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I'm just beginning to learn about group cohomology and I have a question about calculating the group of cocycles. My reasoning may be wrong at any of the following steps or perhaps all of them.

So, I'd like to calculate $Z^2(\mathbb{Z}/2, \mathbb{Z}/2)$ with respect to the trivial action. This group is given by the kernel of the map $$d_2: C^2\rightarrow C^3,$$ which takes $f(g_1,g_2)$ to $$d_2f(g_1,g_2,g_3)=g_2+g_3 -f(g_1+g2, g_3)+f(g_1,g_2+g_3) - f(g_1,g_2),$$ assuming I've done everything up until this point correctly. Thus, I am looking for the collection of functions from $(\mathbb{Z}/2)^2$ to $\mathbb{Z}/2$ that are taken to the identity of $C^3$. Since $C^3$ is an abelian group under pointwise addition, its identity element should be the function that sends every $g_i$ to $0\in \mathbb{Z}/2$.

But, suppose $g_1=g_2=g_3=0$. Then, for $f$ to be in the kernel of the coboundary homomorphism, we require $$d_2f(0,0,0)=f(0,0)=0.$$ Similarly, since we want $d_2f$ to be zero for all $g_i$, if $g_1=g_2=1$ and $g_3=0$, then we need $$d_2(1,1,0)=f(0,0)=1,$$ a contradiction. I know that I must be missing something fundamental, so if anyone could clarify the picture, I'd be very thankful.

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The formula you wrote for $d_2f(g_1,g_2,g_3)$ is wrong. It should be $f(g_2,g_3)-f(g_1+g_2,g_3)+f(g_1,g_2+g_3)-f(g_1,g_2)$ – Mariano Suárez-Alvarez Oct 25 '12 at 2:14
Isn't the first term of the formula given by the action of $g_1$ on $(g_2+g_3)$? – Alexander Sibelius Oct 25 '12 at 2:25
Ah, I see. It's given by the action of $g_1$ on $f(g_2,g_3)$. I'll take a look at it again. – Alexander Sibelius Oct 25 '12 at 2:35

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