Is there a specific set of steps that should be taken in order to the $n$-th root of unity in a cyclic group. To be more specific, I am trying to find the $8$th root of unity for $\mathbb{Z}_{17}^*$. I have already found that $3$ is a generator/primitive root of $\mathbb{Z}_{17}^*$ but I cannot see exactly how this helps find the primitive root.
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$\begingroup$ For any cyclic group or just $(\mathbb{Z}/n)^x$? $\endgroup$– AnotherPersonDec 14, 2015 at 0:18
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$\begingroup$ I was looking at $\mathbb{Z}_{17}^*$ or $\mathbb{Z}_{p}^*$ specifically. Is there are generalized approach? $\endgroup$– dreaminDec 14, 2015 at 0:21
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$\begingroup$ And you are looking to find the generators of the group? $\endgroup$– AnotherPersonDec 14, 2015 at 0:22
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$\begingroup$ I know how to find the generators. I don't know how to find the nth roots of unity or exactly what it is. $\endgroup$– dreaminDec 14, 2015 at 0:23
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$\begingroup$ Well if you don't know what it is then how can you know what to look for? $\endgroup$– AnotherPersonDec 14, 2015 at 0:23
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1 Answer
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You are trying to find the elements $x\in\mathbb{Z}_{17}^*$ such that $x^8=1$. In general if you have the equation $x^n=1$ in $\mathbb{Z}_p$ with $p$ prime, there are exactly $d=gcd(n,p−1)$ solutions. One solution is $s=g^{(p-1)/d}$ where $g$ is a primitive element. Then the complete system of solutions is $$\{s^0,s^1,s^2,\cdots,s^{d-1}\}$$ In this case, $p=17$, $d=gcd(8,17-1)=gcd(8,16)=8$ and $g=3$. Then $s=3^2=9$ and the solutions are
- $s^0=1$
- $s^1=9$
- $s^2=9^2=13$
- $s^3=9^3=15$
- $s^4=9^4=16$
- $s^5=9^5=8$
- $s^6=9^6=4$
- $s^7=9^7=2$
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$\begingroup$ So if I am understanding correctly, the nth root of unity can be given by s = g^{p-1/d} once you have a primitive root of Z and also that for this particular problem there are 8 solutions? $\endgroup$– dreaminDec 14, 2015 at 1:07
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