6
$\begingroup$

A subset $S$ of an abelian group is a subgroup if and only if it is closed under taking differences; that is, the difference of any two elements of $S$ is in $S$. Suppose, however, that we only know that among any three elements of $S$, there are two the difference of which is in $S$; what can $S$ be in this case?

With a specific application in mind, I assume that $S$ is finite, symmetric, and contains $0$: $$ 0\in S=-S,\ |S|<\infty. $$

As an example, $S$ can be a symmetric coset progression, which is a set of the form $$ -ng+K,\dotsc,-g+K,K,g+K,\dotsc,ng+K, $$ where $K$ is a subgroup, and $g$ is a group element of order $2n+1$ at least.

A slightly different construction (suggested by Thomas Bloom's comment below): $$ S = \{0,g_1,g_2\}+H, $$ where $\mathrm{ord}(g_1)=\mathrm{ord}(g_2)=2$, $g_1\ne g_2$, and $H$ is a finite subgroup with $g_1,g_2,g_1+g_2\notin H$.

As an easy exercise, if in this case the underlying group is torsion-free, then $S$ is a symmetric arithmetic progression; hence, a coset progression.

The general problem is: What can the structure of $S$ be given that among any $N$ elements of $S$, there are two elements the difference of which is in $S$? (A multidimensional coset progression?)

Improve this question
$\endgroup$
4
  • $\begingroup$ Do you know of any construction in e.g. $\mathbb{F}_2^n$ which is not a coset progression? (I assume not or you would have mentioned it, just to clarify). $\endgroup$
    – Thomas Bloom
    Commented Sep 25, 2019 at 17:03
  • 1
    $\begingroup$ @ThomasBloom: not that I thought of it much, but I think something special can be invented for $\mathbb F_2^n$; say, the set of all vectors with at least one of the first two coordinates equal to $0$. $\endgroup$
    – Seva
    Commented Sep 25, 2019 at 17:45
  • $\begingroup$ Just to clarify: when you test on a given triple $(s_1,s_2,s_3)$, are you assuming that these elements are distinct? $\endgroup$
    – Yemon Choi
    Commented Sep 25, 2019 at 21:58
  • 1
    $\begingroup$ @YemonChoi: not necessarily, but if they are not pairwise distinct, then the property holds in a trivial way in view of the assumption $0\in S$. $\endgroup$
    – Seva
    Commented Sep 26, 2019 at 4:33

0

Reset to default

You must log in to answer this question.