nLab
free product of groups

Given groups $G_i$, $i\in I$, their free product $G_1\star G_2\star \dots ={\star }_i{G}_i$ is their coproduct in Grp.

Given presentations for the $G_i$, it is straightforward to find a presentation of ${\star }_i{G}_i$; if $G_i=⟨S_i\mid R_i⟩=F_i/N_i$, where each $F_i=⟨S_i⟩$ is the free group on the set $S_i$ and $N_i\subset F_i$ is the normal subgroup of $F_i$ generated by the subset $R_i\subset F_i$, then the free product

$${\star }_i{G}_i:=⟨\coprod _i{S}_i\mid \coprod _i{R}_i⟩=({\star }_i{F}_i)/⟨{\cup }_i{N}_i⟩$$\star_i G_i := \langle \coprod_i S_i | \coprod_i R_i \rangle = (\star_i F_i)/\langle\cup_i N_i\rangle

is presented by the disjoint unions of the $S_i$ and the $R_i$. As with anything satisfying a universal property, the result (up to a unique coherent isomorphism) does not depend on the presentations.

The fact that free products always exist now follows from the fact that any group has a presentation; we can always take $S_i$ to be the underlying set of $G_i$ and take $R_i$ to be the set of all words in $F_i$ that equal the identity in $G_i$. The value of the more general construction above is that one often has much smaller $S_i$ and $R_i$ to work with. Even if $G_i$ is infinite, the $S_i$ and $R_i$ might be finite (in the strictest sense), making this part of finite mathematics and directly subject to the methods of combinatorial group theory.