The Free Product of Groups
 Definition: Let $G_1 = \langle A_1, \mathcal R_1 \rangle$ and $G_2 = \langle A_2, \mathcal R_2 \rangle$. The Free Product of $G_1$ and $G_2$ denoted $G_1 * G_2$ is group with group presentation $G_1 * G_2 = \langle A_1 \cup A_2 : \mathcal R_1 \cup \mathcal R_2 \rangle$. More generally, if $G_j = \langle A_j, \mathcal R_j \rangle$ for all $j \in J$ then the free product of all of the $G_j$s denoted $\displaystyle{*_{j \in J} G_j = \biggr \langle \bigcup_{k \in J} A_j : \bigcup_{j \in J} \mathcal R_j \biggr \rangle}$.

In the above definition, for the free product of groups to be well-defined we must always assume that the generators of the groups we're looking at are given different symbols. We can always do this by Tietze transformations.

For example, consider the following groups:

(1)
\begin{align} \quad G_1 = \langle a : a^2 = 1 \rangle \end{align}
(2)
\begin{align} \quad G_2 = \langle b : b^3 = 1 \rangle \end{align}

Then the free product of $G_1$ and $G_2$ is:

(3)
\begin{align} \quad G_1 * G_2 = \langle a, b : a^2 = 1, b^3 = 1 \rangle \end{align}
 Definition: Let $G_1 = \langle A_1 : \mathcal R_1 \rangle$ and $G_2 = \langle A_2 : \mathcal R_2 \rangle$, and $H$ be groups, and let $f_1 : H \to G_1$ and $f_2 : H \to G_2$ be group homomorphisms. The Free Product of $G_1$ and $G_2$ by Amalgamation of $H$ by $f_1$ and $f_2$ denoted $G_1 *_H G_2$ is the group with group presentation $G_1 *_H G_2 = \langle A_1 \cup A_2 : \mathcal R_1 \cup R_2 \cup \{ f_1(h) = f_2(h) : h \in H \} \rangle$. More generally, if $G_j = \langle A_j : \mathcal R_j \rangle$ for all $j \in J$, $H_{ij}$ are groups, and $\varphi_{ij}^{i} : H_{ij} \to G_i$ and $\varphi_{ij}^j : H_{ij} \to G_j$ are group homomorphisms then the free product of the $G_j$s by amalgamation of the $H_{ij}$s by the $\varphi_{ij}^{i}$s and $\varphi_{ij}^j$s is $\displaystyle{*_{H_{ij}s} G_j = \biggr \langle \bigcup_{j \in J} A_j : \bigcup_{j \in J} \mathcal R_j \cup \bigcup_{i, j \in J} \{ \varphi_{ij}^{i}(h) = \varphi_{ij}^{j}(h) : h \in H_{ij} \} \biggr \rangle}$.

For example, consider the following groups:

(4)
\begin{align} \quad G_1 = \langle x : x^2 = 1 \rangle \end{align}
(5)
\begin{align} \quad G_2 = \langle a, b : ab = ba \rangle \end{align}
(6)
\begin{align} \quad H = \langle y : \emptyset \rangle \end{align}

And let $f_1 : G_1 \to H$ and $f_2 : G_2 \to H$ be group homomorphisms defined by:

(7)
\begin{align} \quad f_1(y) = x \end{align}
(8)
\begin{align} \quad f_2(y) = ab \end{align}

Then the free product of $G_1$ and $G_2$ by amalgamation of $H$ by $f_1$ and $f_2$ is:

(9)
\begin{align} \quad G_1 *_H G_2 &= \langle x, a, b : x^2 = 1, ab = ba, \{ f_1(h) = f_2(h) : h \in H \} \rangle \\ &= \langle x, a, b : x^2 = 1, ab = ba, \{ x = ab, x^2 = (ab)^2, ..., \} \rangle \\ &= \langle x, a, b : x^2 = 1, ab = ba, x = ab \rangle \end{align}