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Can we find finite metabelian (ie with derived length 2) groups with arbitrarily many distinct degrees of irreducible complex characters?

If we cannot, can we somehow find a bound of the form $|cd(G)|\leq f(dl(G))$ for some "interesting" function $f$ (linear would be very cool for instance).

The motivation is that we of course have the bound $dl(G)\leq 2|cd(G)|$ for solvable groups (and conjectured to actually be $dl(G)\leq |cd(G)|$), so I was wondering if a bound in the other direction also existed.

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2 Answers 2

up vote 3 down vote accepted

Sure, you just take direct products of metabelian groups with different character degrees: cd(G×H) = cd(G)×cd(H) and (G×H)′ = G′×H′.

I suggest taking G(p) = AGL(1,p) = Hol(p) to be the normalizer of a Sylow p-subgroup in the symmetric group of degree p, for each prime p, but there are lots of examples. For instance:

  • G(3) = Sym(3) has character degrees {1,2},
  • G(5) = F20 has character degrees {1,4},
  • G(7) = F42 has character degrees {1,6},
  • and G(3) × G(5) × G(7) has character degrees { 1, 2, 4, 6, 8, 12, 24, 48 }.
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Ahh, of course. I guess I should have seen that. It is also easy to find a group whose character degrees are exactly 1 and $p$ for any given prime p, since you just take any non-abelian group of order $p^3$. –  Tobias Kildetoft Jun 21 '10 at 13:47

Much more is true. Let $S$ be an arbitrary finite set of powers of some fixed prime $p$, subject only to the condition that $1 \in S$. Then there exists a class 2 $p$-group (which, of course is metabelian) such that $S$ is exactly the set of degrees of its irreducible characters. This theorem appears in a paper of mine in the AMS Proceedings of 1986 (Volume 96, pages 51--52.)

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