n torsion groups of quadratic twists of elliptic curves If $E$ is an elliptic curve over a number field $K$ and $E^{F}$ is a quadratic twist of $E$. Then it is stated in ``Ranks of twists of elliptic curves and Hilbert’s tenth problem" due to Mazur and Rubin, Remark 2.4, that as $Gal(\bar{K}/K)=G_{K}$-module $E[2]=E^{F}[2]$. I do not quite see the proof.
More generally, if $p$ is a prime, is it true that as $G_{K}$-module $E[p^n] = E^{F}[p^n]$? It would be really helpful to me if someone can explain this to me.
 A: The conceptual way to look at it is to use the fact that there is an isomorphism
$$ f : E \xrightarrow{\;\sim\;} E^F $$
defined over $KF$ that intertwines the representation $\chi : G_K\to\mbox{Gal}(KF/K)\to\{\pm1\}\subset\mbox{End}(E)$ via
$$  f(P)^\sigma = \chi(\sigma)f(P^\sigma). $$
If you restrict $f$ to the 2-torsion points $P\in E[2]$, then $\chi(\sigma)P=P$ for all $\sigma$, since $-P=P$. Hence $f$ induces a $G_K$-isomorphism
$$ f : E[2] \xrightarrow{\;\sim\;} E^F[2]. $$
It does not induce a $G_K$-isomorphism $E[m]\to E^F[m]$ for $m\ge3$, since (assuming it's a non-trivial twist) one can always find a $\sigma$ with $\chi(\sigma)=-1$, and then
$f(P)^\sigma\ne f(P^\sigma)$ for all points $P\in E[m]\setminus E[2]$.
A similar analysis can be done for quartic and sextic twists for $y^2=x^3+x$ and $y^2=x^3+1$, and more generally for twists of abelian varieties. In particular, the result that $A[2]\cong A^F[2]$ as $G_K$-modules holds for quadratic twists of abelian varieties. 
A: Let's write down equations. $y^2=x^3+ax+b$ is $E$, and $dy^2=x^3+ax+b$ is $E^{F}$. Now we know that two torsion points have zero $y$ coordinate, and so the identity map is an isomorphism of $E[2]$ with $E^{F}[2]$.
