Suppose that $E$ an elliptic curve defined over $\mathbb{Q}$ and $p$ an odd prime. Let $G=\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q})$. I am wondering whether the cohomology group $H^1(G, E[p])$ can be nontrivial. If $G=GL\_2(\mathbb{F}\_p)$ (which is the case for all but finitely many primes $p$ if $E$ does not have complex multiplication) then $H^1(G, E[p])$ is trivial. This can be shown by considering the homothety subgroup $Z \le G$ which has order $p-1>1$. One easily sees that $H^i(Z, E[p])=0$ for all $i \geq 0$ and so the result follows from the Hochschild-Serre spectral sequence.


Now suppose that $G$ is a proper subgroup of $GL\_2(\mathbb{F}\_p)$. Can $H^1(G, E[p])$ be nontrivial?



For an elliptic curve $E/\mathbb{Q}$ can the cohomology group $H^1(\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q}), E[p])$ be nontrivial?
Fix elements $\zeta$ and $\alpha$ with $\zeta$ a primitive third root of unity and $\alpha^3 = -4$. These generate a field $K = \Bbb Q(\zeta,\alpha)$ which is the splitting field of $x^3 + 4$, with Galois group $G$ the symmetric group on three letters.


Consider the elliptic curve $y^2 = x^3 + 1$. Unless I have miscalculated, the $3$-torsion points on this curve are the points $(x,y)$ with $x^4 + 4x = 0$. In particular, the points $(0,1)$ and $(\alpha,2\zeta+1)$ are independent 3-torsion points on this curve, so $K = \Bbb Q(E[3])$. Taking these as a basis, the resulting image of the Galois group into $GL\_2(\Bbb F\_3)$ must be
$$\begin{bmatrix}1 & \* \\ 0 & \*\end{bmatrix}$$
because $(0,1)$ is fixed and the map must be injective.


Let $H < G$ be the subgroup of order three. Since the coefficient group $E[3]$ is $3$-torsion, a transfer argument implies that the restriction $H^1(G;E[3]) \to H^1(H;E[3])$ is injective with image the invariants under $G/H \cong \Bbb Z/2$.


If $$A = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$$
represents the generator $\tau$ of $H$, then the group $H^1(H;E[3])$ is $ker(1 + A + A^2) / Im(1 - A)$, which is generated by the column vector $\left[\begin{smallmatrix}0 \\ 1\end{smallmatrix}\right]$. (This describes an element of $H^1$ by where an associated $1$-cocycle sends a chosen generator of $H$.)


As a $1$-cocycle, this is represented by the map $f:H \to E[3]$ with
$$f(\tau^k) = (1 + A + \cdots + A^{k-1})\left[\begin{smallmatrix}0 \\ 1\end{smallmatrix}\right].$$
The action of the element $\sigma = \left[\begin{smallmatrix}1 & 0 \\ 0 & -1\end{smallmatrix}\right]$ on this cocycle is given by
$$({}^\sigma f)(\tau) = \sigma \cdot f(\sigma^{-1} \tau \sigma) = \sigma \cdot f(\tau^2) = \left[\begin{smallmatrix}1 \\ 1\end{smallmatrix}\right]$$
which shows that the two $1$-cocycles ${}^\sigma f$ and $f$ represent the same element of $H^1$. Therefore, this element of $H^1(H;E[3])$ is invariant under $G/H$ and lifts to a nontrivial element of $H^1(G;E[3])$.


