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 $\Phi$ ?

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 $\varphi^* \omega = \omega$ ?

**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.