3rd Galois cohomology of $n$th root of unity vanishes

Question

I'm reading Milne's class field theory, and there is a statement about a group cohomology which I cannot understand. Let $K$ be a field and $K^{sep}$ be a separable algebraic closure of $K$, and let $G=Gal(K^{sep}/K)$. Then we can regard $\mu_n$, the group of $n$th root of unity, as $G$-module. The book saids that from the cohomology sequence of $$0\to\mu_n\to (K^{sep})^{\times} \xrightarrow{x\mapsto x^{n}} (K^{sep})^{\times}\to 0$$ we get $$K^{\times}/(K^{\times})^{n}\simeq H^{1}(G, \mu_n),\,\,\,\,H^{2}(G, \mu_n)\simeq H^{2}(G, (K^{sep})^{\times})[n].$$ I understand that the first isomorphism comes from Hilbert's theorem 90, but I don't know why the second holds. I think we need to show $$H^{3}(G, \mu_{n})=0$$ for the second one but the book does not contain any information about it. How to show this? Thanks in advance.

Answer 1

I'll write $L$ for $K^{\text{sep}}$ and $H^n(A)$ for $H^n(G,A)$.

Part of the exact sequence is $$H^1(L^\times)\to H^2(\mu_n)\to H^2(L^\times)\to H^2(L^\times).$$ By Hilbert 90, $H^1(L^\times)=0$ so this is $$0\to H^2(\mu_n)\to H^2(L^\times)\to H^2(L^\times).$$ Therefore $H^2(G,\mu_n)$ is the kernel of multiplication by $n$ on $H^2(G,(K^{\text{sep}})^\times)$.

Not an $H^3$ in sight!