Let $m>2$ be an integer and $k=\varphi(m)$ be the number of $m$-th primitive roots of the unity. Let $\Phi = \{ \xi\_1, \ldots, \xi\_{k/2}\} $ be a set of $k/2$ pairwise distinct primitive $m$-th roots of the unity such that $\Phi \cap \overline{\Phi} = \emptyset$, where
$\overline\Phi = \{ \overline \xi\_1, \ldots, \overline\xi\_{k/2}\}$ is the set of complex conjugate primitive roots.



> 
> **Question 1.** Does there exist an integer $N$, independent of $m$ and $\Phi$, such that $1$ belongs to $\Phi^N$, the set of products of $N$
>  non-necessarily distinct elements in $\Phi$ ?
> 
> 
> 




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The real question behind this concerns automorphisms of complex Abelian varieties
and symmetric differentials. Let $A$ be a complex Abelian variety, and $\varphi : A \to A$ 
be an automorphism of finite order. 



> 
> **Question 2.** Does there exist an integer $N$, independent of $A$ and
>  $\varphi$, such that there exists a non-zero holomorphic section
>  $\omega \in Sym^N \Omega^1\_A$ satisfying $\varphi^\* \omega = \omega$ ?
> 
> 
> 




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**Update (07/05/2013) :** If $m$ is prime then we can take $N=3$ is Question 1 above. Indeed, the product of any two elements in $\Phi$ is still a primitive root of the unity. If we denote the set of all these products by $\Phi^2$ then it must intersect $\overline \Phi$. Otherwise $\Phi^2 \subset\Phi$ and, inductively, we deduce that $\Phi^m\subset\Phi$. Thus $1 \in \Phi$ contradicting the choice of $\Phi$.


In general $N=3$ does not work. If we take $m=6$ then $\varphi(6)=2$ and for $\Phi=\{\xi\_6\}$ we need to take $N=6$. If we take $m=12$ then $\varphi(12)=4$ and for $\Phi=\{\xi\_{12}, \xi\_{12}^7 \}$ we also need to take $N=6$. 


Concerning Question 2 for the particular case $A=Jac(C)$ is the jacobian of a curve and $\varphi$ is induced by an automorphism of $C$ then I know that $N = 2^2 \times 3 \times 5$ suffices. Indeed, there always exists $N\le 6$ that works for a given pair $(A=Jac(C),\varphi)$ with $\varphi$ induced by an automorphism of $C$.The argument in this cases consists in looking at the quotient orbifold and orbifold symmetric differentials on it. 



Products of primitive roots of the unity
**Theorem** If $m$ is relatively prime to $30$, then $N=3$ works. 


I'll use additive notation. The cyclic group of order $n$ is denoted $C(n)$, the generators of this cyclic group are denoted $U(n)$.


**Lemma** Let $p \geq 7$ be prime and let $X$, $Y$ and $Z$ be subsets of $C(p)$ of size at least $(p-1)/2$. Then $0 \in X+Y+Z$.


**Proof** By the Cauchy-Davenport theorem (thanks quid!), we have $|X+Y+Z| \geq \min(3 (p-1)/2 -2,p)$, which equals $p$ once $p \geq 7$. $\square$


We now prove the Theorem by induction on $m$; the case $m=1$ is vacuously true. Let $p$ be a prime dividing $m$ and let $\pi$ be the projection map $C(m) \to C(m/p)$. Let $A \subset U(m)$ so that $U(m)$ is the disjoint union of $A$ and $-A$. We want to show that $0 \in A+A+A$.


For $b \in U(m/p)$, we have $|\pi^{-1}(b) \cap A|+|\pi^{-1}(-b) \cap A| = p-1$ or $p$; the first case occurs when $GCD(p,m/p)=1$ and the second case occurs otherwise. Choose a subset $B$ of $U(m/p)$ so that $U(m/p)$ is the disjoint union of $B$ and $-B$ and, for each $b \in B$, we have $|\pi^{-1}(b) \cap A| \geq (p-1)/2$.


Inductively, we can find $b\_1$, $b\_2$ and $b\_3$ in $B$ (not necessarily distinct) so that $b\_1+b\_2+b\_3=0$. Choose lifts $c\_i$ of the $b\_i$ to $C(m)$ so that $c\_1+c\_2+c\_3=0$. Let $X\_i$ be $(\pi^{-1}(b\_i) \cap A) - c\_i$ for $i=1$, $2$, $3$. Use the Lemma to find $x\_1$, $x\_2$, $x\_3$ with $x\_i \in X\_i$ and $x\_1+x\_2+x\_3=0$. Then $c\_i+x\_i \in \pi^{-1}(b\_i) \cap A \subseteq A$ and $\sum (c\_i+x\_i) = 0$. $\square$




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As noted in comments below, $12$ always works. Indeed, if we want to have a single $N$ for all $\Phi$, then $12$ is the best possible because $m=12$, $\Phi = \{1,7 \}$ forces $6|N$ and $m=12$, $\Phi = \{ 1,5 \}$ forces $4 | N$. However, it seems to me that a better statement (which I will now prove) is that, for any $\Phi$, either $N=4$ or $N=6$ works. 


Once our inductive claim is that either $4$ or $6$ works, the prime $5$ behaves like any other prime, so we may reduce to numbers of the form $2^a 3^b$. Also, if $9 | m$ and the claim holds for $m/3$, then it holds for $m$, as we get to replace the bound $(p-1)/2$ by $p/2$. So we are reduced to $m$ of the form $2^a$ or $2^a \times 3$. 


gcousin has already done $2^a$. For $m=2^a \times 3$, consider the map $\pi: C(2^a \times 3) \to C(2^a)$. If all of the fibers of this map have size $0$ or $2$, then we may reduce to the case of $2^a$ as above, since then we have subsets of $C(3)$ of size $2$ in the lemma. 


Suppose, then, that there is some $x \in U(2^a)$ such $\pi^{-1}(x) \cap A$ and $\pi^{-1}(-x) \cap A$ nonzero; let $a$ and $a'$ occupy these sets. Then $a+a'$ is $3$-torsion, so $N=6$ works. 


