$\newcommand{\Z}{\mathbb Z/p\mathbb Z}$ Can one partition a group of prime order as $A\cup(A-A)$ where $A$ is a subset of the group, $A-A$ is the set of all differences $a'-a''$ with $a',a''\in A$, and the union is disjoint?

As stated, the answer is "yes", at least if the order of the group is $p\equiv 2\pmod 3$, in which case one can take $A$ to be an appropriately located interval of an appropriate length: namely, $A=[n,2n-1]$ where $n=(p+1)/3$. One also can dilate $A$ replacing intervals with arithmetic progressions.

Are there any other examples where $A$ and $A-A$ partition the whole group?

Suppose that $A\cup(A-A)$ is a partition of a group of prime order; does it follow that $A$ is an arithmetic progression?

Added two days later. A set found by Peter Mueller in the comments can be generalized to produce infinitely many counterexamples, essentially answering the original question. Specifically, for a prime $p\equiv 5\pmod 8$ let $m:=(p+3)/8$ and define $A:=[-(2m-1),-m]\cup[m,2m-1]$. It is easily verified that then $A-A=[-(m-1),m-1]\cup[2m,p-2m]$, so that $A-A$ is disjoint from $A$, and $A\cup(A-A)$ is the whole group, while $A$ is not an arithmetic progression.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. It would be good if someone wrote an answer based on this discussion. $\endgroup$
    – Ben Webster
    Dec 12, 2022 at 0:22
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    $\begingroup$ @BenWebster I disagree with this removing of comments, which does not correspond to the usual MO policy and would appreciate if you take this (old) metaMO post in consideration meta.mathoverflow.net/questions/501/cleaning-up-comments $\endgroup$
    – YCor
    Dec 12, 2022 at 1:16
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    $\begingroup$ @YCor I appreciate the feedback. I had missed the more recent meta post about this (meta.mathoverflow.net/questions/5033/…); given the vote results there, I will be more selective about moving to chat in the future. $\endgroup$
    – Ben Webster
    Dec 14, 2022 at 17:57

2 Answers 2


This is a previous comment which was moved to chat by Ben Webster: In fact for every prime $p$ if $A=[-(2m-1),-m] \cup [m, 2m-1]$ for $(p+3)/8\le m<(p+3)/6$, then $\mathbb Z/p\mathbb Z$ is a disjoint union of $A$ and $A-A$, and in addition this set $A$ is not an arithmetic progression if and only if $m\ne(p+1)/6$. There are many more examples though. One peculiar sporadic one happens for $p=41$ with $A$ the multiplicative subgroup of order $10$ of $(\mathbb Z/p\mathbb Z)^\star$.


It seemed to me that the commenters know much more about this problem than they write in the comments. For that reason alone I am posting this answer, which is most likely a long comment.

For brevity, we call a symmetric set $A\subset G$ ($G$ is an abelian group) a partitioning set if $G=A\cup(A+A)$ and $A\cap(A+A)=\varnothing$.

The definition from the question (if it were formulated there) is equivalent to this since $A-A$ is symmetric, and then $A$ is also symmetric.

And we also call two partitioning sets $A$ and $B$ of $G$ equivalent if $B=\alpha(A)$ for some $\alpha\in\operatorname{Aut}(G)$.

Everywhere below $G=\mathbb{Z}_p$ for some simple $p$. The conjecture in one of the last comments can be phrased like this: If $A\subset\mathbb{Z}_p$ is a partitioning set, then some equivalent of it lies in $\{-p'd,\ldots,-d,d,\ldots,p'd\}$ where $p'=[p/4]$ and $d$ is a nonzero group element.

For example, the Peter Mueller set for $p=29$ is equivalent to $[4,7]\cup[-7,-4]$ and the set for $p=13$ is equivalent to $[2,3]\cup[-3,-2]$.

Partitioning sets for some other prime: $p=11$, $[2,3]\cup[-3,-2]$; $p=17$ and $p=19$, $[3,5]\cup[-5,-3]$; $p=23$, $[4,7]\cup[-7,-4]$.

In fact, the following statement is true.

Let $p>7$ be a prime. If $p=8m\pm1$ or $p=8m+3$ or $p=8m+5$ and $P=[m+1,2m+1]$, then $A=P\cup-P$ is a partitioning set in $\mathbb{Z}_p$.

And one more remark. For $p=23$ we have a partitioning set $A=[-7,-4]\cup[4,7]$. This set is equivalent to the set $B=[-7,-5,-3,-1,1,3,5,7]$, which is an arithmetic progression.


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