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For a specified subgroup $H\le G$ and a chosen alphabet for $G$, the membership problem asks whether an input word represents an element of $H$. Solubility of the ordinary word problem for a group does not imply solubility of this problem, even for finitely generated subgroups: a Mihailova subgroup of a product of finite-rank free groups can have undecidable membership.

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.

False, even when the HNN extension also has a finite presentation. Take the base $P=F_m\times F_m$ and its Mihailova subgroup $L$ from Question 4(b). Its word problem for a group is soluble: project a word to each free factor and freely reduce both projections, as in part (iv). The base is finitely presented, but the HNN extensionhas insoluble word problem by the explicit reduction in Question 4(b). Namely, for an input word $w$ in the presentation with insoluble word problem, set $p=(1,w)$; thenThe obstruction is undecidable membership in the associated subgroup, not undecidable equality in the base. Soluble base word problem alone does not let one recognize the pinches needed for a normal form theorem for an HNN extension algorithm.