[seqfan] Re: Smallest groups with an irreducible representation of given dimension

[Neil Sloane]

David, I don't know that we have any group theorists on the Seq Fan list.
So I asked a group theorist friend (not on this list) to comment on your
proposed sequence. Here is part of the reply:
(quote)
1, 6=S3, 12=A4, 20=C5:C4, 55=C11:C5, 42=C7:C6, 56=(C2xC2xC2):C7,
72=C3xC3:(some regular subgroup of L2(3) on 8 points),
171=C19:C9, 110 = C11:C10, 253=C23:C11, 156=C13:C12, 351=(C3xC3xC3):C13, ...

If one has a small cyclic group of order p such that varphi(p) is a
multiple of n,
then take Cp:Cn. For other n it is reasonable to take elementary abelian
groups of order n+1 that admit a subgroup of their automorphism group
acting regularly on the n non-trivial characters.
(end quote)

Can you go ahead and submit the sequence (of course only including the
terms as far as you are certain of them)?

Thanks

Neil


On Sat, Apr 6, 2013 at 5:36 AM, David Harden <sylow2subgroup at gmail.com>wrote:

> The sequence I am considering is the one for which a_n is the order of the
> smallest finite group(s) possessing an irreducible complex representation
> of dimension n.
> a_1 = 1, and n > 1 implies (writing a_n as the sum of the squares of the
> dimensions of the group's irreducible representations) a_n > n^2 -- since
> a_n has to be a multiple of n, this yields a_n >= n^2 + n.
>
> There is a certain upper bound for this sequence which may always equal the
> sequence: this is nq, where q is the smallest prime-power congruent to 1
> mod n.
> A group of order nq possessing an irreducible representation of dimension n
> is given by the group of all functions from the finite field F_q to itself
> of the form f(x) = ax + b, where b is an arbitrary element of F_q and a is
> an element of F_q such that a^n = 1. This group has n 1-dimensional
> representations and (q-1)/n irreducible representations of dimension n.
>
> To see what else might happen to outdo this upper bound, consider the close
> call which occurs when n=5: the upper bound produces 55, which indeed
> works. But the next multiple of 5 is also the order of a group with an
> irreducible 5-dimensional representation, since 60 is the order of the
> alternating group A_5, which has an irreducible representation of dimension
> 5 (the doubly transitive 6-point representation minus the trivial
> representation).
>
> I know it's not easy to estimate the smallest prime congruent to 1 mod n,
> and replacing "prime" with "prime-power" shouldn't help much. Under what
> assumptions can it be proven that the upper bound always equals the
> sequence (if n > 1)?
>
> For the sake of completeness: the first few values of the sequence,
> assuming the upper bound is tight for n > 1: 1, 6, 12, 20, 55, 42, 56, 72,
> 171, 110, 253, 156, 351, 406, 240, 272, 1751, 342, 3629, 820, 903, 506,
> 1081, 600, 2525, 702, 2943, 812, 1711, 930, 992, 3104, 2211, 3502, 2485,
> 1332, 5513, 7258, 3081, 1640, 3403, 1806, 7439, 3916, 8145, 2162, 13301,
> 2352, 9653, 5050, 5253, 2756 (this is the n=52 term)
>
> ---- David Harden
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>



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Dear Friends, I have now retired from AT&T. New coordinates:

Neil J. A. Sloane, President, OEIS Foundation
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