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A modular machine can be encoded by HNN extensions of $K={\mathbb{Z}}^2\ast ⟨t⟩$. The kernel of $K\to {\mathbb{Z}}^2$ has a free basis of a group $t(r,s)=x^{-r}{y}^{-s}t{x}^r{y}^s$. Associated-subgroup maps send these basis elements according to the two arithmetic transition types. In the resulting group, membership of $t(r,s)$ in the subgroup generated by $t$ and the instruction stable letters is equivalent to $(r,s)\in H_0(\mathcal{M})$. One more HNN extension centralizes that finitely generated subgroup, converting membership to equality and hence to the word problem for a group.

For a modular machine whose $(0,0)$ is terminal, $H_0(\mathcal{M})$ is the set of configurations whose forward computation reaches exactly $(0,0)$, including the zero-step computation there. It may differ from the set of configurations stopping at any terminal configuration. Determinism makes membership invariant along each instruction edge.

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.