Let $G$ be a finite group and $R$ the regular representation. That is, as a vector space $R = F(G)$ is the free vector space with basis $G$. If the basis is $\{e_g : g \in G\}$ the action is defined by
$$g \cdot e_{g'}=e_{gg'}$$
and extended by linearity.
Now, in the book I'm studying the author states the following corolary:
Corollary 2.18: Any irreducible representation $V$ of $G$ appears in the regular representation $\dim V$ times.
The "proof" for this is a little argument before the statement:
We know tha the character of $R$ is simply
$$\chi_R(g)=\begin{cases}0, & g\neq e, \\ |G|, & g= e\end{cases}$$
Thus, we see first of all that $R$ is not irreducible if $G\neq \{e\}$. In fact, if we set $R = \bigoplus V_i^{\oplus a_i}$, with $V_i$ distinct irreducibles, then:
$$a_i = (\chi_{V_i},\chi_R)=\dfrac{1}{|G|}\chi_{V_i}(e)|G|=\dim V_i.$$
All I get from that is: if we decompose $R$ into a direct sum of irreducible representations, the multiplicites are the dimensions.
But what guarantees that any irreducible representation of $G$ appears in that decomposition of $R$? Why all irreducible representations of $G$ appear in the direct sum decomposition of the regular representation?
Best Answer
By definition, a simple module is nonzero. Hence its dimension, which equals its multiplicity in the decomposition, is positive. This guarantees that any irreducible representation does appear in it.