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

Definition. A word is called reduced if it contains no string of the form ${\displaystyle aa^{-1},a\in S^{\pm }.}$ 

Definition. 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. 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 ${\displaystyle aa^{-1}}$  with ${\displaystyle a\in S^{\pm }}$ . 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 ${\displaystyle u_{1}=u_{2}}$ .

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 ${\displaystyle w=g_{1}\cdots g_{n}}$  where ${\displaystyle g_{j}\in A{\text{ or }}B}$  for ${\displaystyle 1\leqslant j\leqslant n}$ .

Definition. A reduced sequence is a sequence ${\displaystyle g_{1},\cdots ,g_{n}}$  such that for ${\displaystyle 1\leqslant j\leqslant n}$  we have ${\displaystyle g_{j}\in A{\text{ or }}B,g_{j}\neq e}$  and ${\displaystyle g_{j},g_{j+1}}$  are not in the same factor ${\displaystyle A}$  or ${\displaystyle B}$ . The identity element is represented by the empty set.

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

(1) If ${\displaystyle w=g_{1}\cdots g_{n},n>0}$ , where ${\displaystyle g_{1},\cdots ,g_{n}}$  is a reduced sequence, then ${\displaystyle w\neq 1}$  in ${\displaystyle A*B.}$ 

(2) Each element of ${\displaystyle A*B}$  can be written uniquely as ${\displaystyle w=g_{1}\cdots g_{n}}$  where ${\displaystyle g_{1},\cdots ,g_{n}}$  is a reduced sequence.

Proof edit

Equivalence edit

The fact that the second statement implies the first is easy. Now suppose the first statement holds and let:

${\displaystyle g_{1}\cdots g_{m}=w=h_{1}\cdots h_{n}.}$ 

This implies

${\displaystyle h_{n}^{-1}\cdots h_{1}^{-1}g_{1}\cdots g_{m}=1.}$ 

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 ${\displaystyle 1\leqslant i\leqslant n.}$  This shows both statements are equivalent.

Proof of (2) edit

Let W be the set of all reduced sequences in A ∗ B and S(W) be its group of permutations. Define φ : A → S(W) as follows:

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

Similarly we define ψ : B → S(W).

It is easy to check that φ and ψ are homomorphisms. Therefore by universal property of free product we will get a unique map φ ∗ ψ : A ∗ B → S(W) such that φ ∗ ψ (id)(1) = id(1) = 1.

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

• Lyndon, Roger C.; Schupp, Paul E. (1977). Combinatorial Group Theory. Springer. ISBN 978-3-540-41158-1..