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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.

Alexander Chervov
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  • Googling on "representation theory of p-groups": http://arxiv.org/abs/0911.0637 http://people.maths.ox.ac.uk/craven/docs/lectures/pgroups.pdf http://scholarsresearchlibrary.com/aasr-vol2-iss6/AASR-2010-2-6-15-27.pdf http://grt.epfl.ch/ln.pdf – Alexander Chervov Sep 06 '12 at 18:15
  • 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. –  Sep 06 '12 at 18:22
  • @Vahid Thank you! Googling "representation theory of upper triangular matrices p-groups" http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.100.4729 http://groupprops.subwiki.org/wiki/Kirillov_orbit_method_for_finite_Lazard_Lie_group http://groupprops.subwiki.org/wiki/Linear_representation_theory_of_unitriangular_matrix_group_of_degree_three_over_a_finite_field http://groupprops.subwiki.org/wiki/Unitriangular_matrix_group:UT(3,3) http://arxiv.org/pdf/1105.4935.pdf http://www.scribd.com/doc/88321666/17/Nilpotent-groups-and-p-groups – Alexander Chervov Sep 06 '12 at 18:55
  • http://arxiv.org/abs/1004.2674 Representations of Finite Unipotent Linear Groups by the Method of Clusters Ning Yan http://web.ceu.hu/math/People/Alumni_and_Friends/Alumni/Zoltan_HALASI_Thesis.pdf Zoltan Halasi On the representations of solvable linear groups – Alexander Chervov Sep 07 '12 at 17:23
  • Involutions in $S_n$ and associated coadjoint orbits A. N. Panov http://arxiv.org/abs/0801.3022 Coadjoint orbits of the group $UT(7,K)$ M.V. Ignatev, A.N. Panov http://arxiv.org/abs/math/0603649 – Alexander Chervov Sep 07 '12 at 17:40
  • http://groupprops.subwiki.org/wiki/Unitriangular_matrix_group:UT(4,2) Special page on groupprops for U(4,2) – Alexander Chervov Sep 08 '12 at 19:44
  • Eric Marberg papers in arXiv lots of on rep-theory of UT http://arxiv.org/find/all/1/all:+Marberg/0/1/0/all/0/1 – Alexander Chervov Sep 09 '12 at 17:02
  • http://arxiv.org/abs/math/0501332 Coadjoint orbits for $A_{n-1}^+, B_{n}^{+}$, and $D_{n}^{+}$

    Shantala Mukherjee

     – Alexander Chervov Sep 09 '12 at 17:04
  • http://mathoverflow.net/questions/68207/irreducible-representations-of-the-unitriangular-group – Alexander Chervov Sep 10 '12 at 12:03
  • http://math.stackexchange.com/questions/99126/conjugacy-classes-of-a-p-group?rq=1 – Alexander Chervov Sep 11 '12 at 18:08
  • http://www.math.kth.se/~boij/kandexjobbVT11/Material/pgroups.pdf Nice text about p^3, p^4 groups – Alexander Chervov Sep 19 '12 at 07:40
  • Related question: http://mathoverflow.net/questions/126932/finite-unipotent-groups-references – Alexander Chervov Apr 09 '13 at 12:39

3 Answers3

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(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.

Frieder Ladisch
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  • 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 8dim1 + 16dim2. So in total there should be 24 irreps and same number of conjugacy classes. Does it sounds reasonable ? Or I miscalculated ? – Alexander Chervov Sep 08 '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 08 '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 08 '12 at 13:04
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    $G=U_4(F_q)$ 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 08 '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 08 '12 at 17:32
  • By the way I've found page about UT(4,2) http://groupprops.subwiki.org/wiki/Unitriangular_matrix_group:UT(4,2) its count of conj. classes 2q^3+q^2-2q agrees with yours count of irreps number – Alexander Chervov Sep 08 '12 at 19:48
  • It is a general fact that $\chi(1)^2\leq |G:Z(G)|$ for any irred. character $\chi$ of a group $G$ (see Isaacs' book on character theory, Corollary 2.30). I apply this to $G/Z(G)$. Since $|(G/Z(G)):Z(G/Z(G))|=q^3$, irreps of $G/Z(G)$ have dim at most $q$. Since $|G/Z(G)|=$ sum of squares of the dims of the irreps, and we have $q^3=|G/[G,G]|$ irreps of dim 1, one can compute the number of irreps of dim $q$.
    A monomial group is a group where every character is induced from a linear character, and nilpotent groups are known to be monomial (see, for example, Corollary 6.14 in Isaacs' book).     – Frieder Ladisch Sep 08 '12 at 23:09
  • Do you rely on the fact that all q-dim irreps of G comes from G/Z(G) ? If yes how to justify it? If not I do not quite understand: how do you get " one can compute the number of irreps of dim q " ? – Alexander Chervov Sep 09 '12 at 13:31
  • Yes, in this special family of groups every irrep of dim $q$ comes from $G/Z(G)$, that is, $Z(G)$ is contained in the kernel. I already mentioned this in my second comment to this answer: First, every irred. char. of a $p$-group is induced from a linear char. If $\chi\in \operatorname{Irr}(G)$ has $\chi(1)=q$, then $\chi=\lambda^G$ and $\lambda\in \operatorname{Lin}(N)$, where $|G:N|=\chi(1)=q$. So $|N|=q^5$ and $N$ is non-abelian and one can show that $Z(G)\leq [N,N]$, so $Z(G)\leq \ker(\lambda)$. But then also $Z(G)\leq \ker(\chi)$. – Frieder Ladisch Sep 09 '12 at 19:24
  • Okay. Thank you very much ! Might be interesting: http://groupprops.subwiki.org/wiki/Linear_representation_theory_of_unitriangular_matrix_group:UT(4,p) – Alexander Chervov Sep 10 '12 at 07:16
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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.

Geoff Robinson
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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.

Dan Shved
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  • Спасибо! Welcome to MO. It might be related to http://mathoverflow.net/questions/106543/p-adic-representations-of-groups .It seems that over complex numbers problem is difficult. – Alexander Chervov Sep 07 '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 07 '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 07 '12 at 16:33