Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Finite p-groups - have p^n elements by definition. According to WP there is rich structure theory.

Question: How far is representation theory of p-groups is understood?

In case this question is too broad let me restrict it to upper triangular matrices over F_q. What is known about its irreps ? Dimensions ? Constructions ? Action of Out(G) ? Characters ?

E.g. 4x4 matrices over F_2 - this group has order 64. So dimensions of irreps over C are 1,2,4 by trivial reasons that 1) dim | order of group and 2) \sum dim^2 = order of group. Is it possible to illustrate the general theory (if it exists) on this example to compute number of different irreps of for this group ?


Remarks:

Notation: let me denote U_n(F_q) = upper triangular matrices over F_q with units on the diagonal.

Some trivial examples: U_2(F_2) = Z/2; U_3(F_2) = D_8 - dihedral group with 8 elements.

Remark: U_n(F_q) is p-Sylow subgroup of GL_n(F_q). Over F_2 it will also be Borel subgroup.

share|improve this question
    
    
As a part of my PhD thesis, I have done some research on $F_p[G]$-modules, where $G$ is a cyclic $p$-group. Although it is restricted to a very specific $p$-groups, it may be useful/inspiring for your work. You can find my results in the article titled "MODULE HOMOMORPHISMS OF GROUP ALGEBRAS OF CYCLIC p-GROUPS IN CHARACTERISTIC p". It is freely available in the internet. –  Vahid Shirbisheh Sep 6 '12 at 18:22
    
    
arxiv.org/abs/1004.2674 Representations of Finite Unipotent Linear Groups by the Method of Clusters Ning Yan web.ceu.hu/math/People/Alumni_and_Friends/Alumni/… Zoltan Halasi On the representations of solvable linear groups –  Alexander Chervov Sep 7 '12 at 17:23
    
Involutions in $S_n$ and associated coadjoint orbits A. N. Panov arxiv.org/abs/0801.3022 Coadjoint orbits of the group $UT(7,K)$ M.V. Ignatev, A.N. Panov arxiv.org/abs/math/0603649 –  Alexander Chervov Sep 7 '12 at 17:40
show 7 more comments

3 Answers

up vote 4 down vote accepted

(This answer is only about representation theory over $\mathbb{C}$, that is, character theory.) You should take a look at §26: Characters of $p$-groups of Bertram Huppert's book Character Theory of Finite Groups. In particular, it contains a proof of Isaacs' results that the set of character degrees of a $p$-group is rather arbitrary (Proc. Amer. Math. Soc. 96 (1986), p.551-552, doi: 10.2307/2046302) and that character degrees of groups of the form $1+ \mathbf{J}(A)$, where $A$ is a finite algebra over the field $F$ with $q$ elements, are powers of $q$ (J. Algebra 177 (1995), p. 708-730, doi: 10.1006/jabr.1995.1325), and a proof that the character degrees of $U_n(F_q)$ are the powers $q^i$ for $0\leq i\leq \lfloor \frac{(n-1)^2}{4}\rfloor$. (A result of Huppert, Arch. Math. 59 (1992), p. 313-318, doi: 10.1007/BF01197044.)
There is also an elaborate theory of "supercharacters" and "superclasses" for the upper triangular group (and other groups). A good place to start reading is, I think (but I'm not an expert), the paper of Diaconis and Isaacs: Supercharacters and superclasses for algebra groups, Trans. Amer. Math. Soc. 360 (2008), p.2359-2392, MR2373317 (doi). There are, by now, many papers about the character theory of the upper triangular group and related topics, which is in part motivated by Higman's conjecture that for every $n$, the number of conjugacy classes of $U_n(F_q)$ is a polynomial in $q$ with integer coefficients.

share|improve this answer
    
Thank you very much for yours kind answer ! Let me ask something. You write that dim(irreps)<= floor[ (n-1)^2/4], is it correct, I mean "floor" not "ceil" rounding ? So in particular if n=4, p= 2 we get [9/4] = 2, so this means that dims of irreps are 1,2. So 4 is forbidden. If i understand correctly G/[G,G] in this case contain 8 elements. So we see that there should be 8*dim1 + 16*dim2. So in total there should be 24 irreps and same number of conjugacy classes. Does it sounds reasonable ? Or I miscalculated ? –  Alexander Chervov Sep 8 '12 at 11:15
    
@Alexander Chervov: for $n=4$, you get $\lfloor 9/4 \rfloor = 2$, correct, but this means that the dims of the irreps are $q^0$, $q^1$ and $q^2$, so for $q=2$ they are $1$, $2$ and $4$. There are $q^3$ dim 1's, $q^3-q$ dim $q$'s and $q^2-q$ dim $q^2$'s. –  Frieder Ladisch Sep 8 '12 at 12:33
    
Ooops, too much thinking ))... Thank you ! How did you get "q3−q dim q's and q2−q dim q2's" ? –  Alexander Chervov Sep 8 '12 at 13:04
    
$G=U_4(F_q)$ has center of order $q$ and $G/Z(G)$ has center of order $q^2$ and index $q^3$, so that the irreps of $G/Z(G)$ have dim at most $q$. So $G/Z(G)$ must have $q^3-q$ irreps of dim $q$. Finally, one needs to see that irreps where $Z(G)$ is not in the kernel have dimension $q^2$. If such an irrep had dim $q$, then it would have to be induced from a linear character of a subgroup of order $q^5$ (since $p$-groups are monomial), but $Z(G)$ is contained in the derived subgroup of every subgroup of order $q^5$, contradiction. –  Frieder Ladisch Sep 8 '12 at 14:25
    
Thank you very much ! How to you see "so that the irreps of G/Z(G) have dim at most q" and "So G/Z(G) must have q3−q irreps of dim q." ? Also this seems to appeal to some background which is not known for me "If such an irrep had dim q, then it would have to be induced from a linear character of a subgroup of order q5 --||WHY?||-- (since p-groups are monomial)--||?||--, but Z(G) is contained in the derived subgroup of every subgroup of order q5, contradiction. " –  Alexander Chervov Sep 8 '12 at 17:32
show 5 more comments

It is already a notoriously difficult problem to determine the number of conjugacy classes of elements of the group of upper unitriangular $n \times n$ matrices over the field of $q$ elements, when $n$ gets moderately large.

share|improve this answer
    
Thank you very much ! Indeed it seems I got too much optimism from WP-article... –  Alexander Chervov Sep 8 '12 at 11:16
add comment

Well, I'm only starting to learn representation theory, but here's a thought about irreps. Forgove me if it seems trivial.

I think that a finite $p$-group always has only one irreducible representation over $F_p$ - the trivial one-dimensional representation. Here's why.

Let $F$ be a field with characteristic $p$ and let $G \leqslant \mathrm{GL}\left( n, F \right)$ be a finite irreducible $p$-group of matrices over $F$. Let $g$ be an element from the center of $G$.

Clearly, $g^{p^m}-1 = 0$ for some positive integer $m$. Since we are in characteristic $p$, it follows that $(g-1)^{p^m} = 0$. Therefore $\ker (g-1) \neq 0$. Now, since $g-1$ commutes with every element of $G$, $\ker (g-1)$ is a $G$-invariant subspace. Remember that $G$ is irreducible, which means that $g-1 = 0$.

So, we have proved that the center of $G$ is trivial, therefore $G$ itself is also trivial, qed.

PS: I am not sure, but maybe you could try googling "Modular representation theory" and "Brauer characters": I think I saw these things mentioned somewhere in connection with a similar question.

share|improve this answer
    
Спасибо! Welcome to MO. It might be related to mathoverflow.net/questions/106543/… .It seems that over complex numbers problem is difficult. –  Alexander Chervov Sep 7 '12 at 13:29
    
Now that I've re-read your question, I see that "over $F_p$" in your question means "triangular matrices over $F_p$", not "representations over $F_p$". My answer totally useless then, sorry. –  Dan Shved Sep 7 '12 at 13:32
    
@Dan Shved do not worry, I was also interested about F_p although the main case for me is C. –  Alexander Chervov Sep 7 '12 at 16:33
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.