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A finitely generated group embeds in a finitely presented group if and only if it has a recursive presentation of a group. The necessity follows by enumerating proofs that substituted generator words are trivial in the larger finite group presentation. The sufficiency is the embedding theorem; it also applies to effectively countably generated recursive presentations after an effective two-generator embedding. This is the theorem of Subgroups of finitely presented groups.

If $N◃G$ is a benign subgroup, use its witness $N=G\cap L$ inside a finitely presented group. Enumerate words on the generators of $L$, together with proofs that each equals a substituted word on the generators of $G$. This enumerates exactly the null words of $G/N$, giving a recursive presentation of a group. The Higman embedding theorem then embeds the finitely generated quotient in a finitely presented group.

Take a recursively enumerable set $S\subseteq {\mathbb{N}}_{>0}$ that is not recursive, for example the diagonal halting set from Question 2 shifted by one. To obtain a finitely generated group with recursive presentation but undecidable word problem for a group, first define the countable Right-angled Artin groupTranslation of the indices gives an automorphism $a_i\mapsto a_{i+1}$. The semidirect product with its infinite cyclic shift is generated by $a=a_0,t$ and has presentationTo verify this presentation, the semidirect product satisfies its relators, while the reverse map sends $a_i$ to $t^{i}a{t}^{-i}$; all translated commutation relations follow by conjugation. These maps are inverse on their generators. Thus $B_S$ embeds in $G_S$.

For $n\in S$, the word $[a_0,a_n]$ is trivial. For $n\notin S$, map $a_0,a_n$ to the free generators $x,y$ of $F(x,y)$ and every other $a_i$ to one. Every defining commutator of $B_S$ maps to one, since the only possible relation involving both surviving generators would have separation $n$, which is absent. The resulting group homomorphism sends $[a_0,a_n]$ to the nontrivial reduced word $[x,y]$. HenceThis proves the shift-invariant graph groups encode arbitrary sets construction rather than supposing that a missing relator automatically means a nontrivial word. A word algorithm for $G_S$ would decide $S$, so none exists. Since $S$ is enumerable, the displayed two-generator presentation is a recursive presentation of a group. By the Higman embedding theorem, $G_S$ embeds in a finitely presented group $G_0$. A word algorithm for $G_0$, composed with substitution of the two fixed generator-image words, would solve the word problem for $G_S$. Therefore $\boxed{G_0\text{ is finitely presented with insoluble word problem}}$.

The Higman embedding theorem says that a finitely generated group embeds in a finitely presented group if and only if it has a recursive presentation of a group, that is, a presentation on finitely many generators with a recursively enumerable set of relators. We use its embedding direction as an assumed theorem. The reverse direction follows by expressing subgroup generators as words in the larger group and enumerating finite products of conjugates of its relators and their inverses; comparison after free reduction enumerates all null words in the subgroup. This is the enumerable identities of finitely generated subgroups argument. The theorem does not assume that the recursively presented group's word problem for a group is decidable.

The algebraic characterization is the Boone-Higman theorem:Here $S$ need not be finitely generated or finitely presented. We prove the equivalence using the Higman embedding theorem, Britton's lemma and the normal form theorem for a free product. The original characterization is Theorem I of An algebraic characterization of groups with soluble word problem.

First suppose the embeddings exist. Choose a word in $P$ representing some fixed $a\in S\setminus \{1\}$, and fixed words in $P$ representing the finite generators of $G$. Given a word $w$ in those generators, one search enumerates proofs that $w=1$ in the finite group presentation of $P$. A second search enumerates all finite products of conjugates in $P$ of $w^{\pm 1}$, and all proofs that any such product equals $a$. All these searches are effective: a proof of equality is a finite product of conjugates of defining relators, checked in the free group, and dovetailing handles the countably many candidates.

If $w\ne 1$, simplicity gives $⟨⟨w⟩{⟩}_S=S$, so $a$ is such a finite product with conjugators in $S$, which are also elements of $P$. The second search eventually succeeds. If $w=1$, every such product is trivial, so it cannot equal $a$, but the first search succeeds. Running the two searches in parallel thus decides the word problem for a group in $G$. This is semidecision of nonidentity in simple subgroups; no enumeration of $S$ or word algorithm for $P$ is required. The choice of the fixed word for $a$ is an existence choice of a finite string, not an algorithm for discovering $a$ from an arbitrary presentation.

Conversely suppose the word problem for a group in $G$ is soluble. We construct a recursive simple envelope by cyclic HNN extensions, keeping all effectiveness requirements explicit. Start with $P_0=G\ast ⟨a⟩$, where $a$ has infinite order. Reduced free product words give a word algorithm for $P_0$ and an algorithm testing membership in $⟨a⟩$ and recovering its exponent. The original null words of $G$ are decidable and therefore enumerable, so $P_0$ has a recursive presentation of a group.

Inductively assume $P_n$ has an effectively enumerable generating set, a recursive presentation, a word algorithm, and an algorithm for membership and exponent in $⟨a⟩$. Form $B_n=P_n\ast ⟨z_n⟩$. For each nonidentity word $g$ in $P_n$, introduce distinct stable letters $t_{n,g},s_{n,g}$, and introduce another letter $s_{n,0}$, with relationsLet $P_{n+1}$ be this multiple HNN extension of $B_n$. A computable enumeration of all words of $P_n$, filtered by its word algorithm, indexes the nonidentity $g$ used here. Repeated representatives of the same element cause no difficulty: their stable letters are simply distinct.

Every associated subgroup in these relations is infinite cyclic. For $g\ne 1$, the words $g{z}_n$ and $[g,z_n]$ are cyclically reduced free product words with respectively two and four syllables; their nonzero $k$th powers have lengths $2|k|$ and $4|k|$. Therefore the reduced syllable length of an input bounds the finitely many possible exponents needed to decide membership in each subgroup and find the exponent. Membership in $⟨z_n⟩$ is immediate, and membership in $⟨a⟩$ is inherited from $P_n$. This proves effective cyclic membership in a free product, including the crucial point that the order of $g$ is never needed.

By Britton's lemma, $B_n$ embeds in $P_{n+1}$. To solve its word problem, repeatedly find a pinch in an HNN extension using these membership algorithms, transport the known exponent across its defining isomorphism, and remove two stable letters. With no pinch left, a word retaining stable letters is nontrivial and not in the base; otherwise the base word algorithm applies. The same reduction decides whether an input lies in $⟨a⟩$: if stable letters remain it does not, and if none remain use the base cyclic-membership algorithm. This verifies the inductive effectiveness claims rather than asserting that HNN extensions automatically preserve decidable word problem. Only the finitely many stable letters in the input need inspection.

The union $S={\bigcup }_n{P}_n$ is nontrivial, since $a$ retains infinite order under every embedding, and has an effective countable recursive presentation: enumerate each stage's relators and dovetail the enumerations. We now prove simplicity. Let $D◃S$ contain $g\ne 1$, and take a stage with $g\in P_n$. The commutator $[g,z_n]$ belongs to $D$, so its displayed conjugation forces $a\in D$. Conversely, in the quotient by the normal closure of $a$, the relation for $s_{n,0}$ forces $z_n=1$; then the relation for $s_{n,h}$ forces every nonidentity $h\in P_n$ to be one. Thus the normal closure of $a$ contains each $P_n$. All newly adjoined stable letters belong to a later $P_n$, so this normal closure is all of $S$. Hence $D=S$, proving that $S$ is a simple group containing $G$.

Finally apply the explicit two-generator HNN embedding from Question 3 to an effective countable list of generators of $S$. Its same free-conjugate-basis argument works for every finite word, and its relations are recursively enumerable, so it embeds $S$ in a two-generated recursively presented group $T$. The Higman embedding theorem embeds $T$ in a finitely presented group $P$. Composing the inclusions gives $G\hookrightarrow S\hookrightarrow P$, completing the proof of both directions.

A subgroup $H\le G$ is a benign subgroup when there are an embedding of $G$ into a finitely presented group $E$ and a finitely generated subgroup $L\le E$ with $G\cap L=H$. Finite generation is required of $L$, not of $H$ itself. We use Britton's lemma, the normal form theorem for an amalgamated free product, and, for the final quotient assertion, the Higman embedding theorem.

Choose witnesses $E_H,L_H$ and $E_K,L_K$ for $H,K$. The amalgamated free product $P=E_H{\ast }_G{E}_K$ embeds both witnesses and their common copy of $G$. It is finitely presented: take their two finite presentations and add the finitely many equalities identifying the images of a finite generating set of $G$. These equalities identify all elements of $G$, so they present precisely the amalgam. The original intersections $G\cap L_H=H$ and $G\cap L_K=K$ are preserved by the embeddings.

Now form the finitely presented groupOnly finitely many commutation relators, one for each of the chosen subgroup generators, are needed. The subgroup $Q=⟨G,t,s⟩$ is exactly the multiple HNN extension of $G$ centralizing $H$ and $K$. To prove this claim, take a word in $G,t,s$ reduced for those associated subgroups. It remains reduced in $T$: a $t$-pinch with a coefficient in $G$ occurs precisely when that coefficient lies in $L_H\cap G=H$, and an $s$-pinch precisely when it lies in $K$. Britton's lemma therefore prevents any extra relation, and it also embeds $G$.

Write $G^u=u^{-1}Gu$. The subgroup $L_{\cap }=G^{ts}$ is finitely generated. If $s^{-1}{t}^{-1}gts\in G$, Britton reduction must first pinch $t^{-1}gt$, requiring $g\in H$; its value is then $g$. The remaining $s^{-1}gs$ can represent a base element only if $g\in K$, and then its value is again $g$. Conversely every $g\in H\cap K$ commutes with both letters. Henceso the intersection is benign.

For the join, put $J=⟨H,K⟩$ and $L_{\vee }=⟨G^t,G^s⟩$, again finitely generated. The subgroup generated by $G,G^t,G^s$ in $Q$ isIndeed, expanding conjugate-copy syllables as $t^{-1}gt$ or $s^{-1}gs$ shows that the only possible Britton reductions are exactly the identifications in the indicated amalgamations; a reduced amalgam word therefore has no additional way to become trivial.

We use this elementary consequence of amalgam normal forms: if $C\le D\le A$, then inside $A{\ast }_{C}B$ the subgroup $⟨D,B⟩$ is $D{\ast }_{C}B$ and intersects $A$ exactly in $D$. A reduced expression with a $B\setminus C$ syllable cannot lie in $A$, because multiplying it by an inverse base element still leaves a reduced nontrivial amalgam word. Apply this first to $G{\ast }_H{G}^t$ with $D=J$, obtaining $D_1=⟨J,G^t⟩$ and $D_1\cap G=J$. Since $K\le J\le D_1$, apply it again in $U=(G{\ast }_H{G}^t){\ast }_K{G}^s$ to obtain $⟨D_1,G^s⟩\cap (G{\ast }_H{G}^t)=D_1$.

Each element of $H$ lies in $G^t$, and each element of $K$ lies in $G^s$, since the corresponding stable letter centralizes it. Thus $J\le L_{\vee }$, and $L_{\vee }=⟨D_1,G^s⟩$. Intersecting the previous identities with $G$ givesThis proves the intersection and join of benign subgroups construction with actual finitely generated witnesses inside $T$.

Finally let $N◃G$ be benign, witnessed by $N=G\cap L$ in a finitely presented group $E$. Choose finite generators $X$ for $G$ and express both $X$ and a finite generating set for $L$ as words in $E$. Enumerate every word $w$ on $X$, every word $v$ on those generators of $L$, and every finite proof in the presentation of $E$ that their substituted words are equal. Dovetail these searches and list $w$ when such a proof is found. Exactly the words representing elements of $G\cap L=N$ are listed: membership has a finite expression in the generators of $L$, and every valid equality has a finite relator proof. Since $N$ is normal, these are exactly the null words of $G/N$.

Use this enumerable list as relators on the finite alphabet $X$. Its normal closure is precisely the kernel of $F(X)\to G/N$, so it is a recursive presentation of a group for $G/N$. The Higman embedding theorem now yieldsThis is normal benign subgroup quotient embedding; no word algorithm for $E$ or membership algorithm for $L$ was assumed.

A finitely generated group with soluble word problem for a group embeds in a countably generated simple group with a recursive presentation of a group. Start with $P_0=G\ast ⟨a⟩$, retaining decidable membership in $⟨a⟩$. At stage $n$, add a free infinite cyclic generator $z_n$. For every nonidentity base element $g$, conjugate $[g,z_n]$ to $a$ and conjugate $g{z}_n$ to $a$; also conjugate $z_n$ to $a$, using distinct stable letters. All associated subgroups are infinite cyclic, with decidable membership by effective cyclic membership in a free product. Britton's lemma preserves embeddings, decidable word problem and decidable membership in the inherited $⟨a⟩$.

In the union, any nontrivial normal subgroup contains some nonidentity $g$ and hence $a$, by the commutator-conjugation relation. The other relations force $z_n$ and then every old $g$ into the normal closure of $a$. Every stable letter becomes an old element at a later stage, so the normal closure of $a$ is the whole union. The union is nontrivial and simple. Effective word enumeration at each stage makes all relators recursively enumerable. A two-generator HNN embedding followed by the Higman embedding theorem embeds this simple envelope in a finitely presented group.

For $G=⟨g_i\mid R⟩$, put $B=G\ast F(a,b)$, $u_0=a$, $v_0=b$, $u_i=b^{-i}a{b}^i$ and $v_i=g_i{a}^{-i}b{a}^i$ for $i\ge 1$. The $u_i$ freely generate by free conjugate bases in a rank-two free group. The $v_i$ do too: projection killing $G$ maps them to a free conjugate basis, so no reduced word in them is trivial. The HNN extension adjoining $t^{-1}{u}_{i}t=v_i$ embeds $G$ by Britton's lemma. Since $b=t^{-1}at$ and $g_i=t^{-1}{u}_{i}t(a^{-i}b{a}^i{)}^{-1}$, its generators reduce to $a,t$. For finitely many $g_i$ and $R$ this is a finite group presentation; for effective countably many generators and enumerable $R$ it is a two-generator recursive presentation of a group.