3 Prime spectra and finite descriptions
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The direct sum has a completely transparent subgroup universe. It separates the complexity of selecting prime components from the complexity of performing the group operation.
3.1 The entire subgroup lattice
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For \(B\subseteq{\mathbb N}\), set
A prime spectrum records which prime orders occur. Define \(\sigma_p(M)=1\) when \(M\) contains a nonzero element annihilated by \(p\), and set it to zero otherwise. Since \(p\) is prime, a nonzero such element has order exactly \(p\).
Theorem 3.1: Subgroup and isomorphism classification
The map \(B\mapsto D_B\) is an order isomorphism from \(\mathcal P({\mathbb N})\) onto \({\operatorname{Sub}}(D)\). Every subgroup is fully invariant under all endomorphisms of \(D\), and
Proof
Let \(H\le D\) and let \(x\in H\). Its support is a finite set \(F\) of primes. For \(p\in F\), choose by the Chinese remainder theorem an integer \(k\) that is one modulo \(p\) and zero modulo every other prime in \(F\). Then \(kx\) is supported only at \(p\), where it is nonzero. It generates the entire \(C_p\) coordinate. Thus \(H\) contains every coordinate factor appearing in one of its elements.
Let \(B\) be the set of indices of these factors. Every element of \(H\) lies in \(D_B\), and every finite sum of the factors in \(B\) lies in \(H\). Hence \(H=D_B\). Uniqueness follows by testing membership of \(e_{p_i}\). Proposition 2.2 proves full invariance. Finally \(\sigma_{p_i}(D_B)=1\) exactly when \(i\in B\), and these values are isomorphism invariants.
The lattice operations have literal set-theoretic descriptions:
The infinite support choice is therefore already visible in an elementary algebraic lattice. No exotic group operation is required.
Give \({\operatorname{Sub}}(D)\) the topology inherited from \(\{0,1\}^D\) by its membership functions. Membership of a vector with finite support depends on finitely many selector bits. Conversely, membership of \(e_{p_i}\) recovers the \(i\)th bit. Consequently the order isomorphism above is also a homeomorphism from Cantor space \({\{0,1\}^{{\mathbb N}}}\) to \({\operatorname{Sub}}(D)\).
Corollary 3.2: Finite observations leave continuum many possibilities
Fix finitely many prime indices and their membership values. There are continuum many pairwise nonisomorphic \(D_B\) consistent with those values. A full compatible sequence of finite membership patterns determines one \(B\), but no fixed finite prefix does.
Proof
The unspecified bits form a countably infinite set of independent binary choices. Distinct completions give nonisomorphic groups by Theorem 3.1. A full compatible family fixes every coordinate and hence fixes the subset.
3.2 Enumeration is enough for a computable copy
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A subset of \({\mathbb N}\) is computably enumerable if a program can list exactly its elements. The program need not decide that an element will never appear. All these notions relativize to an oracle \(X\).
Theorem 3.3: The prime-spectrum criterion
For every oracle \(X\),
Proof
Suppose first that a copy is computable in \(X\). For each \(i\), search its domain for a nonzero element \(x\) with \(p_i x=0\). Each individual test is decidable relative to \(X\). Such a witness exists exactly when \(i\in B\). Dovetailing the searches enumerates \(B\).
Conversely, suppose \(B\) is infinite and enumerable in \(X\). List its distinct elements as \(b_0,b_1,\ldots\). To compute \(b_j\), wait for the \((j+1)\)st distinct enumerated value. This procedure terminates for every \(j\). Use canonical finite tuples with the \(j\)th coordinate in \({\mathbb F}_{p_{b_j}}\). Coordinatewise addition, normalization, and equality are computable in \(X\). If \(B\) is finite, \(D_B\) is a finite group and has a computable copy. This last existence statement does not require a uniform way to decide when a finite enumeration has finished.
This construction is classical. Montalbán gives the prime direct-sum example and its enumeration-degree interpretation explicitly [1], Example III.39. It is included here because its contrast with the later common-completion family is mathematically informative.
A noncomputable enumerable selector can therefore produce a group with computable multiplication and equality. The computable copy labels coordinates by their order of discovery. It does not supply a decision procedure for membership in the original selector.
3.3 A finite definition without computable equality
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Let \(K=\{n:\text{program }n\text{ halts on input }n\}\) be the diagonal halting set. It is enumerable. It is not decidable: a decision procedure would permit a program that halts on its own index precisely when the decision procedure says it does not. Its complement is not enumerable either, since simultaneous enumeration of a set and its complement decides membership.
Define
This is an explicit mathematical definition of a countably infinite, locally cyclic abelian group.
Corollary 3.4
The group \(H\) has no computable copy. It nevertheless has an effectively enumerable infinite list of generators and relators.
Proof
If \(B_K\) were enumerable, filtering its odd entries would enumerate the complement of \(K\). Theorem 3.3 therefore excludes a computable copy.
For the relator description, introduce a commuting generator \(x_i\) for each prime index, impose \(p_i x_i=0\), and impose \(x_{2n+1}=0\) whenever program \(n\) is observed to halt on input \(n\). The resulting quotient retains exactly the prime factors indexed by \(B_K\). All relators are effectively enumerable. Deciding whether an odd generator has become trivial is not supplied by this enumeration.
The conclusion is not absolute finite indescribability. The displayed definition is finite. What fails is the stronger contract of a countable copy with computable operations and decidable equality.
3.4 A purely finite-state boundary
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Consider a restricted grammar in which a finite automaton reads the unary word \(1^n\) and selects the factor \(C_{p_n}\) exactly when it accepts.
Theorem 3.5: Unary selector classification
A group \(D_B\) is realizable by this unary-selector grammar if and only if \(B\) is ultimately periodic.
Proof
On a unary input, the state after \(n\) letters is obtained by iterating a single map of a finite state set. The orbit eventually repeats a state, and from that point it is periodic. Its acceptance sequence is therefore ultimately periodic.
Conversely, any finite prefix followed by a periodic binary pattern is recognized by a finite directed path entering a cycle. The prime-spectrum isomorphism criterion ensures that a different accepted set cannot accidentally realize an isomorphic \(D_B\).
Example 3.6: A computable group outside the unary grammar
Let \(B=\{2^k:k\ge0\}\). It is decidable and enumerable, so \(D_B\) has a computable copy. It is not ultimately periodic. Indeed, if \(d>0\) were an eventual period, choose \(2^k\) beyond the threshold with \(2^k>d\). Periodicity would force \(2^k+d\) to be a power of two, although it lies strictly between \(2^k\) and \(2^{k+1}\). Thus \(D_B\) is outside the unary grammar. A binary-input automaton recognizes the strings \(10^k\), so changing the input representation removes this particular obstruction.
3.5 What the first family does not accomplish
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The groups \(D_B\) have distinct prime spectra, and those differences are already detectable by finite quotients. They do not yet show a failure of reconstruction from the full unmarked finite-quotient system. They isolate a simpler issue: which prime factors have been admitted in the first place.
The main family keeps every prime torsion factor present. It then places the selector in the divisibility of infinite-order elements instead. As a result, all finite quotients remain fixed while the effective cost of the source group varies. The rational space \(P/D\) is the mechanism that permits this stronger separation.