# Group Theory/Free products and amalgamated sums

Definition (reduced word):

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

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

Definition (empty word):

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

Proposition (reduction of tuples to reduced words):

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

Proof: This follows immediately since the length of the tuple ${\displaystyle (r_{1},\ldots ,r_{n})}$ 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. ${\displaystyle \Box }$

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

Proposition (the free group is a group):

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

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

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

## Exercises

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