Why all irreducible representations appear in the regular representation? 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?
 A: 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.  
A: There is a theorem that states:
Let $V$ be a linear representation of $G$ s.t. $V = W_1 \oplus \dots\oplus W_k $ (all of these are irreducible representation of $G$) with $\phi$ being the character. If $W$ is the irreducible representation with character $\chi$, then the number of $W_i$ which is isomorphic to $W$ is the scalar product of $(\phi|\chi)$.
[Serre-Linear representation of finite groups-Ch-2-theorem 4]
So, when it says that "Any irreducible representation $V$ of G appears in the regular representation", it takes into consideration the isomorphs, as we normally do in algebra.
A: This is good question which haunted me for a while as a first learner. I think the book "David S. Dummit, Richard M. Foote - Abstract Algebra 3rd ed" answers it well, in Sec.18.2 Example(3).
The concept one needs to be clear is that the regular representation is afforded by CG ring itself when seen as a CG-module. To see what irreducible representation appear in it amounts to decomposing this CG-module. According to Wedderburn's Theorem, this CG seen as a ring is isomorphic to a decomposition of matrix rings (ideals):
$$
\mathbb{C} G \cong M_{n_{1}}(\mathbb{C}) \times M_{n_{2}}(\mathbb{C}) \times \cdots \times M_{n_{r}}(\mathbb{C}).$$
Furthermore, each $ M_{n_{i}}(\mathbb{C})$ decomposes further as direct sum of $n_i$ isomorphic simple left ideals. These left ideals seen as CG-module give a complete set of isomorphism classes of irreducible CG-modules, which again affords the regular representation.
Thus we reach the conclusion: the regular representation (over C) of G decomposes as the direct sum of all irreducible representations of G, each appearing with multiplicity equal to the degree of the irreducible representation.
