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Let G be a group. Let $G^p$ be the completion of G with respect to the mod p lower central series of G.i.e. $G^p=\varprojlim_{q} G/\gamma_qG$, where $\gamma_qG$ is generated by all $\{[x_1,\cdots,x_s]^{p^t}:sp^t\geq q, x_i\in G\}$ and $[x_1,\cdots,x_s]$ is the iterated commutator $[\cdots[x_1,x_2],\cdots,x_s]$.

Is it true that the group ring of the completion ${\mathbb{Z}}/p[G^p]$ is the completion of the group ring ${\mathbb{Z}}/p[G]$ with respect to the products of the augmentation ideal? i.e. ${\mathbb{Z}}/p[G^p]\cong \varprojlim_{q}{\mathbb{Z}}/p[G]/I^q$, where $I$ is the augmentation ideal of the group ring ${\mathbb{Z}}/p[G]$?

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As Simon points out, the answer is no in a simple case and if you think about his argument the answer should probably be no as soon as $G^p$ is infinite. However, there is a statement that is very close to the question which is true: The completion of the group ring in the $I$-adic topology is isomorphic to $\varprojlim_q\mathbb Z/p[G/\gamma_qG]$. This follows from a result of Jennings-Lazard that the topology on $G$ defined by the $p$-lower central series equals that induced by the $I$-adic filtration on the group ring (see Quillen: On the associated graded ring of a group ring. J. Algebra 10 for an even more precise statement).

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  • $\begingroup$ it seems there is a contradiction. If the group $G$ is not abelian and not finitely generated, then $\varprojlim_{q}{\mathbb{Z}}/p[G/\gamma_qG] \rm{not}\cong \varprojlim_{q}{\mathbb{Z}}/p[G]/I^q$. However, for the same $G$, consider $\varprojlim_{q}{\mathbb{Z}}/p[G]/I^q$ as the completion of the group ring $\mathbb{Z}}/p[G]$ in the $I$-adic topology, it is isomorphic to $\varprojlim_{q}{\mathbb{Z}}/p[G/\gamma_qG]$. $\endgroup$
    – user2529
    Commented Sep 12, 2010 at 5:21
  • $\begingroup$ To clarify: By completion of the group ring in the $I$-adic topology do you mean $\varprojlim_{q}{\mathbb{Z}}/p[G]/I^q$ or something else? Thanks. $\endgroup$
    – user2529
    Commented Sep 12, 2010 at 5:33
  • $\begingroup$ Yes, I mean exactly what you propose. I have to admit that I am a little bit queasy about the non-fg case but I do not see the contradiction you suggest, could you be a bit more specific? I am not suggesting (if that is what you are saying) that $\varprojlim_q \mathbb Z/p[G/\gamma_qG] = \mathbb Z/p[G^p]$. $\endgroup$
    – Torsten Ekedahl
    Commented Sep 12, 2010 at 12:15
  • $\begingroup$ Beriefly the question is like this. Let G be a group. Is the $I$ adic completion of the group ring ${\mathbb{Z}}/p[G]$,denoted by $\varprojlim_{q}{\mathbb{Z}}/p[G]/I^q$ is isomorphic to ${\mathbb{Z}}/p[\varprojlim_{q}G/\gamma_qG]\cong\varprojlim_{q}{\mathbb{Z}}/p[G/\gamma_qG]$? $\endgroup$
    – user2529
    Commented Sep 14, 2010 at 6:47
  • $\begingroup$ Your stated isomorphism is not true. To the left you have only finite sums, to the right you have some infinite sums. $\endgroup$
    – Torsten Ekedahl
    Commented Sep 18, 2010 at 21:24
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I don't quite follow your definition of the mod $p$ lower central series as $s$ only seems to appear once in the definition. However whatever it is the answer is no.

If $G=\mathbb{Z}$ then the $I$-adic completion of $\mathbb{Z}/p[G]$ is isomorphic to a power series in one variable $T=x-1$ with coefficients in $\mathbb{Z}/p$ --- here $x$ is a generator of $G$.

Thus this $I$-adic completion is a commutative Noetherian algebra that is not finitely generated over the base field so cannot be a group algebra of any group since commutativity would imply that the group is abelian and Noetherianity would imply the group has no strictly ascending chains of subgroups. Amongst abelian groups only finitely generated groups have this latter property.

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  • $\begingroup$ I should perhaps add that these $I$-adically completed things are studied mostly under the name of Iwasawa algebras. See math.uiuc.edu/documenta/vol-coates/ardakov_brown.html for a survey of algebraic results. $\endgroup$
    – Simon Wadsley
    Commented Sep 10, 2010 at 16:22
  • $\begingroup$ First, thank you for your answer and the reference.:) Then I correct mistake of $\gamma_qG$. In the case of $G$ not abelian group, with what conditions (such as finitely generated not abel group) the answer will be Yes? $\endgroup$
    – user2529
    Commented Sep 11, 2010 at 0:32
  • $\begingroup$ I think Torsten has now basically answered that question. Probably never yes for infinite groups but you can say something. $\endgroup$
    – Simon Wadsley
    Commented Sep 11, 2010 at 9:46
  • $\begingroup$ Is the group $G\gamma_qG$ nilpotent for arbitrary group $G$? I think if $G$ is nilpotent, then $\varprojlim_{q}{\mathbb{Z}}/p[G]/I^q$ is isomorphic to $\varprojlim_{q}{\mathbb{Z}}/p[G/\gamma_qG]$. Am I right? any comments are welcome. thanks $\endgroup$
    – user2529
    Commented Sep 12, 2010 at 8:14
  • $\begingroup$ I think that all these things will be true when $G^p$ is profinite (in fact necessarily pro-$p$ in that case). Moreover $G$ finitely generated will suffice for this. I suggest consulting a book on profinite or pro-$p$ books such as those by J.S. Wilson or Dixon, Du Sautoy, Mann and Segal. $\endgroup$
    – Simon Wadsley
    Commented Sep 12, 2010 at 13:25

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