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For $\mathbb{Z}/13$, I want to find its primitive roots and 4th roots of unity.

For $g$ to be a primitive root, we must have that $g^6 \neq 1 \pmod{13}$ and $g^4 \neq 1 \pmod{13}$. $2$ satisfies this. Do I just go through all the numbers $0,1,2, \cdots, 12$ and check this? I'm not too clear what a primitive root is.

Then for the 4th roots of unity, do we just go through each number raised to the fourth power and see what gets us $1$?

So, $1,5, 8, 12$ are the 4th roots of unity.

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  • $\begingroup$ The command \bmod (or \pmod, depending on your preferences) renders the mod notation nicely. $\endgroup$ – Travis Mar 2 '15 at 6:14
  • $\begingroup$ Thank you. I've edited the post appropriately. $\endgroup$ – aldnoah.Algebra Mar 2 '15 at 6:21
  • $\begingroup$ Note that if $g$ is a primitive root then $g^3$ and its powers are 4th roots of unity. $\endgroup$ – Gerry Myerson Mar 2 '15 at 6:27
  • $\begingroup$ Yes, this is a very small field, you can check the 12 non-zero elements by hand, just multiply and reduce mod 13 as you go. A primitive root of unity for $\mathbb Z/13\mathbb Z$ is one where its powers give you all 12 non-zero elements. It's enough to raise it to the 6th power, if none of those are one then it must be primitive. $\endgroup$ – Gregory Grant Mar 2 '15 at 6:33
  • $\begingroup$ if $(i,\phi(n))=1$and g is a primitive roots $g^i$is primitive roots and they are all primitve roots $\endgroup$ – ali Mar 2 '15 at 18:14

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