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No Markov property of finitely presented groups is decidable by an algorithm taking an arbitrary finite group presentation as input. The proof reduces the word problem for a group to property recognition using an effective construction which collapses when an input word is trivial and embeds the input group when it is nontrivial.

A finitely presented group is a group admitting a finite group presentation. Finite presentability concerns the existence of such a presentation, not the decidability of its word problem for a group.

The Higman group has the finite group presentation $Q=⟨a_1,a_2,a_3,a_4\mid a_i{a}_{i+1}{a}_i^{-1}=a_{i+1}^2(i\mathrm{mod}4)⟩$. Its subgroup $⟨a_1,a_3⟩$ is a rank-two free group, and the normal closure of each $a_i$ is all of $Q$. It has no nontrivial finite quotients of a group: a least prime divisor among the four generator orders in a finite image contradicts the multiplicative-order constraints imposed by conjugation to squares. These properties give finite-presentation constructions embedding arbitrary finitely presented groups into groups with no nontrivial finite quotients.

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.

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.