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The following is Problem 6 of January 2006 algebra qualifying exam from University of Maryland. See here for the problems.

Let $G$ be a finite group. Suppose that for each normal subgroup $K\neq 1$ of $G$, the quotient group $G/K$ is abelian. Let $\rho: G\to GL_n(\mathbb{C})$ be an irreducible representation of $G$ with $n>1$. Prove that $\rho$ is injective.

We need to show $\ker(\rho)=1$. But I don't see how to use the hypothesis.

The fact that $\rho$ is irreducible will be used if we can construct some invariant subspace, and by irreducibility force it to be the trivial subspace. But could somebody explain how does the condition that $G/K$ is abelian comes into play?

Thank you :)

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    $\begingroup$ If $K$ is contained in $\ker \rho$, then $\rho$ gives an irreducible rep of $G/K$. OTOH the irreducible reps of abelian groups are of dimension one. $\endgroup$
    – Jyrki Lahtonen
    Commented Dec 4, 2013 at 22:17
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    $\begingroup$ To state Jyrki's remark a bit different: the commutator subgroup $G' \subseteq ker(\rho)$, so $\rho$ must be linear since it is irreducible. $\endgroup$
    – Nicky Hekster
    Commented Dec 4, 2013 at 23:46

1 Answer 1

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Thanks Jyrki! I am converting his comment into an answer:

Let $K=\operatorname{ker}(\rho)$. Assume, to the contrary, that $K\neq 1$. Then we have a representation $\overline{\rho}: G/K \to GL_n(\mathbb{C})$ defined by $\overline{\rho}(\overline{g})=\rho(g)$ for each equivalence class $\overline{g}$ in $G/K$. By hypothesis, $G/K$ is abelian. It is known that an irreducible representation of an abelian group is necessarily one-dimensional (see proof here). It follows that $GL_n(\mathbb{C})$ has dimension $1$. But $n>1$ in the hypothesis. This is absurd. We conclude that $K=\ker(\rho)=1$, i.e. $\rho$ is injective.

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