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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.

Form the HNN extension which centralizes $L$:This is the extension associated to the identity isomorphism $L\to L$; relations on the displayed finite generating set imply $k^{-1}hk=h$ for all $h\in L$. Since $K_{\mathcal{M}}$ is finitely presented and $I,J$ are finite, this is a finite group presentation.

For any $g\in K_{\mathcal{M}}$,The forward implication follows from Britton's lemma: if $g\notin L$, the word $k^{-1}gk{g}^{-1}$ is reduced with two stable letters and cannot be the identity. The reverse implication is the defining centralization relation. Consequently the computable wordsatisfiesAn algorithm for the word problem for a group $G_{\mathcal{M}}$ would therefore decide the nonrecursive set $H_0(\mathcal{M})$, a contradiction. Hence $\boxed{G_{\mathcal{M}}\text{ has unsolvable word problem}}$.

We construct the output presentation explicitly, using HNN extensions and an amalgamated free product. Throughout, $P$ also denotes the group given by the input finite group presentation, and $X=\{x_1,\dots ,x_n\}$. Every word introduced below is a literal computable word in the displayed generators; testing whether $w=1$ is never part of the construction.

First fix the auxiliary Higman groupWe need two facts: $⟨a_1,a_3⟩$ is a free group of rank two, and the normal closure of $a_1$ is $Q$. The latter follows immediately from the relations: setting $a_1=1$ forces successively $a_2=a_2^2$, $a_3=a_3^2$, and $a_4=a_4^2$, hence all three are $1$.

Here is a normal-form verification of the first fact. The three-generator groupis an amalgamated free product of two Baumslag-Solitar groups $BS(1,2)$ along their infinite cyclic subgroups generated by $a_2$. Their base cyclic groups embed by Britton's lemma, so this amalgamation is legitimate. Nonzero powers of $a_1$ in the first factor lie outside $⟨a_2⟩$, as shown by the homomorphism recording the exponent of its stable letter $a_1$. Nonzero powers of $a_3$ in the second factor also lie outside $⟨a_2⟩$: the stable-letter exponent map first forces a putative equality $a_3^r=a_2^s$ to have $s=0$, and then injectivity of the base cyclic group forces $r=0$. The normal form theorem for an amalgamated free product therefore makes $⟨a_1,a_3⟩$ a rank-two free group in $Q_{123}$.

The same argument applies to $Q_{341}$, formed from the other two cyclic conjugation relations. The full $Q$ is $Q_{123}{\ast }_{⟨a_1,a_3⟩}{Q}_{341}$, so that rank-two free group embeds in $Q$ as well. In particular $a_1$ has infinite order.

Now take the free productand setIntroduce $n+2$ letters $u_0,\dots ,u_{n+1}$, and define the finite presentationFinally impose the Higman relations and identifyThus the algorithm outputsThe symbols $c,C,g_j$ in this display are abbreviations for the explicit words above, not extra generators. Renaming the finitely many new generators to avoid $X$ makes this a uniform effective construction.

An isomorphism-invariant Markov property of finitely presented groups $\rho$ has two finitely presented witnesses: a group $A_{+}$ with $\rho$, and a group $A_{-}$ which cannot embed in any finitely presented group having $\rho$. In symbols,The second condition is an obstruction to embedding, stronger than merely saying that $A_{-}$ itself fails the property.

Let $U$ be a fixed finitely presented group with unsolvable word problem for a group. Form the finitely presented free product $P=U\ast A_{-}$. Given a word $w$ in the generators of $U$, apply part (a) to this $P$ and $w$, and output the finite presentation ofIf $w=1$ in $U$, it is also $1$ in $P$, so $P(w)$ is trivial and $B(w)\cong A_{+}$ has $\rho$. If $w\ne 1$ in $U$, its image stays nontrivial in the free product $P$, and $P$ embeds in $P(w)$. Therefore $A_{-}$ embeds in $B(w)$, which cannot have $\rho$. We have the effective equivalenceAn algorithm recognizing whether an arbitrary finite group presentation has $\rho$ would decide the unsolvable word problem for a group $U$. This contradiction proves the Adian–Rabin theorem: no Markov property of finitely presented groups is algorithmically decidable.

Let $G=⟨S\mid R⟩$ and $H=⟨T\mid U⟩$ be group presentations, and let $\varphi :H\to \mathrm{Aut}(G)$ be a group homomorphism. Choose a word $w_{t,s}(S)$ representing $\varphi (t)(s)$ for each pair of generators. Then the semidirect product has presentationThe cross-relations allow every word to be written in factor order $gh$, and the multiplication rule matches the prescribed action. The natural homomorphisms to and from the semidirect product are inverse on the generators. Finite factor presentations yield a finite group presentation.

A finite group presentation has a recursive upper bound on its Dehn function exactly when its word problem for a group is decidable. If the maximal relator length is $\ell$ and a word has length $n$, a diagram of area at most $N$ can be chosen with at most $\ell N+n$ edges. Enumerating all finite labelled planar diagrams within that bound gives a terminating test. Conversely, a word decider identifies the finitely many null words of length at most $n$; exhaustive finite diagram searches find their minimal areas, whose maximum computes the Dehn function.

Enumerate all nonidentity words $w_i$ in the rank-two free group and impose relations $w_i^{p^{i+2}}=1$. The resulting two-generated group hasIt is infinite by p-deficiency at least one implies infinitude. Each element has order a power of $p$, so it is a torsion group. The infinitely many relators are essential to this particular construction; finite generation does not imply a finite group presentation.

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.