# Issue with $p$-th root of unity

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.

## 1 Answer

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.

• 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. – John Smith Feb 8 '17 at 23:37
• As in the first displayed equation, $k$ is an integer from $0$ to $p-1$. – David Feb 8 '17 at 23:47
• Okay just double-checking.Other than that, understand it now. – John Smith Feb 8 '17 at 23:48