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Let $p$ be an odd prime. Is it true that there exists a permutation $\sigma$ of the set $$ \{1,\ldots,2p-1\}\setminus \{p\} $$ such that $$ \{\sigma(1),\ldots,(p-1)\sigma(p-1)\}=\{\sigma(p+1),\ldots,(p-1)\sigma(2p-1)\}=\{1,\ldots,p-1\} $$ in $\mathbf{Z}/p\mathbf{Z}$? [The answer is positive for $p \in \{3,5\}$]

An easier version of the problem asks for the existence of a permutation $\mu$ of $\{1,\ldots,p-1\}$ such that $\{\mu(1),\ldots,(p-1)\mu(p-1)\}=\{1,\ldots,p-1\}$ in $\mathbf{Z}/p\mathbf{Z}$.

In this case, the answer is negative since the products of the elements of each set would be the same modulo $p$, which is false by Wilson's theorem (indeed we would have $1\equiv -1\pmod{p}$ for an odd prime $p$). Note that the same method does not apply to the above problem.

On the other hand, an attempt would be: the product of those three sets are the same modulo $p$, hence (here we don't even need Wilson's theorem) we ask: "Does there exist a partition of $\{1,\ldots,2p-1\}\setminus \{p\}$ in two sets with $p-1$ elements such that the product of the elements of each set is $1$ modulo $p$? [Edit: the answer is positive for $p \in \{3,5,7\}$]

The motivation for this question is related to this thread and this post.

Edit: Another variant here.

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Paolo Leonetti
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  • In other words, when $C_n$ denotes a cyclic group, is there a permutation $f$ of $C_{p-1}\times{1,2}$ such that $$ x \mapsto x + f(x,a) $$ is a permutation of $C_{p-1}$ for each $a\in{1,2}$? – hmakholm left over Monica Aug 20 '16 at 10:07
  • Do you mean $x\mapsto xf(x,a)$? – Paolo Leonetti Aug 20 '16 at 10:13
  • No, I'm writing the cyclic group additively. But I'm missing a projection function (and now I've run out of time to edit the comment). – hmakholm left over Monica Aug 20 '16 at 10:14
  • Written additively, yes; anyway, don't worry, it is clear what you mean – Paolo Leonetti Aug 20 '16 at 10:17
  • We could also generalize to: Given a finite group $G$, let $h(G)$ be the smallest (nonzero) size of a family $\mathscr F$ of functions $G\to G$ such that $g\mapsto gf(g)$ is a permutation of $G$ for every $f\in\mathscr F$, and $\sum_{f\in\mathscr F}|f^{-1}{g}| = |\mathscr F|$ for every $g\in G$. Your argument is then that $h(C_{p-1}) > 1$, and you conjecture that it is $2$. – hmakholm left over Monica Aug 20 '16 at 10:28
  • If I understood correctly, you mean $|\bigcup_{f \in \mathscr{F}}f^{-1}{g}|=|\mathscr{F}|$, right? About the underlying group, isn't $((\mathbf{Z}/p\mathbf{Z})^\star,\cdot)$? – Paolo Leonetti Aug 20 '16 at 11:42
  • x @Paolo: no, because if two of the $f$s map the same $g_1$ to $g_2$, that should count for two, so the preimages need to be counted before you sum the counts. As for the underlying group, the multiplicative group modulo a prime is always cyclic, and it seems to be easier to think of the isomorphic group $(\mathbb Z/(p-1)\mathbb Z, +)$, such that one is not distracted by the irrelevant additive structure of $\mathbb Z/p\mathbb Z$. – hmakholm left over Monica Aug 20 '16 at 11:53
  • Ok I see now, you are right. Intuitively, the answer should be positive even if we replace that $p-1$ with an arbitrary even (positive) integer. – Paolo Leonetti Aug 20 '16 at 12:00

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Based on the above comments, we can prove the answer is always positive. Indeed the question simplifies to the existence of a permutation $\sigma$ of $\mathbf{Z}_{p-1} \times \{1,2\}$ (with $\mathbf{Z}_{p-1}$ written additively) such that $x\mapsto x+\sigma(x,a)$ is a permutation of $\mathbf{Z}_{p-1}$ for every $a \in \{1,2\}$ (with the notation of Henning).

The answer is positive also replacing $p-1$ with an even positive integer $2n$. Indeed it is enough to set $$ \sigma(i,1)=(i\bmod{n}) \text{ }\text{ }\text{ and }\text{ }\text{ }\sigma(i,2)=n+(i\bmod{n}) $$ for all $i=1,\ldots,2n$.

Paolo Leonetti
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