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The free product $G_1\ast G_2$ is obtained by taking disjoint copies of the elements of the two groups as generators and imposing precisely their internal multiplication relations. Equivalently, if $G_i=⟨X_i\mid R_i⟩$ with disjoint $X_i$, thenA 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 expressionwhere 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 $L_h{L}_k=L_{hk}$ 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 $G_1\ast G_2$ 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 $f:G_1\ast G_2\to H$ extending both. Explicitly,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.

Eliminate $t$ using $t=xs{x}^{-1}$. The remaining relation $xt{x}^{-1}=s$ becomes $x^{2}s{x}^{-2}=s$, which follows from $x^2=1$. These Tietze transformations giveBoth factors are nontrivial, so this is a nontrivial free product. Conversely, in $C_2\ast ⟨s⟩$ the definition $t=xs{x}^{-1}$ satisfies both conjugation relations, confirming that eliminating $t$ loses no relation.

$\boxed{{\mathbb{Z}}^4\text{ is not a nontrivial free product}.}$ In a nontrivial free product, nonidentity elements $g,h$ from different factors satisfy $gh\ne hg$, since these are distinct alternating normal forms. Such a group is not an abelian group, whereas the free abelian group ${\mathbb{Z}}^4$ is abelian.

$\boxed{A_5\text{ is not a nontrivial free product}.}$ If $g\ne 1$ and $h\ne 1$ belong to different factors, the powers $(gh{)}^r$ have distinct normal forms of length $2r$. Every nontrivial free product is therefore an infinite group, whereas the alternating group $A_5$ is a finite group of order $60$.