A first-order theory of Ulm type

TL;DR

This paper introduces a first-order theory, $T_p$, that characterizes abelian $p$-groups up to bi-interpretability, providing a new example of a theory of 'Ulm type' with specific computability and classification properties.

Contribution

It constructs an elementary first-order theory of Ulm type that captures abelian $p$-groups and addresses a question by Knight about such theories.

Findings

Models are bi-interpretable with abelian $p$-groups.

Any two low models with same infinitary theory are isomorphic.

The isomorphism problem is $oldsymbol{ ext{Σ}}^1_1$-complete but not Borel.

Abstract

The class of abelian $p$-groups are an example of some very interesting phenomena in computable structure theory. We will give an elementary first-order theory $T_p$ whose models are each bi-interpretable with the disjoint union of an abelian $p$-group and a pure set (and so that every abelian $p$-group is bi-interpretable with a model of $T_p$) using computable infinitary formulas. This answers a question of Knight by giving an example of an elementary first-order theory of "Ulm type": Any two models, low for $\omega_1^{CK}$, and with the same computable infinitary theory, are isomorphic. It also gives a new example of an elementary first-order theory whose isomorphism problem is $\mathbf{\Sigma}^1_1$-complete but not Borel complete.

Equations

$$(\mathcal{D}o{m}_{\mathcal{A}}^{\mathcal{B}}/\sim ;R_0/\sim ,R_1/\sim ,...)\cong (A;P_0^{\mathcal{A}},P_1^{\mathcal{A}},...),$$

$$f_{\mathcal{B}}^{\mathcal{A}}\circ {\tilde{f}}_{\mathcal{A}}^{\mathcal{B}}:\mathcal{D}o{m}_{\mathcal{B}}^{(\mathcal{D}o{m}_{\mathcal{A}}^{\mathcal{B}})}\to \mathcal{B}\text{mbox}and{f}_{\mathcal{A}}^{\mathcal{B}}\circ {\tilde{f}}_{\mathcal{B}}^{\mathcal{A}}:\mathcal{D}o{m}_{\mathcal{A}}^{(\mathcal{D}o{m}_{\mathcal{B}}^{\mathcal{A}})}\to \mathcal{A}$$

$$\mathbb{Z}(p^{\infty })=\mathbb{Z}[1/p,1/p^2,1/p^3,\dots ]/\mathbb{Z}.$$

$$\big{(}p^{\alpha}G\big{)}[p]\;/\;\big{(}p^{\alpha+1}G\big{)}[p]$$

$$\mathcal{M}\cong \mathcal{N}⟺\Phi (\mathcal{M})\cong \Phi (\mathcal{N}).$$

$$G\models {\phi }_{\alpha ,n}\iff u_{\alpha }(G)=n.$$

$$\frac{c_1}{p^{k_1}}{x}_1+\cdots +\frac{c_n}{p^{k_n}}{x}_n\ne 0,$$

$$y=\frac{c_1}{p^{k_1}}{x}_1+\cdots +\frac{c_n}{p^{k_n}}{x}_n.$$

$$\frac{c_1}{p^{k_1}}{x}_1+\cdots +\frac{c_m}{p^{k_m}}{x}_m\ne 0.$$

$$(\forall x\forall y\forall z)\left[P_{\ell ,m}^n(x,y,z)\to \left(R_{\ell }(x)\wedge R_m(x)\wedge R_n(z)\right)\right].$$

$$(\forall x)[R_0(x)\leftrightarrow x=0].$$

$$(\forall x)\left[x\in R_n\leftrightarrow (\exists x_2\cdots \exists x_{p-1})\left[P_{n,n}^n(x,x,x_2)\wedge P_{n,n}^n(x,x_2,x_3)\wedge \cdots \wedge P_{n,n}^{n-1}(x,x_{p-1},x_p)\right]\right].$$

$$(\forall x\forall y\forall z\forall z^{\prime})\left[\big{(}P_{\ell,m}^{n}(x,y,z)\wedge P_{\ell,m}^{n^{\prime}}(x,y,z^{\prime})\big{)}\rightarrow z=z^{\prime}\right].$$

$$(\forall x\forall y)\left[\big{(}R_{\ell}(x)\wedge R_{m}(y)\big{)}\rightarrow\bigvee_{n\leq\max(\ell,m)}(\exists z)P_{\ell,m}^{n}(x,y,z)\right].$$

$$(\forall x)[R_{\ell}(x)\rightarrow\big{[}P_{0,\ell}^{\ell}(0,x,x)\wedge P_{\ell,0}^{\ell}(x,0,x)]\big{]}.$$

$$(\forall x)(\exists y)\big{[}R_{\ell}(x)\rightarrow[P_{\ell,\ell}^{0}(x,y,0)\wedge P_{\ell,\ell}^{0}(y,x,0)]\big{]}.$$

$$\begin{split}&(\forall x\forall y\forall z)\bigg{[}\Big{[}R_{\ell}(x)\wedge R_{m}(y)\wedge R_{n}(z)\Big{]}\longrightarrow\\ &\bigvee_{\begin{subarray}{c}r\leq\max(\ell,m)\\ s\leq\max(m,n)\\ t\leq\max(r,n),\max(\ell,s)\end{subarray}}(\exists u\exists v\exists w)\Big{[}P_{\ell,m}^{r}(x,y,u)\wedge P_{r,n}^{t}(u,z,w)\wedge P_{m,n}^{s}(y,z,v)\wedge P_{\ell,s}^{t}(x,v,w)\Big{]}\bigg{]}.\end{split}$$

$$(\forall x\forall y\forall z)\big{[}[R_{\ell}(x)\wedge R_{m}(y)\wedge R_{n}(z)\wedge P_{\ell,m}^{n}(x,y,z)]\rightarrow P_{m,\ell}^{n}(y,x,z)\big{]}.$$

$$\text{x+y=z in G(M) and x∈RℓM, y∈RmM, and z∈RnM}.$$

$$\hat{g}=\sum _{b\in \mathcal{B}}{x}_b\hat{b}$$

$$h=g-\sum _{b\in \mathcal{B}}{x}_{b}b\in pG$$

$$h+\sum _{b\in \mathcal{B}}{x}_{b}b=h^{\prime }+\sum _{b\in \mathcal{B}}{y}_{b}b.$$

$$\sum _{b\in \mathcal{B}}{x}_b\hat{b}=\sum _{b\in \mathcal{B}}{y}_b\hat{b}.$$

$$h+\sum _{b\in \mathcal{B}}{x}_b{a}_b=h^{\prime }+\sum _{b\in \mathcal{B}}{y}_b{a}_b$$

$$ph+\sum _{b\in \mathcal{B}}{x}_{b}b=p{h}^{\prime }+\sum _{b\in \mathcal{B}}{y}_{b}b.$$

$$p(g^{\prime }+\sum _{b\in \mathcal{B}}{x}_b{a}_b)=g+\sum _{b\in \mathcal{B}}{x}_{b}b.$$

$$\begin{split}\mathfrak{H}(G,m)[p]/(p\mathfrak{H}(G,m))[p]&\cong\big{(}G^{*}[p]/(pG^{*})[p]\big{)}\oplus\big{(}(\mathbb{Z}_{p})^{m}[p]/(p(\mathbb{Z}_{p})^{m})[p]\big{)}\\ &\cong(G^{*}[p]/G[p])\oplus(\mathbb{Z}_{p})^{m}.\end{split}$$

$$0=pg=p{g}^{\prime }+\sum _{b\in \mathcal{B}}p{y}_b{a}_b=p{g}^{\prime }+\sum _{b\in \mathcal{B}}{y}_{b}b.$$

$$p\mathfrak{H}(G,m)=p{G}^{\ast }\oplus p({\mathbb{Z}}_p{)}^m\cong p{G}^{\ast }=G$$

Full text

1. Introduction

The class of abelian $p$-groups is a well-studied example in computable structure theory. A simple compactness argument shows that abelian $p$-groups are not axiomatizable by an elementary first-order theory, but they are definable by the conjunction of the axioms for abelian $p$-groups (which are first-order $\forall\exists$ sentences) and the infinitary $\Pi^{0}_{2}$ sentence which says that every element is torsion of order some power of $p$.

Abelian $p$-groups are classifiable by their Ulm sequences [Ulm33]. Due to this classification, abelian $p$-groups are examples of some very interesting phenomena in computable structure theory and descriptive set theory. We will define a theory $T_{p}$ whose models behave like the class of abelian $p$-groups, giving a first-order example of these phenomena. In particular, Theorem 1.6 below answers a question of Knight.

1.1. Infinitary Formulas

The infinitary logic $\mathcal{L}_{\omega_{1}\omega}$ is the logic which allows countably infinite conjunctions and disjunctions but only finite quantification. If the conjunctions and disjunctions of a formula $\varphi$ are all over computable sets of indices for formulas, then we say that $\varphi$ is computable. We use $\Sigma_{\alpha}^{\mathtt{in}}$ and $\Pi_{\alpha}^{\mathtt{in}}$ to denote the classes of all infinitary $\Sigma_{\alpha}$ and $\Pi_{\alpha}$ formulas respectively. We will also use $\Sigma_{\alpha}^{\mathtt{c}}$ and $\Pi_{\alpha}^{\mathtt{c}}$ to denote the classes of computable $\Sigma_{\alpha}$ and $\Pi_{\alpha}$ formulas, where $\alpha<\omega_{1}^{CK}$ the least non-computable ordinal. See Chapter 6 of [AK00] for a more complete description of computable formulas.

1.2. Bi-Interpretability

One way in which we will see that the models of $T_{p}$ are essentially the same as abelian $p$-group is using bi-interpretations using infinitary formulas [Mon, HTMMM, HTMM]. A structure $\mathcal{A}$ is infinitary interpretable in a structure $\mathcal{B}$ if there is an interpretation of $\mathcal{A}$ in $\mathcal{B}$ where the domain of the interpretation is allowed to be a subset of $\mathcal{B}^{<\omega}$ and where all of the sets in the interpretation are definable using infinitary formulas. This differs from the classical notion of interpretation, as in model theory [Mar02, Definition 1.3.9], where the domain is required to be a subset of $\mathcal{B}^{n}$ for some $n$, and the sets in the interpretation are first-order definable.

Definition 1.1**.**

We say that a structure ${\mathcal{A}}=(A;P_{0}^{\mathcal{A}},P_{1}^{\mathcal{A}},...)$ (where $P_{i}^{\mathcal{A}}\subseteq A^{a(i)}$) is infinitary interpretable in $\mathcal{B}$ if there exists a sequence of relations $(\mathcal{D}om_{\mathcal{A}}^{\mathcal{B}},\sim,R_{0},R_{1},...)$, definable using infinitary formulas (in the language of $\mathcal{B}$, without parameters), such that

1. (1)

$\mathcal{D}om^{\mathcal{B}}_{\mathcal{A}}\subseteq\mathcal{B}^{<\omega}$, 2. (2)

$\sim$ is an equivalence relation on $\mathcal{D}om^{\mathcal{B}}_{{\mathcal{A}}}$, 3. (3)

$R_{i}\subseteq(B^{<\omega})^{a(i)}$ is closed under $\sim$ within $\mathcal{D}om_{\mathcal{A}}^{\mathcal{B}}$,

and there exists a function $f^{\mathcal{B}}_{\mathcal{A}}\colon\mathcal{D}om^{\mathcal{B}}_{{\mathcal{A}}}\to{\mathcal{A}}$ which induces an isomorphism:

[TABLE]

where $R_{i}/\sim$ stands for the $\sim$-collapse of $R_{i}$.

Two structures $\mathcal{A}$ and $\mathcal{B}$ are infinitary bi-interpretable if they are each effectively interpretable in the other, and moreover, the composition of the interpretations—i.e., the isomorphisms which map $\mathcal{A}$ to the copy of $\mathcal{A}$ inside the copy of $\mathcal{B}$ inside $\mathcal{A}$, and $\mathcal{B}$ to the copy of $\mathcal{B}$ inside the copy of $\mathcal{A}$ inside $\mathcal{B}$—are definable.

Definition 1.2**.**

Two structures $\mathcal{A}$ and $\mathcal{B}$ are infinitary bi-interpretable if there are infinitary interpretations of each structure in the other as in Definition 1.1 such that the compositions

[TABLE]

are definable in ${\mathcal{B}}$ and ${\mathcal{A}}$ respectively. (Here, we have $\mathcal{D}om_{\mathcal{B}}^{(\mathcal{D}om_{\mathcal{A}}^{\mathcal{B}})}\subseteq(\mathcal{D}om_{\mathcal{A}}^{\mathcal{B}})^{<\omega}$, and $\tilde{f}^{\mathcal{B}}_{\mathcal{A}}\colon(\mathcal{D}om_{\mathcal{A}}^{\mathcal{B}})^{<\omega}\to{\mathcal{A}}^{<\omega}$ is the obvious extension of $f^{\mathcal{B}}_{\mathcal{A}}\colon\mathcal{D}om_{\mathcal{A}}^{\mathcal{B}}\to{\mathcal{A}}$ mapping $\mathcal{D}om_{\mathcal{B}}^{(\mathcal{D}om_{\mathcal{A}}^{\mathcal{B}})}$ to $\mathcal{D}om_{\mathcal{B}}^{\mathcal{A}}$.)

If we ask that the sets and relations in the interpretation (or bi-interpretation) be (uniformly) relatively intrinsically computable, i.e., definable by both a $\Sigma^{\mathtt{c}}_{1}$ formula and a $\Pi^{\mathtt{c}}_{1}$ formula, then we say that the interpretation (or bi-interpretation) is effective. Any two structures which are effectively bi-interpretable have all of the same computability-theoretic properties; for example, they have the same degree spectra and the same Scott rank. See [Mon, Lemma 5.3].

Here, we will use interpretations which use (lightface) $\Delta^{\mathtt{c}}_{2}$ formulas. It is no longer true that any two structures which are $\Delta^{\mathtt{c}}_{2}$-bi-interpretable have all of the same computability-theoretic properties, but it is true, for example, that any two such structures either both have computable, or both have non-computable, Scott rank.

Theorem 1.3**.**

Each abelian $p$-group is effectively bi-interpretable with a model of $T_{p}$. Each model of $T_{p}$ is $\Delta^{\mathtt{c}}_{2}$-bi-interpretable with the disjoint union of an abelian $p$-group and a pure set.

This theorem will follow from the constructions in Sections 3 and 4. Given a model $\mathcal{M}$ of $T_{p}$, $\mathcal{M}$ is bi-interpretable with an abelian $p$-group $G$ and a pure set. The domain of the copy of $G$ inside of $\mathcal{M}$ is definable by a $\Sigma^{\mathtt{c}}_{1}$ formula but not by a $\Pi^{\mathtt{c}}_{1}$ formula. This is the only part of the bi-interpretation which is not effective.

1.3. Classification via Ulm Sequences

Let $G$ be an abelian group. For any ordinal $\alpha$, we can define $p^{\alpha}G$ by transfinite induction:

• •

$p^{0}G=G$;

• •

$p^{\alpha+1}G=p(p^{\alpha}G)$;

• •

$p^{\beta}G=\bigcap_{\alpha<\beta}p^{\alpha}G$ if $\beta$ is a limit ordinal.

These subgroups $p^{\alpha}G$ form a filtration of $G$. This filtration stabilizes, and we call the smallest ordinal $\alpha$ such that $p^{\alpha}G=p^{\alpha+1}G$ the length of $G$. We call the intersection $p^{\infty}G$ of these subgroups, which is a $p$-divisible group, the $p$-divisible part of $G$. Any countable $p$-divisible group is isomorphic to some direct product of the Prüfer group

[TABLE]

Denote by $G[p]$ the subgroup of $G$ consisting of the $p$-torsion elements. The $\alpha$th Ulm invariant $u_{\alpha}(G)$ of $G$ is the dimension of the quotient

[TABLE]

as a vector space over $\mathbb{Z}/p\mathbb{Z}$.

Theorem 1.4** (Ulm’s Theorem, see [Fuc70]).**

Let $G$ and $H$ be countable abelian p-groups such that for every ordinal $\alpha$ their $\alpha$th Ulm invariants are equal, and the $p$-divisible parts of $G$ and $H$ are isomorphic. Then $G$ and $H$ are isomorphic.

1.4. Scott Rank and Computable Infinitary Theories

Scott [Sco65] showed that if $\mathcal{M}$ is a countable structure, then there is a sentence $\varphi$ of $\mathcal{L}_{\omega_{1}\omega}$ such that $\mathcal{M}$ is, up to isomorphism, the only countable model of $\varphi$. We call such a sentence a Scott sentence for $\mathcal{M}$. There are many different definitions [AK00, Sections 6.6 and 6.7] of the Scott rank of $\mathcal{M}$, which differ only slightly in the ranks they assign. The one we will use, which comes from [Mon15], defines the Scott rank of $\mathcal{A}$ to be the least ordinal $\alpha$ such that $\mathcal{A}$ has a $\Pi^{\mathtt{in}}_{\alpha+1}$ Scott sentence. We denote the Scott rank of a structure $\mathcal{A}$ by $\operatorname{SR}(\mathcal{A})$. It is always the case that $\operatorname{SR}(\mathcal{A})\leq\omega_{1}^{\mathcal{A}}+1$ [Nad74]. We could just as easily use any of the other definitions of Scott rank; for all of these definitions, given a computable structure $\mathcal{A}$:

1. (1)

$\mathcal{A}$ has computable Scott rank if and only if there is a computable ordinal $\alpha$ such that for all tuples $\bar{a}$ in $\mathcal{A}$, the orbit of $\bar{a}$ is defined by a computable $\Sigma_{\alpha}$ formula. 2. (2)

$\mathcal{A}$ has Scott rank $\omega_{1}^{CK}$ if and only if for each tuple $\bar{a}$, the orbit is defined by a computable infinitary formula, but for each computable ordinal $\alpha$, there is a tuple $\bar{a}$ whose orbit is not defined by a computable $\Sigma_{\alpha}$ formula. 3. (3)

$\mathcal{A}$ has Scott rank $\omega_{1}^{CK}+1$ if and only if there is a tuple $\bar{a}$ whose orbit is not defined by a computable infinitary formula.

Given a structure $\mathcal{M}$, define the computable infinitary theory of $\mathcal{M}$, $\operatorname{Th}_{\infty}(\mathcal{M})$, to be collection of computable $\mathcal{L}_{\omega_{1}\omega}$ sentences true of $\mathcal{M}$. We can ask, for a given structure $\mathcal{M}$, whether $\operatorname{Th}_{\infty}(\mathcal{M})$ is $\aleph_{0}$-categorical, or whether there are other countable models of $\operatorname{Th}_{\infty}(\mathcal{M})$. For $\mathcal{M}$ a hyperarithmetic structure:

1. (1)

If $\operatorname{SR}(\mathcal{M})<\omega_{1}^{CK}$, then $\operatorname{Th}_{\infty}(\mathcal{M})$ is $\aleph_{0}$-categorical. Indeed, $\mathcal{M}$ has a computable Scott sentence [Nad74]. 2. (2)

If $\operatorname{SR}(\mathcal{M})=\omega_{1}^{CK}$, then $\operatorname{Th}_{\infty}(\mathcal{M})$ may or may not be $\aleph_{0}$-categorical [HTIK]. 3. (3)

If $\operatorname{SR}(\mathcal{M})=\omega_{1}^{CK}+1$, then $\operatorname{Th}_{\infty}(\mathcal{M})$ is not $\aleph_{0}$-categorical as $\mathcal{M}$ has a non-principal type which may be omitted.

In the case of abelian $p$-groups, we can say something even when we replace the assumption that $\mathcal{M}$ is hyperarithmetic with the assumption that $\omega_{1}^{G}=\omega_{1}^{CK}$.

Definition 1.5** (Definition 6 of [FKM*+*11]).**

A class of countable structures has Ulm type if for any two structures $\mathcal{A}$ and $\mathcal{B}$ in the class, if $\omega_{1}^{\mathcal{A}}=\omega_{1}^{\mathcal{B}}=\omega_{1}^{CK}$ and $\operatorname{Th}_{\infty}(\mathcal{A})=\operatorname{Th}_{\infty}(\mathcal{B})$, then $\mathcal{A}$ and $\mathcal{B}$ are isomorphic.

It is well-known that abelian $p$-groups are of Ulm type; however, we do not know of a good reference with a complete proof, so we will give one in Section 2. We also note that there are indeed non-hyperarithmetic abelian $p$-groups $G$ with $\operatorname{SR}(G)<\omega_{1}^{CK}$.

Knight asked whether there was a (non-trivial) first-order theory of Ulm type. By a non-trivial example, we mean that the elementary first-order theory should have non-hyperarithmetic models which are low for $\omega_{1}^{CK}$. Our theory $T_{p}$ is such an example.

Theorem 1.6**.**

The models of $T_{p}$ are of Ulm type. Moreover, given $\mathcal{M}\models T_{p}$ with $\omega_{1}^{CK}=\omega_{1}^{\mathcal{M}}$ and $\operatorname{SR}(\mathcal{M})<\omega_{1}^{CK}=\omega_{1}^{\mathcal{M}}$, $\operatorname{Th}_{\infty}(\mathcal{M})$ is $\aleph_{0}$-categorical.

Proof.

Let $\mathcal{M}$ be a model of $T_{p}$. Now $\mathcal{M}$ is bi-interpretable, using computable infinitary formulas, with the disjoint union of an abelian $p$-group $G$ and a pure set. Thus $\mathcal{M}$ inherits these properties from $G$ (see Theorem 2.1). ∎

Of course, there will be non-hyperarithmetic models of $T_{p}$ with Scott rank below $\omega_{1}^{CK}$.

1.5. Borel Incompleteness

In their influential paper [FS89], Friedman and Stanley introduced Borel reductions between invariant Borel classes of structures with universe $\omega$ in a countable language. Such classes are of the form $\operatorname{Mod}(\varphi)$, the set of models of $\varphi$ with universe $\omega$, for some $\varphi\in\mathcal{L}_{\omega_{1}\omega}$. A Borel reduction from $\operatorname{Mod}(\varphi)$ to $\operatorname{Mod}(\psi)$ is a Borel map $\Phi\colon\operatorname{Mod}(\varphi)\to\operatorname{Mod}(\psi)$ such that

[TABLE]

If such a Borel reduction exists, we say that $\operatorname{Mod}(\varphi)$ is Borel reducible to $\operatorname{Mod}(\psi)$ and write $\varphi\leq_{B}\psi$. If $\varphi\leq_{B}\psi$ and $\psi\leq_{B}\varphi$, then we say that $\operatorname{Mod}(\varphi)$ and $\operatorname{Mod}(\psi)$ are Borel equivalent and write $\varphi\equiv_{B}\psi$. Friedman and Stanley showed that graphs, fields, linear orders, trees, and groups are all Borel equivalent, and form a maximal class under Borel reduction.

If $\operatorname{Mod}(\varphi)$ is Borel complete, then the isomorphism relation on $\operatorname{Mod}(\varphi)\times\operatorname{Mod}(\varphi)$ is $\mathbf{\Sigma}^{1}_{1}$-complete. The converse is not true, and the most well-known example is abelian $p$-groups, whose isomorphism relation is $\mathbf{\Sigma}^{1}_{1}$-complete but not Borel complete. Until very recently, they were one of the few such examples, and there were no known examples of elementary first-order theories with similar properties. Recently, Laskowski, Rast, and Ulrich [URL] gave an example of a first-order theory which is not Borel complete, but whose isomorphism relation is not Borel. Our theory $T_{p}$ is another such example.

Theorem 1.7**.**

The class of models of $T_{p}$ is Borel equivalent to abelian $p$-groups.

Because abelian $p$-groups are not Borel complete, but their isomorphism relation is $\mathbf{\Sigma^{1}_{1}}$-complete, we get:

Corollary 1.8**.**

The class of models of $T_{p}$ is not Borel complete but the isomorphism relation is $\mathbf{\Sigma^{1}_{1}}$-complete.

Theorem 1.7 is a specific instance of the following general question asked by Friedman:

Question 1.9**.**

Is it true that for every $\mathcal{L}_{\omega_{1}\omega}$ sentence there is a Borel equivalent first-order theory?

2. Abelian $p$-groups are of Ulm type

In this section we will describe a proof of the following well-known theorem, which shows that abelian $p$-groups are of Ulm type.

Theorem 2.1**.**

Let $G$ be an abelian $p$-group with $\omega_{1}^{CK}=\omega_{1}^{G}$. Then:

1. (1)

$G$* is the only countable model of $\operatorname{Th}_{\infty}(G)$ with $\omega_{1}^{G}=\omega_{1}^{CK}$, and* 2. (2)

if $\operatorname{SR}(G)<\omega_{1}^{CK}=\omega_{1}^{G}$, then $\operatorname{Th}_{\infty}(G)$ is $\aleph_{0}$-categorical.

The proof of Theorem 2.1 consists essentially of expressing the Ulm invariants via computable infinitary formulas.

Definition 2.2**.**

Let $G$ be an abelian $p$-group. For each ordinal $\alpha<\omega_{1}^{CK}$, there is a computable infinitary sentence $\psi_{\alpha}(x)$ which defines $p^{\alpha}G$ inside of $G$:

• •

$\psi_{0}(x)$ is just $x=x$;

• •

$\psi_{\alpha+1}(x)$ is $(\exists y)[\psi_{\alpha}(y)\wedge py=x]$;

• •

$\psi_{\beta}(x)$ is $\bigdoublewedge_{\alpha<\beta}\psi_{\alpha}(x)$ for limit ordinals $\beta$.

Definition 2.3**.**

For each ordinal $\alpha<\omega_{1}^{CK}$ and $n\in\omega\cup\{\omega\}$, there is a computable infinitary sentence $\varphi_{\alpha,n}$ such that, for $G$ an abelian $p$-group,

[TABLE]

For $n\in\omega$, define $\varphi_{\alpha,\geq n}$ to say that there are $x_{1},\ldots,x_{n}$ such that:

• •

$\psi_{\alpha}(x_{1})\wedge\cdots\wedge\psi_{\alpha}(x_{n})$,

• •

$px_{1}=\cdots=px_{n}=0$, and

• •

for all $c_{1},\ldots,c_{n}\in\mathbb{Z}/p\mathbb{Z}$ not all zero, $\neg\psi_{\alpha+1}(c_{1}x_{1}+\cdots+c_{n}x_{n})$.

Then for $n\in\omega$, $\varphi_{\alpha,n}$ is $\varphi_{\alpha,\geq n}\wedge\neg\varphi_{\alpha,\geq n+1}$, and $\varphi_{\alpha,\omega}$ is $\bigdoublewedge_{n\in\omega}\varphi_{\alpha,\geq n}$.

Lemma 2.4** (Theorem 8.17 of [AK00]).**

Let $G$ be an abelian $p$-group. Then:

1. (1)

the length of $G$ is at most $\omega_{1}^{G}$, and 2. (2)

if $G$ has length $\omega_{1}^{G}$ then $G$ is not reduced (in fact, its $p$-divisible part has infinite rank) and $\operatorname{SR}(G)=\omega_{1}^{G}+1$.

We are now ready to give the proof of Theorem 2.1.

Proof of Theorem 2.1.

Since $\omega_{1}^{CK}=\omega_{1}^{G}$, $G$ has length at most $\omega_{1}^{CK}$. Note that $\operatorname{Th}_{\infty}(G)$ contains the sentences $\varphi_{\alpha,u_{\alpha}(G)}$ for $\alpha<\omega_{1}^{CK}$. Thus any model of $\operatorname{Th}_{\infty}(G)$ has the same Ulm invariants as $G$, for ordinals below $\omega_{1}^{CK}$.

If $\operatorname{SR}(G)<\omega_{1}^{CK}$, let $\lambda$ be the length of $G$. Then $\operatorname{Th}_{\infty}(G)$ includes the computable formula $(\forall x)[\psi_{\lambda}(x)\leftrightarrow\psi_{\lambda+1}(x)]$, so that any countable model of $\operatorname{Th}_{\infty}(G)$ has length at most $\lambda$. Note that in such a model, $\psi_{\lambda}$ defines the $p$-divisible part. Let $n\in\omega\cup\{\omega\}$ be such that $p^{\infty}G$ is isomorphic to $\mathbb{Z}(p^{\infty})^{n}$. Then, if $n\in\omega$, $\operatorname{Th}_{\infty}(G)$ contains the formula which says that there are $x_{1},\ldots,x_{n}$ such that

• •

$\psi_{\lambda}(x_{1})\wedge\cdots\wedge\psi_{\lambda}(x_{n})$,

• •

for all $c_{1},\ldots,c_{n}<p$ not all zero and $k_{1},\ldots,k_{n}\in\omega$,

[TABLE]

• •

for all $y$ with $\psi_{\lambda}(y)$, there are $c_{1},\ldots,c_{n}<p$ and $k_{1},\ldots,k_{n}\in\omega$ such that

[TABLE]

If $n=\omega$, then $\operatorname{Th}_{\infty}(G)$ contains the formula which says that for each $m\in\omega$, there are $x_{1},\ldots,x_{m}$ such that

• •

$\psi_{\lambda}(x_{1})\wedge\cdots\wedge\psi_{\lambda}(x_{m})$, and

• •

for all $c_{1},\ldots,c_{m}<p$ not all zero and $k_{1},\ldots,k_{m}\in\omega$,

[TABLE]

Any countable model of $\operatorname{Th}_{\infty}(G)$ has $p$-divisible part isomorphic to $\mathbb{Z}(p^{\infty})^{n}$. So any countable model of $\operatorname{Th}_{\infty}(G)$ has the same Ulm invariants and $p$-divisible part as $G$, and hence is isomorphic to $\operatorname{Th}_{\infty}(G)$. Hence $\operatorname{Th}_{\infty}(G)$ is $\aleph_{0}$-categorical. This gives (2), and (1) for the case where $\operatorname{SR}(G)<\omega_{1}^{CK}$.

If $\operatorname{SR}(G)=\omega_{1}^{CK}+1$, let $H$ be any other countable model of $\operatorname{Th}_{\infty}(G)$ with $\omega_{1}^{H}=\omega_{1}^{G}=\omega_{1}^{CK}$. Thus $G$ and $H$ both have length $\omega_{1}^{CK}$ and their $p$-divisible parts have infinite rank. As remarked before, they have the same Ulm invariants, and so they must be isomorphic. This completes the proof of (1). ∎

3. The Theory $T_{p}$

Fix a prime $p$. The language $\mathcal{L}_{p}$ of $T_{p}$ will consist of a constant [math], unary relations $R_{n}$ for $n\in\omega$, and ternary relations $P_{\ell,m}^{n}$ for $\ell,m\in\omega$ and $n\leq\max(\ell,m)$. The following transformation of an abelian $p$-group into an $\mathcal{L}_{p}$-structure will illustrate the intended meaning of the symbols.

Definition 3.1**.**

Let $G$ be an abelian $p$-group. Define $\mathfrak{M}(G)$ to be $\mathcal{L}_{p}$-structure obtained as follows, with the same domain as $G$, and the symbols of $\mathcal{L}_{p}$ interpreted as follows:

• •

Set $0^{\mathfrak{M}(G)}$ to be the identity element of $G$.

• •

For each $n$, let $R_{n}^{\mathfrak{M}(G)}$ be the elements which are torsion of order $p^{n}$.

• •

For each $\ell,m\in\omega$ and $n\leq\max(\ell,m)$, and $x,y,z\in G$, set $P_{\ell,m}^{n,\mathfrak{M}(G)}(x,y,z)$ if and only if $x+y=z$, $x\in R_{\ell}^{\mathfrak{M}(G)}$, $y\in R_{m}^{\mathfrak{M}(G)}$, and $z\in R_{n}^{\mathfrak{M}(G)}$.

One should think of such $\mathcal{L}_{p}$-structures as the canonical models of $T_{p}$. The theory $T_{p}$ will consist of following axiom schemata:

1. (A1)

For all $\ell,m,n\in\omega$:

[TABLE] 2. (A2)

($R_{n}$ contains the elements which are torsion of order $p^{n}$.)

[TABLE]

and, for all $n\geq 1$:

[TABLE] 3. (A3)

($P$ defines a partial function.) For all $\ell,m,n,n^{\prime}\in\omega$:

[TABLE] 4. (A4)

($P$ is total.) For all $\ell,m\in\omega$:

[TABLE] 5. (A5)

(Identity.) For all $\ell\in\omega$:

[TABLE] 6. (A6)

(Inverses.) For all $\ell\in\omega$:

[TABLE] 7. (A7)

(Associativity.) For all $\ell,m,n\in\omega$:

[TABLE] 8. (A8)

(Abelian.) For all $\ell,m\in\omega$ and $n\leq\max(\ell,m)$:

[TABLE]

We must now check that the definition of $T_{p}$ works as desired, that is, that if $G$ is an abelian $p$-group, then $\mathfrak{M}(G)$ is a model of $T_{p}$.

Lemma 3.2**.**

If $G$ is an abelian $p$-group, then $\mathfrak{M}(G)\models T_{p}$.

Proof.

We must check that each instance of the axiom schemata of $T_{p}$ holds in $\mathfrak{M}(G)$.

1. (A1)

Suppose that $x$, $y$, and $z$ are elements of $G$ with $P_{m,\ell}^{n,\mathfrak{M}(G)}(x,y,z)$. Then, by definition, $x+y=z$, $x\in R_{\ell}^{\mathfrak{M}}(G)$, $y\in R_{m}^{\mathfrak{M}(G)}$, and $z\in R_{n}^{\mathfrak{M}(G)}$. 2. (A2)

$R_{0}^{\mathfrak{M}(G)}$ contains the elements of $G$ which are torsion of order $p^{0}=1$, so $R_{0}$ contains just the identity. For each $n>0$, $R_{n}^{\mathfrak{M}(G)}$ contains the elements of order $p^{n}$. An element $x$ has order $p^{n}$ if and only if $px$ has order $p^{n-1}$. It remains only to note that if $x$ has order $p^{n}$, then $x,2x,3x,\ldots,(p-1)x$ all have order $p^{n}$ as well. The existential quantifier is witnessed by $x_{2}=2x$, $x_{3}=3x$, and so on. 3. (A3)

If, for some $x$, $y$, $z$, and $z^{\prime}$, $P^{n,\mathfrak{M}(G)}_{\ell,m}(x,y,z)$ and $P_{\ell,m}^{n^{\prime},\mathfrak{M}(G)}(x,y,z^{\prime})$, then $x+y=z$ and $x+y=z^{\prime}$, so that $z=z^{\prime}$. 4. (A4)

Given $x$ and $y$ in $G$ which are of order $p^{m}$ and $p^{\ell}$ respectively, $x+y$ is of order $p^{n}$ for some $n\leq\max(m,\ell)$, and so we have $P_{m,\ell}^{n,\mathfrak{M}(G)}(x,y,x+y)$. 5. (A5)

If $x\in G$ is of order $p^{\ell}$, then $x+0=0+x=x$ and so we have $P_{\ell,0}^{\ell,\mathfrak{M}(G)}(x,0,x)$. 6. (A6)

If $x\in G$ is of order $p^{\ell}$, then $-x$ is also of order $p^{\ell}$, and $x+(-x)=0=(-x)+x$. So we have $P_{\ell,\ell}^{0,\mathfrak{M}(G)}(x,-x,0)$. 7. (A7)

Given $x,y,z\in G$ of order $p^{\ell}$, $p^{m}$, and $p^{n}$ respectively, there are $r\leq\max(\ell,m)$ and $s\leq\max(m,n)$ such that $x+y$ and $y+z$ are of order $p^{r}$ and $p^{s}$ respectively. Then there is $t$ such that $x+y+z$ is of order $p^{t}$; $t\leq\max(r,n)$ and $t\leq\max(\ell,s)$. 8. (A8)

Given $x,y,z\in G$ of order $p^{\ell}$, $p^{m}$, and $p^{n}$ respectively, $n\leq\max(\ell,m)$, and with $x+y=z$, we have $y+x=z$ as $G$ is abelian.

Thus we have shown that $\mathfrak{M}(G)$ is a model of $T_{p}$. ∎

Note that $G$ and $\mathfrak{M}(G)$ are effectively bi-interpretable, proving one half of Theorem 1.3.

4. From a model of $T_{p}$ to an abelian $p$-group

Given an abelian $p$-group $G$, we have already described how to turn $G$ into a model of $T_{p}$. In this section we will do the reverse by turning a model of $T_{p}$ into an abelian $p$-group.

Definition 4.1**.**

Let $\mathcal{M}$ be a model of $T_{p}$. Define $\mathfrak{G}(\mathcal{M})$ to be the group obtained as follows.

• •

The domain of $\mathfrak{G}(\mathcal{M})$ will be the subset of the domain of $\mathcal{M}$ given by $\bigcup_{n\in\omega}R_{n}^{\mathcal{M}}$.

• •

The identity element of $\mathfrak{G}(\mathcal{M})$ will be $0^{\mathcal{M}}$.

• •

We will have $x+y=z$ in $\mathfrak{G}(\mathcal{M})$ if and only if, for some $\ell$, $m$, and $n$, $P_{\ell,m}^{n,\mathcal{M}}(x,y,z)$.

We will now check that $\mathfrak{G}(\mathcal{M})$ is always an abelian $p$-group.

Lemma 4.2**.**

If $\mathcal{M}$ is a model of $T_{p}$, then $\mathfrak{G}(\mathcal{M})$ is an abelian $p$-group.

Proof.

First we check that the operation $+$ on $\mathfrak{G}(\mathcal{M})$ defines a total function. Given $x,y\in\mathfrak{G}(\mathcal{M})$, choose $\ell$ and $m$ such that $x\in R^{\mathcal{M}}_{\ell}$ and $y\in R^{\mathcal{M}}_{m}$. Then by (A3) and (A4), there is a unique $n\leq\max(\ell,m)$ and a unique $z$ such that $P_{\ell,m}^{n,\mathcal{M}}(x,y,z)$. Thus $x+y=z$, and $z$ is unique.

Second, we check that $\mathfrak{G}(\mathcal{M})$ is in fact a group. To see that $0^{\mathcal{M}}$ is the identity, given $x\in\mathfrak{G}(\mathcal{M})$, there is $\ell$ such that $x\in R^{\mathcal{M}}_{\ell}$. By (A5), $P_{\ell,0}^{\ell,\mathcal{M}}(x,0^{\mathcal{M}},x)$ and $P_{0,\ell}^{\ell,\mathcal{M}}(0^{\mathcal{M}},x,0^{\mathcal{M}})$. Thus $x+0^{\mathcal{M}}=0^{\mathcal{M}}+x=x$, and $0^{\mathcal{M}}$ is the identity of $\mathfrak{G}(\mathcal{M})$. To see that $\mathfrak{G}(\mathcal{M})$ has inverses, given $x\in\mathfrak{G}(\mathcal{M})$, there is $\ell$ such that $x\in R^{\mathcal{M}}_{\ell}$, and by (A6) there is $y\in R^{\mathcal{M}}_{\ell}$ such that $P_{\ell,\ell}^{0,\mathcal{M}}(x,y,0^{\mathcal{M}})$ and $P_{\ell,\ell}^{0,\mathcal{M}}(y,x,0^{\mathcal{M}})$. Thus $x+y=y+x=0^{\mathcal{M}}$, and so $y$ is the inverse of $x$. Finally, to see that $\mathfrak{G}(M)$ is associative, given $x,y,z\in\mathfrak{G}(\mathcal{M})$, there are $\ell$, $m$, and $n$ such that $x\in R^{\mathcal{M}}_{\ell}$, $y\in R^{\mathcal{M}}_{m}$, and $z\in R^{\mathcal{M}}_{n}$. Then by (A7) there are $r$, $s$, and $t$, and $u$, $v$, and $w$, such that $P_{\ell,m}^{r,\mathcal{M}}(x,y,u)$, $P_{r,n}^{t,\mathcal{M}}(u,z,w)$, $P_{m,n}^{s,\mathcal{M}}(y,z,v)$, and $P_{\ell,s}^{t,\mathcal{M}}(x,v,w)$. Thus $x+y=u$, $u+z=w$, $y+z=v$, and $x+v=w$. So $(x+y)+z=x+(y+z)$. Thus $\mathfrak{G}(\mathcal{M})$ is associative.

Third, to see that $\mathfrak{G}(\mathcal{M})$ is abelian, let $x,y\in\mathfrak{G}(\mathcal{M})$. There are $\ell$ and $m$ such that $x\in R^{\mathcal{M}}_{\ell}$ and $y\in R^{\mathcal{M}}_{m}$. Let $n\leq\max(\ell,m)$ be such that $z=x+y\in R_{n}^{\mathcal{M}}$. (Such an $n$ and $z$ exist by the arguments above that $+$ is total, via (A3) and (A4).) Then $P_{\ell,m}^{n,\mathcal{M}}(x,y,z)$, and so by (A8), $P_{m,\ell}^{n,\mathcal{M}}(y,x,z)$. Thus $y+x=z$ and so $\mathfrak{G}(\mathcal{M})$ is abelian.

Finally, we need to see that $\mathfrak{G}(\mathcal{M})$ is a $p$-group. We claim, by induction on $n\geq 0$, that $R^{\mathcal{M}}_{n}$ consists of the elements of $\mathfrak{G}(\mathcal{M})$ which are of order $p^{n}$. From this claim, it follows that $\mathfrak{G}(\mathcal{M})$ is a $p$-group. For $n=0$, the claim follows directly from (A2). Given $n>0$, suppose that $x\in R^{\mathcal{M}}_{n}$. Then the witnesses $x_{2},x_{3},\ldots,x_{p}$ to (A2) must be $2x,3x,\ldots,px$. Note that since $P_{n,n}^{n-1,\mathcal{M}}(x,(p-1)x,px)$, $px\in R^{\mathcal{M}}_{n-1}$. Thus $px$ is of order $p^{n-1}$, and so $x$ is of order $p^{n}$. On the other hand, if $x$ is of order $p^{n}$, then $px$ is of order $p^{n-1}$ and so $px\in R^{\mathcal{M}}_{n-1}$. Moreover, $x_{2}=2x,x_{3}=3x,\ldots,x_{p-1}=(p-1)x$ are all of order $p^{n}$. So we have $P_{n,n}^{n,\mathcal{M}}(x,x,x_{2}),P_{n,n}^{n,\mathcal{M}}(x,x_{2},x_{3}),\ldots,,P_{n,n}^{n-1,\mathcal{M}}(x,x_{p-1},x_{p})$. By (A2), $x\in R^{\mathcal{M}}_{n}$. This completes the inductive proof. ∎

We now have two operations, one which turns an abelian $p$-group into a model of $T_{p}$, and another which turns a model of $T_{p}$ into an abelian $p$-group. These two operations are almost inverses to each other. If we begin with an abelian $p$-group, turn it into a model of $T_{p}$, and then that model into an abelian $p$-group, we will obtain the original group. However, if we start with a $\mathcal{M}$ model of $T_{p}$, turn it into an abelian $p$-group, and then turn that abelian $p$-group into a model of $T_{p}$, we may obtain a different model of $T_{p}$. The problem is that the of elements of $\mathcal{M}$ which are not in any of the sets $R_{n}^{\mathcal{M}}$ are discarded when we transform $\mathcal{M}$ into an abelian $p$-group. However, these elements form a pure set, and so the only pertinent information is their size.

Definition 4.3**.**

Given a model $\mathcal{M}$ of $T_{p}$, the size of $\mathcal{M}$, $\#\mathcal{M}\in\omega\cup\{\infty\}$, is the number of elements of $M$ not in any relation $R_{n}$.

Lemma 4.4**.**

Given an abelian $p$-group $G$, $\mathfrak{G}(\mathfrak{M}(G))=G$.

Proof.

Since $\#\mathfrak{M}(G)=0$, we see that $G$, $\mathfrak{M}(G)$, and $\mathfrak{G}(\mathfrak{M}(G))$ all have the same domain. The identity of $\mathfrak{G}(\mathfrak{M}(G))$ is $0^{\mathfrak{M}(G)}$ which is the identity of $G$. If $x+y=z$ in $G$, then, for some $\ell,m,n\in\omega$, we have $P_{\ell,m}^{n,\mathfrak{M}(G)}(x,y,z)$. Thus, in $\mathfrak{G}(\mathfrak{M}(G))$, we have $x+y=z$. So $\mathfrak{G}(\mathfrak{M}(G))=G$. ∎

We make a simple extension to $\mathfrak{M}$ as follows.

Definition 4.5**.**

Let $G$ be an abelian $p$-group and $m\in\omega\cup\{\infty\}$. Define $\mathfrak{M}(G,m)$ to be $\mathcal{L}_{p}$-structure with domain $G\cup\{a_{1},\ldots,a_{m}\}$ with the relations interpreted as in $\mathfrak{M}(G)$. Thus, no relations hold of any of the elements $a_{1},\ldots,a_{m}$.

Lemma 4.6**.**

Given a model $\mathcal{M}$ of $T_{p}$, $\mathfrak{M}(G(\mathcal{M}),\#\mathcal{M})\cong\mathcal{M}$.

Proof.

We will show that if $\#\mathcal{M}=0$, then $\mathfrak{M}(\mathfrak{G}(\mathcal{M}))=\mathcal{M}$. From this one can easily see that $\mathfrak{M}(G(\mathcal{M}),\#\mathcal{M})\cong\mathcal{M}$ in general.

If $\#\mathcal{M}=0$, then $\mathcal{M}$, $\mathfrak{G}(\mathcal{M})$, and $\mathfrak{M}(\mathfrak{G}(\mathcal{M}))$ all share the same domain. It is clear that $0^{\mathcal{M}}=0^{\mathfrak{G}(\mathcal{M})}=0^{\mathfrak{M}(\mathfrak{G}(\mathcal{M}))}$. From the proof of Lemma 4.2, we see that for each $n$, $R_{n}^{\mathcal{M}}$ defines the set of elements of $\mathfrak{G}(\mathcal{M})$ which are torsion of order $p^{n}$, and so $R_{n}^{\mathcal{M}}=R_{n}^{\mathfrak{M}(\mathfrak{G}(\mathcal{M}))}$. Given $\ell,m\in\omega$ and $n\leq\max(\ell,m)$, and $x$, $y$, and $z$ elements of the shared domain, we have $P_{\ell,m}^{n,\mathcal{M}}(x,y,z)$ if and only if

[TABLE]

Since $R_{i}^{\mathcal{M}}=R_{i}^{\mathfrak{M}(\mathfrak{G}(\mathcal{M}))}$ for each $i$, this is the case if and only if $P_{\ell,m}^{n,\mathfrak{M}(\mathfrak{G}(\mathcal{M}))}(x,y,z)$. Thus we have shown that $\mathfrak{M}(\mathfrak{G}(\mathcal{M}))=\mathcal{M}$. ∎

Note that $\mathcal{M}$ and the disjoint union of $\mathfrak{G}(\mathcal{M})$ with a pure set of size $\#\mathcal{M}$ are bi-interpretable, using computable infinitary formulas, completing the proof of Theorem 1.3.

5. Borel Equivalence

In this section we will prove Theorem 1.7 by showing that the class of models of $T_{p}$ and the class of abelian $p$-groups are Borel equivalent. $G\mapsto\mathfrak{G}(\mathfrak{M}(G))=\mathfrak{G}(\mathfrak{M}(G,0))$ is a Borel reduction from isomorphism on abelian $p$-groups to isomorphism on models of $T_{p}$. However, $\mathcal{M}\mapsto\mathfrak{G}(\mathcal{M})$ is not a Borel reduction in the other direction, because two non-isomorphic models of $T_{p}$ might be mapped to isomorphic groups. We need to find a way to turn $\mathfrak{G}(\mathcal{M})$ and $\#\mathcal{M}$ into an abelian $p$-group $\mathfrak{H}(\mathfrak{G}(\mathcal{M}),\#\mathcal{M})$, so that $\mathcal{M}$ and $\#\mathcal{M}$ can be recovered from $\mathfrak{H}(\mathfrak{G}(\mathcal{M}),\#\mathcal{M})$.

We will define $\mathfrak{H}(G,m)$ for any abelian $p$-group $H$ and $m\in\omega\cup\{\infty\}$. It is helpful to think about what this reduction will do to the Ulm invariants: The first Ulm invariant of $\mathfrak{H}(G,m)$ will be $m$, and for each $\alpha$, then $1+\alpha$th Ulm invariant of $\mathfrak{H}(G,m)$ will be the same as the $\alpha$th Ulm invariant of $G$.

Definition 5.1**.**

Given an abelian $p$-group $G$, and $m\in\omega\cup\{\infty\}$, define an abelian $p$-group $\mathfrak{H}(G,m)$ as follows. Let $\hat{\mathcal{B}}$ be a basis for the $\mathbb{Z}_{p}$-vector space $G/{pG}$. Let $\mathcal{B}\subseteq G$ be a set of representatives for $\hat{\mathcal{B}}$. Let $G^{*}$ be the abelian group $\langle G,a_{b}:b\in\mathcal{B}\mid pa_{b}=b\rangle$. Then define $\mathfrak{H}(G,m)=G^{*}\oplus(\mathbb{Z}_{p})^{m}$.

To make this Borel, we can take $\mathcal{B}$ to be the lexicographically first set of representatives for a basis. It will follow from Lemma 5.4 that the isomorphism type of $\mathfrak{H}(G,m)$ does not depend on these choices. First, we require a couple of lemmas.

Lemma 5.2**.**

Each element of $G$ can be written uniquely as a (finite) linear combination $h+\sum_{b\in\mathcal{B}}x_{b}b$ where $h\in pG$ and each $x_{b}<p$.

Proof.

Given $g\in G$, let $\hat{g}$ be the image of $g$ in $G/pG$. Then, since $\hat{\mathcal{B}}$ is a basis for $G/pG$, we can write

[TABLE]

with $x_{b}<p$, where $\hat{b}$ is the image of $b$ in $G/pG$. Thus setting

[TABLE]

we get a representation of $g$ as in the statement of the theorem.

To see that this representation is unique, suppose that

[TABLE]

Then, modulo $pG$,

[TABLE]

Since $\hat{\mathcal{B}}$ is a basis, $x_{b}=y_{b}$ for each $b\in\mathcal{B}$. Then we get that $h=h^{\prime}$ and the two representations are the same. ∎

Lemma 5.3**.**

Each element of $G^{*}$ can be written uniquely in the form $h+\sum_{b\in\mathcal{B}}x_{b}a_{b}$ where $h\in G$ and each $x_{b}<p$.

Proof.

It is clear that each element of $G^{*}$ can be written in such a way. If

[TABLE]

then, in $G$,

[TABLE]

This representation is unique, so $x_{b}=y_{b}$ for each $b\in\mathcal{B}$, and so $h=h^{\prime}$. ∎

Lemma 5.4**.**

The isomorphism type of $\mathfrak{H}(G,m)$ depends only on the isomorphism type of $G$, and not on the choice of $\mathcal{B}$.

Proof.

It suffices to show that if $\mathcal{C}$ is another choice of representatives for a basis of $G/{pG}$, then $G^{*}_{\mathcal{B}}=G^{*}_{\mathcal{C}}$, where the former is constructed using $\mathcal{B}$, and the later is constructed using $\mathcal{C}$. Let $f\colon\mathcal{B}\to\mathcal{C}$ be an bijection.

Given $g\in G^{*}_{\mathcal{B}}$, write $g=g^{\prime}+\sum_{b\in\mathcal{B}}x_{b}a_{b}$ with $g^{\prime}\in G$ and $0\leq x_{b}<p$. This representation of $g$ is unique by Lemma 5.3. Define $\varphi(g)=g^{\prime}+\sum_{b\in\mathcal{B}}x_{b}a_{f(b)}$. It is not hard to check that $\varphi$ is a homomorphism. The inverse of $\varphi$ is the map $\psi$ which is defined by $\psi(h)=h^{\prime}+\sum_{c\in\mathcal{C}}y_{c}a_{f^{-1}(c)}$ where $h=h^{\prime}+\sum_{c\in\mathcal{C}}y_{c}a_{c}$. ∎

The next two lemmas will be used to show that if $G$ is not isomorphic to $G^{\prime}$, or if $m$ is not equal to $m^{\prime}$, then $\mathfrak{H}(G,m)$ will not be isomorphic to $\mathfrak{H}(G^{\prime},m^{\prime})$.

Lemma 5.5**.**

$G=pG^{*}$.

Proof.

Each element of $G$ can be written as $g+\sum_{b\in\mathcal{B}}x_{b}b$ with $g\in pG$. Let $g^{\prime}\in G$ be such that $pg^{\prime}=g$. Then

[TABLE]

Hence $G\subseteq pG^{*}$. Given $h\in G^{*}$, write $h=g+\sum_{b\in\mathcal{B}}x_{b}a_{b}$. Then $ph=pg+\sum_{b\in\mathcal{B}}x_{b}b\in G$. So $pG^{*}\subseteq G$, and so $G=pG^{*}$. ∎

If $G$ is a group, recall that we denote by $G[p]$ the elements of $G$ which are torsion of order $p$.

Lemma 5.6**.**

$\mathfrak{H}(G,m)[p]\;/\;(p\mathfrak{H}(G,m))[p]\cong(\mathbb{Z}_{p})^{m}$.

Proof.

Note that

[TABLE]

We will show that $(G^{*}[p]/G[p])$ is the trivial group by showing that if $g\in G^{*}$, $pg=0$, then $g\in G$. Indeed, write $g=g^{\prime}+\sum_{b\in\mathcal{B}}y_{b}a_{b}$ with $g^{\prime}\in G$. Then

[TABLE]

Since $0\in pG$ has a unique representation (by Lemma 5.2) $0=0+\sum_{b\in\mathcal{B}}0b$, we get that $y_{b}=0$ for each $b\in\mathcal{B}$, and so $g=g^{\prime}\in G$. ∎

By the previous lemma, we can recover $m$ from $\mathfrak{H}(G,m)$. We have

[TABLE]

so that we can also recover $G$.

Thus, using Lemma 4.6, $\mathcal{M}\mapsto\mathfrak{H}(\mathfrak{G}(\mathcal{M}),\#\mathcal{M})$ gives a Borel reduction from $T_{p}$ to abelian $p$-groups. This completes the proof of Theorem 1.7.