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Let $G$ be an abelian group and $A\subseteq G$. Suppose there's a finite set $F\subseteq G$ such that: $$G=FA$$

How can I prove any infinite translation of $A$ is overlapping, that is, there's not any infinite set $C\subseteq G$, such that for each $x,y\in C$,

$$x\ne y \quad \leftrightarrow\quad xA \cap yA\ne \emptyset$$ ?

Edit: (according to comments below) More generally we can say if for some $A_1,...,A_n\subseteq G$, $$A=\bigcup_{k=1}^nA_k$$ then there's some $k$ such that any infinite translation of $A_k$ is overlapping.

The proof suggested below includes existence of a Banach measure. I wonder if there's any elementary proof for this.

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Is $A$ a subgroup or just a subset? –  Derek Holt Apr 17 '13 at 13:30
Do you mean $xA\cap yA\neq\emptyset$ instead of $xA\neq yA$? (Otherwise take $G=\mathbb{Z}$ and $A=\mathbb{Z}\backslash\{0\}$.) –  Sean Eberhard Apr 17 '13 at 13:46

1 Answer 1

up vote 4 down vote accepted

I'll switch to additive notation, because this problem becomes easier if we remind ourselves that $G$ is abelian as often as possible.

Suppose that $G=F+A$, where $G$ is abelian and $F$ is finite. Suppose $|F'|>|F|$ and consider the translates of $A$ by $F'$. I claim they overlap somewhere.

Since $-F'\subset G = F+A$, by the pigeonhole principle there exists $f_1,f_2\in F'$ such that $-f_1,-f_2\in f+ A$ for the same $f\in F$, so $-f\in (f_1+A)\cap(f_2+A)$.

Adding the observation of Alex Ravsky, the proposition clearly extends to the class of all amenable groups, which includes in particular all solvable groups. The proof in this general case is quite obvious and intuitive: if $\mu$ is a left-invariant mean on $G$, and $A$ has $n$ pairwise disjoint translates, then $\mu(A)\leq \frac{1}{n}$, so certainly no fewer than $n$ translates of $A$ suffice to cover $G$.

With this perspective we can answer OP's harder question affirmatively: Given a subset $A=\cup_{k=1}^n A_k$ of an amenable group $G$ and a finite set $F$ such that $G=FA$, there exists $k$ such that no more than $n|F|$ left-translates of $A_k$ can be pairwise disjoint. Indeed, $\mu(A)\geq\frac{1}{|F|}$, so for some $k$ we must have $\mu(A_k)\geq \frac{1}{n|F|}$, and then clearly any $n|F|+1$ translates of $A_k$ are not disjoint.

On the other hand, the claim does not hold in the class of all groups, as the group $G=\mathbf{Z}\ast C_2 = \langle x,y\,|\,y^2=e\rangle$ shows: Take $A$ to be all words beginning with $y$. Then $G=A\cup yA$, but $A,xA,x^2A,\ldots$ are all disjoint.

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I think the assumption that $G$ is abelian is necessary. For instance, consider the free product $G=\mathbf{Z}\ast C_2 = \langle x,y | y^2 = e\rangle$. Let $A$ be the set of all words beginning $y$. Then $A\cup yA = G$, but $A, xA, x^2A,\dots$ are pairwise disjoint. –  Sean Eberhard Apr 17 '13 at 14:37
Could you remind me what the more general question was? –  Sean Eberhard Apr 17 '13 at 14:54
It seems that a natural extension to a more general case, for which the more general question should have a positive answer, can be the case of so-called amenable groups. A group $G$ is amenable if it admits a left-invariant finitely additive probability measure $\mu:\mathcal P(G)\to[0,1]$ defined on the Boolean algebra $\mathcal P(G)$ of all subsets of $G$. It is well-known that each abelian group is amenable. –  Alex Ravsky Apr 17 '13 at 15:23
A subset $A$ of a group $G$ is called large (or left large), if there is a finite subset $F$ of $G$ such that $FA=G$. One can find some results about large sets and partitions of groups in articles by our school. For instance, in the paper by Taras Banakh, Sergiy Slobodianiuk and me “On partitions of G-spaces and G-lattices” and in the paper by Taras Banakh “The Solecki submeasures and densities on groups”. –  Alex Ravsky Apr 17 '13 at 15:23
@CutieKrait True, but still it's perhaps not so hard. Here are some notes giving an introduction to amenability, including a proof that all abelian groups are amenable: dpmms.cam.ac.uk/~bjg23/ATG/Chapter3.pdf. –  Sean Eberhard Apr 17 '13 at 16:28

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