Is $H^{1}(k,E[n])$ a subgroup of $H^{1}(k,E)$?

Question

Let $E/k$ be an elliptic curve. Consider $E[n]$ which is a subgroup of $E$. Is it true that $H^{1}(k,E[n])$ is again a subgroup of $H^{1}(k,E)$ in Galois cohomology? I thought that this was true but I've been told that this is not necessarily the case and I don't really understand why. Can someone explain me the reason? I started to learn Galois cohomology recently, so I'm not completely familiar with the subject. Thanks for your help.

Answer 1

No, this is not true in general. Let me assume that $n$ is prime to the characteristic of $k$. If $k_s$ is a separable closure of $k$, then one has an exact sequence of $G_k=\mathrm{Gal}(k_s/k)$-modules

$$0\to E[n](k_s)\to E(k_s)\xrightarrow{n} E(k_s)\to 0,$$

or if you prefer a short exact sequence of smooth commutative group schemes over $k$

$$0\to E[n]\to E\xrightarrow{n}E\to 0.$$

Taking $G_k$-cohomology (or what amounts to the same, étale cohomology of $\mathrm{Spec}(k)$) gives the exact sequence

$$0\to E[n](k)\to E(k)\xrightarrow{n}E(k)\to H^1(k,E[n])\to H^1(k,E)\xrightarrow{n} H^1(k,E).$$

One thus gets an induced Kummer sequence

$$0\to E(k)/nE(k)\to H^1(k,E[n])\to H^1(k,E)[n]\to 0.$$

In particular, the map $H^1(k,E[n])\to H^1(k,E)$ is injective if and only if $E(k)$ is $n$-divisible. If $k$ is separably algebraically closed, then this will be true (with the assumption on $n$), but for e.g. a number field $k$ with $E(k)$ of positive rank and $n\geq 2$, this can't be true by the Mordell-Weil theorem.

Answer 2

If you have an injection $i:A \to B$ the associated maps in group cohomology is not necessarily injective. Take a look at Theorem 1.2.11 here: http://math.arizona.edu/~sharifi/groupcoh.pdf

In your case, the Kummer exact sequence: $$ 0 \to E[n] \to E \stackrel{[n]}{\to} E \to 0 $$ induces a long exact sequence given by $$ 0 \to H^0(k,E[n]) \to H^0(k,E) \to H^0(k,E) \to H^1(k,E[n]) \to H^1(k,E) \to H^1(k,E) \to \cdots $$ Which can be written as: $$ 0 \to E(k)/[n]E(k) \to H^1(k,E[n]) \to H^1(k,E)[n] \to 0 $$