Let $p \ge 11$ be a prime number, $k,n$ be positive integers such that $n|gcd(p-1,k-1)$ and $p > k > n \ge 5$. Let $s \in \mathbb Z\_p$ such that $ord\_p(s) = n$. Is it possible that the sets $A = \{1,2,3,\dots,k-1\}$ and $B = \{k,k+1,k+2,\dots,p-1\}$ of classes modulo $p$ satisfy $sA = A$ and $sB = B$, simultaneously? I mean, are $A$ and $B$ invariants by multiplication by $s$?


Certainly, if there exist a counter-example then $n$ must be odd. In fact, if $n$ is even then $s^{n/2} \equiv -1 \pmod p$.


We may take $2 \le s \le \min\{k-1,p-k\}$ (otherwise $1\cdot s \not\in A$ or $-1\cdot s \not\in B$. Also, wlog we may assume $k \le (p-1)/2$.


Thanks!



Invariant complementary sets modulo $p$
No. 


I'm going to use $\bar{s} \in \mathbb{Z}\_p, s \in \mathbb{Z}$ its minimal positive representative. 


We can assume WLOG that the initial set we're looking at has size less than $\frac{p - 1}{2}$ as $\{k, k + 1, ..., p - 1\}$ is $s$-invariant if and only if $\{1, 2, ..., p - k\}$ is. Then $s \leq k - 1 \leq \frac{p - 1}{2}$. We also have that $1 = \bar{s}a$ for some $a \in A$. As $\bar{s} \neq 1$, $a \neq 1$, so $a - 1 \in A$. Therefore, $\bar{s} (a - 1) = \bar{s}a - \bar{s} = 1 - \bar{s} \in A$. But $s \leq \frac{p - 1}{2}$, so $1 - \bar{s}$ has a minimal representative that is larger than $\frac{p - 1}{2}$. By the definition of $A$ and the fact that $k - 1 \leq \frac{p - 1}{2}$, this cannot be in $A$ - so we have a contradiction, and are done. 


