Pairing - Is it possible to map two $r$-torsion points to a $r^2$-torsion point? Let $E(\mathbb F_{q^k})$ be an elliptic curve on finite field $\mathbb F_{q^k}$, where $\mathbb F_{q^k}$ is an extension of $\mathbb F_q$ with $k&gt;1$. Let $e: G_1 \times G_2 \rightarrow G_t$ be a bilinear pairing, and $G_1$, $G_2$ are some subgroup of $E(\mathbb F_{q^k})$ of order $r$, where $r| \# E(\mathbb F_q)$ and $r^2 | \# E(\mathbb F_{q^k})$, and $G_t$ is subgroup of order of $\mathbb F^*_{q^k}$.

Is it possible to modify (Weil or Tate) pairing such that the order of $G_t$ be $r^2$? In other word, is it possible to map (with keeping bilinearity property) two $r$-torsion points to a $r^2$-root of unity?
No, it is not possible.

By the definition of bilinearity, we have $e( kG, H ) = k \cdot e( G, H )$.  If the order of $G$ is $r$ (that is, $rG = 0$, we have $e( rG, H ) = e( 0, H ) = r \cdot e(G, H)$.  We know $e(0, H) = 0$ (as bilinear functions maps the identity to the identity), and hence we have $r \cdot e(G, H) = 0$; that is, the order of $e(G, H)$ must be either $r$, or some divisor of $r$.