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I stumbled upon one of these exercises in my textbook and thought it would be a great review for my exam. Here is the question:

Let $n$ be an odd positive integer, and let $\zeta$ be a primitive (complex) $n$-th root of unity. Show that $\zeta^2$ is also a primitive $n$-th root of unity.

This is what I know:

$\zeta$ is a primitive $n$-th root of unity where $O(\zeta) = n$. So in this case,I began to list some of the odd positive integers, where $$n = {1,3,5,7,9,...}$$

Then I considered the number $\sqrt[n]{1}$, where I can derive the appropriate roots based on the value of $n$ I choose. So as an example, $\sqrt[3]{1}$ = {$1,\omega, \omega^2$},

where $\omega$ = $-\frac{1}{2} + \frac {\sqrt{3}}2i$ and $\omega^2$ = $-\frac{1}{2} - \frac {\sqrt{3}}2i$. Is this even the right approach? I want to move on to other problems but not until I figure out what's going on here.

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  • $\begingroup$ Hint: by Bezout's identity, $\zeta^a=1=\zeta^b\implies \zeta^{\gcd(a,b)}=1$ $\endgroup$
    – lulu
    Feb 20, 2017 at 20:26
  • $\begingroup$ If $n$ is odd, $\gcd(n,2)=1$. $\endgroup$
    – Jack D'Aurizio
    Feb 20, 2017 at 20:27
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    $\begingroup$ Interestingly this means that $\zeta^{\large 2^k}$ is a primitive root for any positive integer $k$ $\endgroup$
    – Joffan
    Feb 20, 2017 at 20:27
  • $\begingroup$ Interesting. Bezout's Identity isn't even mentioned in my textbook. $\endgroup$
    – John W. Smith
    Feb 20, 2017 at 20:28

1 Answer 1

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All primitive $n$th roots of unity have the form $\zeta=\exp(\frac{2\pi ik}{n})$ with $k$ an integer such that $(k,n)=1$.

So it is enough to observe that $\zeta^2=\exp(\frac{2\pi i\cdot 2k}{n})$ and that if $k$ is prime to $n$ then so is $2k$, since $n$ is odd.

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  • $\begingroup$ Yes, that is what the others commented below my post. $\endgroup$
    – John W. Smith
    Feb 20, 2017 at 20:31
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    $\begingroup$ If it matters to you, you will note that I posted my answer 23 seconds after the first comment was posted, meaning that I started typing it before the comment appeared. $\endgroup$
    – carmichael561
    Feb 20, 2017 at 20:33
  • $\begingroup$ Ah I couldn't tell. The wifi at my university hasn't been up to par lately :/ $\endgroup$
    – John W. Smith
    Feb 20, 2017 at 20:34

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