Let $G$ be a finite group, and $S\subseteq G$ any non-empty subset. For any $g\in G$, write $Sg:=\{s\cdot g\mid s\in S\}$. I would like to show that there exists some $g\in G$ so that $|G|\cdot |Sg \cap S| \leq |Sg|\cdot |S|$.

I've tried a couple of things, like induction; it is clear when $|S|=1$, so maybe if $T=S\cup\{t\}$ then we may write $Tg\cap T=(Sg\cap S)\cup H$ where $H$ is either empty, contains $t$, contains $tg$ or contains both. But then I get kind of stuck. We have to change this $g$ so that the inequality works for $T$, but I don't really see how we can change it without destroying the nice property we had for $S$.

Another thing I tried was simply out of contradiction. So assume $|G|\cdot |Sg\cap S|>|Sg|\cdot |S|$ holds for all $g\in G$. Then maybe it would be "too much". For example, maybe this gives us some surjective map that should never be surjective. But I can't really think of anything here.

  • $\begingroup$ .....What is $Sg$? $\endgroup$ Feb 26, 2023 at 14:41
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    $\begingroup$ @SouravGhosh I suppose $Sg=\{s\cdot g \mid s \in S\}$ $\endgroup$
    – Desperado
    Feb 26, 2023 at 14:43
  • $\begingroup$ Ok. So i guess it represent cosets. $\endgroup$ Feb 26, 2023 at 14:45
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    $\begingroup$ Yes, @Desperado is right. Allow me to put it in the question $\endgroup$ Feb 26, 2023 at 14:49
  • $\begingroup$ For $S< G$, the inequality holds for every $g\in G\setminus S$. $\endgroup$
    – Kan't
    Feb 26, 2023 at 14:59

1 Answer 1


It's actually not too complicated. Note that

$$\begin{split} \sum_{g\in G}|Sg\cap S|&=\sum_{g\in G}\sum_{s\in S}[sg\in S]\\ &=\sum_{s\in S}\sum_{g\in G}[sg\in S]\\ &=\sum_{s\in S}|sG\cap S|\\ &=\sum_{s\in S}|S|\\ &=|S|^2 \end{split}$$

Now, suppose for the sake of contradiction that $|G|\cdot |Sg\cap S|>|Sg|\cdot |S|$ for all $g\in G$, then

$$|S|^2=\sum_{g\in G}|Sg\cap S|>\frac{1}{|G|}\sum_{g\in G}|Sg|\cdot |S|=\frac{1}{|G|}\sum_{g\in G}|S|\cdot |S|=|S|^2,$$

a contradiction. Thefore, there must exist some $g\in G$ such that $|G|\cdot |Sg\cap S|\leq|Sg|\cdot |S|$, as desired.

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    $\begingroup$ Thanks you so much! $\endgroup$ Feb 26, 2023 at 15:30

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