Algebra
《Algebra》 · 第 68 页 · Chapter I，Section 9「Free Groups, Free Products, and Generators and Relations」
Given a family of groups $\{G_i\mid i\in I\}$ we may assume (by relabeling if necessary) that the $G_i$ are mutually disjoint sets. Let $X=\bigcup_{i\in I}G_i$ and let $\{1\}$ be a one-element set disjoint from $X$. A word on $X$ is any sequence $(a_1,a_2,\ldots)$ such that $a_i\in X\cup\{1\}$ and for some $n\in\mathbf N^*$, $a_i=1$ for all $i\ge n$. A word $(a_1,a_2,\ldots)$ is reduced provided: (i) no $a_i\in X$ is the identity element in its group $G_j$; (ii) for all $i,j\ge1$, $a_i$ and $a_{i+1}$ are not in the same group $G_j$; (iii) $a_k=1$ implies $a_i=1$ for all $i\ge k$. In particular $1=(1,1,\ldots)$ is reduced. Every reduced word $(\ne1)$ may be written uniquely as $a_1a_2\cdots a_n=(a_1,a_2,\ldots,a_n,1,1,\ldots)$, where $a_i\in X$. Let $\prod_{i\in I}^{*}G_i$ (or $G_1*G_2*\cdots*G_n$ if $I$ is finite) be the set of all reduced words on $X$. $\prod_{i\in I}^{*}G_i$ forms a group, called the free product of the family $\{G_i\mid i\in I\}$, under the binary operation defined as follows. $1$ is the identity element and the product of two reduced words $(\ne1)$ essentially is to be given by juxtaposition. Since the juxtaposed product of two reduced words may not be reduced, one must make the necessary cancellations and contractions.