Mihailova subgroup
= Mihailova subgroup
{c}
{title2=$L=\{(u,v)\in F\times F:\pi(u)=\pi(v)\}$}
= Mikhailova subgroup
{c}
{synonym}
For a <finite group presentation> $H=\langle x_1,\ldots,x_m\mid r_1,\ldots,r_n\rangle$, 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.