Let $G$ be a finite group and $A$ a $G$-Module. It is well-known that $H^q(G,A)$ is killed by $|G|$ for all $q \geq 1$. This is usually proved using Restriction-Corestriction (applied with the trivial subgroup).

Is there a way to prove this for $q=1$ without using this machinery? Specifically, is there a computation one can do at the level of cocycles which gives this result for $H^1(G,A)$?


2 Answers 2


Yes, if $c(gh)=c(g)+g\cdot c(h)$ for all $g,h \in G$, then $|G|c(g)= \sum_{h \in G} c(g) = \sum_{h \in G} c(gh)-g \cdot c(h) = a - g \cdot a$ where $a = \sum_{h \in G} c(h)$.


If I remember correctly the proof given in my group theory skript is as follows:

Let f be a derivation with values in $A$. We need to show that $|G|f$ is an inner derivation i.e. it is given by conjugation of the trivial derivation with some element $b \in A $ (conjugation in $A \rtimes G$).

Now let b be $\sum_{g \in G}f(g)$ and we can verify that $|G|f$ is equal to $b0b^{-1}$ where $0$ denotes the trivial derivation (and again I want to think of the conjugation in $A \rtimes G$).

I am aware of the fact that the notation is somewhat difficult to read because sometimes I identify $f(g) \in A$ with $((g,f(g)) \in A \rtimes G$. Nevertheless the basic proof should be valid.


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