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I believe I remember reading the following claim in a paper (the relevant definitions are recalled below):

If $(A,+_A)$ is a torsion-free abelian group and $(\Bbb{Z},+_{\Bbb{Z}})$ is the integers equipped with addition, then for every finite $Z \subset A$ and $r\ge 1$, there exists a Freiman $r$-isomorhpism $\phi:A\to \Bbb{Z}$ (wrt to $Z$).

While the claim makes intuitive sense to me, I'd like to see a formal proof.

Definitions

An abelian group is torsion-free if every non-zero element has infinite order (i.e. if adding $a$ to itself $k$ times is equal to the identity (for some finite $k>0$), then $a$ must be the identity).

Given two groups $(A,*)$, $(B,\circ)$, and a subset $A'\subset A$, we say a homomorphism $\phi:A \to B$ is a Freiman $r$-isomorphism (wrt $A'$), if for $a_1,\dots,a_r,a_1',\dots,a_r' \in A'$, we have

$$\phi(a_1)\circ \ldots \circ \phi(a_r) = \phi(a_1')\circ \ldots \circ\phi(a_r')$$ if and only if $$a_1 * \ldots * a_r = a_1' * \ldots * a_r'.$$

Zach Hunter
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1 Answers1

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That is indeed correct! The first step is to show that every finite subset $A\subseteq \mathbb{Z}^{n+1}$ is Freiman isomorphic to a subset of $\mathbb{Z}$. Since translation is a Freiman isomorphism of all orders, we may assume without loss of generality that $A\subseteq [L]^{n+1}$ for some $L\in\mathbb{N}$. Now fix $s\geqslant 1$ and let $K=sL+1$; I claim that the map $f:[L]^{n+1}\to\mathbb{Z}$ given by $(a_0,\dots,a_n)\mapsto a_0+Ka_1+\dots+K^na_n$ is a Freiman $s$-isomorphism onto its image. This follows from uniqueness of base $K$ expansions; I've written the details below.

Okay, so how about for arbitrary torsion-free abelian groups? The trick we can use here now is the fundamental theorem of finitely generated modules over a PID; in the case of abelian groups this tells us that every finitely generated abelian group $A$ is isomorphic to a group of form $\mathbb{Z}^n\oplus B$ for some torsion group $B$ and some natural $n$. In particular, if $A$ is torsion-free, then $A$ is isomorphic to $\mathbb{Z}^n$ for some $n$.

Now we're essentially done. Given your group $A$ and subset $Z$, we note that the subgroup $A_0\leqslant A$ generated by $Z$ is finitely generated and torsion-free; this puts us in a position to apply the previous two paragraphs.

Fact: Let $K>1$ be a positive integer. For any $n$ and any $\lambda_i,\mu_i\in[K-1]$, we have $\lambda_0+K\lambda_1+\dots+K^n\lambda_n=\mu_0+K\mu_1+\dots+K^n\mu_n$ if and only if $\lambda_i=\mu_i$ for all $i$. $\blacksquare$

Lemma: The map $f$ defined above is a Freiman $s$-isomorphism onto its image.

Proof: We wish to show that, for all $a^1,\dots,a^s$ and $b^1,\dots,b^s$ in $[L]^{n+1}$, we have $$a^1+\dots+a^s=b^1+\dots+b^s$$ if and only if $$f(a^1)+\dots+f(a^s)=f(b^1)+\dots+f(b^s).$$ This follows from the fact: we have $$f(a^1)+\dots+f(a^s)=(a^1_0+\dots+a^s_0)+K(a^1_1+\dots+a^s_1)+\dots+K^n(a^1_n+\dots+a^s_n)$$ and similarly for the sum of the $f(b^i)$. Since $1\leqslant a^i_j\leqslant L$ for all $i$ and $j$, we have $1\leqslant a^1_j+\dots+a^s_j<K$ for all $j$, and similarly for the $b^1_j+\dots+b^s_j$, so we're done. $\blacksquare$

Atticus Stonestrom
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