Given a family of groups $\{G_i:i\in I\}$ we may assume that $G_i$ are pairwise disjoint. Let $X=\bigcup^{}_{i\in I}G_i$ and let $\{1\}$ be one element disjoint from $X.$ A word on on $X$ is any sequence $(a_1, a_2, ...)$ such that $a_i ∈ X \cup \{1\}$ and for some $n ∈ N, a_i = 1$ for all $i ≥ n.$ A word $(a_1, a_2,...)$ is reduced if:
$(a)$ no $a_i ∈ X$ is the identity element in its group $G_i;$
$(b)$ for all $i ≥ 1, a_i$ and $a_{i+1}$ are not in the same group $G_i;$
$(c)$ $a_k = 1$ implies $a_i = 1$ for all $i ≥ k.$
In particular $1=(1,1,...,1)$ is reduced. Show that every reduced word can be written uniquely as $a_1\cdot a_2 \cdot ... \cdot a_n=(a_1,a_2,...,a_n,1,1,...),$where $a_i \in X.$
I dont even know if I understand definition, but here is my attempt:
Assume some word can't be written uniquely, so:
$a_1\cdot a_2 \cdot ... \cdot a_n$=$b_1\cdot b_2 \cdot ... \cdot b_n$
$b_1^{-1}\cdot a_1\cdot a_2 \cdot ... \cdot a_n = e_1 \cdot b_2 \cdot ... \cdot b_n$.
$e_1$ is indentity element in $G_1$. Contradiction with $(a)$ point.