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Let $R$ be an arbitrary associative ring with identity. When does there exist a group $G$ and a field $F$ such that $F[G] = R$?

Do we obtain more rings as $F[G]$ if we loosen the condition that $F$ be a field? When can we get $G$ to be a finite group?

There seem to be some obvious restrictions: The characteristic of $F$ is equal to that of $R$, for instance. Also, when $F$ and $G$ are finite, then so is $R$. Also, all elements of $G$, considered as a subset of $R$, are units in $R$. If $R$ is commutative, $G$ has to be abelian. But I fail to see much more.

Cloudscape
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    A relevant (sufficient) condition might be the additional structure of Hopf algebra on $R$ : every commutative finite dimensional Hopf algebra is the group algebra of some finite group. – Watson Jan 07 '18 at 13:59
  • I don't know anything about Hopf algebras, but I'll be looking them up just now! – Cloudscape Jan 07 '18 at 14:00
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    You might be interested by this MO question. – Watson Jan 07 '18 at 14:00
  • May I inquire whether there exists a good book on these issues? I always like to read things without the screen flickering. – Cloudscape Jan 07 '18 at 14:02
  • By "these issues" I mean Hopf algebras, bialgebras, coalgebras and the like. – Cloudscape Jan 07 '18 at 14:02
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    A necessary condition seems to be the existence of non-trivial idempotent elements in $R$. By the way, I don't know if this could help, but $G \mapsto k[G]$ left adjoint of $A \mapsto A^{\times}$, i.e. $\mathrm{Hom}{k\text{-alg.}}(k[G], A) = \mathrm{Hom}{\mathrm{Grp}}(G, A^{\times})$ (so maybe the question could be reformulated by asking "which groups arise as group of units of some ring?", which has been discussed many times on MSE). – Watson Jan 07 '18 at 14:05
  • This question and maybe that one are discussing the relation between Hopf algebras and group rings. – Watson Jan 07 '18 at 16:04
  • @Watson In your first comment, did you mean cocommutative? Also I believe this statement requires the base field to be algebraically closed. – Julian Rosen Jan 07 '18 at 19:42
  • @JulianRosen : I guess that you are right. I'm a bit confused because for affine algebraic groups, it seems that commutative Hopf algebra give the correct notion (see here). By the way, my first comment was just citing the MO question I referred to. – Watson Jan 07 '18 at 20:30
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    @Watson: your claim about Hopf algebras is just false. Among Hopf algebras, the group algebras are distinguished by being generated by their grouplike elements, and there are many finite-dimensional Hopf algebras, even commutative and cocommutative, with no grouplike elements at all. Also, your suggestion about adjunctions is not a reformulation of the question. – Qiaochu Yuan Jan 07 '18 at 21:27
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    @Watson If $G$ is a finite group, the group ring of $G$ dual to the coordinate ring of (the finite algebraic group associated with) $G$. This is why coordinate rings of groups are commutative while group rings are cocommutative. – Julian Rosen Jan 07 '18 at 21:31
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    (My comment above should read "no nontrivial grouplike elements"; the identity is always grouplike. An explicit example of a commutative and cocommutative finite-dimensional Hopf algebra with no nontrivial grouplike elements given here: https://mathoverflow.net/questions/259307/cartier-kostant-milnor-moore-theorem/259886#259886 ) – Qiaochu Yuan Jan 08 '18 at 00:50
  • Related: https://math.stackexchange.com/questions/2814531/group-like-elements-of-a-hopf-algebra-and-the-group-algebra – Watson Jun 10 '18 at 12:47
  • Related: https://mathoverflow.net/questions/13502/which-r-algebras-are-the-group-ring-of-some-group-over-a-ring-r?rq=1 – Watson Jun 14 '18 at 07:44

2 Answers2

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The simplest case, which is already quite complicated, is if $F$ is algebraically closed of characteristic $0$ and $G$ is finite, or equivalently if $R$ is finite dimensional over $F$ (note that $R$ is necessarily an $F$-algebra). Then $F[G]$ is semisimple by Maschke's theorem, and furthermore splits up as a finite product of matrix algebras $M_n(F)$, one for every irreducible representation of $G$ over $F$ of dimension $n$. So in this special case the question reduces to asking:

Which tuples $(n_1, \dots n_k)$ arise as the dimensions of the irreducible representations of a finite group $G$ over $F$?

I don't think there's any hope of a simple answer to this question. Here are some necessary conditions. Note that $|G| = \sum n_i^2$.

  1. At least one of the $n_i$ must be equal to $1$ (since the trivial representation is always irreducible), and in fact the number of $n_i$ equal to $1$ must divide $|G|$ (since it's the order of the abelianization).
  2. If $|G|$ is prime then each of the $n_i$ must be equal to $1$ (since in this case $G$ is a cyclic group of prime order).
  3. Each of the $n_i$ must divide $|G|$.

For example, $F \times M_2(F)$ is not a group algebra (because $1^2 + 2^2 = 5$ and the only group of order $5$ is $C_5$), but $F \times F \times M_2(F)$ is (it's the group algebra of $S_3 \cong D_3$).

Qiaochu Yuan
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    If $F$ is just a commutative ring but not a field we still get constraints on the structure of $F[G]$ by taking morphisms from $F$ into fields, especially algebraically closed fields as above, and tensoring with them. – Qiaochu Yuan Jan 08 '18 at 20:21
  • I've finally gotten round to learning the relevant representation theory. A simple answer to the problem is perhaps out of reach, but at least for finite simple groups there should be character tables. – Cloudscape Jul 19 '24 at 16:44
  • Also, it seems as though the degree of an irreducible representation divides the index of any Abelian normal subgroup. – Cloudscape Jul 19 '24 at 17:18
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If your group $G$ contains a non-trivial element of finite order then your group ring $R:=SG$ contains a zero-divisor (here, $S$ is just a ring, not necessarily a field): If $g^n=1$ then $$\begin{align*} &(1-g)(1+g+g^2+\cdots+g^{n-2}+g^{n-1})\\ &=(1+g+g^2+\cdots+g^{n-2}+g^{n-1})-(g+g^2+g^3+\cdots+g^{n-1}+1)\\ &=0.\end{align*}$$

Note that in his PhD thesis Higman proved an "opposite" result using locally indictable groups (groups where every non-trivial, finitely generated subgroup maps onto $\mathbb{Z}$): G. Higman, The units of group rings, Proc. London Math. Soc., vol. 46 (1940), pp. 231- 248.

user1729
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