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.
Best Answer
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.