Elements of arbitrary large order in the first Galois cohomology of an elliptic curve Let $E$ be an elliptic curve over $k=\mathbb{Q}$.  Consider $H^1(k,E)$. 
In this answer Daniel Loughran writes: "I'm pretty sure that this cohomology group has elements of arbitrarily large order". I would be happy to have an explanation of this fact and/or a reference.
(I usually work with linear algebraic groups. For an abelian linear algebraic 
group $A$ there exists a natural number $d=d(A_{\bar k})$ such that the order of any element of $H^1(k,A)$ is $\le d$.)
EDIT: Cassels, § 10 and § 27, writes that this result was first proved by Shafarevich in 1957.
 A: Here is the kind of method I had in mind.
We have the elliptic curve Kummer sequence
$$0 \to E[n] \to E \to E \to 0,$$
Here I denote by $E[n]$ the $n$-torsion group scheme of $E$. Applying Galois cohomology we obtain
$$0 \to E(\mathbb{Q})/nE(\mathbb{Q}) \to H^1(\mathbb{Q}, E[n]) \to H^1(\mathbb{Q}, E)[n] \to 0.$$
By the Mordell-Weil theorem, the group  $E(\mathbb{Q})/nE(\mathbb{Q})$ is finite (its cardinality grows roughly like $n^{\mathrm{rank}(E)}$).
Thus it suffices to show that $H^1(\mathbb{Q}, E[n])$ is infinite. I think that this should be some general property of Galois cohomology for non-trivial finite abelian group schemes over number fields, which probably you already know about.
Anyway, the argument should go as follows: Choose a splitting field $k/\mathbb{Q}$ for $E[n]$. We then apply inflation-restriction to obtain
$$0 \to H^1(\mathrm{Gal}(k/\mathbb{Q}), E[n]) \to H^1(\mathbb{Q}, E[n]) \to H^1(k, (\mathbb{Z}/n\mathbb{Z})^2)^{\mathrm{Gal}(k/\mathbb{Q})} \to H^2(\mathrm{Gal}(k/\mathbb{Q}), E[n]).$$
The first and the latter group are finite. The group $H^1(k, (\mathbb{Z}/n\mathbb{Z})^2)$ is clearly infinite, and I think that it is still infinite after taking Galois invariants. Though this last step is the part I did not fully check. Is it clear to you?
A: Theorem 6 of this paper gives a generalization of Shafarevich's Theorem:  

Theorem: Let $K$ be a Hilbertian field.  Let $A_{/K}$ be a nontrivial abelian variety.  If $n > 1$ is indivisible by the characteristic of $K$ and $A(K)/nA(K)$ is finite, then $H^1(K,A)$ has infinitely many elements of order $n$.  

As explained in the paper, the hypothesis that $A(K)/nA(K)$ be finite cannot be omitted, but the hypothesis that $n$ is indivisible by the characteristic of $K$ can be weakened to: $A(K^{\operatorname{sep}})$ contains a point of order $n$.  Via the Kummer sequence, the proof quickly reduces to showing that $H^1(K,A[n])$ contains infinitely many elements of order $n$, which is related to your other question and (yet more closely) to Lior Bary-Soroker's answer to it.
