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Consider group algebra k[G] of finite group G. If k is alg.closed then every irrep lives there with multiplicity equal to dimension. (More conceptually as bimodule over GxG it is multiplicity free and all irreps live there).

Now let k=F_p what is known then ? Is true that all irreps of G live in k[G] ? (Seems obviously yes... but to be sure) But multiplicities ?

If assume order of G does not divide p ?


Example:

G=Z/nZ, the F_p[G] = F_p[x] / (x^p-1) and every irreducible polynom generates ideal and hence subrepresentation, so these are quests about irred. polynoms. E.g. like this:

http://math.stackexchange.com/questions/172534/for-which-values-of-n-is-the-polynomial-px-1xx2-cdotsxn-irreducible/172540#comment396429_172540

http://math.stackexchange.com/questions/172468/for-what-n-k-there-exists-a-polynomial-px-in-f-2x-s-t-degp-k-and


Motivation: "cyclic error correcting codes" are exactly the ideals in F_p[x] / (x^p-1) = F_p[G], why take G= Z/nZ ? may be take other groups give "better" codes ?

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If by irreps you mean indecomposable representations, then not only it is not obvious that all of them are in the group algebra: it is in fact generally false (for there are infinitely many of them) On the other hand, if by irrep you mean simple representations, the situation is more hopeful; for example, in the extreme case where $G$ is a $p$-group, there is only one simple representation, the trivial one. But now you have to decide what exactly you want "live in F[G]" to mean, as the algebra is no longer semisimple. –  Mariano Suárez-Alvarez Jul 18 '12 at 19:51
    
(The statement in your first sentence is only true if $k$ is of characteristic zero) –  Mariano Suárez-Alvarez Jul 18 '12 at 19:52
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Mariano, "irrep" means "irreducible (that is, simple) representation." –  Ben Webster Jul 18 '12 at 20:01
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The keyword you want here (for starters) is "modular representation theory." This is an old and, as I understand it, big subject. –  Qiaochu Yuan Jul 18 '12 at 20:18
    
See also Geoff Robinson answer at MSE math.stackexchange.com/questions/173811/… –  Alexander Chervov Jul 22 '12 at 11:22
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3 Answers

up vote 10 down vote accepted

It seems natural to work with an algebraically closed field of characteristic $p$, or, less restrictively, a splitting field of characteristic $p$ for $G$. For example, any field containing the primitive $m$-th roots of unity, where $|G| = p^{a}m$ and $p$ dos not divide $m$, so I assume now that $k$ is algebraically closed of characteristic $p.$ We have entered the realm of modular representation theory, whose theory was first extensively developed by Richard Brauer. By now the basic theory is covered in numerous texts (eg by Alperin, by Curtis-Reiner).

The indecomposable direct summands of the group algebra $kG$ are the so-called projective indecomposable modules (sometimes called principal indecomposables). The number of isomorphism types of these is the number of conjugacy classes of elements of order prime to $p$ of $G.$ Each of these has a simple socle (the socle of module is its largset semisimple submodule) and a unique maximal submodule (so a simple head, or largest semisimple quotient module). Because the group algebra is a symmetric algebra, the socle and head of $P$ are isomorphic. If $P$ is one of these projective indecomposable modules, and has (simple) socle $S$, then up to isomorphism, $P$ occurs ${\rm dim}_{k}(S)$ times as a summand of the group algebra $k[G].$ Hence $S$ occurs $dim_{k}(S)$ times (up to isomophism) as a summand of the socle of the regular module $kG.$ Also, $S$ occurs ${\rm dim}{k}(P)$ times as a composition factor of the regular module $k[G].$ Every simple module for the group algebra $k[G]$ has a projective cover, and these all occur as direct summands of the group algebra. As Mariano remarks in his comment, the theory of non-projective indecomposable modules is much less transparent. The results over non-splitting field can be recovered using Galois theory and Clifford theory. However, modular representation theory is much richer and diverse than this brief description allows for. These basic facts (and much much more) were all known to Brauer, perhaps sometimes in different language, and these are just the beginnings. If $|G|$ is not divisible by $p,$ the projective indecomposable $P$ and its socle $S$ are the same module, and the theory degenerates to the semisimple case, which is much like the characterisic zero situation.

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Thank you for the answer! What if field is not alg. closed e.g. just F_2={0,1} ? (This is main example to think if keep in my error-correcting codes). –  Alexander Chervov Jul 19 '12 at 5:57
    
In tht case, as I said, Galois theory comes into play. A simple module over the prime field breaks up when the field is extended to a splitting field, into a direct sum of a number of Galois conjugate modules, all of the same dimension, which can be easily calculated. Since the regular module is realizable over the prime field, the simple modules over the prime field all still appear in the socle of the regular module, but the multiplicity as a summand is the dimension of one of the absolutely irreducible simple module (ie over the algebraically closed field). –  Geoff Robinson Jul 19 '12 at 7:18
    
Thank you very much ! –  Alexander Chervov Jul 19 '12 at 19:34
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The basic roadblock for a finite group over a finite prime field (whether or not the characteristic divides the group order) is clear-cut: you rarely get all of the absolutely irreducible representations of the group over the prime field, which is equally a problem when working over $\mathbb{Q}$. You always get various irreducible reprsentations living in the group algebra, but typically these split up further if you extend the field as Geoff indicates. Then it can get arbitrarily complicated to sort everything out, but the tools are there. Trying to work exclusively over a prime field is seldom productive, without reference to a splitting field.

Geoff has given a broad sketch of the subject, which is treated thoroughly in the 1962 book of Curtis-Reiner and (in a more modern style) in their later two-volume work. Alperin's book is short, readable, and module-oriented, while the third part of Serre's also short book is devoted to the basics of Brauer theory (without block theory) following an introduction to ordinary character theory, etc.

There's plenty of literature on coding theory, some of which uses finite group theory in an essential way, but it's hard to do anything insightfully without at least some theoretical background in modular representation theory.

P.S. I don't understand the parenthetic remark in your first paragraph.

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Thank you for the answer. Does the standard construction of embedding End(V) to k[G] work over F_p ? I mean irrep (p,V) take a basis e_i in V. Consider matrix elements as elements for each element p(g)_ij - this gives some elements in k[G]. Over complex numbers this is subalgebra in k[G] isomorphic to End(V). It seems it works over any field. Am I wrong ? –  Alexander Chervov Jul 19 '12 at 7:06
    
End_{F}(V) embeds in the socle of FG (as rightmodule) when F is a simple FG-module, but in general if F is not a splitting field it will not be isomorphic a full matix ring over F. –  Geoff Robinson Jul 19 '12 at 7:32
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Adding a few words from the coding theory side. Abelian groups without $p$-torsion are somewhat more natural in coding theory, because then we get the machinery of discrete Fourier transform (which is, of course, just representation theory) to play with. Dihedral groups have been used as symmetry groups of cyclic codes (add order reversal symmetry), but for the most part symmetries of codes just aid proving things about their properties.

When the group has $p$-torsion, the theory is less clean. Nevertheless, the topic has been studied. Check out papers by Karl-Heinz Zimmermann (http://www.tu-harburg.de/ti6/mitarbeiter/khz/pub.html). In his papers from the 90s a lot of representation theoretical concepts appear. I don't know, if they are very hot from the point of view of constructing new and better codes, though.

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Notifier: May I kindly ask you to look at mathoverflow.net/questions/117505/… PS Happy New year ! –  Alexander Chervov Dec 29 '12 at 12:31
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