# $n$-th root of unity implies $d$-th primitive root of unity for some $d\big|_n$

Let $F$ be a field and let $n \in \mathbb{N}$ and let $\varepsilon \in \overline{F}$ ($\overline{F}$ is an algebraic clausure of $F$) be a $n$-th root of unity ($\varepsilon$ is a root of $X^n - 1 \in F[X]$). Can I state that there exists $d \in \mathbb{N}$, with $d\big|_n$, such that $\varepsilon$ is a $d$-th primitive root of unity? ($\varepsilon$ is $d$-th primitive root of unity if $\varepsilon$ is $d$-th root of unity and $\varepsilon$ generates all the elements in the group of $d$-th roots of unity, which is a subgroup in $\overline{F} \setminus \{0\}$). I am going to denote by $W_k$ the group of $k$-th roots of unity and $W_k^*$ the elements in $W_k$ which are primitive.

I have not found any counterexample. For instance, if we take $F = \mathbb{C}$, then $W_4 = \{1 , i , - 1 , i\}$. In this group, the only elements which do not belong to $W_4^*$ are $1$ and $- 1$; however, $- 1 \in W_2^*$ and $1 \in W_1^*$, so no counterexample can be found here. On the other hand, if we take $F$ as the finite field of $9$ elements, $F = GF(9)$ (whose operations between the elements can be found here: http://www.cs.miami.edu/home/burt/learning/Csc609.032/notes/gf9example.html), we can see that $W_4 = \{1 , 2 , x , 2 x\}$. Here, $W_4^* = \{x , 2 x\}$, but $2 \in W_2^*$ and $1 \in W_1^*$. Finally and it is my last example, if we take $F = GF(4)$, we have that $X^4 - 1 = {(X - 1)}^4$ and $X^2 - 1 = {(X - 1)}^2$, so $1$ is a primitive element, as $W_4 = W_2 = \{1\}$.

My attempt for the proof: let $d$ the order for $\varepsilon$ as element in $W_n$ (each element in a group generates a cyclic group), which is a cyclic and finite group. Then ${\varepsilon}^d = 1$ and $\langle \varepsilon \rangle = \{1 , \varepsilon , \ldots , {\varepsilon}^{d - 1}\}$. Appearently, we are done because ${\varepsilon}^d = 1$, which shows that $\varepsilon \in W_d$, but ${({\varepsilon}^{j})}^d = 1^j = 1$ for all $j = 0 , 1 , \ldots , d - 1$, which shows that $\langle \varepsilon \rangle \subset W_d$ and $|W_d| \leq d$, which shows that $\langle \varepsilon \rangle = W_d$.

The main problem which I have found in the proof is that I cannot state that the group $W_n$ has $n$ elements (see my last example), it happens in finite groups or in groups which characteristic is not zero. Generally, I can state that $m \leq d$ (it is an equality in $\mathbb{C}$ por example), where $m = |W_d|$ but I cannot state that $m = d$.

• What is $d\mid_n?$ – 伽罗瓦 Aug 22 '18 at 16:46
• I am puzzled about where you think your proof uses the statement that $W_n$ has $n$ elements. – Eric Wofsey Aug 22 '18 at 16:47
• It means that $d$ divides to $n$, so in this case there exists $k \in \mathbb{N}$ such that $n = d k$. – joseabp91 Aug 22 '18 at 16:47
• If $g^n$ is trivial then the order of $g$ divides $n$. This is true in any group. – Daniel Mroz Aug 22 '18 at 16:49
• I have never seen your notation, $d\mid_n$, to mean that "$d$ is a factor of $n$". I have thought that this relation is always (?) written $d\mid n$, in other words $n$ should not be in the subscript position. Is your notation actually used in some book? I find it very strange. – Jyrki Lahtonen Aug 22 '18 at 17:17

The proof looks fine -- you never assume that $W_d$ has $d$ elements (which, as you noted, is not true in general), rather you derive it from the following facts:
• $\langle\varepsilon\rangle$ has $d$ elements
• $\langle\varepsilon\rangle \subset W_d$
• $|W_d| \leq d$
None of these facts require $|W_d|=d$. Together they imply it: the first two taken together imply $d \leq |W_d|$, and together with the third we have $|W_d|=d$.