Does H2 depends only on abelian quotient? Consider a finite group $G$ and an abelian group $N$. Let $G$ act trivially on $N$. Is $H^{2}(G,N)\cong H^{2}(G^{ab},N)$? ($G^{ab}=G/[G,G]$ the abelianization of $G$)
I don't get group cohomology so I have no intuitions about this, and computing $H^{2}$ is a pain, so I am not sure how to even look for examples.
 A: For trivial action, the relationship between $H^2(G, N)$ and the abelianization of $G$ is given by the universal coefficient theorem, which says that there is a split exact sequence
$$0 \to \text{Ext}^1(H_1(G), N) \to H^2(G, N) \to \text{Hom}(H_2(G), N) \to 0.$$
Here $H_1(G)$ is the abelianization. $H_2(G)$ is a group called the Schur multiplier, which can be computed using Hopf's formula, and it is generally nontrivial, so it's possible for the second term to be nontrivial even if the first term vanishes. 
Explicitly, take $G = A_5$. Since the abelianization of this group is trivial, the first term vanishes, and we have
$$H^2(A_5, N) \cong \text{Hom}(H_2(A_5), N).$$
Now we need to know $H_2(A_5)$. Fortunately this was computed by our ancestors long ago: it turns out to be $\mathbb{Z}_2$. So if we take $N = \mathbb{Z}_2$ we get a nontrivial cohomology group. 
If you're familiar with the relationship between $H^2(G, N)$ and central extensions of $G$ by $N$, you can also exhibit a nontrivial element of $H^2(A_5, \mathbb{Z}_2)$ by exhibiting a nontrivial $\mathbb{Z}_2$ central extension of $A_5$. There is a unique such extension called the binary icosahedral group $\widetilde{A}_5$, and it's the universal central extension of $A_5$. 
Not getting group cohomology is a typical experience, I think. I expect that for many people it's a long sequence of unmotivated definitions and computations. The real meaning behind the subject is in homotopy theory and higher category theory but this requires time, experience, and the right background to appreciate. 
A: No, these are not the same.  For example, if we choose $G=D_{4n}$ then its Abelianization is $(\Bbb Z/2)^2$, and so $H^2(D_{4n}^{ab}, \Bbb Z)=(\Bbb Z/2)^2$.  However, as shown here, the $H^2(D_{4n}, \Bbb Z)$ group is only $\Bbb Z/2$.
Side remark: in fact, non-commutative cohomology is one of the harder subjects that I've studied.  Sorry, I have to go to sleep, so it's faster to give another example than to go through the details of my initial example.
