In the book Galois Cohomology by J-P. Serre, page no. 95, in the proof of proposition 18, it is written that $H^0(\hat{\mathbb{Z}},A)$ and $H^0(\hat{\mathbb{Z}},A)$ have the same cardinality when $A$ is a finite module. I am trying to prove this claim in the following way:
We know that $\hat{\mathbb{Z}}$ is topologically generated by $1$. Let $F$ be the automorphism of $A$ corresponding to the action of $1$ on $A$. It is known (cf. Local Fields, Serre, p. 188-189) that $$H^0(\hat{\mathbb{Z}},A)=A^F\qquad H^1(\hat{\mathbb{Z}},A)=A/(F-1)A$$
The most (possibly) natural map $$A^F\longrightarrow A/(F-1)A$$ one can think of is given by $$a\mapsto[a]$$
I am trying to show that this map is bijective. But I am stuck at this.
My progress (not really):
Let $a\in A^F$ be such that $[a]=0$. Then, there exists $b\in A$ such that $a=F(a)-a$. Since $A$ if finite, there are smallest natural numbers $m,n$ such that $na=0$ and $F^m(b)=b$. Then we get that
$$a=F(b)-b\\a=F(a)=F^2(b)-F(b)\\\vdots\\a=F^{m}(b)-F^{m-1}(b)$$ summing all these we get $ma=F^m(b)-b=0$ and also for $m$ replaced by $n$, we get the same and conclude that $m=n$. We have to prove $m=n=1$ in order to show injectivity. How can I prove this claim? Any help would be appreciated. Thanks in advance!
Update:
for any $[a]\in A/(F-1)A$, an obvious choice of an element $b\in A^F$ in terms of $a\in A$ (when $a$ is not in $A^F$) is as follows $$b=a+F(a)+F^2(a)+\cdots+F^{n-1}(a)$$ where $F^n(a)=a$. It's easy to see that $F(b)=b$ and hence $b\in A^F$. But how can I show that $[a]=[b]$, in order to show surjectivity?
We have the following exact sequence of Abelian groups
$$0\longrightarrow A^F\longrightarrow A\overset{F-1}{\longrightarrow} A\longrightarrow {A/(F-1)A}\longrightarrow 0$$
This shows that $$|A^F|=|A/(F-1)A|$$
Please feel free to comment if this answer has any glitches.