$$.

It's obvious that $$(FM(X),*)$$ has the identity $$1$$ as $$1 * x_1 \ldots x_n = x_1 \ldots x_n$$ by the definition of $$1$$ and $$*$$. $$\square$$

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

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

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

Free group spanned by a set
Now let us construct the more "free" group spanned by a set $$X$$. Informally, what we will do is insert in the monoid $$FM(X)$$ the inverse elements that are missing in it for it to be a group. In a more precise way, we will have a set $$\bar X$$ equipotent to $$X$$, choose a bijection from $$X$$ to $$\bar X$$ and in this way achieve a "association" between the elements of $$X$$ and the elements of $$\bar X$$. Then we face $$x_1 \ldots x_n \in FM(X)$$ (with $$x_1,\ldots,x_n \in X$$) as having the inverse element $$\overline{x_n} \ldots \overline{x_1}$$ (with $$\overline{x_1},\ldots,\overline{x_n} \in X$$), where $$x_1,\ldots,x_n \in X$$ is is associated with $$\overline{x_1} \ldots \overline{x_n}$$, respectively. Let us note that the order of the elements in $$\overline{x_n} \ldots \overline{x_1}$$ is "reversed" because the inverse of the product $$x_1 \ldots x_n = x_1 * \cdots * x_n$$ must be $$x_n^{-1} * \cdots * x_1^{-1}$$, and the $$x_1^{-1},\ldots,x_n^{-1}$$ are, respectively, $$\overline{x_1},\ldots,\overline{x_n}$$. The way we do that $$\overline{x_n} \ldots \overline{x_1}$$ be the inverse of $$x_1 \ldots x_n$$ is to take a congruence relation $$R$$ that identifies $$x_1 \ldots x_n \overline{x_n} \ldots \overline{x_1}$$ with $$1$$, and pass to the quotient $$FM(X \cup \bar X)$$ by this relation (defining then, in a natural way, the binary operation of the group, $$[u]_R \star [v]_R = [u * v]_R$$). By taking the quotient, we are formalizing the intuitive idea of identifying $$x_1 \ldots x_n \overline{x_n} \ldots \overline{x_1}$$ with $$1$$, because in the quotient we have the equality $$[x_1 \ldots x_n \overline{x_n} \ldots \overline{x_1}]_R = [1]_R$$. Let us give the formal definition.

Definition ''Let $$X$$ be a set. Let us take another set $$\overline{X}$$ equipotent to $$X$$ and disjoint from $$X$$ and let $$f : X \rightarrow \overline{X}$$ be a bijective application.''
 * 1) For each $$x \in X$$ let us denote $$f(x)$$ by $$\bar x$$, for each $$x \in \overline{X}$$ let us denote $$f^{-1}(x)$$ by $$\bar x$$ and for each $$x_1 \ldots x_n \in FM(X \cup \overline{X})$$ let us denote $$\overline{x_n} \ldots \overline{x_1}$$ by $$\overline{x_1 \ldots x_n}$$.
 * 2) Let $$R$$ be the congruence relation of $$FM(X \cup \overline{X})$$ spanned by $$G = \{(u * \bar u,1) : u \in X \cup \overline{X}\}$$, this is, $$R$$ is the intersection of all the congruence relations in $$FM(X \cup \overline{X})$$ wich have $$G$$ as a subset. We denote the quotient set $$FM(X \cup \overline{X})/R$$ by $$FG(X)$$.

Frequently, abusing the notation, we represent an element $$[u]_R \in FG(X)$$ simply by $$u$$.

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

Lemma ''Let $$X$$ be a set. It is well defined in $$FG(X)$$ the binary operation $$\star$$ by $$[u]_R \star [v]_R = [u_r * v_r]_R$$ (where $$R$$ is the congruence relation of the previous definition).''

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

Because the definition is valid, we present it.

Definition ''Let $$X$$ be a set. We define in $$FG(X)$$ the binary operation $$\star$$ by $$[u]_R \star [v]_R = [u_r * v_r]_R$$.''

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

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

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

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

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

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

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

Given $$u \in FM(X \cup \bar X)$$, let us denote the exact number of times that the "letter" $$x$$ appears in $$u$$ by $$|u|_x$$ and let us denote the exact number of times the "letter" $$\bar x$$ appears in $$u$$ by $$|u|_\bar x$$. Then "cutting" $$x$$'s with $$\bar x$$'s it remains a reduced word word with $$|u|_x - |u|_\bar x$$ times the letter $$x$$ (if $$|u|_x - |u|_\bar x < 0$$, let us us consider that there aren't any letters $$x$$ and remains $$-(|u|_x - |u|_\bar x)$$ times the letter $$\bar x$$). Let us denote $$|u|_x - |u|_\bar x$$ by $$|u|_{x - \bar x}$$. We have In this way, each element $$[u]_R \in FG(X)$$ is determined by the integer number $$|u|_{x - \bar x}$$ and the product $$\star$$ of two elements $$[u]_R,[v]_R \in FG(X)$$ correspondent to the sum of they associated integers numbers $$|u|_{x - \bar x}$$ and $$|v|_{x - \bar x}$$. Therefore, it seems that the group $$(FG(X),\star)$$ is "similar" to $$(\mathbb{Z},+)$$. indeed $$(FG(X),\star)$$ is isomorph to $$(\mathbb{Z},+)$$ and the application $$|\cdot|_{x - \bar x} : FG(X) \rightarrow \mathbb{Z}$$ is a group isomorphism.
 * 1) $$[u]_R = [v]_R$$ if and only if $$|u|_{x - \bar x} = |v|_{x - \bar x}$$ and
 * 2) $$\forall [u]_R,[v]_R \in FG(X), \, |uv|_{x - \bar x} = |u|_{x - \bar x} + |v|_{x - \bar x}$$.

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

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

Let us see some examples of presentations of the free group $$FG(X)$$ and the groups $$\mathbb{Z}_n$$, $$\mathbb{Z} \oplus \mathbb{Z}$$, $$\mathbb{Z}_m \oplus \mathbb{Z}_n$$ and $$S_3$$. 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 $$X$$ be a set. $$$$ is a presentation of $$FG(X)$$ because $$FG(X) \simeq FG(X)/\{1\}$$, where $$\{1\}$$ is the normal subgroup of $$FG(X)$$ spanned by $$\emptyset$$. In particular, $$<\{x\}:\emptyset >$$ is a presentation of $$(\mathbb{Z},+) \simeq FG(\{x\})$$, more commonly denoted by $$$$. Another presentation of $$(\mathbb{Z},+)$$ is $$<\{x,y\}:\{xy^{-1}\}>$$, more commonly denoted by $$$$. Informally, in the presentation $$$$ we insert a new element $$y$$ in the spanning set, but we impose the relation $$xy^{-1} = 1$$, this is, $$x = y$$, which is the same as having not introduced the element $$y$$ and have stayed by the presentation $$$$.
 * 2) Let $$X = \{x\}$$. $$$$ (where $$x^n = x \star \cdots \star x \in FG(X)$$ $$n$$ times) is a presentation of $$\mathbb{Z}_n$$. Indeed, the subgroup of $$FG(X)$$ spanned by $$\{x^n\}$$ is $$N = \{\ldots,\bar{x}^{2n},\bar{x}^n,1,x^n,x^{2n},\ldots\} \simeq n\mathbb{Z}$$ and $$FG(X) \simeq \mathbb{Z}$$, therefore$$\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z} \simeq FG(X)/N$$. Is more common to denote $$<\{x\} : \{x^n\}>$$ by $$$$.
 * 3) Let $$X = \{x,y\}$$ (with $$x$$ and $$y$$ distinct) and $$R = \{xyx^{-1}y^{-1}\}$$. $$$$ is a presentation of $$\mathbb{Z} \oplus \mathbb{Z}$$. Informally, what we do is impose comutatibility in $$FG(X)$$, this is, $$xy = yx$$, this is, $$xyx^{-1}y^{-1} = 1$$, obtaining a group isomorph to $$\mathbb{Z} \oplus \mathbb{Z}$$. It's more usually denote $$<\{x,y\} : \{xyx^{-1}y^{-1}\}>$$ by $$$$.
 * 4) Let $$X = \{x,y\}$$ and $$R = \{xyx^{-1}y^{-1},x^m,y^n\}$$. $$$$ is a presentation of $$\mathbb{Z}_m \times \mathbb{Z}_n$$. Informally, what we do is impose commutability in the same way as in the previous example, and we impose $$x^m = 1$$ and $$x^n = 1$$ to obtain $$\mathbb{Z}_m \times \mathbb{Z}_n$$ instead $$\mathbb{Z} \oplus \mathbb{Z}$$. It's more common denote$$<\{x,y\}:\{xyx^{-1}y^{-1},x^m,y^n\}>$$ by $$$$.
 * 5) $$<\{a,b,c\}:\{aa,bb,cc,abac,cbab\}>$$, more commonly written $$$$, is a presentation of $$S_3$$, the group of the permutations of $$\{1,2,3\}$$ with the composition of applications. To verify this, one can verify that any group with presentation $$$$ as exactly six elements $$id$$, $$a$$, $$b$$, $$c$$, $$a$$, $$ab$$ and $$ac$$, and that the multiplication of this elements results in the following Cayley table that is equal to the Cayley table of $$S_3$$. Just to give an idea how this can be achieved, a group with presentation $$$$ as exactly the elements $$id$$, $$a$$, $$b$$, $$c$$, $$a$$, $$ab$$ and $$ac$$ because none of this elements are the same (the relations $$a^2 = b^2 = c^2 = abac = cbab = 1$$ don't allow us to conclude that two of the elements are equal) and because "another" elements like $$bc$$ are actually one of the previous elements (for example, from $$cbab = id$$ we have $$cb = ab$$, and taking inverses of both members, we have $$b^{-1}c^{-1} = b^{-1} a^{-1}$$, which, using $$a^2 = b^2 = c^2 = id$$, this is, $$a = a^{-1}$$, $$b = b^{-1}$$ and $$c = c^{-1}$$, results in $$bc = ba$$). Then, using the relations of the presentation, one can compute the Cayley table. For example, $$a (ab) = b$$ because we have the relation $$a^2 = 1$$. Another example: we have $$b(ac) = a$$ because we can multiply both members of the relation $$abac = id$$ by $$a$$ and then use $$a^2 = id$$. One could have suspected of this presentation by taking $$a = (1 \ 2)$$, $$b = (1 \ 3)$$ and $$c = (2 \ 3)$$ and then, trying to construct the Cayley table of $$S_3$$, found out that it was possible if one know that $$a^2 = b^2 = c^2 = abac = cbab = 1$$.

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 $$(G,\times)$$ be a group.
 * 1) The application $$\varphi : FG(G) \rightarrow G$$ defined by $$\varphi([x_1]_R \star \cdots \star [x_n]_R) = x_1 \times \cdots \times x_n$$ (where $$x_1,\ldots,x_n \in G$$) is an epimorphism of groups.
 * 2) $$$$ is a presentation of $$(G,\times)$$.

Proof
 * 1) $$\varphi$$ is well defined because every element of $$FG(X)$$ as a unique representations in the form $$[x_n] \star \cdots [x_n]_R$$ with $$x_1,\ldots,x_n \in G$$, with the exception of $$[1]_R$$ appears several times in the representation, but that doesn't affect the value of $$x_1 \times \cdots \times x_n$$ Let $$[x_1]_R \star \cdots \star [x_m]_R,[y_1]_R \star \cdots \star [y_n]_R \in FG(X)$$ be any elements, where $$x_1,\ldots,x_m,y_1,\ldots,y_n \in G$$. We have $$\varphi(([x_1]_R \star \cdots \star [x_m]_R) \star ([y_1]_R \star \cdots \star [y_n]_R)) = (x_1 \times \cdots \times x_m) \times (y_1 \times \cdots \times y_n) =$$ $$\varphi([x_1]_R \star \cdots \star [x_m]_R) \times \varphi([y_1]_R \star \cdots \star [y_n]_R)$$, therefore $$\varphi$$ is a morphism of groups. Because $$\forall x \in G, \, \varphi ([x]_R) = x$$, then $$G$$ is a epimorphism of groups.
 * 2) Using the first isomorphism theorem (for groups), we have $$FG(G)/\textrm{ker} \, \varphi \simeq \varphi(G) = G$$, therefore $$<G:\textrm{ker} \, \varphi>$$ is a presentation of $$(G,\times)$$. $$\square$$

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