5
$\begingroup$

Show that the primitive n-th roots of unity have the form $e^{2ki\pi/n}$ for $k,n$ coprime for $0\leq k\leq n$.

Since all primitive n-th roots of unity are n-th roots of unity by definition they all have that form, the question is, how to show $k$ and $n$ are coprime.

$\endgroup$
2
  • 4
    $\begingroup$ Put another way: if $k$ and $n$ are not coprime, then is $e^{2\pi i k/n}$ a primitive $n$-th root of $1$? What does being a primitive $n$-th root mean? $\endgroup$ – Dylan Moreland Jan 21 '12 at 21:29
  • 1
    $\begingroup$ please say the definition of primitive root. thanks. $\endgroup$ – user97295 Sep 27 '13 at 14:14
5
$\begingroup$

Primitivity means that no positive power of $\zeta=e^{2\pi i k/n}$ less than $n$ will achieve unity. If $k$ is not coprime to $n$ and $\gcd(k,n)=m$, then observe $\zeta^{(n/m)}=e^{2\pi i (k/m)}=1$ but $n/m<n$ if $m>1$.

$\endgroup$
2
$\begingroup$

More generally, if $G$ is a cyclic group of order $n$ and $g$ is one of its generators, then the order of $g^k$ is $n/(k,n)$. In particular, the generators are $g^k$ with $(k,n)=1$.

This is easy to prove: $t$ is the order of $g^k$ iff $t$ is the smallest number such that $1=(g^k)^t=g^{kt}$ iff $kt$ is the smallest multiple of $k$ that is also a multiple of $n$ iff $kt=\operatorname{lcm}(k,n) = kn/(k,n)$ iff $t=n/(k,n)$.

$\endgroup$
1
$\begingroup$

Just to help your intuition in support of @lhf's and @anon's answers...

Are you aware of the formula (for positive integers $n$) $$ n = \sum_{0<d|n}\phi(d) $$ where $\phi(n)$ is Euler's totient function? It's basically giving you a breakdown of how many $n$th roots of unity there are which are primitive $d$th roots of unity for each $d$ dividing $n$ (there are $\phi(d)$).

For example, of the six sixth roots of unity, one has order $1$, one order $2$, two have order $3$, and two have order $6$: $\phi(1)=\phi(2)=1$ & $\phi(3)=\phi(6)=2$. Try drawing them on the unit circle.

So with this intuition, we need to show that the order $d$ of $\zeta=e^{2\pi ik/n}$ is $n$ iff $k$ & $n$ are relatively prime, i.e. iff their gcd $g=(k,n)$ is $1$. However, note that for $t=n/g$, $\zeta^t=e^{2\pi ik/g}=1$ since $g|k$, so that $g>1$ implies that the $ord(\zeta)|t<n$, i.e. that $\zeta$ is not a primitive $n$th root but rather a primitive $t$th root of unity, where $t$ is a proper divisor of $n$.

$\endgroup$
2
  • 1
    $\begingroup$ $n = \sum_{0<d|n}\phi(d)$ comes from considering the reduced forms of the fractions $1/n, 2/n, ... n/n$. $\endgroup$ – lhf Jan 22 '12 at 0:21
  • $\begingroup$ Yes, of course. Put another way, we're counting subsets of $\mathbb{Q}\cap(0,1]$ represented as reduced fractions $p/q$ having denominator $q\leq n$ in two ways -- by row, and by column (I think). $\endgroup$ – bgins Jan 22 '12 at 0:33

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.