I run, somewhat indirectly, into the following problem and I have no hints where to look in the literature in search for answers or clues.
Let $K$ be a number field, which we may assume Galois if it helps, $\cal O$ its ring of integers and for each prime number $p$ let $R_p={\cal O}/p{\cal O}$ (a finite product of finite fields of characteristic $p$ for almost all $p$). Fix $\lambda\in{\cal O}$, $\lambda\neq0$ or a root of $1$. Then $\bar\lambda\in R_p^\times$ for almost all $p$ and the period $\pi_p=\pi_p(\lambda)$ of $\lambda$ is defined as the smallest positive $d$ such that $\bar\lambda^d=1$.
It is obvious that if we fix an integer $n$ the number of $p$'s such that $\pi_p\leq n$ is finite, since $\pi_p=d$ implies that $p|(\lambda^d-1)$ and there are only finitely many of those.
On the other hand, if $n\geq2$ the number of $p$'s such that $\max\{\text{Supp}(\pi_p)\}\leq n$ (the support $\text{Supp}(N)$ of an integer $N$ is the set of prime divisors of $N$) is infinite. For instance, the set of elements $\lambda^{2^k}-1$ has an infinite set of rational prime divisors because $\lambda^{2^{k+1}}-1=(\lambda^{2^k}-1)(\lambda^{2^k}+1)$ and the only common prime divisors in $\cal O$ to the 2 factors are primes of residual characteristic 2. Thus, as k grows, a new set of primes adds up at each step, so to speak.
Now the question is: fix an arithmetic progression ${\cal P}:a,a+d,a+2d,\dots$ with $(a,d)=1$, is it true that there are infinitely many primes in $\cal P$ such that $\max\{\text{Supp}(\pi_p)\}\leq n$? Conditionally on $n$?
In particular: suppose $K$ quadratic, and let $\ell>2$ a prime. Are there infinitely many primes $p\equiv 1\bmod\ell$ such that $\max\{\text{Supp}(\pi_p)\}\leq\ell-1$?
