Edit: alright, I guess I will provide the complete quote of the page, and will not leave out a single detail so that my question is comprehensive.
I am reading Abstract Algebra: 3rd Edition by Dummit and Foote.
On page 216, they make a couple definitions. Note that $S$ is a set:
Let $S^{-1}$ be any set disjoint from $S$ such that there is a bijection from $S$ to $S^{-1}$. For each $s\in S$ denote its corresponding element in $S^{-1}$ by $s^{-1}$ and similarly for each $t\in S^{-1}$ let the corresponding element of $S$ be denoted by $t^{-1}$ (so $(s^{-1})^{-1}=s$). Take a singleton set not contained in $S\cup S^{-1}$ and call it $\{1\}$. Let $1^{-1}=1$ and for any $x\in S\cup S^{-1}\cup \{1\}$, let $x^1=x$.
Next we describe the elements of the free group on a set $S$. A word on $S$ is by definition a sequence $$(s_1, s_2, s_3, …)\text{ where }s_i\in S\cup S^{-1}\cup \{1\}\text{ and }s_i=1\text{ for all }i\text{ sufficiently large} $$(that is, for each sequence there is an $N$ such that $s_i=1$ for all $i\geq N$). Thus we can think of a word as a finite product of elements of $S$ and their inverses (where repetitions
are allowed). Next, in order to assure uniqueness of expressions we consider only words which have no obvious "cancellations" between adjacent terms (such as $baa^{-1}b=bb$).The word $(s_1,s_2,s_3,…)$ is said to be reduced if $$(1)\ s_{i+1}\neq s_i^{-1}\text{ for all }i\text{ with }s_i\neq 1\text{, and} $$ $$(2)\text{ if }s_k=1\text{ for some }k\text{, then }s_i=1\text{ for all }i\geq k.$$
The reduced word $(1, 1, 1, … )$ is called the empty word and is denoted by $1$. We now simplify the notation by writing the reduced word $(s_1^{\epsilon_1}, s_2^{\epsilon_2}, …, s_n^{\epsilon_n}, 1, 1, 1, …)$, $s_i\in S$, $\epsilon_i=\pm 1$, as $s_1^{\epsilon_1}s_2^{\epsilon_2}…s_n^{\epsilon_n}$. Note that by definition, the reduced words $r_1^{\delta_1}r_2^{\delta_2}…r_m^{\delta_m}$ and $s_1^{\epsilon_1}s_2^{\epsilon_2}…s_n^{\epsilon_n}$ are equal if and only if $n=m$ and $\delta_i=\epsilon_i$, $1\leq i\leq n$.
Original question:
Wait, what? Why is this true? Nowhere in the definition did it mention that the $s_i$s had to be distinct, for example. Does it have to do with the fact that the $\epsilon_i$s are equal to $\pm 1$?
Additional details:
Perhaps I'm asking about what a word is. It's not always easy to ask a precise question when I don't have sufficient knowledge of the subject. How should I think of words? Like strings? The sentence "…we can
think of a word as a finite product of elements of $S$ and their inverses" makes me want to view them as actual elements of $S$ undergoing some operation with each other. Why are the words $r_1^{\delta_1}r_2^{\delta_2}…r_m^{\delta_m}$ and $s_1^{\epsilon_1}s_2^{\epsilon_2}…s_n^{\epsilon_n}$ equal if and only if they are equal component-wise? What is the intuition behind this, and how should I think about it?
Best Answer
First you have the set (which I will denote by $W$) of all sequences of elements of $S\cup S^{-1}\cup\{1\}$.
Two sequences are equal if and only if they are identical, term by term. We call these sequences "words".
Then we will define the free group. The underlying set (its elements) will be a specific subset of $W$. That subset is the subset of "reduced words", defined as in the text quoted, with the described shorthand fora reduced word.
These are elements of $W$. So the reduced words represented by the shorthand $s_1^{\epsilon_1}\cdots s_n^{\epsilon_n}$ and $r_1^{\delta_1}\cdots r_m^{\delta_m}$ are equal if and only if the sequences they represent are equal (identical). You can the verify that this happens if and only if $n=m$, and $s_i=r_i$ and $\epsilon_i=\delta_i$ for $i=1,\ldots,n$. Because this is the only way that the sequences represented by these two shorthands are identical.
This only defines the set. We still need to define the operation before we have a group, and that happens later.
The intuition is that we are going to construct a group that has every element of $S$ as an element, but with no assumptions other than those we get from this being a group. So we know that if $s\in S$, we will also need to have its inverse in our group, $s^{-1}$. And products and inverses of things we have, etc. But we will not do any simplification that does not follow from just the axioms of a group. So we know things like $s_1s_2(s_3s_2)^{-1}$ are the same as $s_1s_3^{-1}$ (because we can show these are equal using only the fact that we have a group), but we will not claim that $s_1$ and $s_1^{-1}$ are equal (even though they might be equal in some groups) because that equality does not follow from just the axioms of a group.
This "no simplifications unless they follow from the axioms of a group" is coded into the definition of "reduced word" (one in which there are no simplifications that can be done from just the axioms of a group, except for a bunch of "trailing ones", analogous to having a bunch of leading zeros in a number), and in the multiplication rule that you will see right after the definition you quote.
Saying they are equal if and only if they are identical is just like saying that two sequences of digits with trailing zeros, interpreted as a natural number written "backwards" are the same number if and only if they are identical.