[proofplan] We first replace the ambient torsion-free abelian group by the subgroup generated by $A$, which is finitely generated and therefore isomorphic to $\mathbb{Z}^r$ for some $r \geq 0$. Under this isomorphism the doubling hypothesis is preserved. We then invoke the Freiman-Ruzsa theorem for finite subsets of torsion-free finitely generated abelian groups to obtain a proper generalized arithmetic progression with rank and size controlled only by $K$, and transport that progression back to $G$. [/proofplan] [step:Reduce the ambient group to a finitely generated torsion-free subgroup] Let $H := \langle A \rangle$ denote the subgroup of $G$ generated by $A$. Since $A$ is finite and nonempty, $H$ is a finitely generated abelian group. Because $G$ is torsion-free, every subgroup of $G$ is torsion-free, so $H$ is torsion-free. By the [Structure Theorem for Finitely Generated Abelian Groups](/theorems/18), there exists an integer $r \geq 0$ and a group isomorphism \begin{align*} \varphi: H &\to \mathbb{Z}^r. \end{align*} Define \begin{align*} B := \varphi(A) \subset \mathbb{Z}^r. \end{align*} Since $\varphi$ is bijective, $B$ is finite and nonempty, and $|B| = |A|$. Since $\varphi$ is a group homomorphism, it maps sumsets bijectively: \begin{align*} B + B &= \{\varphi(a_1) + \varphi(a_2) : a_1, a_2 \in A\} \\ &= \{\varphi(a_1 + a_2) : a_1, a_2 \in A\} \\ &= \varphi(A + A). \end{align*} Hence $|B+B| = |A+A| \leq K|A| = K|B|$. [guided] The first obstacle is that $G$ may have infinite rank, while Freiman-type inverse theorems are normally stated for finite-rank torsion-free groups. The set $A$ itself is finite, so all additive relations inside $A$ live in the subgroup it generates. Define $H := \langle A \rangle$, the smallest subgroup of $G$ containing $A$. Because $A$ is finite, $H$ is generated by finitely many elements. Because $G$ is torsion-free, every subgroup of $G$ is torsion-free: if $h \in H$ and $mh = 0$ for some $m \in \mathbb{N}$, then the same equation holds in $G$, so $h = 0$. Thus $H$ is a finitely generated torsion-free abelian group. The [Structure Theorem for Finitely Generated Abelian Groups](/theorems/18) applies and gives an integer $r \geq 0$ together with a group isomorphism \begin{align*} \varphi: H &\to \mathbb{Z}^r. \end{align*} Now define the modeled set \begin{align*} B := \varphi(A) \subset \mathbb{Z}^r. \end{align*} Since $\varphi$ is a bijection, $B$ is finite and nonempty and $|B| = |A|$. The doubling hypothesis is also preserved. Indeed, since $\varphi$ is a homomorphism, \begin{align*} B + B &= \{\varphi(a_1) + \varphi(a_2) : a_1, a_2 \in A\} \\ &= \{\varphi(a_1 + a_2) : a_1, a_2 \in A\} \\ &= \varphi(A + A). \end{align*} Because $\varphi$ is injective, the last map is a bijection from $A+A$ to $B+B$. Therefore \begin{align*} |B+B| = |A+A| \leq K|A| = K|B|. \end{align*} This reduction uses all structural hypotheses on $G$: commutativity makes the sumset theory applicable, torsion-freeness removes finite cyclic components, and finiteness of $A$ makes the generated subgroup finitely generated. [/guided] [/step] [step:Apply the finite-rank torsion-free Freiman-Ruzsa theorem] By the [Freiman-Ruzsa Theorem in Finitely Generated Torsion-Free Abelian Groups](/theorems/19), applied to the finite nonempty set $B \subset \mathbb{Z}^r$ with $|B+B| \leq K|B|$, there exist constants $d(K)$ and $C(K)$ depending only on $K$ and a proper generalized arithmetic progression $Q \subset \mathbb{Z}^r$ such that \begin{align*} B \subset Q, \qquad \operatorname{rank}(Q) \leq d(K), \qquad |Q| \leq C(K)|B|. \end{align*} Here $Q$ has the form \begin{align*} Q = \left\{q_0 + n_1 q_1 + \cdots + n_m q_m : 0 \leq n_i < L_i \text{ for } 1 \leq i \leq m\right\}, \end{align*} where $m \leq d(K)$, $q_0,q_1,\dots,q_m \in \mathbb{Z}^r$, and $L_1,\dots,L_m \in \mathbb{N}$. Properness means that the parameter map \begin{align*} \Theta: \prod_{i=1}^m \{0,1,\dots,L_i-1\} &\to \mathbb{Z}^r \\ (n_1,\dots,n_m) &\mapsto q_0 + n_1q_1 + \cdots + n_mq_m \end{align*} is injective. [guided] We now use the inverse theorem in the finite-rank torsion-free setting. The [Freiman-Ruzsa Theorem in Finitely Generated Torsion-Free Abelian Groups](/theorems/19) requires a finite nonempty subset of a torsion-free finitely generated abelian group with small doubling. We have exactly that: $B \subset \mathbb{Z}^r$ is finite and nonempty, $\mathbb{Z}^r$ is torsion-free and finitely generated, and the previous step proved \begin{align*} |B+B| \leq K|B|. \end{align*} Therefore the theorem gives constants $d(K)$ and $C(K)$ depending only on $K$, together with a proper generalized arithmetic progression $Q \subset \mathbb{Z}^r$, such that \begin{align*} B \subset Q, \qquad \operatorname{rank}(Q) \leq d(K), \qquad |Q| \leq C(K)|B|. \end{align*} Writing out the progression, there are an integer $m \leq d(K)$, elements $q_0,q_1,\dots,q_m \in \mathbb{Z}^r$, and lengths $L_1,\dots,L_m \in \mathbb{N}$ such that \begin{align*} Q = \left\{q_0 + n_1 q_1 + \cdots + n_m q_m : 0 \leq n_i < L_i \text{ for } 1 \leq i \leq m\right\}. \end{align*} The word proper is important: it means the parameter map \begin{align*} \Theta: \prod_{i=1}^m \{0,1,\dots,L_i-1\} &\to \mathbb{Z}^r \\ (n_1,\dots,n_m) &\mapsto q_0 + n_1q_1 + \cdots + n_mq_m \end{align*} is injective. This injectivity will be preserved when we transport the progression back through the group isomorphism. [/guided] [/step] [step:Pull the progression back to the original group] Define \begin{align*} p_i := \varphi^{-1}(q_i) \in H \subset G \qquad \text{for } 0 \leq i \leq m, \end{align*} and define \begin{align*} P := \varphi^{-1}(Q) = \left\{p_0 + n_1p_1 + \cdots + n_mp_m : 0 \leq n_i < L_i \text{ for } 1 \leq i \leq m\right\} \subset G. \end{align*} Since $A = \varphi^{-1}(B)$ and $B \subset Q$, we have $A \subset P$. The rank of $P$ is $m$, so $\operatorname{rank}(P) \leq d(K)$. Since $\varphi$ is bijective from $P$ to $Q$, $|P| = |Q|$, and therefore \begin{align*} |P| = |Q| \leq C(K)|B| = C(K)|A|. \end{align*} Finally, the parameter map for $P$ is injective because it is the composition $\varphi^{-1} \circ \Theta$. Hence $P$ is a proper generalized arithmetic progression in $G$. This proves the theorem. [guided] It remains to undo the isomorphism. For each generator of $Q$, define its pullback \begin{align*} p_i := \varphi^{-1}(q_i) \in H \subset G \qquad \text{for } 0 \leq i \leq m. \end{align*} Then define the corresponding progression in $G$ by \begin{align*} P := \varphi^{-1}(Q) = \left\{p_0 + n_1p_1 + \cdots + n_mp_m : 0 \leq n_i < L_i \text{ for } 1 \leq i \leq m\right\} \subset G. \end{align*} Because $B \subset Q$ and $A = \varphi^{-1}(B)$, applying $\varphi^{-1}$ gives $A \subset P$. The rank is unchanged: $P$ is parametrized by the same $m$ step directions, and $m \leq d(K)$. The size estimate is also unchanged because $\varphi$ is a bijection. Its restriction maps $P$ bijectively onto $Q$, so \begin{align*} |P| = |Q| \leq C(K)|B| = C(K)|A|. \end{align*} Finally, we verify properness. The parameter map for $P$ is \begin{align*} \Psi: \prod_{i=1}^m \{0,1,\dots,L_i-1\} &\to G \\ (n_1,\dots,n_m) &\mapsto p_0 + n_1p_1 + \cdots + n_mp_m. \end{align*} For every parameter tuple, $\Psi = \varphi^{-1} \circ \Theta$. Since $\Theta$ is injective and $\varphi^{-1}$ is injective, $\Psi$ is injective. Thus $P$ is proper. We have constructed a proper generalized arithmetic progression $P \subset G$ containing $A$, with rank at most $d(K)$ and size at most $C(K)|A|$, which is exactly the desired conclusion. [/guided] [/step]