Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 104 / 6

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.

If $w=1$ in $P$, its image is the identity in the output group, so $c=1$. The equations $u_j^{-1}c{u}_j=g_j$ then give $z=d=1$ and $x_{i}z=1$ for every $i$, hence $x_i=1$. Therefore $C=1$, and the relation $C=a_1$ gives $a_1=1$. Since the normal closure of $a_1$ is $Q$, every $a_i$ is $1$. The remaining identifications make every $u_j$ equal to $1$ as well. All output generators vanish, provingThis proves collapse even when the original group $P$ itself is nontrivial.

Suppose $w\ne 1$ in $P$. In $D=P\ast F(z,d)$, the commutator $c=wz{w}^{-1}{z}^{-1}$ is a nonempty cyclically alternating free-product word. Every positive power remains reduced, so $c$ has infinite order. Each $g_j$ also has infinite order: this is immediate for $z,d$; if $x_i=1$, then $x_{i}z=z$, and otherwise $x_{i}z$ is a cyclically alternating word of length two.

Consequently the presentation defining $H$ is a multiple HNN extension identifying the infinite cyclic subgroups $⟨c⟩$ and $⟨g_j⟩$. By the normal form theorem for an HNN extension, $D$ embeds in $H$.

The subgroupis a free group on exactly these $n+3$ elements. For a proof, view $D$ as $(P\ast ⟨z⟩)\ast ⟨d⟩$. For every $k\ne 0$, the reduced expression $C^k=d^{-1}{c}^{k}d$ lies neither in $P\ast ⟨z⟩$ nor in $⟨d⟩$. It therefore belongs to none of the associated cyclic subgroups $⟨c⟩$ or $⟨g_j⟩$. Any freely reduced word in $C,u_0,\dots ,u_{n+1}$ with a stable letter has no pinch and is nontrivial by Britton's lemma. A word with no stable letters is a nonzero power of $C$, also nontrivial. This proves the claimed free basis of a group.

Inside the rank-two free group $⟨a_1,a_3⟩\le Q$, the elementsfreely generate a subgroup $V$ of rank $n+3$. This is the same free-product normal-form calculation as for the conjugate basis in Question 5: between conjugates with distinct indices, a nonzero power of $a_3$ remains. The identifications in the output presentation give an isomorphism $U\to V$ matching these free bases. HenceThe normal form theorem for an amalgamated free product makes both factors embed. Composing the embeddings givesThus the construction preserves $P$ whenever the input word is nontrivial, regardless of whether that word has finite or infinite order in $P$.

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.

The auxiliary Higman group $Q$ has no nontrivial finite quotients of a group. One can use this permitted property directly, but there is a short verification. In any finite quotient, let $n_i$ be the order of the image of $a_i$. Since conjugation sends $a_{i+1}$ to its square, $n_{i+1}$ must be odd. If some $n_i>1$, choose the least prime number $p$ dividing any $n_i$. It is odd. Conjugation by the preceding generator givesThe multiplicative order of $2$ modulo $p$ is therefore a divisor of $n_{i-1}$. It is greater than one and divides $p-1$ by the Fermat little theorem, so it has a prime divisor smaller than $p$. That prime also divides $n_{i-1}$, contradicting the choice of $p$. Hence every $n_i=1$ and the quotient is trivial.

Given any finitely presented group $P$, form $\hat{P}=P\ast ⟨v⟩$, and apply part (a) to the word $w=v$. This word is nontrivial by the normal form theorem for a free product, so the output $J=\hat{P}(v)$ contains $P$ by the embedding already proved.

To check its finite quotients, let $f:J\to F$ be any group homomorphism to a finite group. The image of the copy of $Q$ is trivial by the preceding property. The identifications $u_j=a_3^{j+1}{a}_1{a}_3^{-(j+1)}$ make all $f(u_j)$ trivial. Thus the equations $u_j^{-1}c{u}_j=g_j$ give, in the image,Since $f(c)=f(z)$, this forces $f(x_i)=1$ for all generators of $\hat{P}$, including $v$. The defining word $c=[v,z]$ then has trivial image, so $f(z)=f(d)=1$ as well. Every generator of $J$ has trivial image. ThereforeThis uses the concrete construction of part (a), not an assumption that arbitrary quotients preserve an embedding of $P$.

Every finitely presented group $P$ embeds in $P\times C_2$ by $p\mapsto (p,1)$. This direct product of groups is finitely presented: combine finite presentations of the factors and add the finitely many cross-commutation relations between their generators. Projection onto $C_2$ gives a nontrivial finite quotient of a group.

Thus no finitely presented $P$ can serve as the forbidden witness $A_{-}$ for this property. Hence

Part (c) shows that every finitely presented group embeds in a finitely presented group with no nontrivial finite quotients of a group. Consequently this property also has no possible forbidden witness $A_{-}$. ThereforeBoth properties have positive witnesses, for example $C_2$ and the trivial group respectively. It is the required embedding obstruction that fails in each case.