The discrete periodic tiling conjecture

Let $G=(G,+)$ be a finitely generated discrete abelian group. For subsets $A,F\subseteq G$ , write $A\oplus F=G$ when every element of $G$ has a unique representation $a+f$ with $a\in A$ and $f\in F$ ; then $F$ tiles $G$ by translations. Call such a tiling periodic when $A$ is a finite union of cosets of a finite-index subgroup of $G$ . Discrete periodic tiling conjecture. Let $F$ be a finite non-empty subset of $G$ . If $F$ tiles $G$ by translations, then $F$ periodically tiles $G$ by translations. This conjecture would imply algorithmic decidability of whether a given finite set tiles an explicitly presented finitely generated abelian group, while the general case remains open.

The conjecture asserts that every finite tile in a finitely generated discrete abelian group has a periodic tiling. On September 19, 2022, Rachel Greenfeld and Terence Tao announced a counterexample.

Known results

$G=\mathbb{Z}$ and $G=\mathbb{R}$ : established before the counterexample.
$G=\mathbb{Z}\times G_0$ , with $G_0$ finite abelian: established.
$G=\mathbb{Z}^2$ : established, including work of Siddhartha Bhattacharya.
In higher dimensions, the result was known when $|F|$ is prime or $|F|=4$ .

September 2022 counterexample

The announced construction gives a finite abelian group $G_0$ and finite $F\subseteq\mathbb{Z}^2\times G_0$ with an aperiodic tiling; a reduction yields a finite tile in $\mathbb{Z}^d$ for sufficiently large $d$ that has no periodic tiling. The claim is presented as a complete disproof, but remains unverified here.

Current status (as of September 2026): The conjecture is claimed false by the Greenfeld--Tao counterexample in sufficiently high-dimensional free abelian groups; lower-dimensional cases and independent verification of the reported construction remain unsettled.

The discrete periodic tiling conjecture

Primary source

Progress summary

Solutions 0