# Coset Progression is Freiman Isomorphic to Bohr Set

• For an abelian group $$G$$, $$H$$ a finite subgroup of $$G$$, $$x_1, \dots, x_r \in G$$ and $$L_1, \dots, L_r \in \mathbb N$$, let: $$P(x ; L) = P(x_1, \dots, x_r ; L_1, \dots, L_r) = \{l_1x_1 + \dots + l_rx_r \mid |l_i| \leq L_i \; \forall i\}$$ be a progression of rank $$r$$. Then: $$H + P(x;L) = \{h + p \mid h \in H, \; p \in P(x;L)\}$$ is a coset progression of rank r in $$G$$.

• Given $$x \in \mathbb R / \mathbb Z$$, define $$|| x ||_{\mathbb R / \mathbb Z} = |\hat x|$$, where $$\hat x$$ is the representative of $$x$$ in $$\mathbb R / \mathbb Z$$ with $$\hat x \in [-\frac{1}{2}, \frac{1}{2}]$$

• For a finite abelian group $$G$$, let $$\Gamma = \{\gamma_1, \dots, \gamma_r\} \subset \hat G = Hom(G, \;\mathbb R / \mathbb Z)$$ and $$\rho \in [0, \frac{1}{2}]$$. Define the Bohr Set of rank $$r$$: $$B(\Gamma, \rho) = \{g \in G \mid ||\gamma_i(g)||_{\mathbb R / \mathbb Z} \leq \rho, \; \forall i\}$$.

• Given two subsets $$A,B$$ of some abelian groups $$G,H$$, a function $$f:A \rightarrow B$$, we say that $$f$$ is a Freiman $$m$$-Homomorphism, if whenever $$a_1, \dots, a_m, a_{m+1}, \dots, a_{2m} \in A$$ are such that $$a_1 + \dots+ a_m = a_{m+1} + \dots +a_{2m}$$, then we have $$f(a_1) + \dots + f(a_m) = f(a_{m+1}) + \dots + f(a_{2m})$$

I am asked to show that every coset progression of rank $$r$$ is Freiman isomorphic to a Bohr set of rank $$r$$.

My immediate thought is that since the ranks are related, we must be "constructing" the $$\gamma_i$$'s using the $$x_i$$'s.

Additionally, if we had some condition similar to linear independence between the $$x_i$$'s, then I can see how we may be able to step through a product of finite, cyclic groups to achieve the isomorphism. However, in a situation where there is some sort of "linear relation" with the $$x_i$$'s, how does this not affect the rank of our final Bohr set?

I am struggling to see how we might be able to construct a finite group, with which we may find some $$\Gamma$$ and $$\rho$$ to give us the desired result. Given this, perhaps we can instead start with an arbitrary Bohr Set of rank r, and show that the Freiman isomorphic image is a coset progression of rank $$r$$, but I believe this is actually false, though I can't seem to find that in my notes currently.

How should I approach this question? How can I effectively construct a finite abelian group, even though it is possible that some of the $$x_i$$'s have infinite order? Would it be sensible instead then to use $$H$$ as the starting point, in which case how might I be able to deal with the fact that our $$\gamma_i$$'s may no longer have the $$x_i$$'s in their domain?

Any help that may be offered would be very much appreciated, thank you.