Group Theory/Free products and amalgamated sums

Definition (reduced word):

Let $S$ be any set, and define the set $S^{-1}$ to be the set of formal inverses to the elements of $S$; that is, $S^{-1}:=\{s^{-1}|s\in S\}$, so that $S\cap S^{-1}=\emptyset$; for example, we could define $s^{-1}:=\{\emptyset ,s\}$. Let $\varepsilon$ denote the empty tuple. Then a reduced word over $S$ is either

1. the empty tuple $\varepsilon$, or
2. a finite tuple $(r_{1},\ldots ,r_{n})$ of elements of $S\cup S^{-1}$ such that whenever $r_{j-1},r_{j}$ are two adjacent elements, then neither $r_{j-1}\in S$ and $r_{j-1}=r_{j}^{-1}$ nor $r_{j-1}\in S^{-1}$ and $r_{j-1}^{-1}=r_{j}$.

Definition (empty word):

The empty tuple $\varepsilon =()$ is also called the empty word

Proposition (reduction of tuples to reduced words):

Let $S$ be any set, and let $S^{-1}$ be the set of formal inverses. Suppose that $(r_{1},\ldots ,r_{n})$ is any tuple (not necessarily a reduced word). Then in finitely many steps, one may obtain a reduced word from $(r_{1},\ldots ,r_{n})$ by removing adjacent elements $r_{j-1},r_{j}$ such that either $r_{j-1}\in S$ and $r_{j-1}=r_{j}^{-1}$ or $r_{j-1}\in S^{-1}$ and $r_{j-1}^{-1}=r_{j}$.

Proof: This follows immediately since the length of the tuple $(r_{1},\ldots ,r_{n})$ is an integer, which is reduced by 2 whenever adjacent elements that contradict the definition of a reduced word are eliminated. Doing this elimination repeatedly until it is no longer possible will hence lead to a reduced word in a finite number of steps. $\Box$

Note that when $n$ is odd, then the reduced word obtained in this way will not be the empty tuple. Otherwise, the empty tuple may result.

Definition (free group):

Let $S$ be any set. Then the free group over $S$ is defined to be the group $F\langle S\rangle$ whose elements are the reduced words over $S$ and whose group operation is given by first concatenation and then reduction to a reduced word.

Proposition (the free group is a group):

Let $S$ be a set. Then $F\langle S\rangle$ is a group.

Proof: The empty tuple $\varepsilon$ serves as an identity. Associativity holds because if $(r_{1},\ldots ,r_{n}),(t_{1},\ldots ,t_{m}),(u_{1},\ldots ,u_{k})$ are three reduced words, then

Finally, whenever $(r_{1},\ldots ,r_{n})$ is a reduced word, we claim by induction on $n$ that it has an inverse. Certainly the empty word has Indeed, suppose that $r_{n}\in S$; then $(r_{1},\ldots ,r_{n})(r_{n}^{-1})=(r_{1},\ldots ,r_{n-1})$, which has an inverse $(t_{1},\ldots ,t_{m})$ by the induction hypothesis, so that by associativity $(r_{n}^{-1})(t_{1},\ldots ,t_{m})$ is an inverse of $(r_{1},\ldots ,r_{n})$. $\Box$

Exercises

1. Prove that when $S$ is a set such that $|S|\geq 2$, then $F\langle S\rangle$ is not an abelian group.