# Does $f(gh)=gf(h)+f(g)$ make $H^1(G,A)$ small enough?

I'm currently reading the text Beginnings of Group Cohomology, trying to understand $H^1(G,A)$ for $G$ a group and $A$ a $G$-module. In the article, the space $H^1(G,A)$ is motivated as a space of obstructions to lifting fixed points, as I now briefly explain (for fixing notation and to get my own thoughts straight). Feel free to skip to the questions at the bottom.

Suppose $B$ is an abelian group on which $G$ acts and $A$ is a subgroup of $B$ which is stable under the action of $G$. Then $G$ acts on $A$ and also on $B/A$. Let $\bar{b}\in B/A$ be a fixed point for the $G$-action. We can then try to lift this point to a fixed point in $B$.

Pick any representative $b\in B$ of $\bar{b}$. Then we can look at a function $$f_b : G\to A : g\mapsto gb-b ,$$ and this function is identically zero exactly if $b$ is a fixed point. Moreover, picking another representative $b-a\in B$ (with $a\in A$) gives rise to the function $$f_{b-a} : G\to A : g\mapsto g(b-a)-(b-a) = (gb-b)-(ga-a) .$$ Noting that $\bar{b}$ can be lifted to a fixed point precisely if $f_{b-a}=0$ for some $a\in A$, we conclude that $\bar{b}$ can be lifted to a fixed point exactly if $f_b$ is of the form $G\to A : g\mapsto ga-a$ for some $a\in A$. Write $$B^1(G,A) = \{ G\to A : g\mapsto ga-a \mid a\in A \} .$$ Then $\bar{b}$ can be lifted to a fixed point exactly if $f_b\in B^1(G,A)$ for some (and hence every) representative $b$ of $\bar{b}$.

Now for any point $b\in B$ the function $f_b$ satisfies the condition $$\tag{\ast} f_b(gh) = gf_b(h)+f_b(g) .$$ Write $Z^1(G,A)$ for all maps $G\to A$ satisfying ($\ast$). We then define the space $$H^1(G,A) = \frac{Z^1(G,A)}{B^1(G,A)}$$ (this is a quotient of abelian groups). We can now associate to any point $\bar{b}\in B/A$ an element of $H^1(G,A)$ by $\bar{b}\mapsto \bar{f_b}$ and a fixed point $\bar{b}$ can be lifted exactly if we associate $0\in H^1(G,A)$ to it. This is very nice. I want to know that $H^1(G,A)$ is not "too big" in the following sense.

Why do we define $Z^1(G,A)$ the way we do? What is special about $(\ast)$? In particular, is this condition strong enough to ensure that $f_b$ is of the form $g\mapsto gb-b$ for some $b$ in some $G$-module $B$ that has $A$ as a $G$-invariant subgroup?

Another way to state my question would be: is every element of $H^1(G,A)$ associated to some $\bar{b}$ in some $G$-module $B$ that has $A$ as a $G$-invariant subgroup? If the answer is "yes", then the matter is settled. If the answer is "no", then why do people choose to work with $(\ast)$?

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I previously answered a similar question. – Zhen Lin Feb 4 '13 at 17:48
@ZhenLin: A lot of the math in your answer is too advanced for me (for now). Nevertheless, I'll take a careful look at it and make the best of it. I suspect your answer shows that the answer to my question is "yes". Am I correct? – Daan Michiels Feb 5 '13 at 12:52
Actually, if you're just interested in whether it is true that every element of $H^1 (G, A)$ is associated with some $\bar{b}$, then you only need to know that every $G$-module $A$ can be embedded in an injective $G$-module $B$. Then the long exact sequence in cohomology becomes $\cdots \to H^0 (G, B/A) \to H^1 (G, A) \to 0$. – Zhen Lin Feb 5 '13 at 13:31