# Existence and Uniqueness of Reduced Form of Group Word

## Contents

## Theorem

Let $X$ be a set.

Let $w$ be a group word on $X$.

Then $w$ has a unique reduced form.

## Outline of proof

For existence, we proceed by induction on the length.

For uniqueness, we again proceed by induction on the length, by observing that two distinct elementary reductions have a common elementary reduction.

## Proof

### Existence

By induction on the length of $w$.

#### Basis for the induction

Let $w$ have length $0$.

Then $w$ is the empty word.

By Empty Group Word is Reduced, $w$ is reduced.

#### Induction step

Let $w$ have length $n \geq 1$.

If $w$ is reduced, we are done.

Otherwise, let $k \in \{1, \ldots, n-1\}$ be such that $w_{k+1} = w_k^{-1}$.

Then $w$ has an elementary reduction of length $n-2$.

By the induction hypothesis, $w$ has a reduction.

The result follows by induction.

$\Box$

### Uniqueness

By induction on the length of $w$.

#### Basis for the induction

Let $w$ have length $0$.

By Length of Reduced Form of Group Word is at most Length of Word, a reduced form of $w$ must have length $0$.

Thus a reduced form of $w$ can only be $w$ itself.

$\Box$

#### Induction step

Let $w$ have length $n \geq 1$.

Let $u$ and $v$ be reduced forms of $w$.

Let $u_1$ and $v_1$ be elementary reductions of $w$ with reduced forms $u$ and $v$.

By Distinct Elementary Reductions of Group Word have Equal Elementary Reduction, there exists a word $t$ which is an elementary reduction of $u_1$ and $v_1$.

Let $s$ be a reduced form of $t$.

Then $s$ is a reduced form of $u_1$ and $v_1$.

By the induction hypothesis, $s=u=v$.

The result follows by induction.

$\blacksquare$