Why an isogeny induces a surjection between points over maximal unramified extension? Let $E$ and $E'$ be elliptic curves over $\mathbb Q$, and let $\phi\colon E\to E'$ be an isogeny defined over $\mathbb Q$. Let $p$ be a prime relatively prime to the degree of $\phi$. Let $\mathbb Q_p^{nr}$ be the maximal unramified extension of $\mathbb Q$. Why is is true that $\phi$ induces a surjection $E(\mathbb Q_p^{nr})\to E'(\mathbb Q_p^{nr})$?
My attempt is as follows. We have an exact sequence $0\to E(\mathbb Q_p^{nr})[\phi] \to E(\mathbb Q_p^{nr})\to E'(\mathbb Q_p^{nr})\to H^1(\operatorname{Gal}(\overline{\mathbb Q_p}/\mathbb Q_p^{nr}), E[\phi])$.
But how to show that the cohomology group is zero? If I could show that every element of $\operatorname{Gal}(\overline{\mathbb Q_p}/\mathbb Q_p^{nr})$ has order prime to $\#E[\phi]$ then maybe that would work.
 A: Assume that $E$ (and hence $E'$) has good reduction at $p$.
First version: Let $P\in E'(K)$ where $K$ is the maximal unramified extension of $\mathbb{Q}_p$.
The reduction $\tilde P$ is in $E'(\overline{\mathbb{F}_p})$ and since that field is algebraically closed there is a point $R$ in $E(\overline{\mathbb{F}_p})$ such that the reduction of $\phi$ maps $R$ to $\tilde {P}$.
Now lift $R$ to a point $Q$ in $E(K)$ which can be done by Hensel's lemma as it is non-singular.
Then consider $P-\phi(Q)$.
Since it reduces to zero it belongs to the formal group $\hat {E'}(\mathfrak m)$ for the maximal ideal $\mathfrak m$ in $K$.
However, since the degree of $\phi$ is coprime to $p$, the map $\phi:\hat E(\mathfrak m ) \to \hat {E'}(\mathfrak m)$ is surjective. Hence $P-\phi(Q)$ and $P$ are in the image of $\phi$.
Second version: Pick a point $S$ in $E(\overline{\mathbb{Q}_p})$ in the preimage of $P$ under $\phi$.
Let $\sigma$ be an element of the Galois group of $\overline{\mathbb{Q}_p}/K$ and consider $\sigma(S)-S$.
It is in the kernel of $\phi$.
Since $\sigma$ is in the inertia group, the reduction of $\sigma(S)$ is equal to the reduction of $S$ in the non-singular reduction $\tilde E$.
Hence $\sigma(S)-S$ also belongs to the formal group.
However, since the degree of $\phi$ is coprime to $p$, the intersection of $\hat{E}(\mathfrak m)$ and $E[\phi]$ is trivial.
Therefore $\sigma(S)-S=O$, which shows that $S\in E(K)$.
(And that the connecting Kummer homomorphism to $H^1\bigl(K,E[\phi]\bigr)$ is zero.)
Third version: Since the degree of $\phi$ is coprime to $p$ and $E$ has good reduction, Proposition VIII.1.5 in Silverman's "The arithmetic of ellitpic curves" shows that $K\bigl( \phi^{-1} E(K) \bigr)$ is unramified and hence trivial. (But its proof is the second verison.)
Will's comment above shows that good reduction is needed.
