Let $(\pi_1, V_1), ... , (\pi_t,V_t)$ be all the irreducible representations of a finite group $G$ over the complex numbers with respective degrees $n_i$ and characters $\chi_i$. If $(\rho,V)$ is the left regular representation of $G$, then the number of times $\pi_i$ occurs in $V$ is $n_i$. I'm a bit confused on the following problem:

Suppose $(\phi,W)$ is an $n < \infty$-dimensional representation of $G$ with character $\chi$, and that $\chi(g) = 0$ for $g \neq 1_G$. Show that $\phi$ is an integral multiple of the character $\lambda$ of $\rho$.

Well, the number of times $\pi_i$ occurs in $W$ is $$(\phi, \pi_i) = \frac{1}{|G|} \sum\limits_{g \in G} \chi(g) \chi_i(g^{-1}) = \frac{n}{|G|}n_i$$ and so $$ \chi = \sum\limits_{i=1}^t (\phi,\pi_i) \chi_i = \frac{n}{|G|} \sum\limits_{i=1}^t n_i \chi_i = \frac{n}{|G|} \lambda$$ That was easy enough, but how can I be sure that $\frac{n}{|G|}$ is actually an integer? Is there something obvious about this? Certainly $\frac{n}{|G|}n_i$ has to be an integer, hence so does $\frac{n^2}{|G|^2}n_i^2$, and hence so does $$\sum\limits_{i=1}^t \frac{n^2}{|G|^2}n_i^2 =\frac{n^2}{|G|^2}\sum\limits_{i=1}^t n_i^2=\frac{n^2}{|G|^2}|G| = \frac{n^2}{|G|}$$ That's close, but not quite what I need to prove.

  • 1
    $\begingroup$ For the trivial representation, $n_i = 1$. $\endgroup$ – Joey Zou Jan 1 '16 at 2:03
  • 1
    $\begingroup$ I conjecture that Joey Zou was hinting at the following argument. The trivial representation has multiplicity 1 in the regular representation, so it has multiplicity $n/|G|$ in your representation with character $\chi$. But the multiplicity of any irreducible representation in any (actual) representation is an integer. $\endgroup$ – Andreas Blass Jan 1 '16 at 2:25
  • $\begingroup$ Thanks, I forgot about the trivial representation. $\endgroup$ – D_S Jan 1 '16 at 2:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.