Factor elements act consistently by permutations on amalgamated normal words

Given: The objects and hypotheses in the statement.

Let $K$ be embedded in $G$ and $H$ as in def-free-product-with-amalgamation. By def-axiom-of-choice, choose left-coset transversals $S_G,S_H$ containing the identity. A normal word is $s_1\cdots s_{n}k,$ where $n\in \mathbb{N}$ (def-natural-numbers), $k\in K$, every $s_j$ is a nonidentity representative from $S_G$ or $S_H$, and consecutive representatives come from different factors. Length zero means the word is just $k$. The written form depends on the transversals. (Transversal normal-form data for an amalgamated free product).

For every set $X$, the triple $(\mathrm{Sym}(X),\circ ,{\mathrm{i}\mathrm{d}}_X)$ of def-symmetric-group is a group (def-group); the inverse of a permutation $f$ is its inverse function $f^{-1}$. If $X$ contains three distinct elements $a$, $b$, $c$, then $\mathrm{Sym}(X)$ is not abelian: the transpositions $\tau =(ab)$ and $\rho =(bc)$ satisfy $\tau \circ \rho \mathrm{\ne }\rho \circ \tau$. ($\mathrm{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

Let $(M,\cdot ,e)$ and $(M^{\mathrm{\prime }},{\cdot }^{\mathrm{\prime }},e^{\mathrm{\prime }})$ be monoids (def-semigroup-and-monoid). A monoid homomorphism from $M$ to $M^{\mathrm{\prime }}$ is a function $f:M\to M^{\mathrm{\prime }}$ such that - (H1) $f(x\cdot y)=f(x){\cdot }^{\mathrm{\prime }}f(y)$ for all $x,y\in M$; - (H2) $f(e)=e^{\mathrm{\prime }}$. Let $G$ and $G^{\mathrm{\prime }}$ be groups (def-group). A group homomorphism from $G$ to $G^{\mathrm{\prime }}$ is a function $f:G\to G^{\mathrm{\prime }}$ satisfying (H1) alone: $f(xy)=f(x)f(y)\text{for all }x,y\in G\mathrm{.}$ Condition (H2) is not imposed for groups because it follows: a group homomorphism automatically satisfies $f(e)=e^{\mathrm{\prime }}$ and $f(x^{-1})=f(x{)}^{-1}$ (lem-group-homomorphism-basic-properties). For monoids it does not follow and must be assumed, which is why the two definitions differ. A homomorphism from a structure to itself is an endomorphism. The identity map of $M$ is a monoid homomorphism, and a composite of monoid homomorphisms is one, since $(g\circ f)(xy)=g(f(x)f(y))=g(f(x))g(f(y))$ and $(g\circ f)(e)=g(e^{\mathrm{\prime }})=e^{\mathrm{\prime }\mathrm{\prime }}$; the same computation, without the second clause, shows a composite of group homomorphisms is a group homomorphism. (Monoid homomorphism and group homomorphism).