Let $E$ be an elliptic curve over $k=\mathbb{Q}$. Consider $H^1(k,E)$. 
In [this answer](https://mathoverflow.net/a/239404/4149) 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](http://jlms.oxfordjournals.org/content/s1-41/1/193), § 10 and § 27, writes that this result was first proved by Shafarevich in 1957.



Elements of arbitrary large order in the first Galois cohomology of an elliptic curve
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?


