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Statement

Setting (pp. 1-2). A subset $S$ of an abelian semigroup is a basis (of order two) if every element is $s_1+s_2$ with $s_1,s_2\in S$. The basis is perfect if each element has exactly one such representation, up to the order of the summands. The representation function counts ordered representations. For a subset $C$ of an abelian group and an integer $n\ge1$, $nC=\{nc:c\in C\}$, so $2G$ is the set of doubles of elements of $G$.

Theorem 1 (p. 2, quoted). "Let $G$ be an infinite abelian group with $|2G|=|G|$. (i) If $G$ is not the direct sum of a group of exponent 3 and the group of order 2, then $G$ has a perfect basis. (ii) If $G$ is the direct sum of a group of exponent 3 and the group of order 2, then $G$ does not have a perfect basis, but has a basis such that every element of $G$ has at most two representations (distinct under permuting the summands) as a sum of two elements of the basis."

The hypothesis $|2G|=|G|$ is needed: the paper observes (p. 2) that when $G$ is infinite with $|2G|<|G|$, every basis $S$ (indeed every subset with $|S|=|G|$) gives some element $|G|$ representations of the form $2s$ with $s\in S$; this covers every infinite group of exponent 2. The abstract calls the theorem a complete solution of the Erdős-Turán problem for infinite groups. The proofs assume the axiom of choice (p. 3).

Source. Sergei V. Konyagin and Vsevolod F. Lev, The Erdős-Turán problem in infinite groups, arXiv:0901.1649v1 (2009); published in Additive Number Theory, Springer, New York, 2010, 195--202. Labels and pages here are those of arXiv v1: the definitions on p. 1, Theorem 1 on p. 2, Lemmas 1 and 2 on pp. 3-4, the proof on pp. 5-6. The edition read is identified on the source card.

Read depth. Claims checked: the definitions and the statement were read clause by clause on the printed pages. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pages 5-6, in three parts. For $G$ of exponent 3, Lemma 1 (p. 3: an infinite abelian group of prime exponent $p$ is isomorphic to $\mathbb F\times\mathbb F$ for an algebraically closed field $\mathbb F$ of characteristic $p$) reduces to $\mathbb F\times\mathbb F$, where the parabola $\{(x,x^2):x\in\mathbb F\}$ is a perfect basis, since a representation of $(u,v)$ solves a quadratic in $x$ with one or two roots. For $G=F\oplus\{0,h\}$ with $F$ of exponent 3 and $h$ of order 2, a perfect basis $S$ of $F$ together with its translate $h+S$ gives at most two representations, and a perfect basis of $G$ is ruled out by taking the unique representation of $h$ and doubling. In general the perfect basis is built by transfinite recursion along a well-ordering of $G$ indexed by the initial ordinal of $|G|$, adding, at each successor step whose element is not yet represented, a pair $s,t$ with that element as their sum, subject to conditions (a)-(e) on p. 6, of which (b)-(e) keep representations unique. Lemma 2 (p. 4: if $2G$ is infinite and $\max\{|A|,|B|\}<\min\{|2G|,|3G|\}$, some $s\in G$ has $2s\notin A$ and $3s\notin B$) supplies the pair when $|3G|=|G|$; a coset argument handles $3\le|3G|<|G|$; $|3G|=1$ means $G$ has exponent 3 (the first part), and $|3G|=2$ makes $G$ the direct sum of a group of exponent 3 and the group of order 2, the case excluded in (i).

Dependencies

Lemma 1 (p. 3) and Lemma 2 (p. 4) of the same paper, and the standard facts of linear algebra and set theory listed on p. 3.

Bears on

• Problem 1192: the problem asks, for each $r\ge2$, for a basis $A\subset\mathbb N$ of order $r$ with $\sum_{n\le x}f_r(n)^2\ll x$. Theorem 1 concerns infinite abelian groups with $|2G|=|G|$ and order two only; it is a group analogue of the case $r=2$ and says nothing about bases of $\mathbb N$.