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| numb3rs3 |
 Posted: Mon Feb 20, 2006 6:42 pm |
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It is well known that if a finite group has only 1 dim irreducible representationss then is is abelian.
However, if it has one or two dimensional representations (irreducibe) is it tre that G must contain a normal subgroup that has index 2 in it? |
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| Guest |
| Posted: Tue Feb 21, 2006 12:00 am |
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In article <23550029.1140496961153.JavaMail.jakarta@nitrogen.mathforum.org>,
numb3rs3 <mathrulezzzzz333@hotmail.com> writes:
[quote:97a4b52114]It is well known that if a finite group has only 1 dim irreducible representationss then is is abelian.
However, if it has one or two dimensional representations (irreducibe) is it tre that G must contain a normal subgroup that has index 2 in it?
[/quote:97a4b52114]
Presumably you mean *only* one or two-dimensional representations.
Yes, this follows from a theorem of Isaacs (in his book "The Character Theory
of Finite Groups", p. 201), that, if the set of character degrees of a finite
group G is {1,m}, then either
a) G has an abelian normal subgroup of index m, or
b) m=p^e (p prime) and G is a direct product of a p-group and an abelian group.
There is probably a more elementary proof for the case m=2.
Derek Holt. |
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| Guest |
| Posted: Tue Feb 21, 2006 7:46 am |
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