Talk:Group ring

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Some of the edits I just made were based on memory and could thus be slightly wrong. In particular, I'm not sure about the norm used to define the non-reduced C* algebra. Prumpf 00:48, 10 Sep 2004 (UTC)

It looks right to me. Also various relations between representations weakly contained in the left regular rep and the reduced C*-algebra should be put in at some point. I also think the name group algebra is preferable to group ring.CSTAR 01:02, 10 Sep 2004 (UTC)

There used to be separate group ring and group algebra articles; then they were merged. We need group rings such as Z[G] for abstract algebra. So, I think we should probably go back, now, to separate pages. Charles Matthews 07:47, 10 Sep 2004 (UTC)

My proposal was moving this page to group algebra and keeping only that one article as long as it doesn't get too long. I guess we could turn it into a disambiguation page as well, but I think I'd prefer some common text. For example, all group algebras are commutative if G is abelian (with the converse being true in the non-pathological cases), and for finite groups, all group algebras coincide (if considered over the complex numbers).

Maybe you want to fix the group ring section? Some previous author seems to have focussed on group rings over a field, but in my experience, the most common case is ZG, so maybe at least that case should be explained in a bit more detail. Prumpf 11:14, 10 Sep 2004 (UTC)

A very common case is CG, since this is the fundamental object of study in complex group representations. This probably explains the bias. Ben 11:26, 11 August 2006 (UTC)

The section 'Representations of a group ring' (the one after 'Representations of a group algebra') only talks about representations of group algebras, and doesn't say too much not said earlier. I'm not sure if the sections should merged, or if the second section should be rewritten to talk about modular systems or just integral representations. JackSchmidt 01:58, 8 July 2007 (UTC)Reply

Group ring definition does not imply usually that the ring is commutative. Jean-Louis Margot 12:12, 30 Sep 2005 (UTC)

Is this assumption required for the adjunction? I.e., is GrpRing an adjoint to ${\displaystyle GrpUnits:(R\downarrow Ring)\to Grp}$? Thehotelambush (talk) 19:00, 9 May 2009 (UTC)Reply

This could use a clean up - at the moment it's a bit of a hodge podge of facts and statements. Leland McInnes 21:16, 28 January 2006 (UTC)Reply

Perhaps the page could use a clean-up in terms of consistency with the PNG-style equations (small/large) and in terms of R[G] versus RG? If no-one objects, I'll go ahead and do this. Xantharius 17:18, 4 June 2007 (UTC)Reply

Sounds like a plan. Go for it. And while I can get in a preference: I think R[G] is nicer. ;-) Leland McInnes 18:37, 4 June 2007 (UTC)Reply

Okay. And on R[G] vs. RG, I concur. :) Xantharius 18:59, 4 June 2007 (UTC)Reply

In section 'group rings over an infinite group' one may take into account that the statement that C[G] is free of non-trivial idempotents if G is torsion-free is proved for all groups which satisfy the Baum-Connes conjecture. In fact, in that case even the reduced C*-algebra of G is free of nontrivial idempotents. The class of groups which are known to satisfy Baum-Connes is much larger than that of abelian, free, or elementary amenable groups, for instance, it includes amenable groups. —Preceding unsigned comment added by 134.76.82.127 (talk) 15:19, 17 September 2007 (UTC)Reply

Hi, Is this section even correct? Is it really known that if KG has no nontrivial idempotent elements then it has no zero-divisors? I think this is wrong, ie I think there is not any known proof that nonexistence of nontrivial idempotent elements implies the nonexistence of other types of zero-divisors. There are results along those lines for nilpotent elements in one of D Passman's books, but not for idempotent elements. Also Ithink that Kaplansky's conjecture mayp possibly be solved for some more cases than listed here. 137.205.56.18 (talk) 11:26, 17 February 2011 (UTC)Reply

This has been partially refuted by Giles Gardam, who found a meta-abelian torsion-free group with zero-divisors in its group algebra. — Preceding unsigned comment added by Mecciu (talk • contribs) 12:19, 15 October 2021 (UTC)Reply

That last comment is wrong, Gardam found a counterexample (in positive characteristic) to the unit conjecture, which is stronger than the zero-divisor conjecture. The group in which Gardam found a nontrivial unit is knwon to satisfy the zero-divisor conjecture. jraimbau (talk) 10:47, 16 October 2021 (UTC)Reply

The second paragraph makes it sound like all Group rings are R-Modules. This is not the case. Some are, but some are not. For example, a group ring with a non-abelian group is not an R-module; a group ring with a non-additive group could potentially not be an R-module, depending on the definition of "suitable multiplication" between the ring and the group. —Preceding unsigned comment added by 129.92.250.41 (talk) 19:19, 9 July 2008 (UTC)Reply

Howdy, you may be talking about a different thing. In this article, a "group ring" is a bit like a polynomial ring, but instead of being formal sums of monomials xⁿ, it is a formal sum of group elements. The precise definition is given in the article. A group ring R[G] is a very specific R-module with a multiplication that makes it into an R-algebra (assuming R is commutative, otherwise, just a ring extension). In particular, even for non-abelian groups the group ring R[G] is an R-module. This is the definition from Curtis, Passman, and the Springer EoM entry.

This is not talking about the general idea of a group that is a subset of a ring. JackSchmidt (talk) 19:36, 9 July 2008 (UTC)Reply

I understand your definition of the group ring, perhaps it's the definition of the module we're not agreeing on. I'm using MacLane & Birkhoff, Algebra, which defines an R-module specifying an additive abelian group. —Preceding unsigned comment added by 129.92.250.41 (talk) 15:25, 15 July 2008 (UTC)Reply

I checked just to be sure, and the Wikipedia definition also requires that the R-module be an additive abelian group. —Preceding unsigned comment added by 129.92.250.41 (talk) 15:30, 15 July 2008 (UTC)Reply

The group ring is an additive abelian group. For instance, choosing the coefficient ring R to be the integers, and the group G to be the nonabelian group with six elements, then the group ring R[G] is just Z⁶ = Z × Z × Z × Z × Z × Z, as an abelian group. JackSchmidt (talk) 16:21, 15 July 2008 (UTC)Reply

Oh, I think I see now. It is the group ring itself which is acting as the additive abelian group (rather than the group used to construct the group ring) which is then mapped with the ring back into the group ring. Thank you for clearing that up.

Yup, exactly. No problem. JackSchmidt (talk) 17:07, 15 July 2008 (UTC)Reply

I worked over the definition, added a couple of canonical properties, and removed the "confusing" tag. Hope it's okay. LDH (talk) 02:00, 20 November 2008 (UTC)Reply

What if the ring doesn't have any units?=r peterson216.86.177.36 (talk) 22:32, 24 February 2012 (UTC)Reply

The article says group rings satisfy ``a universal property, citing Polcino & Sehgal (2002), p. 131. Can somebody with a copy of this text (or somebody who otherwise knows) fill in what the property is?71.36.206.9 (talk) 16:00, 23 August 2012 (UTC)Reply

You can have a look in Google Books: http://books.google.com/books?id=7m9P9hM4pCQC&pg=PA131 --Kompik (talk) 06:38, 15 December 2012 (UTC)Reply

The "Hopf algebra" material in this article is inconsistent with the article group Hopf algebra, which says that the notions of "group algebra" and "group Hopf algebra" coincide only when the order of the group is finite. Thoughts? Thatsme314 (talk) 05:42, 9 October 2023 (UTC)Reply

I think it's best to remove the second part of this sentence in the lede (to wit : "A group algebra over a field has a further structure of a Hopf algebra; in this case, it is thus called a group Hopf algebra." <- "A group algebra over a field has a further structure of a Hopf algebra." The article on "group hopf algebra" doesn't have much additional content compared to that on Hopf algebras. jraimbau (talk) 06:46, 9 October 2023 (UTC)Reply