Proposition 5

Under hypothesis (A) every element $x$ of $M$ admits a unique reduced decomposition $\sigma$. Every decomposition $\sigma' \neq \sigma$ of $x$ satisfies $l(\sigma') > l(\sigma)$.

(A) Uniqueness of a reduced decomposition:

Let $\Sigma$ denote the set of sequences $\sigma = (a; i_1, \ldots, i_n; p_1, \ldots, p_n)$ with $n \geq 0$, $a \in A, i_\alpha \in I$ and $p_\alpha \in P_{i_\alpha} - \{e_{i_\alpha}\}$ for $1 \leq \alpha \leq n$, such that $i_\alpha \neq i_{\alpha+1}$ for $1 \leq \alpha < n$. Let $\Phi$ denote the mapping of $\Sigma$ into $M$ defined by

A reduced decomposition of $x \in M$ is an element $\sigma$ of $\Sigma$ such that $\Phi(\sigma) = x$.

For all $i \in I$, let $\Sigma_i$ be the subset of $\Sigma$ consisting of the sequences $(e; i_1, \ldots, i_n; p_1, \ldots, p_n)$ with $i \neq i_1$ when $n > 0$. Let

be in $\Sigma_i$ and $\xi$ in $M_i$; let $\xi = h_i(a) \cdot p$ with $a \in A$ and $p \in P_i$, and

It is immediate that $\Psi_i$ is a bijection of $M_i \times \Sigma_i$ onto $\Sigma$.

Let $i \in I$ and $x \in M_i$; as $\Psi_i$ is bijective, a mapping $f_{i,x}$ of $\Sigma$ into itself is defined by

Further, for $a \in A, f_a$ denotes the mapping of $\Sigma$ into itself defined by

Clearly $f_{i,e_i}$ is the identity mapping of $\Sigma$ and $f_{i,xx'} = f_{i,x} \circ f_{i,x'}$ for $x, x'$ in $M_i$ and $f_{i,h_i(a)} = f_a$ for $a \in A$ and $i \in I$.

Then Proposition 4 may be applied to the case where $M'$ is the monoid of mappings of $\Sigma$ into itself with law of composition $(f,f') \mapsto f \circ f'$ and where f_i is the homomorphism $x \mapsto f_{i,x}$ of $M_i$ into $M'$; then there exists a homomorphism $f$ of $M$ into $M'$ such that $f_{i,x} = f(\phi_i(x))$ for $i \in I$ and $x \in M_i$. Let

be in $\Sigma$. Formulae (5) to (7) imply by induction on $n$ the relation that is $\sigma = f(\Phi(\sigma))(e)$. This proves that $\Phi$ is injective.

(B) Existence of a decomposition:

Let $D$ be the set of elements of $M$ admitting a decomposition. Then $e \in D$ and $M$ is generated by $\bigcup_{i \in I} \phi_i(M_i)$ and hence by $h(A) \cup \bigcup_{i \in I} \phi_i(P_i)$. Then $D.\phi_i(P_i) \subset D$ for all $i \in I$; to prove that $D = M$, it thus suffices to prove the relation $D.h(A) \subset D$. This follows from the following more precise lemma: