most recent 30 from http://mathoverflow.net 2013-05-22T05:02:34Z http://mathoverflow.net/feeds/question/38289 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38289/comparing-lower-central-series-and-augmentation-ideal-completions Colin Tan 2010-09-10T10:00:16Z 2010-09-11T06:01:50Z

<p>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 $[\cdots[x_1,x_2],\cdots,x_s]$.</p> <p>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 ${\mathbb{Z}}/p[G]$?</p>

http://mathoverflow.net/questions/38289/comparing-lower-central-series-and-augmentation-ideal-completions/38318#38318 Simon Wadsley 2010-09-10T14:43:57Z 2010-09-10T14:43:57Z

<p>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.</p> <p>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 $G$. </p> <p>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.</p>

http://mathoverflow.net/questions/38289/comparing-lower-central-series-and-augmentation-ideal-completions/38393#38393 Torsten Ekedahl 2010-09-11T06:01:50Z 2010-09-11T06:01:50Z

<p>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).</p>