Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 104 / 1 / iii

Eliminate $t$ using $t=xs{x}^{-1}$. The remaining relation $xt{x}^{-1}=s$ becomes $x^{2}s{x}^{-2}=s$, which follows from $x^2=1$. These Tietze transformations giveBoth factors are nontrivial, so this is a nontrivial free product. Conversely, in $C_2\ast ⟨s⟩$ the definition $t=xs{x}^{-1}$ satisfies both conjugation relations, confirming that eliminating $t$ loses no relation.