# Composition of Irreducible Representation and Surjective Homomorphism

Let $\varphi:G\to H$ be a epimorphism and let $\psi:H\to GL(V)$ be an irreducible representation. We wish to show that $\psi\circ\varphi$ is an irreducible representation of $G$. I have started this problem by considering the orthogonal relationship of the characters, namely by trying to show that $\langle\chi_{\psi\circ\varphi},\chi_{\psi\circ\varphi}\rangle=1$, and thereby showing that ${\psi\circ\varphi}$ is in fact irreducible. I have the following steps, but unsure how to continue (or if this is in the right direction): \begin{align} \langle\chi_{\psi\circ\varphi},\chi_{\psi\circ\varphi}\rangle & =\frac{1}{|G|}\sum_{g\in G}\chi_{\psi\circ\varphi}(g)~\overline{\chi_{\psi\circ\varphi}(g)} \\ &=\frac{1}{|G|}\sum_{g\in G}\chi_{\psi\circ\varphi}(g)~\chi_{\psi\circ\varphi}(g^{-1}). \end{align}

Any suggestions, or if you can explain it using group actions, would be ideal.

• This is more easily proved using the definition of irreducibility. – Eric O. Korman Apr 5 '14 at 19:12

Since $\chi_{\psi\circ\varphi}(g)=\chi_\psi(\varphi(g))$, we have $$\frac{1}{|G|}\sum_{g\in G}\chi_{\psi\circ\varphi}(g)\overline{\chi_{\psi\circ\varphi}(g)}=\frac{1}{|G|}\sum_{h\in H}\frac{|G|}{|H|}\chi_{\psi}(h)\overline{\chi_{\psi}(h)}=\frac{1}{|H|}\sum_{h\in H}\chi_{\psi}(h)\overline{\chi_{\psi}(h)}=1$$
• Do you mean $\chi_\psi(\varphi(g))$? – smithe Apr 5 '14 at 19:26
• No worries, I'm just glad you saw where to use the subjectiveness of $\varphi$. Thanks. I also think proving this by the definition would be equally quick as the Eric O. Korman suggested. But thank you both. – smithe Apr 5 '14 at 19:29
Suppose $W$ is a $G$-invariant subspace of $V$. Then, for each $g$ $\psi \circ \rho(g)|_{W}$ is an element of $GL(W)$. Since, for every element $h$ in $H$, there exists a $g$ such that $h = \rho(g)$, we must have $\psi(h)|_{W}$ is an element of $GL(W)$. This implies that $W$ is an $H$-invariant subspace of $V$ and is hence either $0$ or $1$. Thus, $V$ is $G$-irreducible as well.