Proof that the order of a cyclic rotation of $p$ elements is always $1$ or $p$ for $p$ prime

Question

By a "cyclic rotation" $F$ of a $p$-tuple I mean: $$ F(a_1, a_2, ..., a_p) = (a_2, a_3, ..., a_p, a_1) $$ This definition is motivated by McKay's proof of Cauchy's theorem (e.g. Gone's answer). A consequence of that lemma in that answer is that if $p$ is prime then the order of $F$ on a given tuple is always $1$ or $p$. I was wondering if there was another proof of that fact, using non group-theoretic results preferably. Because to prove that, it was needed the lemma above, and that lemma above uses $\langle a^k \rangle = \langle a^{gcd(n,k)} \rangle$ which makes the result 'unintuitive' at least to me. There just 'seems' that a right combination of $a_i = a_j$ should make it work for some $F^k$, $k<p$.

Answer 1

If you think of this cyclic permutation as adding 1 in modular arithmetic/mod p, then this is a field $\mathbb{F}_p$ if and only if $p$ is prime. If $p$ is not prime, then there is a nontrivial ‘zero divider’ (a number which multiplies to zero, eg $2\cdot 3=6\cong 0\in\mathbb{F}_p$), which then leads to a smaller number of permutations to get back to the initial state than $p$. Thus $p$ being prime means there are no non-trivial zero-dividers, so the only ways to permits back to the original tuple is to permute no times (order 1) or to permute $p$ times and cycle all the way back round (order p). Since the order of a cyclic rotation is the number of times to return to the original tuple, this explains why. I hope this helped! Stay safe