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Given a finite group and a normal subgroup, does there always exist an irreducible complex representation, whose kernel is this normal subgroup?

Sorry, just it was just mentioned that this is a duplicate. See http://mathoverflow.net/questions/57129/which-finite-groups-have-faithful-complex-irreducible-representations.

edited Apr 12 2011 at 15:40

Charles Matthews
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asked Apr 12 2011 at 13:19

Marc Palm
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I think the answer is clearly no. Irreducible complex representations of abelian groups are 1-dimensional, while every subgroup is normal. So no proper normal subgroup can be the kernel of an irreducible representation. – David Hill Apr 12 2011 at 13:46

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Well, for finite abelian groups the kernels of irreducible complex representations would be the subgroups such that the quotient is cyclic. – Charles Matthews Apr 12 2011 at 13:53

Kate Juschenko · Accepted Answer · 2011-04-12 13:34:07Z UTC

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take the canonical quotient $q:G\rightarrow G/H$ and $\lambda :G/H\rightarrow U(n)$ to be the left-regular representation of $G/H$, where $n=|G/H|$, then the composition is the map that you are looking for, unless I am missing something...

Edit: if you consider irreducible representations, then the answer is no for the group $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$.

edited Apr 12 2011 at 13:34

answered Apr 12 2011 at 13:27

Kate Juschenko
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Sorry forgot the adjective "irreducible" before. – Marc Palm Apr 12 2011 at 13:28

for the irreducible case, the answer is no, see above – Kate Juschenko Apr 12 2011 at 13:36

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the sum of $3$ terms of $\mathbb{Z}_2$ is just for having proper normal subgroup. But if you don't want to assume this, the answer of BS is completely describes the situation. – Kate Juschenko Apr 12 2011 at 13:52

BS · Answer 2 · 2011-04-12 13:36:17Z UTC

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In the new formulation (irreducible rep), the question is wether any finite group has a faithful irreducible complex representation. This is false for noncyclic abelian groups. A more complete answer is here

answered Apr 12 2011 at 13:36

BS
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unknown (google) · Answer 3 · 2011-04-12 13:23:26Z UTC

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Yes, consider the quotient by this normal subgroup, embeds it into symmetric group of $n$ letters with $n$ large enough, identify symmetric group with a subgroup of $GL(n)$, you get the representation.

answered Apr 12 2011 at 13:23

unknown (google)
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Good answer, forgot the adjective "irreducible". Is it still true? – Marc Palm Apr 12 2011 at 13:27

Does every normal subgroup appear as a kernel of an irreducible representation?

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