# $\zeta_n^k$ is primitive if and only if $(k,n) = 1$ [duplicate]

Show that for $$k \in \mathbb{Z}$$:

Is $$\zeta_n$$ a primitive $$n$$-th root of unity, then $$\zeta_n^k$$ is primitive if and only if $$(k,n) = 1$$.

I only need the backwards direction:

$$\zeta_n^k$$ is primitive $$\Rightarrow$$ $$(k,n) = 1$$

## marked as duplicate by Dietrich Burde abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 30 at 11:39

First of all notice that $$\zeta_n^k$$ is of the form $$e^{2\pi i\frac kn}$$ and the hypotesis of being primitive means that $$n$$ is the smallest (positive) integer such that $$(\zeta_n^k)^n=1$$. If by contradiction $$(k,n)=d>1$$ then you can write $$k=ad$$ and $$n=bd$$, with $$a,b\in\Bbb Z$$ and $$a,b\ge1$$. Thus $$\zeta_n^k=e^{2\pi i\frac {ad}{bd}}=e^{2\pi i\frac d{b}}.$$ From here we deduce that $$(\zeta_n^k)^b=1$$ and since $$b (in fact $$0\le k\le n-1$$) our root cannot be primitive.
• $(3,9)=3$ so whats $a,b$ here? It should be $1 \leq a,b$ but is your argument then still working? – user625682 Jan 30 at 12:09
• The key fact is that, since $k\le n-1$ then the common divisor between $k$ and $n$ strictly divides $n$, that is $n=bd$ with both $b$ and $d$ (wlog positive) and $<n$. In your case $k=3$ and $n=9$, thus we can take $b=d=3$ (I have edited $a,b\ge1$ and in this case $a=1$). – Joe Jan 30 at 13:29
• The question is tagged abstract-algebra and nothing about it suggests that the field in question is a subfield or an extension of $\mathbb{C}$, so an approach which argues on the basis of primitive roots in $\mathbb{C}$ doesn't really answer the question. – Peter Taylor Jan 30 at 14:40