Let $G$ be a finite group and $\rho_1,\rho_2,\cdots,\rho_m$ be pairwisely non-isomorphic irreducible representations of $G$. Then is it right that $\bigcap\limits_{i=1}^m\ker\rho_i=1$?


This is true if you mean that the list above exhausts all irreducible representations.

To see this, note that by extending a representation $\rho:G\to GL_n(\mathbb{C})$ linearly, we get a representation $\tilde{\rho}: \mathbb{C}G\to M_n(\mathbb{C})$. Since elements of $G$ form a basis for $\mathbb{C}G$, we see that $\ker(\rho)=\{1\}$ if, and only if, $\ker(\tilde{\rho})=\{0\}$. Therefore, we are left to check that $\bigcap_i \tilde{\rho_i}=\{0\}$.

Well, there is an isomorphism $$\mathbb{C}G\cong M_{n_1}(\mathbb{C})\oplus M_{n_2}(\mathbb{C})\oplus\cdots\oplus M_{n_m}(\mathbb{C})$$ and $\rho_i$ is the projection onto the $i$th factor under this isomorphism. Thus, $x\in\bigcap_i\ker(\tilde{\rho_i})$ if and only if $x\mapsto 0$ under the isomorphism above. That is, $x=0$.

  • $\begingroup$ Why is the map $g\mapsto(\rho_1(g),\rho_2(g),\cdots,\rho_m(g))$ an isomorphism? $\endgroup$ – Li Li Dec 7 '16 at 9:48
  • $\begingroup$ Are you asking for a proof that the group algebra is isomorphic to a direct sum of matrix groups? That is a long answer that can be found by searching this site. The key words are "Maschke's Theorem" and "Wedderburn-Artin Theorem" $\endgroup$ – David Hill Dec 7 '16 at 16:50

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