2

One can see that $$\mathbb{Z}/19\mathbb{Z}=\{[0],[1],\ldots,[18]\}=\{[0],[3^1],[3^2],\ldots,[3^{18}]\}.$$ If I have $\mathbb{Z}/p\mathbb{Z}$, $p>2$ prime, when is there a $k\in\{2,3,\ldots,p-1\}$ such that $[k ^n]$ and $[0]$ recover $\mathbb{Z}/p\mathbb{Z}$? Further, when is there a prime $k$?

I'm a little removed from my last algebra course, but this interested me and I couldn't come up with a solution offhand.

Edit: nonessential to answer, but appreciated: Regarding when $k$ is prime, are there any partial results for any special classes of primes $p$?

GPhys
  • 1,570
  • For your last question, see Greg Martin's paper linked in this answer. As long as $p$ is large enough and none of the L-functions of Dirichlet characters mod $p$ have zeroes of small height far away from the critical line, then there exists a rather small (less than some power of $\log p$) prime $k$ which works. In particular, GRH would imply this for all sufficiently large primes. – Erick Wong Jun 21 '17 at 16:33

2 Answers2

3

The group $(\mathbb{Z}/p\mathbb{Z})^\times$ of non-zero elements of $\mathbb{Z}/p\mathbb{Z}$ is cyclic, which means there always exists an element $k\in\{1,\ldots,p-1\}$ whose powers, along with $[0]$, give every element of $\mathbb{Z}/p\mathbb{Z}$. Such a $k$ is called a primitive root modulo $p$. The number of primitive roots is $\varphi(p-1)$, the number of integers between $1$ and $p-1$ that are coprime to $p-1$.

Regarding trying to find a prime $k$ that does the job, this question gives an argument that the answer is probably yes, but that proving this might be very hard.

Community
  • 1
User
  • 792
2

What you are really asking is "Can I find a generator for $(\Bbb Z/p\Bbb Z)^{\ast}$?"

The answer is "always", but there is no "general formula" for "how", and such a generator need not be some prime less than $p$.

The question of "if" there is always such a prime, is to the best of my knowledge, still open, but there "probably is".

As strange as it seems, trial-and-error is still the best method for investigating this question for any given $p$.

David Wheeler
  • 15,842
  • 1
  • 27
  • 39
  • Sure, but this prime is not necessarily less than $n$. – David Wheeler Sep 24 '15 at 01:15
  • That there is such a prime (between $1$ and $p$ whose residue class generates the units modulo $p$). Feel free to edit my response if you feel this needs clarification. – David Wheeler Sep 24 '15 at 01:19
  • Ah, your second line is saying a given generator needn't be prime, and the third is saying it might be open if there is always a generator that is prime ($<n$). Nevermind. – anon Sep 24 '15 at 01:22
  • I understand that the sequence $(a_k): a_k = a_0 + kp$ contains infinitely many (and thus at least one, right?) primes. This is the theorem you refer to yes? – David Wheeler Sep 24 '15 at 01:25
  • Yes, that is Dirichlet's theorem that I referred to. – anon Sep 24 '15 at 03:20