Prove that any finite group with exactly two minimal normal subgroups has a faithful irreducible $\mathbb{C}$-character.

What I have tried:

Let $N_1$ and $N_2$ be two minimal normal subgroups of $G$. If $G$ is abelian the result is straightforward. So let's suppose that $G'$ is nontrivial and then, one of the minimal normal subgroups, say $N_1$, is contained in $G'$. Now if $N_2 \cap G'=1$, there exist a $\lambda \in Lin(G)$ with $N_2 \cap Ker(\lambda)=1$ and also a nonlinear irreducible character $\phi$ such that $N_2 \subset Ker(\phi)$ and $N_1 \cap Ker(\phi)=1$. In this situation $\lambda \phi$ will be a faithful irreducible character of $G$. But what if both $N_1$ and $N_2$ are contained in $G'$? How should I deal with this situation?

  • 1
    $\begingroup$ If not, then every irreducible character would arise from an irreducible character of at least one of $G/N_1$ and $G/N_2$ and the trivial character would be both, so the sums of the squares of their degrees would be less than $|G|/|N_1| + |G|/|N_2|$ and hence could not equal $|G|$. $\endgroup$ – Derek Holt Nov 14 '15 at 20:09
  • $\begingroup$ Exactly! Thanks a lot. $\endgroup$ – user97635 Nov 14 '15 at 20:16

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