Let k be a field and let G be a finite group. I would like to know if there is any nice description of the group of units in the group algebra kG. (If there is no nice answer in this generality, assume that the characteristic of k is p and that G is a p-group.)
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3 Answers
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If $k$ has characteristic $p$ and $G$ is a finite $p$-group, then the augmentation ideal $J$ of $kG$ is nilpotent and of codimension one, hence coincides with the Jacobson radical. This means that the units of $kG$ are precisely the complement $kG \setminus J = \{ \sum_g c_g g \mid \sum_g c_g \neq 0 \}$ of the augmentation ideal.
Is this the description you asked for, or do you rather want a procedure for determining the isomorphy type of the group of units (which is the product of $k^{\times}$ and a $p$-group of order $|k|^{|G|-1}$) from that of $G$? When $G$ is abelian, this problem is completely solved in R. Sandling's paper Units in the modular group algebra of a finite abelian $p$-group (J. Pure Appl. Algebra 1984). In the non-abelian case there are many papers in the literature devoted to particular $p$-groups $G$, but as noted in Stefan Kohl's answer, the problem is not understood in full generality.
answered Sep 22, 2013 at 21:22
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$\begingroup$ Vesselin:We were writing our answer in the same time...I will remove mine one if you prefer. $\endgroup$ Commented Sep 22, 2013 at 21:30
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$\begingroup$ I was just about to remark the same thing :-). But please do not remove your answer as it contains valuable references to the literature! (My answer gives, in turn, the complementary reference to Sandling's paper.) $\endgroup$ Commented Sep 22, 2013 at 21:37
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$\begingroup$ Thank you Vesselin; This is helpful. I will look at these references, too. $\endgroup$– CheboluCommented Sep 23, 2013 at 16:14
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If $k$ is a field of characteristic $p>0$ and $G$ is a finite $p$-group then it is well-known that the Jacobson radical of $kG$ coincides with the augmentation ideal $\omega(kG)$ of $kG$. The group of units of $kG$ is given by $kG\backslash \omega(kG)$ and is isomorphic to the direct product $k^\times \times (1+\omega(kG))$, where $k^\times=k\backslash\{0\}$. Some other results on the group of units of group algebras can be found, e.g., in the books "S.K. Sehgal: Topics in group rings", "D. Passman: The algebraic structure of group rings", "I. Passi: group rings and their augmentation ideal", and in many papers of the existing literature on the subject.
answered Sep 22, 2013 at 21:25
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$\begingroup$ Sure. -- But what you don't say is what is the isomorphism type of the normalized unit group $1+\omega(kG)$ -- and that is the difficult thing. $\endgroup$– Stefan Kohl ♦Commented Sep 22, 2013 at 21:28
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2$\begingroup$ Yes, I agree, but a discussion about that problem can be found in some of the books I mentioned, too. $\endgroup$ Commented Sep 22, 2013 at 21:33
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Even if $k = \mathbb{F}_p$ for some prime $p$ and $G$ is a $p$-group, it is certainly not easy to describe the group of units of the modular group algebra $kG$. -- For example the Modular Isomorphism Problem which asks whether non-isomorphic $p$-groups have always non-isomorphic $p$-modular group algebras is still open. Note that the problems of deciding isomorphism of the group algebras and of their groups of units are essentially equivalent. For a discussion of this problem, see e.g.
Inger Christin Borge: A cohomological approach to the modular isomorphism problem, 2002, https://www.duo.uio.no/handle/10852/10683
Bettina Eick, Alexander Konovalov: The modular isomorphism problem for the groups of order 512, Proceedings of Groups St. Andrews 2009, 375--383 (2011), http://www.icm.tu-bs.de/~beick/publ/mip512.pdf.
Czesław Bagiński, Alexander Konovalov: The modular isomorphism problem for finite $p$-groups with a cyclic subgroup of index $p^2$, http://arxiv.org/abs/math/0607292
Alexander Konovalov: Computer investigations of the modular isomorphism problem, 2005, http://alexk.host.cs.st-andrews.ac.uk/papers/Bedlew05.pdf.
In 2003, Inger Christin Borge and Olav Arnfinn Laudal announced a solution of the modular isomorphism problem. In the sequel, the problem was thought to be solved until about a year later, a gap in the proof was found. Cf. http://homepages.abdn.ac.uk/mth192/pages/html/archive/borge-laudal.html. For a discussion of what the valid parts of the proof can still be used for, see:
Martin Hertweck, Marcos Soriano: Central $p$-group extensions; Laudal obstruction spaces revisited, http://www.igt.uni-stuttgart.de/LstDiffgeo/Hertweck/preprints/HeSo_Lobs.pdf
answered Sep 22, 2013 at 20:43
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$\begingroup$ very interesting! Thank you for these references. $\endgroup$– CheboluCommented Sep 23, 2013 at 16:14