Let $p$ be a prime and $n$ a positive integer not divisible by $p$. When working on a fixed field in the cyclotomic field $Q(e^{2i\pi/n})$, I tumbled into the condition: $p$ does not divide $\frac{\phi(n)}{\mathrm{ord}(n)},$ where $\phi(n)$ is Euler's totient function, and $\mathrm{ord}(n)$ is the multiplicative order of $p$ mod $n$. That is, $\mathrm{ord}(n)$ is the smallest positive integer $k$ such that $p^k \equiv 1 (\bmod n)$. I would appreciate very much if you could provide some examples for this condition to be satisfied. A family (in either $p$ or $n$) of examples is even nicer.



Euler's totient phi and a prime
Below I write $\mathrm{ord}(x) $ for $\mathrm{ord}\_x(p)$. As usual, $\nu\_p(t)$ denotes the maximal $m$ for which $p^m$ divides $t$.


Let $n=\prod q\_i^{\alpha\_i}$ be a prime factorization of $n$. Then $\nu\_p(\phi(n))=\sum\_i \nu\_p(q\_i-1)$. On the other hand, $\mathrm{ord}(n)=\mathrm{lcm} \{\mathrm{ord}(q\_i^{\alpha\_i}):i=1,2,\ldots\}$. Since $\mathrm{ord}(q\_i^{\alpha\_i})$ divides $\phi(q\_i^{\alpha\_i})=(q\_i-1)q\_i^{\alpha\_i-1}$, we get $\nu\_p(\mathrm{ord}(n))\leqslant \max\_i \nu\_p(q\_i-1)$.


Therefore, if there exist at least two indices $i$ for which $p$ divides $q\_i-1$, we get $\nu\_p(\mathrm{ord}(n))<\nu\_p(\phi(n))$, and your condition does not hold.


If $p$ divides no $q\_i-1$, then your condition obviously holds as $p$ does not divide even $\phi(n)$ itself.


So, the interesting case to consider is when there exists unique $i$ for which $p$ divides $q\_i-1$. Say, $q\_i=1+p^Ab$, where $p$ does not divide $b$ (and $A>0$). Then your condition holds if and only if $p^A$ divides $\mathrm{ord}(q\_i^{\alpha\_i})$. Note that $\mathrm{ord}(q\_i)$ divides
$\mathrm{ord}(q\_i^{\alpha\_i})$, and the ratio is a certain power of $q\_i$ (indeed, $q\_i$ divides $p^r-1$, where $r=\mathrm{ord}(q\_i)$, hence $q\_i^2$ divides $p^{rq\_i}-1=(p^r-1)(1+p^r+\ldots+p^{r(q\_i-1)})$ and therefore $\mathrm{ord}(q\_i^2)$ divides $rq\_i$, and so on. This argument is an easy version of the so called Lifting the Exponent Lemma).


Thus, your condition holds if and only if $p^A$ divides $\mathrm{ord}(q\_i)$, in other words, if $q\_i$ does not divide $p^{(q\_i-1)/p}-1$. Or, in yet other words, if $p$ is not a $p$-th power modulo $q\_i$.


I doubt that there is a good characterization of such pairs $(p,q\_i)$. If $p=2$, this holds if and only if $2$ is not a quadratic residues modulo $q\_i$, that is, $q\_i$ is congruent to 3 or 5 modulo 8.


