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

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.