Because the generating sets are disjoint, the <free product> has the presentation
$$
\boxed{G*H=\langle X\sqcup Y\mid R\sqcup S\rangle.}
$$
This is immediate from the <universal property of a group presentation>: a map out of this presented group is exactly a pair of homomorphisms from $G$ and $H$ to the target group.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 104 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-104/1/a/solution