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.

1$\begingroup$ The representation $\rho^F$ of $G_K$ on $E^F[m]$ for any $m$ is the product of the representation $\rho$ on $E[m]$ times the (scalar) character $\chi$ corresponding to the extension $F/K$. If $m>2$ , there will be $\sigma$ in $G_K$ with $\chi(\sigma)=1$. Since $1$ and $+1$ are not congruent modulo $m>2$, we have $E^F[m]\neq E[m]$. $\endgroup$ – Chris Wuthrich May 5 '15 at 15:07
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 2torsion 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 nontrivial 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.

$\begingroup$ Thank you for the clarification. For $p^{n}$ torsion case is it true that $H^{1}(K, E^{F}[p^{n}])$ can be cannonically identified with $H^{1}(K,E[p^{n}])$? $\endgroup$ – dipramit majumdar May 5 '15 at 17:03

$\begingroup$ For example, Selmer Companion Curves, authors seems to identify these groups in the remarks following definition 6.2. $\endgroup$ – dipramit majumdar May 5 '15 at 17:08

$\begingroup$ I can't comment on the paper, but the statement seems pretty unlikely. Even for $p^n=3$, the Galois modules $E[3]$ and $E^F[3]$ are different, so I see no reason why their first cohomology should be the same. If you want to construct an example, I'd suggest taking $E[3]\subset E(K)$. $\endgroup$ – Joe Silverman May 5 '15 at 17:11
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 $E^{F}[2]$.

$\begingroup$ What about the group of $p^n$ torsion points? Also is there a more conceptual way to look at it? $\endgroup$ – dipramit majumdar May 5 '15 at 14:58