Intuition for decomposition of regular representation Working over $\mathbb{C}$ it is well-known that if $G$ is a finite group and $V$ is its regular representation then every irreducible representation $V_i$ of $G$ occurs as a summand of $V$ with multiplicity $\dim V_i$.  This is often proved using character theory, but I find this argument slightly unenlightening from the prespective of understanding why the result is true.
I recently came across a nicer proof in Ed Segal's online notes in which he shows that for any representation $W$ of $G$ there is a natural isomorphism of vector spaces
$$ \operatorname{Hom}_G (V, W) \cong W $$
by evaluating a homomorphism at the basis element of $V$ corresponding to the identity in $G$.  But this is still a little indirect, going via Schur's theorem.
Don't get me wrong, I think both proofs are lovely pieces of mathematics (and they're obviously essentially equivalent), however I don't feel like either of them makes the result seem obvious.  Perhaps this is just not possible, given the dependence on the ground field (although I guess the second proof works over any algebraically closed field whose characteristic does not divide $|G|$), but I'd be grateful to hear of any intuitive explanations for this decomposition.
 A: Intuitively, for $H$ a subgroup of $G$ and $\sigma$ a representation of $H$, the induced representation $Ind_H^G \sigma$ is "built up from" all of the irreducible representations of $G$ which, when restricted to $H$, contain a copy of $\sigma$.
Essentially by definition, the regular representation is the representation $Ind_1^G 1$ induced from the trivial representation of the trivial group. So you'd expect that every irreducible representation occurs in the regular representation.
As for the multiplicity, for an irreducible representation $\pi$ of $G$ you can use Frobenius reciprocity to calculate
$$\dim Hom_G(\pi,Ind_1^G 1)=\dim Hom_1(\pi|_1,1)=\dim Hom_1(1^{\oplus \dim \pi},1)=\dim \pi.$$
A: Suppose $V$ is an irreducible $\mathbb{C}G$-module and $v_0\in V$ is nonzero. Then, regarding $\mathbb{C}G$ as the left regular module, the map
$$\mathbb{C}G\to V,\;\;\;g\mapsto g.v_0$$
is a surjective $\mathbb{C}G$-module homomorphism. The kernel of this map is $$\mathrm{Ann}(V)=\{x\in\mathbb{C}G\mid xv=0\mbox{ for all }v\in V\}$$ 
It follows that $V\cong \mathbb{C}G/\mathrm{Ann}(V)$ as $\mathbb{C}G$-modules. 
The annihilator $\mathrm{Ann}(V)$ is a left ideal in $\mathbb{C}G$, and is therefore a left $\mathbb{C}G$-module. But, we know by Maschke's Theorem that $\mathbb{C}G$ is completely reducible, so $\mathbb{C}G=X\oplus\mathrm{Ann}(V)$ for some $\mathbb{C}G$-submodule $X\subset\mathbb{C}G$. Finally, we get
$$X\cong \mathbb{C}G/\mathrm{Ann}(V)\cong V$$
as desired.
