# Are there finite metabelian groups with arbitrarily many character degrees?

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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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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