I'm trouble in understanding the proof of Lemma X.4.3 in Silverman's Arithmetic of Elliptic Curves (2nd edition), that claims $H^1(G_{\bar{K}/K}, M; S)$ is finite. In page 334, the book state the map $$Hom(G_{L/K}, M; S) \rightarrow Hom(G_{\bar{K}/K}, M; S)$$ is an isomorphism. But I cannot understand why this is true.
The inflation-restriction sqeuence of Galois cohomology $$0 \rightarrow Hom(G_{L/K}, M; S) \rightarrow Hom(G_{\bar{K}/K}, M; S) \rightarrow Hom(G_{\bar{K}/L}, M; S)$$ says if the above map is isomorphism then the image of the right map in the exact sequence is {0}. This means the restrictions of $Hom(G_{\bar{K}/K}, M; S)$ to $G_{\bar{K}/L}$ is trivial map. The book claims that this is true because $mM = 0$ and $G_{L/K}$ has exponent $m$.
If $\sigma \in (G_{\bar{K}/K})^m$ then $f(\sigma) = 0$ for all $f \in Hom(G_{\bar{K}/K}, M; S)$. And $(G_{\bar{K}/K})^m \subset G_{\bar{K}/L}$.
But I cannot deduce the following, which is that I want to prove.
If $\sigma \in G_{\bar{K}/L}$ then $f(\sigma) = 0$ for all $f \in Hom(G_{\bar{K}/K}, M; S)$.
How can I prove the above fact?
I've straighten out this problem. I appreciate Ferra's informative comments.
Let $f \in Hom(G_{\bar{K}/K},M;S)$ and $K_f$ the fixed field of $\ker(f)$. Then $K_f$ is an abelian extension of $K$ having exponent $m$ that is unramified outside of $S$.
$\because$ $K_f$ is abelian because $M$ is abelian,
$K_f$ is unramified outside of $S$ because $f$ is trivial on inertia groups $I_v$ ($v \not\in S$),
and $K_f$ has exponent $m$ because $mM = 0$ so $G_{\bar{K}/K}^m \subset \ker(f)$.
Therefore, by the definition of $L$, $K_f \subset L$. This means $f$ is trivial on $G_{\bar{K}/K}$, which is what I want to say.