In a recent preprint arXiv:1811.10503, I proved that if $a_1,\ldots,a_n$ are distinct elements of a torsion-free additive abelian group $G$, then there is a permutation $\pi\in S_n$ such that all those elements $ka_{\pi(k)}\ (k=1,\ldots,n)$ are pairwise distinct. I also showed that for any odd prime $p$ there is no permutation $\pi$ such that all the numbers $k\pi(k)\ (k=1,\ldots,p-1)$ are pairwise incongruent modulo $p$. In view of this, it is natural for me to pose the following new combinatorial problem for finite groups.

Conjecture. Let $G$ be a finite multiplicative group with $|G|>1$, and let $p(G)$ be the smallest prime factor of $|G|$. If $A$ is a subset of $G$ with $0<|A|=n<p(G)-1$, then we may write $A=\{a_1,\ldots,a_n\}$ such that all those $$a_1,\ a_2^2,\ \ldots,\ a_n^n$$ are pairwise distinct.

I have confirmed this for $n\le3$. For $n=4,5,6,7,8,9$, I have verified the conjecture for cyclic groups $G$ with $|G|$ not exceeding $$100,\ 100,\ 70,\ 60,\ 30,\ 30$$ respectively.

I'm confident that my above conjecture should be true (at least for finite abelian groups). The problem looks quite challenging. Any ideas towards its solution?

PS: If we want to extend the conjecture to include infinite groups, then I think it suffices that $G$ has no non-identity element of order not exceeding $n+1$.

  • $\begingroup$ Since $n<p$, the element $g^i$ generates the same cyclic group as does $g$ for all $g\in G$ and $1\leq i\leq n$. Thus, the set $A$ above can be partitioned into sets each of which is a set of generators of a cyclic group. Hence, the problem reduces to cyclic groups only. In this case, the result is true when $n/p$ is sufficiently small (since it is true for the infinite cyclic group). $\endgroup$ – M. Farrokhi D. G. Dec 8 '18 at 10:22

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