Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $H^1(G,A)$?

share|cite|improve this question
up vote 7 down vote accepted

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 $a = \sum_{h \in G} c(h)$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.