Coproducts in $\text{Grp}$

Question

Background:

Let $G, H$ be groups. I am currently trying to show that $G*H$, the free product of $G$ and $H$, is the coproduct of $G$ and $H$ in $\text{Grp}$.

I have no resources for this, so I am probably making a great many mistakes, but my idea was to try and use the universal property of free groups to show $F(G\amalg H)=G*H$ satisfies the universal property of the coproduct in $\text{Grp}$. I'm not sure if this can work though. The problem I am having is that the universal property of free groups involves set functions from $\{G,H\}$ to groups, whereas the univeral property for coproducts involves homomorphisms from $G$ and $H$.

As Thomas Andrews points out in the comments $F(G\amalg H)\not \cong G*H$, so this approach is not a really approach at all. As such my question is:

How does one see that $G*H$ is the coproduct in group?

Answer 1

Assume that we have a group $T$ and morphisms $\phi_G:G\to T$ and $\phi_H:H\to T$. Denote by $\iota_G:G\hookrightarrow G\ast H$ the inclusion, same with $\iota_H$. For a letter $g\in G\mathop{\dot\cup} H$, let $$D(g) = \left\{ \begin{array}{lcl} G &;& g \in G \\ H &;& g \in H \end{array}\right. $$ and define the map $\phi$ given by $$\begin{align*} G\ast H &\longrightarrow T \\ g_1\ldots g_r &\longmapsto \phi_{D(g_1)}(g_1)\cdots\phi_{D(g_r)}(g_r). \end{align*}$$ The relations of $G\ast H$ make this well-defined and a morphism of groups with the property $\phi\circ\iota_G=\phi_G$ and also $\phi\circ\iota_H=\phi_H$.

On the other hand, given any morphism $\psi:G\ast H \to T$ with $\psi\circ\iota_G=\phi_G$ and $\psi\circ\iota_H=\phi_H$, it must map a letter $g\in G\mathop{\dot\cup} H$ to $\phi_{D(g)}(g)$, forcing $\psi=\phi$. This should prove existence and uniqueness.