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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.

Write $R=⟨⟨r_j:j\in J⟩{⟩}_F$ and let $\pi :F\to G_0=F/R$ be the quotient group homomorphism. The subgroup in the question is exactly the fiber product of groupsEvery specified generator belongs to the right-hand side, so one inclusion is immediate. For the reverse inclusion, diagonal generator pairs produce $(u,u)$ for every $u\in F$. Alsoso their products and inverses produce $(1,r)$ for every $r\in R$. If $\pi (u)=\pi (v)$, then $u^{-1}v\in R$ and $(u,v)=(u,u)(1,u^{-1}v)\in H$, proving equality.

In particular $\boxed{(1,w)\in H⟺w\in R⟺w=1\text{ in }{G}_0}$. Constructing the input pair $(1,w)$ is effective, so a solution of the membership problem for a subgroup $H\le F\times F$ would solve the insoluble word problem for a group in $G_0$. The subgroup membership problem is therefore undecidable. This is the Mihailova subgroup construction; its finite generating set follows from the finiteness of $X,J$.

Let $\pi :F\to H$ send the free generators to the given presentation generators, and let $N=\ker \pi$, the normal closure of the relators $w_j$. The Mihailova subgroup here is exactly the fiber product of groupsTo prove this equality, each specified generator belongs to the displayed fiber product, so $L$ is contained in it. The diagonal generators produce every $(a,a)$. Conjugating $(1,w_j{)}^{\pm 1}$ by a diagonal element gives $(1,a^{-1}{w}_j^{\pm 1}a)$; products give all $(1,n)$ for $n\in N$. Conversely, if $\pi (u)=\pi (v)$, then $u^{-1}v\in N$ andIn particular $(1,w)\in L$ if and only if $w=1$ in $H$. Thus membership in this finitely generated subgroup is undecidable, despite the soluble word problem for a group in the ambient direct product of groups $F\times F$.

For an explicit finite presentation, use $a_i$ and $b_i$ for the first and second copies of the free basis. The base has presentation $⟨a_1,\dots ,a_m,b_1,\dots ,b_m\mid [a_i,b_j]=1(1\le i,j\le m)⟩$. The pairs $(x_i,x_i)$ and $(1,w_j(x))$ are $a_i{b}_i$ and $w_j(b)$. Consequently the requested HNN extension has the finite presentationCommutation with the displayed subgroup generators implies commutation with all their products and inverses, so these finitely many relations are exactly the specified centralizing extension. The associated subgroup is $L$ on both sides, and the associated isomorphism is the identity.

For an input word $w$ in the generators of $H$, form $p=w(b)\in P$ and the effectively constructed wordIf $w=1$ in $H$, then $p=(1,w)\in L$ and $W=1$ by centralization. If $w\ne 1$, then $p\notin L$; the displayed word contains stable letters and no pinch in an HNN extension, so Britton's lemma gives $W\ne 1$. This is the centralizing HNN extension detects subgroup membership equivalenceA solution of the word problem for a group in $G$ would therefore solve that in $H$, a contradiction. The finitely presented extension $G$ has insoluble word problem.

The fiber product of groupsis a group under componentwise multiplication. The maps $m\mapsto (m,1)$ and $(e,g)\mapsto g$ give an exact sequenceThe section $g\mapsto (s(f(g)),g)$ has extension cocycleTherefore this pullback extension represents $f^{\ast }([\varphi ])\in H^2(G,M)$.