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 $s_1\cdots s_nk,$ 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 $(\operatorname{Sym}(X), \circ, \mathrm{id}_X)$ 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 $\operatorname{Sym}(X)$ is not abelian: the transpositions $\tau = (a\,b)$ and $\rho = (b\,c)$ satisfy $\tau \circ \rho \ne \rho \circ \tau$. ($\operatorname{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

Let $(M,\cdot,e)$ and $(M',\cdot',e')$ be monoids (def-semigroup-and-monoid). A monoid homomorphism from $M$ to $M'$ is a function $f : M \to M'$ such that - (H1) $f(x \cdot y) = f(x) \cdot' f(y)$ for all $x, y \in M$; - (H2) $f(e) = e'$. Let $G$ and $G'$ be groups (def-group). A group homomorphism from $G$ to $G'$ is a function $f : G \to G'$ satisfying (H1) alone: $f(xy) \;=\; f(x)\, f(y) \qquad \text{for all } x, y \in G .$ Condition (H2) is not imposed for groups because it follows: a group homomorphism automatically satisfies $f(e) = e'$ 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 $(g \circ f)(e) = g(e') = e''$; the same computation, without the second clause, shows a composite of group homomorphisms is a group homomorphism. (Monoid homomorphism and group homomorphism).