In Cassels' paper Arithmetic on curves of genus 1. IV introducing the Cassels–Tate pairing the following lemma is stated.
Lemma 5.1: Let $q$ be a rational prime and $\Gamma$ the Galois group of the algebraic closure $\bar k$ over $k$. Let $A$ be a $\Gamma$-module which when considered only as a $\mathbb{Z}$-module is isomorphic to $\mathbb{Z}/q\mathbb{Z}\oplus \mathbb{Z}/q\mathbb{Z}$. Then an element of $H^2(\Gamma,A)$ is trivial if and only if it is trivial locally everywhere.
Here, $k$ is a number field. The statement of the lemma in more modern language is
$$Ш^2(k,A):=\ker\left(H^2(k,A)\to \prod_{v} H^2(k_v,A)\right)=0$$
where the product is taken over the places $v$ of $k$, the map is restriction, and for a field $K$, $H^i(K,A)=H^i(\operatorname{Gal}(\bar K/K),A)$. In the proof of this lemma, the following argument is made:
Let $\Gamma_1$ be the subgroup of $\Gamma$ that leaves $A$ pointwise fixed. Then $\Gamma/\Gamma_1$ is finite and there exists (by Sylow's theorem) a subgroup $\Gamma_2$ of $\Gamma$ of index prime to $q$ such that $\Gamma_1$ is of index a power of $q$ in $\Gamma_2$. The restriction-inflation sequence shows that the restriction map $$H^2(\Gamma,A)\to H^2(\Gamma_2,A)$$ is an isomorphism.
I don't see the justification for inflation–restriction. First off, I don't see why $\Gamma_2$ is normal in $\Gamma$, which would be necessary for inflation–restriction. Since the automorphism group of $A$ as a $\mathbb{Z}$-module is $\operatorname{GL}_2(\mathbb{F}_q)$, $\Gamma/\Gamma_1$ is guaranteed to be a subgroup of $\operatorname{GL}_2(\mathbb{F}_q)$. If it is the entire group, then $\Gamma_2$ is not guaranteed to be normal as $$\left\{\begin{pmatrix} 1 & x \\ 0 & 1\end{pmatrix}:x\in \mathbb{F}_q\right\}\subseteq \operatorname{GL}_2(\mathbb{F}_q)$$ is a Sylow-$q$ subgroup that is not normal.
Even assuming that $\Gamma_2$ is normal in $\Gamma$, I am not sure how this is sufficient. First of all, the inflation-restriction sequence only says something about $H^2$s if $H^1(\Gamma_2,A)$ vanishes, and I'm not sure why that would be the case. Using the full Hochschild–Serre spectral sequence does not seem to help.
This isomorphism does not seem to be essential in the proof. Indeed, as long as the restriction map is an injection it should suffice to show $Ш^2(\bar k^{\Gamma_2},A)=0$. I can show this fact: since $\Gamma_2$ is an open subgroup of $\Gamma$, the map $$\operatorname{cor}\circ\operatorname{res}:H^2(\Gamma,A)\to H^2(\Gamma_2,A)\to H^2(\Gamma,A)$$ acts as multiplication by $[\Gamma:\Gamma_2]$. Since $A$ is $q$-torsion, its cohomology groups are also $q$-torsion, so this map is an isomorphism. This then implies that $\operatorname{res}$ must be an injection.
Although I can justify this reduction with just an injection, I am still curious how we obtain an isomorphism between the cohomology groups, because it is entirely unclear to me how Cassels has done this.
