 Mihailova subgroup

For a finite group presentation $H=⟨x_1,\dots ,x_m\mid r_1,\dots ,r_n⟩$, let $F$ be the free group on its generators and $\pi :F\to H$. The corresponding fiber product of groups is generated by $(x_i,x_i)$ and $(1,r_j)$. Conjugation by the diagonal pairs supplies every pair $(1,r)$ with $r$ in the normal closure of the relators, and $(u,v)=(u,u)(1,u^{-1}v)$ proves the generating claim. Membership of $(1,w)$ is equivalent to $w=1$ in $H$, so this finitely generated subgroup can have undecidable membership even though $F\times F$ has soluble word problem.