A question on the injectivity of a canonical map between galois cohomology groups I'm currently reading the book "Galois theory of $p$-extensions" by Helmut Koch.
There, we calculate the cohomological dimension of the galois group $G(K/k)$ where $K$ is the maximal (normal) $p$-extension of $k$.
(Here $p$ is a prime and $k$ is a local field or global field of finite type, i.e finite extension of $\mathbb{Q}$ of $\mathbb{Q}_p$.)
As $G(K/k)$ is a pro-p group, we study $H^2(G(K/k), \mathbb{F}_p)$ where $\mathbb{F}_p$ is the finite field with $p$ elements with trivial group action.
Let $k'$ be the field generated by $k$ and the $p$'th roots of unity, and let $K'$ be the maximal $p$-extension of $k'$. Then there is a canonical group homomorphism from $G(K'/k')$ to $G(K/k)$ (restriction map).
This induces a homomorphism from $H^2(G(K/k),\mathbb{F}_p)$ to $H^2(G(K'/k'),\mathbb{F}_p)$.
The question is, is this map injective?
 A: I believe this map is always injective. Here is a quick argument: first note that $K'$ is normal over $k$ (because it is invariant under any automorphism of the algebraic closure of $k$ which preserves $k'$, and $k'$ is normal over $k$). This means that we can view the map $G(K'/k') \to G(K/k)$ as a composition of two maps
$$ G(K'/k') \stackrel{f}{\to} G(K'/k) \stackrel{g}{\to} G(K/k) .$$
It will hence suffice to show that both $f^*:H^2(G(K'/k),\mathbb{F}_p) \to H^2(G(K'/k'),\mathbb{F}_p)$ and $g^*:H^2(G(K/k),\mathbb{F}_p) \to H^2(G(K'/k),\mathbb{F}_p)$ are injective. Now $f: G(K'/k') \to G(K'/k)$ is an inclusion of a subgroup of finite index $[k':k]$, and hence there exists a corestriction map $f_!:H^2(G(K'/k'),\mathbb{F}_p) \to H^2(G(K'/k),\mathbb{F}_p)$ such that $f_! \circ f^*$ is multiplication by $[k':k]$. Since $[k':k]$ is coprime to $p$ and $H^2(G(K'/k),\mathbb{F}_p)$ is a $p$-torsion group it follows that $f^*$ is injective. To show that $g^*$ is injective, we note that $g: G(K'/k) \to G(K/k)$ is a surjective group homomorphism with kernel $G(K'/K)$, and so the Hochschild-Serre spectral sequence gives an exact sequence
$$ H^1(G(K'/K),\mathbb{F}_p)^{G(K/k)} \to H^2(G(K/k),\mathbb{F}_p) \to H^2(G(K'/k),\mathbb{F}_p) .$$
To show that $g^*$ is injective it will hence suffice to show that $H^1(G(K'/K),\mathbb{F}_p)^{G(K/k)} = 0$, i.e., that $K$ does not admit any non-trivial cyclic $p$-extensions inside $K'$ which are normal over $k$. But this is just because $K$ does not admit any non-trivial cyclic $p$-extensions which are normal over $k$: indeed, it is the maximal $p$-extension of $k$.
