Let $G_{\mathbb{Q}_\infty}$ be the absolute Galois group of $\mathbb{Q}_\infty$, where $\mathbb{Q}_\infty$ is the cyclotomic $\mathbb{Z}_p$- extension of $\mathbb{Q}$ and let $A$ be a discrete $G_{\mathbb{Q}}$-module. Consider the cohomology group $H^1(G_{\mathbb{Q}_\infty}, A)$ on which $G_{\mathbb{Q}_\infty}$ acts trivially.
Why did this action allow us to regard the global cohomology group $H^1(G_{\mathbb{Q}_\infty}, A)$ as a discrete Gal$(\mathbb{Q}_\infty/\mathbb{Q})$-module?
This is a very general fact. Let $G$ be a profinite group and $H \triangleleft G$ be a closed subgroup. Let $A$ be a discrete $G$-module.
Let $C^{\cdot}$ be the continuous cochain complex for $H$: for $n \geq 0$, $C^n$ is the space of continuous (ie locally constant) $H$-equivariant maps $H^{n+1} \rightarrow A$. For $n \geq 0$, we have a $H$-linear differential $d: C^n \rightarrow C^{n+1}$ such that $df(h_1,\ldots,h_{n+2})=\sum_{i=1}^{n+2}{(-1)^{i-1}f(h_1,\ldots,\hat{h_i},\ldots,h_{n+2})}$.
Then the cohomology of $C^{\cdot}$ computes the groups $H^{\cdot}(H,A)$.
The goal is to show that $G$ acts naturally on every $H^i(H,A)$, and that the $H$-action is trivial.
To do that, we let $G$ act on every $C^n$, by $g \cdot f(h_0,\ldots,h_n)=g \cdot f(g^{-1}h_0g,\ldots,g^{-1}h_ng)$. It’s easy to see that $d$ commutes with this action of $G$, so that $G$ acts on the cohomology of the complex, which is $H^{\cdot}(H,A)$.
Note that as every $f$ is locally constant (and has finite image), its stabilizer in $G$ is open. Thus the action of $G$ on the cohomology is discrete.
So we want to show now that through this action, $H$ acts trivially.
In fact, we are going to show that the action of any $h \in H$ on $C^n$ is homotopic to the identity.
By definition, the action of $h \in H$ is by right translation of the arguments of $f$.
For $f \in C^n$, define $P(f) \in C^{n-1}$ by $Hf(h_1,\ldots,h_n)=\sum_{i=1}^n{(-1)^nf(h_1,\ldots,h_i,h_ih,\ldots,h_nh)}$.
It’s easy to see that $P$ is well-defined, and if I’m not mistaken $(Pd+dP)f(h_0,\ldots,h_n)=\sum_{i=0}^n{f(h_0,\ldots,h_{i-1},h_ih,\ldots,h_nh)-f(h_0,\ldots,h_i,h_{i+1}h,\ldots,h_nh)}$, so that $(dP+Pd)(f)=h \cdot f-f$. This concludes.
There’s also a much more pleasant construction with a little more work on the homological algebra: take an injective resolution $A \rightarrow I^{\cdot}$ in the category of discrete $G$-modules (it’s standard, but not completely obvious that they exist). The $I^n$ are injective as discrete $H$-modules, (again, standard but not completely obvious).
Thus $H^n(H,A)$ is computed by the cohomology of $(I^n)^H$. But $(I^n)^H$ is a $G$-submodule of $I^n$, so you have a $G$-action on the complex $(I^{\cdot})^H$, hence on its cohomology. But by definition of the complex, $H$ acts trivially!