Hi, I know that over $\mathbb{C}$ the dimension of an irreducible representation of a finite group $G$ must divide the order of the group. I've read somewhere that if $p$ does not divide the order of the group then the representations over $\tilde{F_p}$ are "essentially the same" (Is this true? What does it mean other than the dimensions are of the same size and number?)

My question is if anything can be said in general about the dimesions of the irreducible representations when $p$ does divide $|G|$? Can anything be said about dimensions of faithful irreducible representations (if such exist)?

Also, can someone recomend a good "friendly" (I am from computer science dept.) refrence for such statements?


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    $\begingroup$ They are "essentially the same" in a much better sense: you choose a basis of the representation so that the group elements are matrices with (algebraic) integer entries; then reducing these modulo p (essentially) produces the irreps over $\bar{\mathbb{F}}_p$. $\endgroup$ – Ben Webster Jan 8 '13 at 19:04
  • $\begingroup$ That's great! Thanks! Do you know where I could find a reference for this? I'm sorry if the next question is very basic, but how can I always find a basis in which all the entries are algebraic integers? It's clearly enough to show this on a generating set, and I think I can show this for a single group element. But how do I show that the same basis works on the entire generating set simultaneously? $\endgroup$ – A.B. Jan 8 '13 at 20:34

I'm not sure where you are starting in terms of background and references, but the standard short book for such questions is Serre's Linear Representations of Finite Groups (Springer GTM 42, a good English translation by Len Scott of older lecture notes dating back in their first version to the 1960s). Part III mixes the ordinary and modular theories (where the characteristic is 0 or prime).

For the "easy" case where the prime doesn't divide the group order, see Serre's 15.5. Of course, this is based on a lot of previous general theory, so you may prefer a more direct argument along the lines suggested in Ben's comment. Either way, you have to start in characteristic 0 with a matrix representation, then find a suitable basis for reduction modulo $p$. Starting with an irreducible representation (over a big enough field such as $\mathbb{C}$) then yields an irreducible representation in characteristic $p$. (But there is some theory required for this process.)

The case when $p$ does divide the group order is fairly subtle, which shows up in the more sophisticated tone of Serre's third part. (The old 1962 book by Curtis and Reiner is more down-to-earth, but doesn't get to this topic until the very end.) The short answer to your question here is that lots of things can happen, as Richard Brauer and his followers have shown in detail. The dimensions of irreducible representations in characteristic $p$ usualy won't divide the group order and have nothing obvious to do with dimensions in characteristic 0. But there are subtle connections, still being worked out for simple groups in particular. Only narrowly focused questions are likely to have clear answers.

  • $\begingroup$ Thank-you. Well,I am narrowingly interested in faithful irreducible representations of $\mathbb{Z}_q\rtimes \mathbb{Z}_{p^2}$ where $p,q$ are primes, $q=np+1$ and the action is that of $\mathbb{Z}_p$ on $\mathbb{Z}_q$. Over $\mathbb{C}$ (and therefore as I understand over $\tilde{F}_u,u\neq p,q$) by the statement every faithful irreducible representation is of dimension $\geq p$ and it must be exactly $p$ because the order of an abelian subgroup is $pq$. My question is if I can bound the dimensions of the irreducible faithful representations (if such exist) in characteristic $p$ as well? $\endgroup$ – A.B. Jan 8 '13 at 20:56
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    $\begingroup$ @A.B. It's probably useful to post a follow-up question about your particular solvable (nonabelian) group, stated as precisely as possible. It may challenge experts more than the generalities. I'm not a specialist but for instance can see a natural faithful representation of degree $p$ in characteristic $q$. But then it gets tricky to sort out (faithful) degree possibilities for all relevant $p, q$, etc. Anyway, a tag finite-groups might work better than characteristic-p. $\endgroup$ – Jim Humphreys Jan 9 '13 at 14:22
  • $\begingroup$ Thankyou for the advice. I guess I thought it was something well known, like in the complex case. I don't see any reason that the faithful irreducible representation I already have won't be a faithful representation with every charactaristic, but I am not sure how to check if it is irreducible when the characteristic is $p,q$. I'll try to ask in a new post. $\endgroup$ – A.B. Jan 10 '13 at 12:56

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