# The Free Group on a Set X, F(X)

Let $X$ be a set. We aim to construct what is called the Free Group on the set $X$ which we will denote by $F(X)$.

First suppose that $X \neq \emptyset$. Let $X^{-1}$ be any set that is disjoint from $X$ for which there exists a bijection $f : X \to X^{-1}$. For each point $x \in X$ let $x^{-1} \in X^{-1}$ be defined to be $x^{-1} = f(x)$. Let $\{ 1 \}$ be a set containing a single element that is disjoint from both $X$ and $X^{-1}$.

Definition: Let $X$ be a set, $X^{-1}$ a nonempty set disjoint from $X$ and in bijection with $X$, and $\{ 1 \}$ a singleton set disjoint from both $X$ and $X^{-1}$. A Word on $X$ is a sequence $(a_n) \in X \cup X^{-1} \cup \{ 1 \}$ for which there exists an $N \in \mathbb{N}$ such that $a_n = 1$ for all $n > k$. The Empty Word on $X$ is the constant sequence $(1, 1, ...)$. |

For example, if $X = \{ a, b, c, ..., z \}$, $X^{-1} = \{a^{-1}, b^{-1}, ..., z^{-1} \}$, then $(a, g, t^{-1}, l, 1, 1, 1, ... )$ is a word on $X$. It is conventional to write words in $X$ NOT in sequence notation, but instead writing them as concentation of non-$1$ terms. We write the empty word to be $1$, and the word $(a, g, t^{-1}, l, 1, 1, 1, ... )$ to be:

(1)Definition: Let $X$, $X^{-1}$, and $\{ 1 \}$ be sets as described in the previous definition. A Reduced Word on $X$ is a word $(a_n)$ in $X$ with the properties that:1) $x$ and $x^{-1}$ are never adjacent terms in $(a_n)$.2) If $a_i = 1$ then $a_j = 1$ for all $j \geq 1$. |

Note that every nonempty word reduced word is of the form:

(2)Where $x_1, x_2, ..., x_n \in X \cup X^{-1}$ and $\lambda_1, \lambda_2, ..., \lambda_n \in \mathbb{N}$ where for each $1 \leq i \leq n$ we define $x_i^{\lambda_i} = \underbrace{x_ix_i...x_i}_{\lambda_i \: \mathrm{symbols}}$.

Furthermore, two nonempty reduced words $x_1^{\lambda_1}x_2^{\lambda_2}...x_n^{\lambda}$ and $y_1^{\delta_1}y_2^{\delta_2}...y_m^{\delta_m}$ in $X$ are equal if and only if $n = m$, $x_i = y_i$ for each $1 \leq i \leq n$ and $\lambda_i = \delta_i$ for each $1 \leq i \leq n$.

We will let $F(X)$ denote the set of all reduced words on $X$. We aim to make $F(X)$ a group with the operation of concatentation of words. Unfortunately, the concatenation of two reduced words need not be a reduced word. To fix this, we just define the product of two reduced words to be the reduction of the concatenation of these two words.

Definition: Let $X$ be a set. The Free Group on $X$ denoted by $F(X)$ is defined as follows.a) If $X = \emptyset$ then $F(X)$ is the trivial group.b) If $X \neq \emptyset$ then $F(X)$ is the set of reduced words on $X$ with the operation of (reduced) concatenation and furthermore, $F(X)$ is generated by the set $X$, i.e., $F(X) = \langle X \rangle$. If $X$ is a finite set with $|X| = n$ it is conventional to write $F_n$ instead, and call $F_n$ the Free Group on $n$ Generators. |

*Sometimes the term "Free Group Generated by the set $X$" is used to describe $F(X)$.*

*If $X$ and $Y$ are finite sets and $|X| = |Y| = n$, then $F(X)$ and $F(Y)$ are clearly isomorphic, so the notation $F_n$ above is unambiguous.*

By definition, the empty word is the identity of $F(X)$ and $1$ has inverse $1$. If $x \in X$ then the inverse of $x$ in $x^{-1} \in X^{-1}$. Conversely, the inverse of $x^{-1} \in X^{-1}$ is $x$. All that really remains to verify in showing that $F(X)$ is a group is that the operation of reduced concatenation is associative. This proof is VERY cumbersome and tedious, so we will omit it.