At pages 25 and 26 of his notes (http://www.math.umn.edu/~webb/oldteaching/Year2010-11/8246CohomologyNotes.pdf), Peter J. Webb calculates $H^2(C_2 \times C_2, \mathbb{F}_2)$ (with $\mathbb{F}_2$ a trivial $\mathbb{Z}[C_2 \times C_2]$-module) explicitly using the presentation $\langle a, b \mid a^2 = b^2 = [a,b] = 1 \rangle$ of $C_2 \times C_2$.
Let $G = C_2 \times C_2$ and $IG$ the augmentation ideal of the group ring $\mathbb{Z}G$. The idea is basically to find a description of $\mathrm{Hom}_{\mathbb{Z}G}(R/R',\mathbb{F}_2)$ where $R$ is the subgroup generated by the relations $\{a^2,b^2,[a,b]\}$ of the free subgroup generated by $a$ and $b$.
As a first step, he observes that $$\mathrm{Hom}_{\mathbb{Z}G}(R/R',\mathbb{F}_2) \cong \mathrm{Hom}_{\mathbb{Z}G}(\mathbb{F}_2 \otimes_{\mathbb{Z}} (R/R')/(IG\cdot R/R'),\mathbb{F}_2)\cong \mathrm{Hom}_{\mathbb{Z}}(\mathbb{F}_2 \otimes_{\mathbb{Z}G} R/R',\mathbb{F}_2).$$
The way I see this, is as follows. We have that $$\mathrm{Hom}_{\mathbb{Z}G}(R/R',\mathbb{F}_2) \cong \mathrm{Hom}_{\mathbb{Z}G}((R/R')/(IG\cdot R/R'),\mathbb{F}_2)$$ since every element of $\mathrm{Hom}_{\mathbb{Z}G}(R/R',\mathbb{F}_2)$ factors through $(R/R')/(IG\cdot R/R')$ because the action of $G$ on $\mathbb{F}_2$ is trivial. Next, we see that $$\mathrm{Hom}_{\mathbb{Z}G}((R/R')/(IG\cdot R/R'),\mathbb{F}_2) \cong \mathrm{Hom}_{\mathbb{Z}G}((R/R')/(IG\cdot R/R'),\mathrm{Hom}_\mathbb{Z}(\mathbb{F}_2,\mathbb{F}_2)) \cong \mathrm{Hom}_{\mathbb{Z}}(\mathbb{F}_2 \otimes_{\mathbb{Z}G} R/R',\mathbb{F}_2)$$ because $\mathrm{Hom}_\mathbb{Z}(\mathbb{F}_2,\mathbb{F}_2) \cong \mathbb{F}_2$ and by the tensor-Hom duality. Do you guys think this is the reasoning the writer intended or is there a more logical approach? Is this a trick often used?
The calculations that follow in the notes, I do understand (more or less). Only his last conclusion is somewhat of a mystery to me. He seems to use the fact that $$\mathrm{Hom}_{\mathbb{Z}}(\mathbb{F}_2 \otimes_{\mathbb{Z}G} R/R',\mathbb{F}_2) \cong (\mathbb{F}_2 \otimes R/R')/(IG \cdot \mathbb{F}_2 \otimes R/R').$$ Is this true or do I misinterpret his conclusion? And if it's true, why?
If I put $C = R/R' $ then in Hom$_G(\mathbb{F}_2 \otimes_\mathbb{Z} C/I_GC, \mathbb{F}_2)$ all of the groups have trivial $G$ action, So you can put Hom$_\mathbb{Z}(\mathbb{F}_2 \otimes_\mathbb{Z} C/I_GC, \mathbb{F}_2)$ instead of it and You can easily see that: $$ \mathbb{F}_2 \otimes_\mathbb{Z} C/I_GC = (\mathbb{F}_2 \otimes_\mathbb{Z} C)_G = \mathbb{F}_2 \otimes_G C $$