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p=7
G = DirichletGroup(p); G

m=3; n=ZZ((p-1)/m); print m,n

c=G[1]

c1=c^n;c1

The output is:

Dirichlet character modulo 7 of conductor 7 mapping 3 |--> zeta6 - 1

Can anyone explain what zeta6 is? Is this the Riemann-Zeta function? Is this the whole group of units? Is there a relation to the Eisenstein primes? I'm still a bit weak in this material and am having trouble grasping some of these sage outputs. Thank you in advance!

EDIT: If I take the log and then exponentiate I get, for p=7;

$$ -\frac{3i}{2}\sqrt{3} +\frac{1}{2}$$

Not sure what this is exactly.

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    $\begingroup$ the best place to ask this question is here $\endgroup$ – Masacroso Dec 5 '18 at 17:12
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    $\begingroup$ $3$ is a generator of $(\mathbb{Z}/7\mathbb{Z})^\times$ and the Dirichlet character is defined by "$\chi(3) = \zeta_6,\chi(n+7) = \chi(n)$" which implies $\chi(7n) = 0, \chi(3^l+7n) = \zeta_6^l$ where $\zeta_6 = e^{2i \pi /6} $ or any primitive $6$-th root of unity (and it is an element of the ring of Eisenstein integers). Because $(\mathbb{Z}/7\mathbb{Z})^\times$ is cyclic so is the group of Dirichlet characters modulo $7$ and $\chi$ is a generator of it. $\endgroup$ – reuns Dec 5 '18 at 18:12
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    $\begingroup$ Indeed, there is an accepted answer to this question at ask.sagemath.org/question/44593/… $\endgroup$ – kcrisman Dec 7 '18 at 18:14
  • $\begingroup$ I know. I posted the question :) $\endgroup$ – Nicklovn Dec 7 '18 at 23:20
  • $\begingroup$ I suppose you could post that answer here ... apparently you aren't supposed to close your own questions for this reason though. $\endgroup$ – kcrisman Dec 20 '18 at 12:42
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The question has been answered on Ask SageMath (by me) and also in the comments above.

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