Jae Choon Cha
POSTECH
Free groups
Consider a collection $S=\{x_i\}_{i\in I}$ of symbols $x_i$. For each $i\in I$, let $x_i^{1}$ be another formal symbol, and let $\overline S=\{x_i^{1}\}_{i\in I}$. Recall that a word on $S\cup\overline S$ is an expression of the form $w=x_{i_1}^{\epsilon_1}\cdots x_{i_\ell}^{\epsilon_\ell}$ where $\epsilon_k\in \{1,1\}$. We allow $\ell=0$, that is, the empty word, which we denote by $e$, is a word too. Define the concatenation, or product, of words as follows: for $w =x_{i_1}^{\epsilon_1}\cdots x_{i_r}^{\epsilon_r}$ and $v =x_{j_1}^{\delta_1}\cdots x_{j_s}^{\delta_s}$,
Define a relation $\sim$ on the set of words on $S$ as follows: $w\sim v$ if there is a finite sequence of words $w=w_0,w_1,\ldots,w_n=v$ such that for each $i=1,\ldots,n$, there are words $u,u'$ such that $\{w_{i1},w_i\}=\{uu', ux_j^\epsilon x_j^{\epsilon}u'\}$ for some $j$ and $\epsilon$. It is easily seen that $\sim$ is an equivalence relation.
Theorem. The set
Proof. The welldefinedness of the operation and its associativity is straightforward to verify. Also, it is easy to check that the identity is the empty word $e$, and the inverse of (the equivalence class of) $w=x_{i_1}^{\epsilon_1}\cdots x_{i_\ell}^{\epsilon_\ell}$ is (represented by) $w^{1}=x_{i_\ell}^{\epsilon_\ell}\cdots x_{i_1}^{\epsilon_1}$. $\quad\square$
The group $F\langle S\rangle$ is called the free group on the generators $x_i$, or the free group generated by $x_i$. The symbols $x_i$ are called generators.
Note that $F\langle S\rangle$ and $F\langle T\rangle$ are isomorphic if there is a bijection between $S$ and $T$. When $S$ is a finite set with $n$ elements, $F\langle S\rangle$ is called a free group of rank $n$.
For brevity, we denote the equivalence class of a word $w$ by $w$. Also, as done before, we write the product of $n$ copies of $w$ by $w^n$.
The function $i\colon S\to F\langle S\rangle$ defined by $i(x_i)=x_i$ is onetoone. (Exercise: prove it!) So, we may regard $S$ as a subset of $F\langle S\rangle$, by viewing $x_i\in S$ as a oneletter word representing an element of $F\langle S\rangle$.
The following fundamental property characterizes a free group.
Theorem (Universal property of the free group). For any function $f\colon S \to G$ to a group $G$, there is a unique group homomorphism $h\colon F\langle S\rangle \to G$ such that $h(x_i)=f(x_i)$ for all generators $x_i$.
Furthermore, if $H$ is another group equipped with a function $j\colon S\to H$ such that $(H,j)$ has the above property of $(F\langle S\rangle, i)$, then there is an isomorphism $\phi\colon F\langle S\rangle \to H$ such that $\phi i= j$.
Proof. For a given $f\colon S\to G$, define $h\colon F\langle S\rangle \to G$ by $h(x_{i_1}^{\epsilon_1}\cdots x_{i_\ell}^{\epsilon_\ell}) = f(x_{i_1})^{\epsilon_1}\cdots f(x_{i_\ell})^{\epsilon_\ell}$. It is straightforward to verify that $h$ is a welldefined group homomorphism. If $h'\colon F\langle S\rangle \to G$ is another homomorphism satisfying $h'(x_i)=f(x_i)$, then $h'(x_{i_1}^{\epsilon_1}\cdots x_{i_\ell}^{\epsilon_\ell})$ must be equal to $f(x_{i_1})^{\epsilon_1}\cdots f(x_{i_\ell})^{\epsilon_\ell}$, by the homomorphic property of $h'$. This proves the uniqueness of $h$.
The last statement is proven by a standard argument using the universal property. Details are left as an exercise. $\quad\square$
Group presentations
We will present a method to express a group as a quotient of a free group. For this purpose, we want to describe a normal subgroup of a free group. First, recall that the subgroup generated by a given subset $A$ of a group is defined to be the smallest subgroup of $G$ containing $A$. It is equal to the intersection of all subgroups containing $A$. (Exercise: prove this!) Alternatively, consider the following subset of $G$:
Since $H$ is closed under product and inversion, $H$ is a subgroup. Also, if $K$ is another subgroup of $G$ which contains $A$, then for each $a_i\in A$, we have $a_1^{\epsilon_1} \cdots a_r^{\epsilon_r} \in K$. So the above $H$ is contained in $K$. It follows that $H$ is the smallest subgroup containing $A$. That is, $H$ is the subgroup generated by $A$.
Now, given a subset $A$ of a group $G$, the normal subgroup generated by $A$ is defined to be the smallest normal subgroup in $G$ containing $A$. Similarly to the above paragraph, it is equal to the intersection of all normal subgroups in $G$ containing $A$, and equal to the following subgroup:
In particular, the normal subgroup generated by $A$ is equal to the subgroup generated by conjugates of elements in $A$.
Definition. For a subset $R\in F\langle S\rangle$, define
The expression $\langle S\mid R\rangle$ is called a group presentation. When $S=\{x_1,\ldots,x_n\}$ and $R=\{r_1,\ldots,r_m\}$ with each $r_i$ a word, we write
instead of $\langle S\mid R\rangle$. We call an element $x_i$ of $S$ a generator, and an element $r_j$ of $R$ a relator. An element in the normal subgroup generated by $R$ is called a relation.
We often view a word in $x_i$ as an element in $\langle S\mid R\rangle$.
The following is a fundamental property of a group presentation.
Theorem. Suppose $S$ is a set and $G$ is a group. Suppose $f\colon S\to G$ is a function such that $f(x_{i_1}^{\epsilon_1})\cdots f(x_{i_\ell}^{\epsilon_\ell}) = e$ for each $x_{i_1}^{\epsilon_1}\cdots x_{i_\ell}^{\epsilon_\ell}\in R$. Then there is a unique homomorphism $h\colon \langle S\mid R\rangle \to G$ satisfying $h(x_i)=f(x_i)$ for each $x_i\in S$.
Proof. By the universal property of a free group, there is a homomorphism $g\colon F\langle S\rangle \to G$ satisfying $g(x_i)=f(x_i)$ for all $x_i\in S$. It follows that $g(r)=e$ for each $r\in R$, by the hypothesis. Therefore $g$ induces a homomorphism $h\colon\langle S \mid R \rangle \to G$. Since the elements $x_i$ generate $\langle S \mid R\rangle$, the property $h(x_i)=f(x_i)$ determines the homomorphism $h$ uniquely. $\quad\square$
The group presentation is a very useful method to describe a group.
Theorem. Every group $G$ has a group presentation. That is, $G\cong \langle S\mid R\rangle$ for some set $S$ and some subset $R$ of $F\langle S\rangle$.
Proof. First, observe that there exists an epimorphism $F\langle S\rangle \to G$ for some set $S$. For instance, take $G$ itself as $S$, and define a homomorphism $h\colon F\langle S\rangle \to G$ by $g\mapsto g$ for each $g\in G$. This homomorphism $h$ exists by the universal property of the free group $F\langle S\rangle$, and $h$ is obviously surjective. Let $N$ be the kernel of $h$. Choose a set $R$ which normally generates $N$. For instance, one may take $N$ itself as $R$. Then, by the first isomorphism theorem, $h$ induces an isomorphism $F\langle S\rangle / N \cong G$. This shows that $G\cong \langle S\mid R\rangle$.
Note that there are more than one group presentation which represent the same group. In general, it is a very difficult problem to determine when two group presentations give the same group.
Observe that a group presentation given by the construction in the proof of the above theorem is not very efficient in general. For example, for the infinite cyclic group, the proof gives us a group presentation with infinitely many generators. In practice, we often prefer working with simpler group presentations.
Definition. A group is finitely generated if it has a presentation with finitely many generators. A group is finitely presented if it has a presentation with finitely many generators and finitely many relators.
Examples.

The presentation $\langle x \mid \cdot \rangle$ gives an infinite cyclic group. To give a proof, define a homomorphism $\phi\colon \langle x \mid \cdot \rangle \to \Z$ by $\phi(x)=1$, using the above theorem. Since $\phi(x^n)=n$ for any integer $n$, $\phi$ is 11 and onto. It follows that $\phi$ is an isomorphism.

A similar argument shows that $\langle x \mid x^n \rangle \cong \Z_n$, the cyclic group of order $n$.

Let $D_{2n}$ be the dihedral group of order $2n$. That is, $D_{2n}$ is the group of plane isometries preserving the set $\{e^{2\pi k i/n}\mid k=0,1,\dots,n1\}$. Recall that an element of $D_{2n}$ is either the $(2\pi k/n)$rotation, or a reflection about the line $\theta=\pi k/n$.
We claim that $D_{2n}$ has the presentation $\langle x,y\mid x^2,\, y^n,\, xyxy\rangle$. To verify this, let $G=\langle x,y\mid x^2,\, y^n,\, xyxy\rangle$, and define a homomorphism $\phi\colon G \to D_{2n}$ by letting $\phi(x)$ be the reflection about the $x$axis, and $\phi(y)$ be the rotation by $2\pi/n$. Our $\phi$ is welldefined since $\phi(x)^2$, $\phi(y)^n$ and $\phi(x)\phi(y)\phi(x)\phi(y)$ are equal to the identity. (Check!) Furthermore, $\phi$ is onto since the subset $\{\phi(x),\phi(y)\}$ generates $D_{2n}$. Finally, to show that $\phi$ is 11, choose an element of $G$, which is represented by a word $w$ in $x$ and $y$. Since $xy=y^{1}x^{1}$ in $G$, we may assume $w$ is of the form $w=x^a y^b$. Furthermore, since $x^2=e$ and $y^n=e$, we may assume $0\le a\le 1$ and $0\le b \le n1$. This shows that $G$ has at most $2n$ elements. Since $D_{2n}$ has $2n$ elements and $\phi$ is onto, it follows that $G$ has exactly $2n$ elements and $\phi$ is 11. This completes the proof that $\phi$ is an isomorphism.