Let $A$ be any bialgebra (associative, unital, etc.) over a ring $k$. Then among other things it has a counit $\epsilon : A \to k$, and hence an *augmentation ideal* $I = \ker \epsilon$, which is a Hopf ideal. Any ideal determines a filtration
$$ A \supseteq I \supseteq I^2 \supseteq \dots$$
and hence an *associated graded* vector space
$$ \operatorname{gr} A = \frac A I \oplus \frac I {I^2} \oplus \frac {I^2} {I^3} \oplus \dots $$
Since $\epsilon: A \to A/I = k$ is a morphism of bialgebras, $I$ is a "Hopf ideal", and hence $\operatorname{gr} A$ is a bialgebra. But it is graded with zero part a Hopf algebra, hence $\operatorname{gr} A$ is Hopf.

Moreover, $\operatorname{gr} A$ is generated by $I/I^2$ as an algebra, and each element of $I/I^2$ is primitive. So in particular $\operatorname{gr} A$ is generated by its primitive part, and hence is a quotient of some universal enveloping algebra of some graded Lie algebra $\mathfrak a$; moreover, $\mathfrak a$ is generated as a Lie algebra by its degree-$1$ part, which is precisely $I/I^2$. (Generically the surjection is not an isomorphism, as $A$ might be finite-dimensional but ${\rm U}\mathfrak a$ never is.)

For now, I am interested in the following special case. The ground ring $k$ is a field of characteristic $0$. $G$ is a discrete group, and $A = k\cdot G$ is its group algebra. Then I can calculate $I / I^2$ has as its basis the non-identity elements of the abelianization $G_{\rm ab}$ of $G$. $I^2/I^3$ is spanned by pairs $(g,h) \in G\times G$, modulo $(g,1) = 0$ and $(g,hk) = (g,h) + (g,k)$, and the same relations on the other side (and maybe more relations?), and the multiplication $(I/I^2)^{\otimes 2} \to (I^2/I^3)$ is $gh = (g,h)$.

Anyway, $\operatorname{gr} (k\cdot G)$ feels a lot like some homological construction, but I don't know much homology theory. So:

Question:What's a hands-on description of $\operatorname{gr} (k\cdot G)$? How does it relate to other constructions I might have met?

**Update:** I definitely made an error in the above, which just means that I understand less than I thought. I'd like to explain an example, and then ask a second, more precise version of the above question.

Suppose that $G$ is a abelian. Then $k\cdot G$ is a commutative cocommutative Hopf algebra, and so is the algebra of functions on some affine algebraic group, which for want of a better name I'll call $G^\vee$ --- it's the "dual group" to $G$. The augmentation ideal then corresponds to the identity element $e\in G^\vee$, and the filtration is the filtration of the algebra of functions on $G^\vee$ in Taylor series. Then $\operatorname{gr}(k G)$ is the symmetric algebra of ${\rm T}^*_e G^\vee$. (If you complete at the augmentation ideal, you're writing down "formal power series near $e$".)

For example, when $G$ is the group with two elements and $\operatorname{char}(k) = 0$, then $I^n = I$ for $n>0$, and so $\operatorname{gr}(kG) = k$ is one-dimensional, not two-dimensional like $kG$. I was confused in my original question, because there are two kinds of filtrations on a vector space --- going up and going down --- and in one of them $\operatorname{gr}$ preserves dimensions, and in the other it may not.

So this shows that I may have been wrong when I wrote "Generically the surjection is not an isomorphism". Note that in the abelian case, $\operatorname{gr}(kG)$ is a symmetric algebra, and in particular is a universal enveloping algebra (the Hopf structure is the right one).

In the nonabelian case, I can't use quite as much geometric language, but I expect something similar should still be true:

Updated question:Is $\operatorname{gr}(kG)$ a universal enveloping algebra? If so, how is the corresponding Lie algebra related to $G$ (which remember is just a discrete group)?

elementof a group...oy vey!). A more modern account of the same material, where the results just to by "Jennings's Theorem", is given in Dixon et al, "Analytic pro-p Groups." $\endgroup$ – Cam McLeman Feb 16 '11 at 13:49