E an elliptic curve over Z[1/N], how many p such that E(Z/p^2) = (Z/p)^2?

Question

For a fixed elliptic curve $E$ over $\mathbb{Z}[1/N]$, is there a non-trivial upper bound in terms of $x$ for the number of primes $p \leq x$ with $p \nmid N$ for which $E(\mathbb{Z}/p^2\mathbb{Z}) \cong (\mathbb{Z}/p\mathbb{Z})^2$? Anything less than $\frac{x}{(\log x)^{2}}$ would be very useful for me.

Some thoughts. The condition $E(\mathbb{Z}/p^2\mathbb{Z}) \cong (\mathbb{Z}/p\mathbb{Z})^2$ implies that $E(\mathbb{F}_p)\cong \mathbb{Z}/p\mathbb{Z}$. Conversely, given an elliptic curve $E$ over $\mathbb{Z}[1/N]$ and a prime $p$ not dividing $N$ such that $E(\mathbb{F}_p) \cong \mathbb{Z}/p\mathbb{Z}$, there are two possibilities for $E(\mathbb{Z}/p^2\mathbb{Z})$: either the extension \[ 0 \rightarrow E_1(\mathbb{Z}/p^2\mathbb{Z}) \rightarrow E(\mathbb{Z}/p^2\mathbb{Z}) \stackrel{\textrm{reduction mod $p$}}{\longrightarrow} E(\mathbb{F}_p) \rightarrow 0 \] doesn't split, in which case we have $E(\mathbb{Z}/p^2\mathbb{Z}) \cong \mathbb{Z}/p^2\mathbb{Z}$, or it does split, so that $E(\mathbb{Z}/p^2\mathbb{Z}) \cong (\mathbb{Z}/p\mathbb{Z})^2$.

Still assuming that $E(\mathbb{F}_p)\cong \mathbb{Z}/p\mathbb{Z}$, the second case occurs if and only if $E_{\mathbb{Z}/p^2\mathbb{Z}} = E \otimes_{Z[1/N]} \mathbb{Z}/p^2 \mathbb{Z}$ is the canonical lifting of $E_{\mathbb{F}_p}$ to $\textrm{Spec}(\mathbb{Z}/p^2 \mathbb{Z})$. At least, this is what I've been told by some people working with canonical liftings, so I'm willing to go along with this. Anyway, I don't know if it helps, but somehow, the second case seems to be less "likely" to happen than the first.

Motivation. My question might seem a strange one, so I'll tell you why I am interested in this. Given an $E$ and $p$ as above, if $E(\mathbb{F}_p)$ is cyclic, we may immediately conclude that $E(\mathbb{Q}_p)$ is a procyclic topological group (i.e. topologically generated by a single element), except when $E(\mathbb{Z}/p^2\mathbb{Z}) \cong \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$. I am interested in finding, for a given $E$, infinitely many $p$ such that $E(\mathbb{Q}_p)$ is procyclic. (I actually want this condition to hold for all quadratic twists of $E$, but I don't think I need to go into the specifics any further.) Since there are all sort of results in the literature concerning cyclicity of $E(\mathbb{F}_p)$, the only real obstacle is the occurrence of the groups $(\mathbb{Z}/p\mathbb{Z})^2$ as above.

Answer 1

I don't know how easy it will be to prove anything along these lines, but the condition is that the Serre-Tate parameter of $E$ at $p$ satisfies $q \equiv 1 \mod p^2$. It's natural to assume that $(q-1)/p \mod p$ is random as $p$ varies, just because I know of no reason why it should be otherwise. If that's the case the number of your primes up to $x$ would be $O(\log \log x)$ kind of like Wieferich primes.