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?

  • $\begingroup$ In case it helps, an explicit isomorphism is $g[G,G]\mapsto (g-1)+I^2$. The point here is that the elements $\lbrace g-1\rbrace_{g\in G}$ form a basis for $I$ as a $\mathbb{Z}$-module, noting that $I$ is (by definition) the kernel of $\mathbb{Z}[G]\to\mathbb{Z}$ given by $g\mapsto 1$ for $g\in G$. $\endgroup$ – Chris Gerig Dec 27 '15 at 23:12

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.


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