The first cohomology group of automorphism group

Question

I want to know $H^1(K(\mu_p)/K,\mu_p)$ where $K$ is a number field.

Since $K(\mu_p)/K$ is cyclic, $H^1(K(\mu_p)/K,\mu_p)=H^{-1}(K(\mu_p)/K,\mu_p)$.

Using calculation, $H^{-1}(K(\mu_p)/K,\mu_p) =0$.

So, I hope to know if the question is true.

Q :"If $A$ is finite abelian group and if $G:=Aut(A)$, then $H^1(G,A)=0$."

Answer 1

This is false. Take $A=\Bbb Z/4\Bbb Z$. I claim $H^1(G;A)\cong\Bbb Z/2\Bbb Z$. If you don't believe my manual calculations below, here's one way to see that this couldn't possibly be zero:

Consider the automorphism group $G$ of $A$ is $\Bbb Z/2\Bbb Z$; let $0\to\Bbb Z/4\Bbb Z\to\Bbb Z/8\Bbb Z\to\Bbb Z/2\Bbb Z\to0$ be the short exact sequence of $G$-modules where $G$ acts by negation in all cases. The LES in cohomology begins with $0\to\Bbb Z/2\Bbb Z=\Bbb Z/2\Bbb Z\overset{0}{\to}\Bbb Z/2\Bbb Z\to H^1(G;A)$; $H^1(G;A)$ cannot be zero, as that would imply $0:\Bbb Z/2\Bbb Z\to\Bbb Z/2\Bbb Z$ is surjective, which is obviously false.

$G\cong\Bbb Z/2\Bbb Z$ with generator $\gamma:A\cong A$, $1\mapsto3$.

$H^1(G;A)=0$ iff. every derivation is principal. I claim that $\psi:G\to A$, $\gamma\mapsto1$ is a derivation but is not principal.

Firstly, if it were principal then $1=\gamma(a)-a$ for some $a\in A$; however you can check that $\gamma(a)-a=0,2,0,2$ for $a=0,1,2,3$ so none of these options work. Therefore it remains to check $\psi$ is a derivation.

That is, we must check: $$\forall f,g\in G:\psi(f\circ g)=f(\psi(g))+\psi(f)$$When one of $f,g$ is the identity, this is trivial as $\psi(\mathrm{id}):=0$. Therefore the only nontrivial check is: $$\psi(\gamma\circ\gamma)\overset{?}{=}\gamma(\psi(\gamma))+\psi(\gamma)$$But the left hand side is zero and the right hand side is $\gamma(1)+1=3+1=0$, so it all works out.

There is another nonprincipal derivation $\psi':\gamma\mapsto3$ but in fact $\psi$ is homologous with $\psi'$. We conclude $H^1(G;A)\cong\Bbb Z/2\Bbb Z\neq0$.