For Galois cohomology, one uses the cohomology constructed for profinite groups instead of the usual group cohomology. In other words, one also uses/takes into account the Krull topology we have on the Galois groups.
Is there a significant reason why we are using this definition of Galois cohomology?
For infinite Galois theory, I know that the topology is needed to construct a generalization of the fundamental theorem for finite Galois theory. But is there also a similar reason why we want to consider the topology when talking about cohomology for Galois groups?
Let $G$ be a group, and $M$ be a $G$-module. Then in terms of group cohomology, we have $H^{0}(G,M):=M^{G}$, and $H^{1}(G,M)=\{\text{1-cocycles}\}/\{\text{1-coboundaries}\}$.
When considering profinite groups $G$, we consider discrete $G$-modules $M$. These are $G$-modules such that the group action is continuous with respect to the discrete topology. Then the cohomology groups are given by $H_{c}^{0}(G,M):=M^{G}$, and $H_{c}^{1}(G,M)=\{\text{cont. 1-cocycles}\}/\{\text{cont. 1-coboundaries}\}$. In general, $H_{c}^{1}(G,M)$ does not have to be equal to $H^{1}(G,M)$.
Now a profinite group (for instance a Galois group), is obiously also just a group. Meaning that we could also just consider the usual group cohomology, i.e. $H^{1}(G,M)$ instead of $H^{1}_{c}(G,M)$. Or in other words, just skip the continuity condition.
So I wonder, why do we want also want to take into account the continuity condition. Wouldn't it be a lot easier if we just use the basic group cohomology? Or are there any significant results for which we need the extra continuity condition.