The following was taken from Hungerford. This is how the author constructed a set, $F$, of reduced words on an arbitrary set $X$. Then, he proceeded to show that $F$ is a free group in the category of groups.
Given a set $X$ we shall construct a group $F$ that is free on the set $X$ in the sense of Definition 7.7. If $X = \emptyset$, $F$ is the trivial group $\langle e \rangle$. If $X \neq \emptyset$ let $X^{-1}$ be a set disjoint from $X$ such that $|X| = |X^{-1}|$. Choose a bijection $X \to X^{-1}$ and denote the image of $x \in X$ by $x^{-1}$. Finally choose a set that is disjoint from $X \cup X^{-1}$ and has exactly one element; denote this element by $1$. A word on $X$ is a sequence $(a_1, a_2, \dotsc)$ with $a_i \in X \cup X^{-1} \cup \{1\}$ such that for some $n \in \mathbb{N}^*$, $a_k = 1$ for all $k \geq n$. The constant sequence $(1, 1, \dotsc)$ is called the empty word and is denoted by $1$. (This ambiguous notation will cause no confusion.) A word $(a_1, a_2, \dotsc)$ on $X$ is said to be reduced provided that
- for all $x \in X$, $x$ and $x^{-1}$ are not adjacent (that is, $a_i = x \implies a_{i+1} \neq x^{-1}$ and $a_i = x^{-1} \implies a_{i+1} \neq x$ for all $i \in \mathbb{N}^*$, $x \in X$) and
- $a_k = 1$ implies $a_i = 1$ for all $i \geq k$.
In particular, the empty word is reduced.
To show that every group $G$ is the homomorphic image of a free group, we let $X$ be the generating set of $G$. So, every element in $G$ takes the form $a_1^{c_1}a_2^{c_2} \ldots a_n^{c_n}$ where each $a_i \in X$. Rewrite the element as $a_1a_2 \ldots a_s$ where $s = c_1 + c_2 \ldots + c_n$. Is it possible to have $a_ia_k = 1$ for some $i, k$? Based on the author's definition, the answer seems to be no. With the bijection between $X$ and $X^{-1}$:
$$a_i \mapsto a_i^{-1} = a_k \text{ and } a_k \mapsto a_k^{-1} = a_i$$
This contradicts that $X \cap X^{-1} = \emptyset$. Hence, if $a \in X$, you will not find its inverse in $X$ …. Did I understand this correctly?
Now, forget that $X$ is the generating set of $G$. What if I have a set $Y \subset G$ such that $1 \in Y$, or $a, a^{-1} \in Y$? In that case, then I cannot construct $Y^{-1}$. If $Y$ contains $1$, then $Y^{-1}$ also contains $1$. If $Y$ contains $a, a^{-1}$, so does $Y^{-1}$. Therefore, in the construction of $F$, the beginning set $Y$ is not arbitrary. Specifically, $Y$ is devoid of inverses and $1$.
Best Answer
When constructing the free group on a set of generator, you should only see it as a set, and forget that it is also a subset of the group. So for example, the elements $x^{-1}$ obtained using some bijection $X\to X^{-1}$ as in your text are not the inverses of the elements of $X$, but rather new elements that need not even be in $G$. Similarly the element $1$ is not the unit of $G$, but again some new element not necessarily in $G$.
Your free group is constructed as explained in your text, and can be mapped onto $G$ by sending every element of $X$ to itself, every $x^{-1}$ to the inverse of $x$ in $G$, $1$ to the unit in $G$, and every word to the corresponding product, computed in $G$.