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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 Which finite groups have faithful complex irreducible representations?.

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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 '11 at 13:46
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 '11 at 13:53
up vote 5 down vote accepted

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 $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$.

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Sorry forgot the adjective "irreducible" before. – Marc Palm Apr 12 '11 at 13:28
for the irreducible case, the answer is no, see above – Kate Juschenko Apr 12 '11 at 13:36
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 '11 at 13:52

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

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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.

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Good answer, forgot the adjective "irreducible". Is it still true? – Marc Palm Apr 12 '11 at 13:27

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