Given an abelian variety $A$ over the rational integers $\mathbb{Q}$, and a prime $p$, we know that $\mathbb{Q}(\zeta_p)$ is contained in $\mathbb{Q}(A[p])$, the $p$-division field of $A$, and where $\zeta_p$ is the primitive $p$-th root of unity.

More generally, for a finite set $S\subset A(\bar{\mathbb{Q}})$, let denote by $\mathbb{Q}(S)$ the field obtained by adjoining to $\mathbb{Q}$ the coordinates of the points in $S$.

We know also that $A[p](\bar{\mathbb{Q}})\cong (\mathbb{Z}/p\mathbb{Z})^{2g}$, where $g$ is the dimension of $A$.

Can we have a simple abelian variety $A$ over $\mathbb{Q}$ such that for each odd prime $p$, we have at most one subgroup $G\subset A[p](\bar{\mathbb{Q}})$ of order $p$ such that $\zeta_p\notin \mathbb{Q}(G)$?

Answers and comments are welcome. Thanks in advance.