I have seen the definition of a free group go like this:
Let $S = \{s_i : i\in \mathbb{N} \}$ be a countable set. Let $S^{-1}$ be the set $\{s_i^{-1}: i\in \mathbb{N}\}$. Here one is to understand $s_i^{-1}$ as a notation for an element of this other set. Let $W(S)$ be the set of words in $S$ and $S^{-1}$. That is, the elements in $W(S)$ are strings of elements from $S$ and $S^{-1}$. One then defines a function $W(S) \times W(S) \to W(S)$. One makes $W(S)$ into a group (free group) be defining an operation with certain cancellation operations (so $s_is_i^{-1}$ is the empty word).
All of this makes much sense intuitively. My question is:
Is there is a more precise way to define this?
What I mean is, talking about strings, empty word, and cancellation properties doesn't seem very clear (precise). I understand that a more precise definition would be harder to grasp intuitively, but I was just wondering if there, for example, was some extremely precise way to define a free group only relying on language from, for example, set theory.
Edit: I am in beginning Abstract Algebra.
Best Answer
Sure, you can make this totally precise. There is no reason to assume $S$ is countable; just let $S$ be any set. Let $T=S\times\{0,1\}$; we think of elements of the form $(s,0)\in T$ as representing "$s$" and $(s,1)$ as representing "$s^{-1}$". Write $\pi_0:T\to S$ for the projection map onto the first coordinate and $\pi_1:T\to \{0,1\}$ for the projection map onto the second coordinate. If $n\in\mathbb{N}$, write $[n]=\{i\in\mathbb{N}:i<n\}$ and say that a word (on $T$) of length $n$ is a function $w:[n]\to T$. Say that a word $w$ is reduced if for all $i<n-1$, if $\pi_0(w(i))=\pi_0(w(i+1))$ then $\pi_1(w(i))=\pi_1(w(i+1))$ (intuitively, this means that your word never has $s$ adjacent to $s^{-1}$). Given a word $w:[n]\to T$, say that another word $w':[n-2]\to T$ is a one-step reduction of $w$ if there exists $i<n-1$ such that $\pi_0(w(i))=\pi_0(w(i+1))$, $\pi_1(w(i))\neq \pi_1(w(i+1))$, and $w'$ is given by the formula $$w'(j)=\begin{cases} w(j) & \text{ if $j<i$} \\ w(j+2) & \text{ if $j\geq i$.}\end{cases}$$ (Intuitively, you are obtaining $w'$ from $w$ by cancelling an adjacent $s$ and $s^{-1}$ appearing in the word $w$).
Write $W$ for the set of all words on $T$ (of any length). Say that a word $w'$ is a reduction of a word $w$ if there exists a natural number $n$ and function $v:[n]\to W$ such that $v(0)=w$, $v(n-1)=w'$, and $v(i+1)$ is a one-step reduction of $v(i)$ for each $i<n-1$. It is then possible to prove the following theorem:
Theorem: Let $w\in W$. Then there is a unique reduced word that is a reduction of $w$.
The proof is a bit messy but not too hard, and uses induction on the length of $w$. Write $r(w)$ for the unique reduced word that is a reduction of $w$.
We need one more ingredient before we can define the free group. Given two words $w:[n]\to T$ and $w':[m]\to T$, define their concatenation $w*w':[n+m]\to T$ by the formula $$(w*w')(i)=\begin{cases} w(i) & \text{ if $i<n$} \\ w'(i-n) & \text{ if $i\geq n$.}\end{cases}$$
We can now define the free group. Let $F(S)$ (the "free group on $S$") be the set of all reduced words on $T$, and define a multiplication $\cdot:F(S)\times F(S)\to F(S)$ by $w\cdot w'=r(w*w')$, i.e. the reduced word obtained by reducing the concatenation of $w$ and $w'$. You can then prove that $(F(S),\cdot)$ is a group.