Let $M$ be a $G$-module, and let $f \in Z^2(G,M)$. Show that $f(1,1)=f(1,\sigma)=f(\sigma,1),\forall \sigma \in G$. I'm doing the problems about group cohomology in Morandi,Field and Galois Theory.
The problem 3 in page 102,
Let $M$ be a $G$-module, and let $f \in Z^2(G,M)$. Show that $f(1,1)=f(1,\sigma)=f(\sigma,1),\forall \sigma \in G$.
By the $2$-cocycle condition:  
$\sigma_1f(\sigma_2,\sigma_3)+f(\sigma_1,\sigma_2\sigma_3)=f(\sigma_1\sigma_2,\sigma_3)+f(\sigma_1,\sigma_2)$
i can easily show $f(1,1)=f(1,\sigma)$ by taking $(\sigma_1,\sigma_2,\sigma_3)=(1,1,\sigma)$ to the equation above,
but how could i get $f(1,1)=f(\sigma,1)$?
 A: I don't think that the claim holds for all 2-cocycles (I guess you need to do some normalizing for it to work out, but I'm too rusty to recall the details).
Recall that any function $\phi:G\to M$ gives rise to a 2-coboundary $f$ via the recipe
$$
f(\sigma,\tau):=\sigma\cdot \phi(\tau)-\phi(\sigma\tau)+\phi(\sigma).
$$
It is an easy standard exercise to check that a 2-coboundary is always a 2-cocycle.
Consider the following example. Let $G=\langle\sigma\rangle\simeq C_2$ be a cyclic group of order two, acting on $M=\Bbb{Z}_3$ via $1\cdot x=x$, $\sigma\cdot x=-x$ (so $M$
becomes a $G$-module). Let us define $\phi:G\to M$ by declaring $\phi(1)=1$,
$\phi(\sigma)=2=-1$. Then
$$
\begin{aligned}
f(1,1)&=1\cdot\phi(1)-\phi(1)+\phi(1)=1,\\
f(1,\sigma)&=1\cdot\phi(\sigma)-\phi(\sigma)+\phi(1)=1,\\
f(\sigma,1)&=\sigma\cdot\phi(1)-\phi(\sigma)+\phi(\sigma)=-1,\\
f(\sigma,\sigma)&=\sigma\cdot\phi(\sigma)-\phi(1)+\phi(\sigma)=-1.
\end{aligned}
$$
You see that $f(1,g)=f(1,1)$ for all $g\in G$, but $f(\sigma,1)\neq f(1,1)$.
So the claim does not hold for all 2-coboundaries, and hence cannot hold for all 2-cocycles either.
