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The <free product> $G_1*G_2$ is obtained by taking disjoint copies of the elements of the two <groups> as generators and imposing precisely their internal multiplication relations. Equivalently, if $G_i=\langle X_i\mid R_i\rangle$ with disjoint $X_i$, then
$$
G_1*G_2=\langle X_1\sqcup X_2\mid R_1\sqcup R_2\rangle.
$$
A <nontrivial free product> has both factors different from the <trivial group>.
The <normal form theorem for a free product> says that every element has a unique expression
$$
\boxed{g_1g_2\cdots g_r,\quad g_j\in G_{\epsilon_j}\setminus\{1\},\quad \epsilon_j\ne\epsilon_{j+1},}
$$
where the empty expression represents the identity. The $g_j$ are called <syllables in a free product>. Thus nonempty alternating products cannot be the identity, and the natural maps of the factors into the <free product> are injective.
For existence, multiply adjacent syllables from the same factor and delete identity syllables until the word alternates. Each change uses a defining relation and decreases the number of syllables. To establish uniqueness without assuming that these reductions are confluent, let $\mathcal W$ be the set of alternating nonidentity syllable sequences, including the empty sequence. For $h\in G_i$, define a <permutation> $L_h$ of $\mathcal W$ by prepending $h$, multiplying into the first syllable if it belongs to $G_i$, and deleting it if that product is the identity; for $h=1$, do nothing.
The multiplication law inside $G_i$ gives $L_hL_k=L_{hk}$ for $h,k\in G_i$, and $L_h^{-1}=L_{h^{-1}}$. This can be checked at the first syllable: if a product deletes it, the next syllable belongs to the other factor and is treated as a new first syllable. Hence the defining relations give a <group action> of $G_1*G_2$ on $\mathcal W$. An alternating word $g_1\cdots g_r$ sends the empty sequence to its own syllable sequence. Two such words representing the same group element induce the same <permutation>, and therefore have identical sequences. This proves the <normal form theorem for a free product>.
The <universal property of a free product> states that, for <group homomorphisms> $f_i:G_i\to H$, there is a unique <group homomorphism> $f:G_1*G_2\to H$ extending both. Explicitly,
$$
\boxed{f(g_1\cdots g_r)=f_{\epsilon_1}(g_1)\cdots f_{\epsilon_r}(g_r).}
$$
The internal multiplication relations are respected because each $f_i$ is a <group homomorphism>, so this assignment descends from words to the presented <group>. Alternatively, multiplication of two normal forms is concatenation followed by precisely those internal relations, which do not change its value in $H$. Uniqueness holds because the two factors generate the <free product>.
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