Let $K$ be a finite field with $p^n$ elements. Then the group of unit is cyclic of order $p^n-1$. What are the primes which are dividing $p^n-1$? What are the Sylow-subgroups of the group of units of $K$? Of course they are cyclic but of which order?
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1$\begingroup$ Sometimes (p-1)/2 is a prime, sometimes it is not. Conjecturally, the former happens infinitely often. There is no universal answer. Probably you should explain what you are looking for, and why, so that we can help. Are you interested in a worst case? Then define "worst". Are you interested in average-case behavior, then what are you averaging over, p or n, etc. $\endgroup$– Boris BukhCommented Oct 31, 2014 at 14:58
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$\begingroup$ I think both questions are well formulated. So the answer is what Peter has written already: no answer in general. Thanks:-) $\endgroup$– Sven WirsingCommented Oct 31, 2014 at 16:15
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$\begingroup$ Zsigmondy's theorem shows that, for fixed p and any value of n, there exists a prime dividing $p^n-1$ which does not divide $p^m-1$ for any $m < n$. en.wikipedia.org/wiki/Zsigmondy's_theorem $\endgroup$– Padraig Ó CatháinCommented Nov 1, 2014 at 1:51
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$\begingroup$ Open problem: so the question is answered! $\endgroup$– Sven WirsingCommented Nov 3, 2014 at 8:01
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1 Answer
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What kind of answer do you expect? Of course, not much about the prime factorization of $p^n-1$ can be said, for otherwise one could settle questions about Mersenne primes $2^q-1\in\mathbb P$ ($q$ necessarily a prime). One could even settle the existence question of Fermat primes $F_k=2^{2^k}+1$, if we knew the prime factorization of $\prod_{k=0}^nF_k=2^{2^{n+1}}-1$.
As to your general question, note that $p^n-1=\prod_{k\mid n}\Phi_k(p)$, where $\Phi_k$ is the $k$-th cylotomic polynomial. So one gets some little information about prime factors, as each prime divisor $q$ of $\Phi_k(p)$ divides $k$ or fulfills $q\equiv1\pmod{k}$.
answered Oct 31, 2014 at 9:24
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$\begingroup$ Thank you Peter, I was thinking about this topic and thought I did not see something obvoius but thanks than for this background information. So its still an open topic in algebra:-) $\endgroup$ Commented Oct 31, 2014 at 11:32