How to show that if $p$ is a prime number, then every imaginary $p$-th root of unity is necessarily a primitive $p$-th root of unity.

I am really clueless at how to even approach this problem. I've tried selecting a prime number, say $5$, and listing it's primitive elements but that idea isn't going anywhere.


The $p$th roots of unity are solutions of $z^p=1$, that is, $$z=e^{2k\pi i/p}\ ,\quad k=0,1,2,\ldots,p-1\ .$$ If $p$ is odd, then all of these are imaginary except the one with $k=0$. The root $z$ is a primitive $p$th root of unity if the smallest positive integer $n$ such that $z^n=1$ is $n=p$. If $k=1,2,\ldots,p-1$, we have $$\eqalign{z^n=1\quad &\Leftrightarrow\quad e^{2kn\pi i/p}=1\cr &\Leftrightarrow\quad \frac{kn}p\ \hbox{is an integer}\cr &\Leftrightarrow\quad p\mid kn\cr &\Leftrightarrow\quad p\mid n\quad \hbox{since $p$ is prime and $p\not\mid k$}\ .\cr}$$ So the smallest such $n$ is $n=p$, and this shows that $z$ is a primitive $p$th root of unity.

For the case $p=2$ there are no imaginary $p$th roots of unity, so the statement is vacuously true.

  • $\begingroup$ Okay this makes sense but is $k$ an integer in your proof? I haven't had much exposure to the n-th root of unity we just started talking about it today in my class. $\endgroup$ – John Smith Feb 8 '17 at 23:37
  • 1
    $\begingroup$ As in the first displayed equation, $k$ is an integer from $0$ to $p-1$. $\endgroup$ – David Feb 8 '17 at 23:47
  • $\begingroup$ Okay just double-checking.Other than that, understand it now. $\endgroup$ – John Smith Feb 8 '17 at 23:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.