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This question may seem elementary to experts but I am quite confused about it:

According to the entry of Lyndon–Hochschild–Serre spectral sequence on wikipedia, for a group extension $1\to N\to G\to Q\to1$, there is a Lyndon–Hochschild–Serre spectral sequence if $G$ is a profinite group and $N$ is a closed normal subgroup of $G$:

Also in the book Neukirch, Jürgen; Schmidt, Alexander; Wingberg, Kay (2000), Cohomology of Number Fields, the same condition was assumed.

However, according to some other books, for example Brown, Kenneth S. (1972), Cohomology of Groups, and A User's Guide to Spectral Sequences by John McCleary, the profiniteness condition on the group $G$ was NOT assumed.

Why there is such a difference? Do we really need the condition that $G$ is a profinite group and $N$ is a closed normal subgroup of $G$ to construct the Lyndon–Hochschild–Serre spectral sequence?

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What kind of group cohomology are you considering? When talking about profinite groups, you usually work with discrete modules and continuous cochains. – user2035 Apr 25 '12 at 8:04
a-fortiori should expand this comment into an answer and claim his/her reward! – Mark Grant Apr 27 '12 at 9:26
In the context of your question there are two cohomology theories: One for abstract groups and one designed for profinite groups. Each has a LHS spectral sequence (both are formally equal). The difference between the two theories is that for profinite groups one uses continuous cochains. More specifically: In the def. of $X^n$ at the beginning of §2 in the Neukirch book, in the profinite case you use continuous maps $G^{n+1} \to A$ while for abstract groups you use all maps $G^{n+1} \to A$. //And, of course, I agree with Mark that a-fortiori should make his/her comment into an an answer. – Ralph Apr 27 '12 at 9:50
BTW: The wikipedia article is bad: There is no need for "$G$ [to] be a finite group". Better use the other sources you mentioned. – Ralph Apr 27 '12 at 9:54

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