Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 104 / 5

In a direct product of groups, $(a,b{)}^r=(a^r,b^r)$. Thus a finite order for $(a,b)$ makes both coordinate orders finite. Conversely, if their orders are $p$ and $q$, the least positive exponent annihilating both is their least common multiple. Hencewhen both orders are finite; if either coordinate has infinite order, so does $(a,b)$.

Use a reduced sequence in an HNN extension for $\gamma$. By repeatedly conjugating a prefix to the other end and reducing any resulting pinch, one obtains a cyclically reduced sequence in an HNN extension: either an element of the base group, or a sequence of positive stable-letter length with no pinch even across its cyclic junction. For example, after conjugating away the initial base coefficient, write it as $t^{{ϵ}_1}{g}_1\cdots t^{{ϵ}_k}{g}_k$. A pinch across the junction can occur only if ${ϵ}_k=-{ϵ}_1$ and $g_k\in C_{{ϵ}_k}$. Moving the first stable letter to the end then exposes that pinch and reduces the stable-letter length by two. Iteration must terminate.

If the resulting cyclically reduced sequence has $k>0$, every positive power remains reduced and has stable-letter length equal to that power times $k$. By Britton's lemma, none is the identity. Thus a finite-order element must be conjugate into the base group $G$. Since $G$ embeds in the HNN extension, the order of a base-group element is unchanged, and conjugation also preserves order. ThereforeThe argument also applies to several stable letters: cyclic pinches must involve a stable letter and its own inverse.

The base free product for the group encoding of a modular machine isThe letters $x,y$ generate the first factor, and $t$ generates the second. Let $T_K$ be the kernel of the group homomorphism $K\to ⟨x,y⟩\cong {\mathbb{Z}}^2$ that kills $t$. Thenis a free group with this displayed free basis of a group. Indeed every word in $K$ can be rewritten as a product of such conjugates followed by an element of $⟨x,y⟩$, so they generate the kernel. A freely reduced product of these conjugates is nontrivial by the normal form theorem for a free product: after combining adjacent occurrences of the same conjugate, different successive indices give a nonzero intervening $x,y$-syllable. Thus there is no relation among the proposed basis elements.

For clarity, a modular machine of modulus $m>1$ has at most one instruction for each residue pair $(a,b)$, with $0\le a,b<m$ and $0\le c<m^2$. Its right and left transitions are respectivelyIts halting set at a designated terminal configuration $H_0(\mathcal{M})$ consists of the nonnegative pairs whose forward computation reaches $(0,0)$, including $(0,0)$ itself. These formulas explain the exponents in the associated-subgroup maps below.

For $M>a\ge 0$ and $N>b\ge 0$, defineThe normal form theorem for a free product giveswhere $X,Y,T$ map to $x^M,y^N,t(a,b)$. To see injectivity, expand an alternating word in $⟨x^M,y^N⟩$ and powers of $t(a,b)$. The conjugating $x,y$ factors commute with the lattice syllables, so each intervening nonidentity lattice syllable stays nonidentity in the expanded free-product normal form.

Its intersection with $T_K$ is the free group with basisThis follows by moving all $x^M,y^N$ factors to the right; the zero-exponent lattice remainder characterizes membership in the kernel.

The required isomorphisms are specified on the three generators byandEach source and target has the same presentation $⟨X,Y,T\mid [X,Y]=1⟩$, and each displayed assignment identifies its three abstract generators. Hence these are genuine isomorphisms, not merely maps of generating sets.

On the kernel basis they give exactly the machine transitions:For example, rewrite the first input as $x^{-mu}{y}^{-mv}t(a_i,b_i)x^{mu}{y}^{mv}$ and apply ${\varphi }_i$ to each factor. These identities also hold for negative $u,v$, since they are identities in the groups.

Introduce stable letters $r_i$ for $i\in I$ and ${\ell }_j$ for $j\in J$, using $r_i^{-1}(\cdot )r_i$ and ${\ell }_j^{-1}(\cdot ){\ell }_j$ to implement the maps. Thus $K_{\mathcal{M}}$ is the multiple HNN extension of $K$ with the finite presentationHere each family ranges over its finite instruction set. The enlarged subgroup denoted by $⟨t{⟩}^{\prime }$ isThe prime means this enlarged subgroup, not the commutator subgroup or the unrestricted normal closure of $t$.

The key relationship isHere is a justification that also identifies the relevant intersection. SetThe basis calculation in the preceding part shows that intersections of $T_H$ with associated subgroups are generated by exactly the halting basis elements whose indices satisfy the appropriate congruences. Since the machine is deterministic and $(0,0)$ is terminal, halting at $(0,0)$ is equivalent for the two ends of every instruction edge. Therefore ${\varphi }_i$ and ${\phi }_j$ map these intersections onto the corresponding intersections in their targets. The same assertion holds on the integer lattices after declaring pairs with a negative coordinate nonhalting: the bounds on $a,b,c$ imply that each coordinate's sign is preserved by a transition and its inverse.

Apply Britton's lemma to a word in $B_H$ which represents an element of $K$. If it has stable letters, it must have a pinch. The base coefficient of that pinch lies in $T_H$ and an associated subgroup, and the intersection property just proved makes its transported image lie in $T_H$ again. Successive pinch removals therefore keep all base coefficients in $T_H$, and eventually remove every stable letter. It follows thatFinally, $t=t(0,0)$ belongs to $T_H$, so $L\subseteq B_H$. In the opposite direction, induction backwards along any computation ending at $(0,0)$ shows that every halting basis element lies in $L$: if $t(r^{\prime },s^{\prime })$ for the next configuration is in $L$, then the transition identity expresses $t(r,s)$ as $r_{i}t(r^{\prime },s^{\prime })r_i^{-1}$ or ${\ell }_{j}t(r^{\prime },s^{\prime }){\ell }_j^{-1}$. Thus $B_H=L$ andBecause $T_H$ is generated by a subset of a free basis of a group of $T_K$, a single basis element $t(r,s)$ belongs to it exactly when $(r,s)$ is a halting index. This proves the boxed relationship.

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 $\boxed{G_{\mathcal{M}}\text{ has unsolvable word problem}}$.

The factors ${\mathbb{Z}}^2$ and $\mathbb{Z}$ are torsion-free groups, and so is their free product $K$. Indeed every element of a free product is conjugate either into one factor or to an alternating word whose first and last syllables lie in different factors. A word of the latter kind has nonempty reduced powers of every positive length, so has infinite order. A nonidentity element of either factor also has infinite order.

By the finite-order result in part (b), every finite-order element of the multiple HNN extension $K_{\mathcal{M}}$ is conjugate into the torsion-free base $K$, and hence is the identity. Apply the same result once more to $G_{\mathcal{M}}$, whose base is $K_{\mathcal{M}}$. It follows thatThus the failure of a word problem for a group to be decidable does not require torsion.