free product with amalgamated subgroup

It follows by ``general nonsense'' that the free product of $G_1$ and $G_2$ with amalgamated subgroup $G_0$ , if it exists, is ``unique up to unique isomorphism.'' The free product of $G_1$ and $G_2$ with amalgamated subgroup $G_0$ , is denoted by $G_1\bigstar_{G_0} G_2$ . The following theorem asserts its existence.

Proof. [Sketch of proof] Without loss of generality assume that $G_0$ is a subgroup of $G_k$ and that $i_k$ is the inclusion for $k=1,2$ . Let $$G_k=\langle (x_{k;s})_{s\in S}\, |\, (r_{k;t})_{t\in T} \rangle$$ be a presentation of $G_k$ for $k=1,2$ . Each $g\in G_0$ can be expressed as a word in the generators of $G_k$ ; denote that word by $w_k(g)$ and let $N$ be the normal closure of $\{w_1(g)w_2(g)^{-1}\,|\,g\in G_0\}$ in the free product $G_1\bigstar G_2$ . Define $$G_1\bigstar_{G_0} G_2:=G_1\bigstar G_2/N\,$$ and for $k=0,1$ define $j_k$ to be the inclusion into the free product followed by the canonical projection. Clearly (1) is satisfied, while (2) follows from the universal properties of the free product and the quotient group.

Notice that in the above proof it would be sufficient to divide by the relations $w_1(g)w_2(g)^{-1}$ for $g$ in a generating set of $G_0$ . This is useful in practice when one is interested in obtaining a presentation of $G_1\bigstar_{G_0} G_2$ .

In case that the $i_k$ 's are not injective the above still goes through verbatim. The group thusly obtained is called a ``pushout''.

Examples of free products with amalgamated subgroups are provided by Van Kampen's theorem.