irreducible representation contained in regular rep 
Why is every irreducible representation contained in the regular representation?

Suppose $W$ is a irreducible representation. ( i.e. a vector space over $\mathbb{C}$ which $G$ acts on with no $G$-invariant subspace)
Then fix $0\neq w\in W$ and define $\mathbb{C}[G]\rightarrow W, \; \sum\alpha_gg \mapsto\sum\alpha_gg(w)$. Hence by irreducibility of $W$, $W$ is isomorphic to $\mathbb{C}[G]/(M)$ for some maximal ideal $M$ in $\mathbb{C}[G]$.
But I still can't see why W can be regarded a subspace of $\mathbb{C}[G]$? 
 A: Update
Looking back at this old answer of mine, I want to improve it.
For $W$ to be irreducible, we mean it has no nonzero proper subrepresentations. If we take any $0 \neq w \in W$, then its orbit $G w$ is a nonzero subrepresentation of $W$, so we must have $W = G w$.
Now since $W$ is a $G$-module, we can regard it as a $\mathbb{C}[G]$-module if we extend by linearity.  And there's a natural map of $G$-modules $\mathbb{C}[G] \to W$ given by
$$\left(\sum_i a_i g_i\right) \mapsto \sum_i a_i (g_i \cdot w)$$
We've just seen this map is surjective, so if $I$ is its kernel then by the first isomorphism theorem $W \cong \mathbb{C}[G]/I$.
Now since $\mathbb{C}[G]$ is semisimple, any quotient of $\mathbb{C}[G]$ is isomorphic to a submodule of $\mathbb{G}[G]$.

Previous answer
I'm assuming $G$ is a finite group and $\mathbb{C}[G]$ is the group algebra.
The way I've seen this done is via character theory.  Observing that the regular representation is a permutation representation, we can extract its character.  It follows from character theory that any $k$-dimensional irreducible representation of $G$ appears $k$ times in the regular representation.  
