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Algebra by Michael Artin Exer 3.1.7 (Actually Exer 7 of Ch3.2)

By finding primitive elements, verify that the multiplicative group $\mathbb F_p^{\times}$ is cyclic for all primes $p < 20$.

After some highly tedious manual computation (or excel/sheets, wolfram etc), we obtain that for each prime less than 20, we have, resp, the following as possible generators $1,2,3,3,7,11,11,13$. There are other generators for most of the primes less than 20, but are there patterns or properties possible generators? Like every 2p-5 is a generator of $\mathbb F_p^{\times}$ or every generator of $\mathbb F_p^{\times}$ is odd or prime if p > 5 or something.

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    There is, in general, no efficient way to find a generator. – Gerry Myerson Oct 17 '18 at 05:45
  • @GerryMyerson Thanks! ^-^ Post as answer? – BCLC Oct 17 '18 at 06:15
  • @GerryMyerson Wait, do you disagree with P Vanchinathan? Or does your 'in general' refer to all primes $p$ instead of just primes less than 20? – BCLC Oct 17 '18 at 06:35
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    See https://math.stackexchange.com/a/2512103/589 – lhf Oct 17 '18 at 10:54
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    "efficient" means "significantly better than trial and error". For $p<20$, trial and error doesn't take long. – Gerry Myerson Oct 17 '18 at 11:20
  • @lhf Thanks! ^-^ – BCLC Oct 17 '18 at 11:22
  • @GerryMyerson Ok but what is the trial and error here? My thought was to try every number in $\mathbb F_p^{\times}$ and keep multiplying it to itself to see if I can generate the whole $\mathbb F_p^{\times}$, but apparently, I can just do the primes because if a generates $\mathbb F_p^{\times}$, then $a$ is prime? I was thinking this might be relevant, but $\mathbb F_p^{\times}$ doesn't have prime order...Is Fermat's little theorem relevant here? – BCLC Oct 17 '18 at 11:24
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    No, primitive roots don't have to be primes. Probably best to have a look at the link @lhf gives, and the link you'll find there to Wikipedia. – Gerry Myerson Oct 17 '18 at 11:28
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    There are results of the form that the smallest generator is less than some fractional power of $p$, possibly times a constant, possibly including a logarithmic factor – Jyrki Lahtonen Oct 17 '18 at 11:28
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    See Wikipedia. Unconditionally, the smallest primitive root is below $Cp^{\frac14+\epsilon}$ where the factor $C$ depends on $\epsilon>0$. Conditionally to GRH the smallest primitive root is $O(\log^6p)$. – Jyrki Lahtonen Oct 17 '18 at 11:30
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    I hope it is clear that those results are highly non-trivial. – Jyrki Lahtonen Oct 17 '18 at 11:31
  • @GerryMyerson Posted an answer. So Fermat's little theorem IS relevant? I was thinking to multiply both sides of equation in FLT with $a^{-1}$, which exists for $a \not\equiv 0$. Either that or Euler's theorem (at least that's my understanding based on Wiki for Euler's theorem). – BCLC Oct 17 '18 at 11:38
  • @JyrkiLahtonen I saw those just awhile ago. Thanks! – BCLC Oct 17 '18 at 11:40

2 Answers2

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You have said there are "other generators for most primes". If by most you mean "other the primes 2 and 3" it is ok. Actually from $p\geq5$ one is guaranteed to have more than one generator. The number is given by $\phi(p-1)$.

(The calculation for $p<20$ can not be termed tedious. As one can always omit squares from consideration and as there are just 8 primes up to 20 one can get them all without software in less than 5 minutes.)

P Vanchinathan
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  • Thanks, but what about the actual generators? (Ponders more) ah, the generators must be prime? I edited question to clarify. – BCLC Oct 17 '18 at 05:16
  • Wait why can we omit squares? See. I knew there were some shortcuts or properties in doing this instead of testing out every number! – BCLC Oct 17 '18 at 11:40
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    The order being $ p- 1$ is even. So the homomorphism $x \mapsto x^2$ is not onto. So all the elements in the image subgroup consisting of squares can generate only a proper subgroup hence can't be primitive roots. – P Vanchinathan Oct 17 '18 at 11:41
  • Is this the proof of your first 2 statements? Thanks! – BCLC Oct 17 '18 at 11:44
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My paraphrasing based on this section from Wiki as pointed out as by lhf in h comment to h answer.

For a prime $p$, if the multiplicative order of a number m modulo p is equal to $\varphi \left(p\right) = p-1$ (the order of $\mathbb F_p^{\times}$), then it is a primitive root. In fact the converse is true: If m is a primitive root modulo p, where $p$ is prime, then the multiplicative order of m is $\varphi \left(p\right)=p-1$. We can use this to test for primitive roots.

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