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Introduction

In mathematics, particularly in combinatorial group theory, a normal form for a free group over a set of generators or for a free product of groups is a representation of an element by a simpler element, the element being either in the free group or free products of group. In case of free group these simpler elements are reduced words and in the case of free product of groups these are reduced sequences. The precise definitions of these are given below. As it turns out, for a free group and for the free product of groups, there exists a unique normal form i.e each element is representable by a simpler element and this representation is unique. This is the Normal Form Theorem for the free groups and for the free product of groups. The proof here of the Normal Form Theorem follows the idea of Artin and van der Waerden.

Normal form for free groups

Let ${\displaystyle G}$ be a free group with generating set ${\displaystyle S}$. Each element in ${\displaystyle G}$ is represented by a word ${\displaystyle w}$, ${\displaystyle w=a_{1}a_{2}\ldots \,a_{n}}$, where ${\displaystyle a_{j}\in \,S^{\pm }\,\forall \,1\leq \,j\leq \,n.}$

Definition(Reduced Word)

A word ${\displaystyle w}$ is reduced if it contains no part ${\displaystyle aa^{-1},\,a\in \,S^{\pm }}$.

Definition(Normal Form)

A normal form for a free group ${\displaystyle G}$ with generating set ${\displaystyle S}$ is a choice of a reduced word in ${\displaystyle S}$ for each element of ${\displaystyle G}$.

Normal Form Theorem for free groups

Statement A free group has a unique normal form i.e. each element in ${\displaystyle G}$ is represented by a unique reduced word.

Proof An elementary transformation of a word ${\displaystyle w\in \,G}$ consists of inserting or deleting a part of the form ${\displaystyle aa^{-1}\,a\in \,S^{\pm }}$. Two words ${\displaystyle w_{1}}$ and ${\displaystyle w_{2}}$ are equivalent, ${\displaystyle w_{1}\equiv \,w_{2}}$, if there is a chain of elementary transformations leading from ${\displaystyle w_{1}}$ to ${\displaystyle w_{2}}$. This is obviously an equivalence relation on ${\displaystyle G}$. Let ${\displaystyle G_{0}}$ be the set of reduced words. We shall show that each equivalence class of words contains exactly one reduced word. It is clear that each equivalence class contains a reduced word, since successive deletion of parts ${\displaystyle aa^{-1}}$ from any word ${\displaystyle w}$ must lead to a reduced word. It will suffice then to show that distinct reduced words ${\displaystyle u}$ and ${\displaystyle v}$ are not equivalent. For each ${\displaystyle x\in \,S}$ define a permutation ${\displaystyle x\Delta }$ of ${\displaystyle G_{0}}$ by setting ${\displaystyle w(x\Delta )=wx}$ if ${\displaystyle wx}$ is reduced and ${\displaystyle w(x\Delta )=u}$ if ${\displaystyle w=ux^{-1}}$. Let ${\displaystyle P}$ be the group of permutations of ${\displaystyle G_{0}}$ generated by the ${\displaystyle x\Delta ,\,x\in \,S}$. Let ${\displaystyle {\Delta }^{*}}$ be the multiplicative extension of ${\displaystyle \Delta }$ to a map ${\displaystyle {\Delta }^{*}:W\mapsto \,P}$. If ${\displaystyle u_{1}\equiv \,u_{2},}$ then ${\displaystyle u_{1}{\Delta }^{*}=u_{2}{\Delta }^{*}}$; moreover ${\displaystyle 1(u{\Delta }^{*})=u_{0}}$ is reduced with ${\displaystyle u_{0}\equiv \,u.}$ It follows that if ${\displaystyle u_{1}\equiv \,u_{2}}$ with ${\displaystyle u_{1},\,u_{2}}$ reduced, then ${\displaystyle u_{1}=u_{2}}$.

Normal form for free products

Let ${\displaystyle G\,=\,A\,*\,B}$ be the free product of groups ${\displaystyle A}$ and ${\displaystyle B}$. Every element ${\displaystyle w\,\in \,G}$ is represented by ${\displaystyle w\,=\,g_{1}g_{2}...g_{n}}$ where ${\displaystyle g_{j}\in \,A\,\,{\text{or}}\,\,B}$ for ${\displaystyle 1\leq \,j\leq \,n}$.

Definition(Reduced Sequence)

A reduced sequence is a sequence ${\displaystyle g_{1},\,g_{2}\ldots \,g_{n}}$ such that ${\displaystyle g_{j}\,\in \,A\,{\text{or}}\,B\,\forall \,1\leq \,j\leq \,n}$ with the property that ${\displaystyle g_{j}\,\neq \,e\,\forall \,j,}$ and ${\displaystyle g_{j},\,g_{j+1}}$ are not in the same factor ${\displaystyle A}$ or ${\displaystyle B}$.

Definition(Normal Form)

A normal form for a free product of groups is a representation or choice of a reduced sequence for each element in the free product.

Normal Form Theorem for free product of groups

There are two equivalent version of Normal Form Theorem in the case of Free products.

Statement Consider the free product ${\displaystyle A*B}$ of two groups ${\displaystyle A}$ and ${\displaystyle B}$. Then the following two equivalent statements hold.

• If ${\displaystyle w=g_{1}g_{2}\cdots g_{n}\,\,,n>0}$, where ${\displaystyle g_{1},g_{2},\dots ,g_{n}}$ is a reduced sequence, then ${\displaystyle w\neq 1}$ in ${\displaystyle A*B}$.
• Each element ${\displaystyle w}$ of ${\displaystyle A*B}$ can be written uniquely as ${\displaystyle w=g_{1}g_{2}\cdots g_{n}}$ where ${\displaystyle g_{1},g_{2},\dots ,g_{n}}$ is a reduced sequence (for ${\displaystyle w=1}$ we will take the reduced sequence to be empty).

Proof : First consider the equivalence of the above two statements.

The second statement implies the first is easy.

Suppose the first statement holds. Let ${\displaystyle w=g_{1}g_{2}\cdots g_{m}}$ and ${\displaystyle w=h_{1}h_{2}\cdots h_{n}}$, then we have ${\displaystyle g_{1}g_{2}\cdots g_{m}\,=\,h_{1}h_{2}\cdots h_{n}.}$, which implies ${\displaystyle h_{n}^{-1}h_{n-1}^{-1}\cdots h_{1}^{-1}g_{1}g_{2}\cdots g_{m}\,=\,1.}$ Hence by first statement left hand side cannot be reduced. This can happen only if ${\displaystyle h_{1}^{-1}g_{1}\,=\,1}$, i.e ${\displaystyle g_{1}\,=\,h_{1}.}$ Proceeding inductively we have ${\displaystyle m\,=\,n}$ and ${\displaystyle g_{i}\,=\,h_{i}}$ for all ${\displaystyle i\,=\,1,2,\dots ,n.}$ This shows both statements are equivalent.

Now we will show that these statements hold.

Let ${\displaystyle W}$ be the set of all reduced sequences in ${\displaystyle A*B}$. Let ${\displaystyle S(W)}$ be the group of permutations of ${\displaystyle W}$. Define ${\displaystyle \phi :A\rightarrow S(W)}$ as follows. If ${\displaystyle a=id}$, ${\displaystyle \phi (a)=id}$. Otherwise define ${\displaystyle \phi }$ as

${\displaystyle \phi (x)(g_{1},g_{2},\dots ,g_{m})={\begin{cases}(a,g_{1},g_{2},\cdots ,g_{m})&{\text{if }}g_{1}\in B.\\(ag_{1},g_{2},\dots ,g_{n})&{\text{if }}g_{1}\in A\,\,\,and\,\,\,ag_{1}\neq 1.\\(g_{2},g_{3},\dots ,g_{n})&{\text{if}}\,ag_{1}=1.\end{cases}}}$

Similarly we define ${\displaystyle \psi :B\rightarrow S(W)}$.

It is easy to check that ${\displaystyle \phi }$ and ${\displaystyle \psi }$ are homomorphisms. Therefore by universal property of free product we will get a unique map ${\displaystyle \phi *\psi :A*B\rightarrow S(W)}$ and ${\displaystyle \phi *\psi (id)(1)=id(1)=1}$.

Now suppose ${\displaystyle w=g_{1}g_{2}\cdots g_{n}\,\,,n>0}$, where ${\displaystyle g_{1},g_{2},\dots ,g_{n}}$ is a reduced sequence, then ${\displaystyle \phi *\psi (w)(1)=(g_{1},g_{2},\dots ,g_{n}).}$ Therefore ${\displaystyle w=1}$ in ${\displaystyle A*B}$ implies ${\displaystyle n=0}$, a contradiction.

References

• C. Lyndon, Roger; E. Schupp, Paul (1977). Combinatorial Group Theory. Springer. ISBN 978-3-540-41158-1. {{cite book}}: Cite has empty unknown parameter: |1= (help).