# A new combinatorial problem for finite groups

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 $$0<|A|=n, then we may write $$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 $$n+1$$.

• 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). – M. Farrokhi D. G. Dec 8 '18 at 10:22