This question was inspired by Qiaochu's recent question, Which commutative groups are the group of units of some field? - my question is close to being the inverse of it.

As mentioned here, given a ring $R$, the functor $GrpRing:Grp\rightarrow R$-$Alg$ taking a group $G$ to the group ring $R[G]$ is left adjoint to the functor $GrpUnits:R$-$Alg\rightarrow Grp$ taking an $R$-algebra to its group of units. What is the essential image of $GrpRing$, i.e., which $R$-algebras are isomorphic to the group ring of some group over $R$?

One might ask more generally, when is a ring $R$ a group ring over some ring, not fixed at the outset? (Obviously, any ring $R$ is isomorphic to $R[1]$, the group ring of the trivial group over itself, but let's exclude this trivial case.)

  • $\begingroup$ Just a point in terminology: the "image" of a functor as you defined it above is known as its essential image. In the same vein, a functor F: A → B such that every y ∈ B is isomorphic to F(x) for some x ∈ A is called essentially surjective. $\endgroup$ – Alberto García-Raboso Jan 31 '10 at 18:31
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    $\begingroup$ Several answers seem to be implicitly assuming that the group is finite, especially David Treumann's. Is this intended as part of the question? $\endgroup$ – Pete L. Clark Feb 1 '10 at 0:05
  • $\begingroup$ Hmmm - no, that wasn't what I intended, though Wikipedia tells me that much less is known about the infinite case, so I suppose it's okay if answers are only valid for the finite case (as long as the answer mentions which case it is handling). $\endgroup$ – Zev Chonoles Feb 1 '10 at 0:20
  • $\begingroup$ I realize this is an old post, but in view of answers such as Yemon's, I presume more mileage would be gotten if instead of working with rings/algebras, one works with bialgebras. $\endgroup$ – Todd Trimble Nov 26 '16 at 14:08

By the way, there's one very interesting restriction on group algebras that has yet to be mentioned. If $k$ is a field, then even over characteristic p, the group algebra has the remarkable property that all its projective modules are also injective. The proof is simple: the map $k[G]\times k[G]\to k$ given by the value at the identity of the product is a non-degenerate pairing. So $k[G]$ is the dual of a projective and thus injective, so its summands are as well.

This may not seem impressive to those of you who don't study finite dimension algebras, but this is an extraordinary property, and show that most finite dimensional algebras don't show up in group algebras.

  • $\begingroup$ This is for finite groups, of course (for $G$ infinite cyclic, $k[G]$ is a projective and non-injective $k[G]$-module). $\endgroup$ – YCor Nov 26 '16 at 17:01

A group algebra over the complex numbers, like any semisimple algebra, is isomorphic to a product of matrix rings, $$R = M_{d_1 \times d_1} \times M_{d_2 \times d_2} \times \cdots \times M_{d_n \times d_n}$$ The d_i that appear are exactly the dimensions of the irreducible representations of G. I don't know how to classify all sets of numbers that appear in this way (but the answer is not "all of them").

Over nonalgebraically closed fields of characteristic zero, nontrivial divison algebras can appear in the group algebra. I don't know if there's a restriction on the division algebras that can appear.

The situation in characteristic p is more complicated, but we can say something. A finite-dimensional algebra A has a square matrix of numerical invariants called the Cartan matrix. (I am not talking about the Lie-theoretic Cartan matrix, but I would be interested to know if they are named the same for a reason.) There is a one-to-one correspondence between simple A-modules and indecomposable projective A-modules, and the ij entry in the Cartan matrix is the Jordan-Holder index of the ith simple module in the jth projective module.

The theory of Brauer (and the subject of "part 3" of Serre's famous book on representation theory of finite groups) imposes strong conditions on the Cartan matrix when A is the group algebra of a finite group (over a large finite field). It must admit a factorization as D.D^t, where D is another matrix with nonnegative integer entries. (D is the "decomposition matrix" which describes what happens to simple modules in characteristic zero when reduced mod p.) For instance the Cartan matrix must be symmetric.

What if we work over a ring that is not a field, e.g. the integers? Here's a comment on Yemon's point that there are many pairs of groups G and H for which the group rings C[G] and C[H] are isomorphic. It is more difficult to construct such isomorphisms over smaller rings, and whether or not an isomorphism of the form Z[G] = Z[H] implied that G = H was an open problem for a long time (the "isomorphism problem for integral group rings," posed by Brauer in the 60s) A counterexample was found by Hertweck 10 years ago:


(Pete points out above that I am assuming G is finite. When G is infinite C[G] cannot be analyzed by Wedderburn's theorem, there's no such thing as a Cartan matrix, everything breaks down. Is there a counterexample to the isomorphism problem simpler than Hertweck's if we do not require G and H to be finite?)

  • $\begingroup$ Thanks for the in depth answer, and interesting reference - for some reason, group algebras/rings are really capturing my interest lately. $\endgroup$ – Zev Chonoles Jan 31 '10 at 20:31
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    $\begingroup$ Regarding whether or not there are easier infinite counterexamples to the isomorphism problem: I think that Hertweck's example was the first counterexample, period, and that it's even more surprising that the examples are finite. I don't know what's happened in the interim. There's a brief history up to Hertweck here: ime.usp.br/~polcino/group_rings $\endgroup$ – Autumn Kent Feb 3 '10 at 17:05

Reid Barton's very nice answer to Computing the structure of the group completion of an abelian monoid, how hard can it be? contained a pointer to the wikipedia page for the Eilenberg-Mazur swindle. Towards the bottom of that page, one finds the following relevant paragraph.

Example: (Lam 2003, Exercise 8.16) If $A$ and $B$ are any groups then the Eilenberg swindle can be used to construct a ring $R$ such that the group rings $R[A]$ and $R[B]$ are isomorphic rings: take $R$ to be the group ring of $A + B + A + B + \ldots$

  • $\begingroup$ Wow! That's a really clever trick they've got there. $\endgroup$ – Zev Chonoles Feb 5 '10 at 0:37

It doesn't answer your question, but might be worth noting en passant: depending on the choice of $R$, two different groups might give rise to the same group ring. The standard example is that when ${\mathbb C}$ is the field of complex numbers, and $G$ and $H$ are two finite abelian groups with the same cardinality, then the group algebras ${\mathbb C}G$ and ${\mathbb C}H$ are isomorphic. (In fact, even if you equip them with their natural $C^\ast$-algebra norms, the corresponding $C^\ast$-algebras will be *-isomorphic.)

If we work over a field $k$ of characteristic zero, then an old result of Kaplansky tells us that for any group $G$ (not necessarily finite) the group algebra $kG$ is "directly finite", in the sense that every left invertible element is right invertible. So any $k$-algebra which fails to have this property cannot arise as $kG$ for some group $G$.

There may perhaps also be homological obstructions of various sorts, but I don't know too much about such things in this setting.

  • $\begingroup$ Excellent points - out of curiosity, is there a characterization of rings $R$ for which $G\not\simeq G'$ implies $R[G]\not\simeq R[G']$? $\endgroup$ – Zev Chonoles Jan 30 '10 at 21:07
  • $\begingroup$ Pete's posted an answer that there are in fact no such rings. $\endgroup$ – Zev Chonoles Feb 5 '10 at 0:38

A $R$-algebra $A$ is a group algebra over $R$ if and only if there exists a $R$-module basis of $A$, which is also central and a subgroup of $A^*$. Of course, this is trivial, but I don't think that there is a nice characterization.

Observe that some $R$-algebras $A$ don't have any $R$-homomorphism $A \to R$ at all. However, group algebras admit the augmentation map $R[G] \to R$. Besides, the diagonal map $G \to G \times G$ induces $R[G] \to R[G] \otimes R[G]$ and endow $R[G]$ with the structure of a coalgebra.

  • $\begingroup$ It seems like the polynomial ring R[t] (= the monoid ring R[(N,+)]) is a counterexample to your claim. Are you thinking that the R-algebra should be finite? $\endgroup$ – Pete L. Clark Jan 30 '10 at 23:35
  • $\begingroup$ which group do you take? $\endgroup$ – Martin Brandenburg Jan 31 '10 at 0:23
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    $\begingroup$ I believe Pete is saying that the basis {t^k} for k >= 0 is a central submonoid of R[t], yet R[t] is not a group algebra (note the units in R[t] are just the units in R, but a group ring is always spanned over R by invertible elements). This shows your characterization in the first paragraph is not correct. $\endgroup$ – Tom Church Jan 31 '10 at 1:01
  • $\begingroup$ Oh, my remark applies to monoid rings. I've edited it. $\endgroup$ – Martin Brandenburg Jan 31 '10 at 1:16
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    $\begingroup$ Perhaps you meant A^\times instead of R^* $\endgroup$ – S. Carnahan Jan 31 '10 at 3:52

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