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A220470
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Smallest order of a group with an irreducible complex representation of dimension n.
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2
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1, 6, 12, 20, 55, 42, 56, 72, 144, 110, 253, 156, 351, 336, 240, 272, 1751, 342, 3420, 500, 672, 506, 1081, 600, 2525, 702, 1512, 812, 1711, 930, 992, 1440
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OFFSET
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1,2
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COMMENTS
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a(1) = 1 because the trivial group works.
If n > 1, then let G_n be a group of minimal order which has an irreducible complex representation of dimension n.
Since G_n has the n-dimensional and trivial representations, a_n = |G_n| >= n^2 + 1.
Since n divides the order of any finite group with an irreducible complex representation of dimension n, n divides a_n and this allows us to strengthen the lower bound to a_n >= n^2 + n.
An upper bound is given by a_n <= n*q_n, where q_n is the smallest positive integer which is a power of a prime and which is congruent to 1 mod n. This is because the group of affine transformations x --> a*x + b (from the finite field F_(q_n) to itself), where a^n = 1 and b is an arbitrary element of F_(q_n), has order n*q_n and an irreducible complex representation of dimension n.
The upper bound and the lower bound coincide if and only if (n > 1 and) n+1 is a power of a prime. This means that, for any such n, a(n) = n^2 + n.
Upper bound is n*A224503(n), for n > 1.
This function is sub-multiplicative: a(m*n) <= a(m)*a(n) because (any) G_m x G_n has an irreducible complex representation of dimension m*n.
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REFERENCES
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Marshall Hall, Jr., Theory of Groups, AMS Chelsea Publishing, 1999 reprinting, pp. 136-138.
I. Martin Isaacs, Finite Group Theory, AMS, 2008, p. 35.
Gordon James and Martin Liebeck, Representations and Characters of Groups, Cambridge University Press, 1995 reprinting, pp. 217-218.
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LINKS
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EXAMPLE
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For n=5, the lower bound is a(5) >= 30, while the upper bound is a(5) <= 55.
So one only needs to prove that no group of order 30, 35, 40, 45, or 50 has an irreducible complex representation of dimension 5.
All groups of order 35 and 45 are Abelian, so all their irreducible complex representations have dimension 1.
For orders 30, 40, and 50, recall that the dimension of an irreducible complex representation is bounded from above by the index of an Abelian subgroup.
A group of order 50 has an Abelian subgroup of index 2.
A group of order 40 has a normal Sylow 5-subgroup P, which must be centralized by an element t of order 2 in a Sylow 2-subgroup. Then <P,t> is a cyclic subgroup of index 4.
A group of order 30 has a cyclic subgroup of index 2.
All these indices are too low to allow such a group to have an irreducible complex representation of dimension 5, and the a(5) = 55 result is proved.
Comments from Gabriele Nebe (Lehrstuhl D für Mathematik, RWTH Aachen), added by N. J. A. Sloane, Apr 13 2013; extended by David L. Harden, Nov 08 2017: (Start)
The following are examples of groups that realize the given values of a(n).
1=C_1, 6=S_3, 12=A_4, 20=C_5:C_4, 55=C_11:C_5, 42=C_7:C_6, 56=(C_2xC_2xC_2):C_7, 72=C_3xC_3:(some regular subgroup of L_2(3) on 8 points), 144=A_4xA_4, 110 = C_11:C_10, 253=C_23:C_11, 156=C_13:C_12, 351=(C_3xC_3xC_3):C_13, 336 = (C_2xC_2xC_2):C_7 x S_3, 240 = (C_2xC_2xC_2xC_2):C_15, 272 = C_17:C_16, 1751 = C_103:C_17, 342 = C_19:C_18, 3420 = PSL_2(19).
If one has a small cyclic group of order p where phi(p) is a multiple of n, then take C_p:C_n. 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 nontrivial characters. (End)
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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