# If $G$ is a finite group, show that there is some $g\in G$ so that $|G|\cdot |Sg \cap S| \leq |Sg|\cdot |S|$

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.

• .....What is $Sg$? Feb 26, 2023 at 14:41
• @SouravGhosh I suppose $Sg=\{s\cdot g \mid s \in S\}$ Feb 26, 2023 at 14:43
• Ok. So i guess it represent cosets. Feb 26, 2023 at 14:45
• Yes, @Desperado is right. Allow me to put it in the question Feb 26, 2023 at 14:49
• For $S< G$, the inequality holds for every $g\in G\setminus S$. Feb 26, 2023 at 14:59

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