Regular Representation and Irreducible Representation It seems every irreducible representation under certain carrier space $W\subset V$ (which is NOT a subspace of $\mathbb{C}G$), there always exists an isomorphism $g: W\rightarrow W_i$ where $W_i$ is the subspace of $\mathbb{C}G$ yielding an irreducible representation.
It seems to prove such a statement requires showing the existence of $g$ and $W_i$ simultaneously. Can somebody please give me proof of the existence of such an isomorphism?
I don't have any background in the Field and Ring theory.
 A: Usually you proceed as follows: (I assume you speak of finite groups, as you use $\mathbb{C}G$)
You first derive a bit theory about characters, that is, for a (finite-dimensional) representation $\Gamma:G\to\text{GL}(V)$ just the trace. For these you have some sort of orthoginality relation, namely
$$
 \frac{1}{|G|}\sum_{g\in G} \overline{\text{tr}\,\Gamma(g)}~\text{tr}\,\Gamma'(g)=\delta_{\Gamma,\Gamma'}
$$
for two irreducible representations $\Gamma:G\to\text{GL}(V),~\Gamma':G\to\text{GL}(V')$, where the $\delta_{\Gamma,\Gamma'}$ indicates that this sum equals 1 if the irreducible representations are the same (that is, equivalent, recalling the cyclicity of the trace), and 0 otherwise.
An arbitrary representation may now be decomposed into a direct sum of irreducible representations. The character of such a representation is then just summing the characters of the irreducible representations contained in that decomposition. So the above orthogonality representation may be used as a criterion "how many times a certain irreducible representation is contained in a given representation", since we just count "how many times the $\delta_{\Gamma,\Gamma'}$ is non-zero.
To the very end, you can just do that for the regular representation on $\mathbb{C}G$ and find that any irreducible representation with dimension $d$, occurs in the regular representation $d$ times. So you can conclude that for any ( $d$-dimensional) irreducible representation $\Gamma$ the $\mathbb{C}G$ contains $d$ $d$-dimensional subspaces (particularly one, which you had asked for), which pairwise only intersect in $\{0\}$ and on each of them the regular representation acts like $\Gamma$. The last part "acts like $\Gamma$" means that the regular representation, restricted to such a subspace, is equivalent to that given $\Gamma$. This equivalence then yields the isomorphism you had asked for.
A: The above answer, which I had chosen to be the proper answer to my question, turned out to be incomplete. Fortunately, I finally found the answer based on Ed Segal's lecture note on the representation theory.
The proof starts with the following two claims.

*

*Denoting the vector space spanned by $G$-linear map $f:V\rightarrow W$ by $\mathrm{Hom}_G(V,W)$, the $\dim{\mathrm{Hom}_G(V,W)}=0$ counts the multiplicity of the irrep $W$ in $V$. The proof is immediate from Schur's lemma.

*There is a natural isomorphism between $\mathrm{Hom}_G(V,W\oplus U)\approx\mathrm{Hom}_G(V,W)\oplus\mathrm{Hom}_G(V,U)$. The isomorphism is defined by $P: P(f)=(\pi_W\circ f, \pi_U\circ f)$.

Meanwhile, the statement I wanted to prove is as follows.
Every irrep $W$ appears in the regular representation $V_{reg}$ whose multiplicity is determined by $\dim{W}$.
The proof follows from demonstrating $\mathrm{Hom}_G(V_{reg},W)\approx W$. To prove this statement, we investigate the map defined by the evaluation at $e$ , i.e., $T:T(f)=f(e)$. Clearly, this map is linear.
Next, we investigate $\ker{T}$, i.e., some $f$ such that $f(e)=0$. Then by $G$-linearity, $0=\rho_Wf(e)=f(\rho_{V_{reg}}(e))=f(g)$ for any g. Thus $f$ must be zero map, and hence $T$ is injective.
Finally, we investigate $\mathrm{Im}T$. Suppose $w\in W$. Then if we define $f(g)\equiv\rho_W(g)(w)$, $f(\rho_{V_{reg}}(h)(g))=\rho_W(hg)(w)=\rho_W(h)f(g) $ and hence $f\in\mathrm{Hom}_G(V_{reg},W)$ and $T(f)=w$. Thus $T$ is surjective.
Hence $\mathrm{Hom}_G(V_{reg},W)\approx W$ and by the claim 1, $\dim{W}$ is the multiplicity of $W$ in thee regular representation.
