Property of the $p^n$-Selmer group Consider the $p^n$-Selmer group of an elliptic curve $E/\mathbb{Q}$, $\operatorname{Sel}_{p^n}(E)$. Must we always have $\operatorname{Sel}_{p^n}(E) \cong (\mathbb{Z}/p^n\mathbb{Z})^{s}$ for some $s$? It seems that all the examples I've seen with $p^n = 2, 3$ this seems to be true, but why?
 A: Let $p=2$ and $n=2$, and consider $\text{Sel}_4(E/\mathbb{Q})$, the Selmer group associated to the isogeny $[4]:E\to E$. By definition, the Selmer group and the Tate-Shafarevich group fit in an exact sequence:
$$0\to E(\mathbb{Q})/4E(\mathbb{Q})\to \text{Sel}_4(E/\mathbb{Q})\to \text{Sha}(E/\mathbb{Q})[4]\to 0.$$
Suppose that $E(\mathbb{Q})\cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}$. Then, 
$$E(\mathbb{Q})/4E(\mathbb{Q})\cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}.$$
Moreover, suppose that $\text{Sha}(E/\mathbb{Q})[4]$ is trivial. Then, 
$$\text{Sel}_4(E/\mathbb{Q})\cong E(\mathbb{Q})/4E(\mathbb{Q})\cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}.$$
Of course, showing that $\text{Sha}(E/\mathbb{Q})[4]$ is trivial is hard, but if you are willing to assume the Birch and Swinnerton-Dyer conjecture, then you may look for examples where the analytic order of Sha is $1$ (or, more generally, odd analytic order). You can find a list of $100$ examples that satisfy the above properties here. For instance,
$$E: y^2+xy=x^3-x$$
is one such example.
A: The $p^n$-Selmer group is a finite $p^n$-torsion abelian group.  When $n = 1$ such a guy must be of the form $(\mathbb{Z}/p^n \mathbb{Z})^s$ for some $s$, but for $n > 1$ such a guy is allowed to be a direct sum of copies of $\mathbb{Z}/p^m \mathbb{Z}$ for all $1 \leq m \leq n$.  Another way to say this is that a finitely generated $\mathbb{Z}/p^n \mathbb{Z}$-module need not be free unless $n = 1$.  
I'm not sure I have seen an explicit example where the $p^n$-Selmer group is not $\mathbb{Z}/p^n \mathbb{Z}$-free, but nevertheless I am prepared to (virtually!) guarantee you that such a thing exists.  If I were you, I would start looking at $4$-Selmer groups and I'm sure you'll find an example sooner rather than later.  
If you try for a while to find a non-free $4$-Selmer group and fail, let me know and I'll think a little harder on how to search for / construct such a thing.  
