Free Product of Groups with Presentations

The Universal Property of the free product $G*H$ of groups $G$ and $H$ is that there exist homomorphisms $i_G:G \to G*H$ and $i_H:H \to G*H$ such that, for any group $K$ and any homomorphisms $\tau_G:G \to K$ and $\tau_H:H \to K$, there is a unique homomorphism $\phi:G*H \to K$ with $\phi i_G=\tau_G$ and $\phi i_H=\tau_H$.

It is not hard to prove uniqueness of $G*H$ up to isomorphism directly from this definition.

To prove existence, let $G = \langle A \mid R \rangle$ and $H = \langle B \mid S \rangle$ be presentations of $G$ and $H$ , and let $P = \langle A \cup B \mid R \cup S \rangle$. Then by basic properties of group presentations, the identity maps on $X$ induce homomorphisms $i_G:G \to P$ and $i_H:H \to P$.

Also, given $K,\tau_G,\tau_H$ as above, define $\psi:F(A \cup B) \to K$ by $\psi(a) = \tau_G(a)$ and $\psi(b) = \tau_H(b)$ for $a \in A,b \in B$. Then, since $\tau_G$ and $\tau_H$ are homomorphisms of $G$ and 0f $H$, $\psi$ maps all elements of $R \cup S$ to the identity, and so $\phi$ induces a homomorphism $\phi:P \to H$ with $\phi i_G=\tau_G$ and $\phi i_H=\tau_H$. Hence $P \cong G*H$.

This an old question that recently got bumped, but there's another nice proof that uses the fact that colimits commute with one another and left adjoints, so I thought I'd add it for potential future visitors.

Let $G_i =\langle X_i\mid R_i\rangle$ for $1\le i \le n$ be groups given by presentation. Another way of saying this is that we have the coequalizer diagram $$ F(R_i)\rightrightarrows F(X_i)\to G_i, $$ with the top map of the fork being induced by $R_i\subseteq F(X_i)$, and the bottom map being the zero map.

Now the coproduct in the category of groups is the free product, $\newcommand\bigast{\mathop{{\Large *}}}\bigast$, so taking the coproduct of these coequalizers, since colimits commute, we get the coequalizer $$ \bigast_{i=1}^n F(R_i) \rightrightarrows \bigast_{i=1}^n F(X_i)\to \bigast_{i=1}^n G_i, $$ but the free group functor is left adjoint to the forgetful functor, so it also commutes with colimits, so we get a coequalizer $$ F\left(\coprod_{i=1}^n R_i\right) \rightrightarrows F\left(\coprod_{i=1}^n X_i\right) \to \bigast_{i=1}^n G_i. $$

Translating this back into a statement about group presentations, this says that $\bigast_{i=1}^n G_i$ is given by the presentation $\left\langle \coprod_{i=1}^n X_i\mid \coprod_{i=1}^n R_i \right\rangle$, as desired.