Augmentation ideal and the abelianization of $G$ On a qual problem recently, I came across the following fact:

If $G$ is a finite group, and $\mathfrak{a}$ is the augmentation ideal
  of the integral group ring $\mathbb{Z}G$, then
  $$\mathfrak{a}/\mathfrak{a}^2 \cong G/G', \qquad \text{as abelian groups.}$$

I understood the proof as far as it went, but I'm looking to absorb this on a deeper level. What is this "really" saying? Where does this fact come up? In what canonical resource "should" I have read about it already? Is it a simple fact about group cohomology in disguise? Is there a relation with the notion of the tangent space as $(I/I^2)^*$ for the ideal of functions vanishing at a point?
 A: Yes, this is a simple fact about group (co)homology in disguise.
Recall that the abelianization is $H_1(G, \mathbb{Z})$. This suggests that you should be trying to relate the augmentation ideal to this homology group via some long exact sequence. The augmentation ideal $I$, as a $G$-module, by definition fits into a short exact sequence
$$0 \to I \to \mathbb{Z}G \to \mathbb{Z} \to 0$$
which induces a long exact sequence in group homology the end of which goes
$$\dots H_1(G, \mathbb{Z} G) \to H_1(G, \mathbb{Z}) \to H_0(G, I) \to H_0(G, \mathbb{Z}G) \to H_0(G, \mathbb{Z}) \to 0.$$
By freeness $H_1(G, \mathbb{Z} G) = 0$. We also have $H_0(G, \mathbb{Z}G) \cong \mathbb{Z}$, and in fact the natural map to $H_0(G, \mathbb{Z})$ is an isomorphism. By exactness we get
$$H_1(G, \mathbb{Z}) \cong H_0(G, I)$$
and now it remains to verify that $H_0(G, I) \cong I/I^2$. In fact it's generally true that $H_0(G, M) \cong M/IM$. Throughout this argument there is no need to assume that $G$ is finite. 
It's unclear how to interpret this in terms of tangent spaces since $\mathbb{Z}[G]$ is usually noncommutative. 
