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Let $G$ be a group and let $S$ be a subset of $n$ distinct elements of $G$ with the property that $a\in S$ implies $a^{-1}\not\in S$. Consider the $n^2$ products (not necessarily distinct) of the form $ab$, where $a\in S$ and $b\in S$. Prove that at most $n(n-1)/2$ of these products belong to $S$.


I have been thinking about this problem for two days, I have done nothing about the proof, but just exploring some examples.
I consider the group $G=\langle\mathbb Z_{11}, \cdot\rangle$, and let $$S=\{2,3,5,7\}$$ which clearly satisfy the property.
Then I tested it, in $16$ of the numbers, exactly $6$ of them are in $S$.
Any suggestion for how to prove it?

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Let $T$ be the set of pairs $(a,b)\in S^2$ such that $ab\in S$. Let $U$ be the set of pairs $(a,b)\in S^2$ such that $b\in aS$.

Define the permutation $q:G^2\to G^2$ by $q(a,b)=(a,ab)$. Then $q(T)=U$.

By the assumption $S\cap S^{-1}=\emptyset$, we have $(a,a)\notin U$ and $(a,b)\in U$ implies $(b,a)\notin U$. Hence $|U|\le n(n-1)/2$. Since $|U|=|T|$ it follows that $|T|\le n(n-1)/2$.

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  • $\begingroup$ Your solution really amaze me, how can you figure out the definitions of $T$ and $U$? $\endgroup$ – kelvin hong 方 Feb 14 '19 at 7:57
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    $\begingroup$ The definition of $T$ was implicit in your question and explicit in the other tentative answer. The definition of $U$ is quite natural when familiar with Schreier graphs and partial actions. Outputting $n(n-1)/2$ suggested finding an oriented graph structure, and oriented comes naturally with the assumption on $S$. $\endgroup$ – YCor Feb 14 '19 at 8:00

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