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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.

The allowed algebraic characterization is the Boone-Higman theorem:The simple group $S$ need not itself be finitely presented. The primary statement is Theorem I of An algebraic characterization of groups with soluble word problem. In particular, a finitely generated group with insoluble word problem for a group cannot embed in any finitely presented simple group: otherwise take that group as both $S$ and $P$ in the characterization.

Use the following effective form of the permitted Messuage lemma for groups. Given a finite group presentation of $K$ and a word $w$, one can algorithmically write a finite presentation $K_w$ such that $w=1$ in $K$ implies $K_w=1$, while $w\ne 1$ implies that $K$ embeds in $K_w$. The construction is performed without first deciding whether $w$ is trivial. A collapse-and-embedding lemma with these properties is also stated as Lemma 5.13 in Combinatorial Group Theory. No proof of that allowed lemma is needed here.

Fix a finitely presented group $U$ with insoluble word problem for a group, whose existence is allowed. For each input word $w$ in its generators, construct $U_w$ by the lemma and then form the finite presentation ofIf $w=1$, then $P_w\cong C_2$, which is a nontrivial simple group. If $w\ne 1$, the lemma embeds $U$ in $U_w$, and the normal form theorem for a free product embeds $U_w$ in $P_w$. If $P_w$ were simple, it would be a finitely presented simple group containing $U$, contradicting the characterization. HenceAn algorithm deciding simplicity from arbitrary finite presentations would therefore solve the insoluble word problem for a group in $U$. This proves simplicity of finitely presented groups is algorithmically undecidable. Using $C_2$ avoids any convention about whether the trivial group is called simple.