irreducible representation extending as projection onto matrix algebra I want to ask a question about some statement in Peter Webb's "A course in finite group representation theory", p40 (book on his website, not the published version)
Let $\rho_1,\ldots,\rho_r$ be irreducible representations of finite group $G$ over $\mathbb{C}$ with degree $d_1,\ldots,d_r$ and character $\chi_1,\ldots,\chi_r$. It says that after extending by linearity $\rho_i:G\rightarrow M_{d_i}(\mathbb{C})$ to $\mathbb{C}$-algebra homomorphism $\rho_i: \mathbb{C}[G]\rightarrow M_{d_i}(\mathbb{C})$, we idetify $\mathbb{C}[G]$ with $\Pi_{j=1}^rM_{d_j}(\mathbb{C})$ by Artin-Wedderburn, then each $\rho_i:\Pi_{j=1}^rM_{d_j}(\mathbb{C})\rightarrow M_{d_i}(\mathbb{C})$ is a projection onto the $i$th matrix summand.
My question is why he can conclude that $\rho_i$ is a projection onto $M_{d_i}(\mathbb{C})$ or why $\rho_i$ is surjective. The book says it is because of the way we decompose $\mathbb{C}[G]$ as a sum of matrix algebra. But I have spent whole day on this problem and still can't figure it out. 
 A: This argument is from Proposition 7.2, p.382, and Lemma 8.3, p387, in Grillet's Abstract Algebra 2/e.


*

*Recall that
\begin{eqnarray*}
\Bbb{C}G
&\cong& M_{d_1}(\Bbb{C})\oplus M_{d_2}(\Bbb{C})\oplus \cdots\oplus M_{d_r}(\Bbb{C})\\
&\cong&
\overbrace{S_1\oplus S_1\oplus \cdots\oplus S_1}^{d_1\text{-times}}\oplus \\
&& \overbrace{S_2\oplus S_2\oplus \cdots\oplus S_2}^{d_2\text{-times}}\oplus \\
&&\cdots\oplus\\
&& \overbrace{S_r\oplus S_r\oplus \cdots\oplus S_r}^{d_r\text{-times}} \\
&\cong&
S_1^{d_1}\oplus S_2^{d_2}\oplus \cdots \oplus S_r^{d_r}.
\end{eqnarray*}
In fact,
$$S_i\cong \left\{\left(0, ..., 0, \begin{pmatrix}
0 & \cdots & 0 & a_{1k} & 0 & \cdots & 0 \\
0 & \cdots & 0 & a_{2k} & 0 & \cdots & 0 \\
\vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 & \cdots & 0 &  a_{d_i k} & 0 & \cdots & 0 \\\end{pmatrix}, 0, ..., 0\right)\mid a_{1k}, a_{2k}, ..., a_{d_i k}\in \Bbb{C}\right\}$$
for some $k\in \{1, 2, ..., d_i\}$.

*Since $\rho_i$ is an irreducible representation of $G$ over $\Bbb{C}$,
let $L_i$ be the corresponding simple $\Bbb{C}G$-module.
That is, $\rho_i:G\to GL(L_i)$.

*$L_i$ is isomorphic to a simple $\Bbb{C}G$-submodule $S_i$ of $\Bbb{C}G$.
(See Theorem C-2.33 in Rotman's Advanced Modern Algebra 3/e Part II.)
So $\rho_i:G\to GL(S_i)$.
In rigorous,
we should write $L_i\cong 0\oplus \cdots \oplus 0\oplus S_i\oplus 0\oplus\cdots \oplus 0$.

*By the correspondence between the representations and modules,
for any $\sum_{g\in G}z_g g\in \Bbb{C}G$ and $s\in S_i$,
\begin{equation}
\tag{1}
\left(\sum_{g\in G}z_g g\right)\cdot s
=\sum_{g\in G}z_g (\rho_i(g)(s)).
\end{equation}

*We define $\overline{\rho_i}:\Bbb{C}G\to \text{End}_{\Bbb{C}}{(S_i)}$
by $$\overline{\rho_i}\left(\sum_{g\in G}z_g g\right)
=\sum_{g\in G}z_g \rho_i(g).$$

*For any
\begin{eqnarray*}
s
&\in& S_i\\
&\cong& 0\oplus \cdots \oplus 0\oplus S_i\oplus 0\oplus\cdots \oplus 0\\
&\subseteq & 0\oplus \cdots \oplus 0\oplus M_{d_i}(\Bbb{C})\oplus 0\oplus\cdots \oplus 0\\
&\subseteq & \Bbb{C}G
\end{eqnarray*}
and
\begin{eqnarray*}
x=\sum_{g\in G}z_g g
&\in& M_{d_j}(\Bbb{C}) \text{ (Notice the subscript!)}\\
&\cong& 0\oplus \cdots \oplus 0\oplus M_{d_j}(\Bbb{C})\oplus 0\oplus \cdots \oplus 0\\
&\subseteq & \Bbb{C}G,
\end{eqnarray*}
where $j\neq i$,
we have
$x\cdot s=0$ because they are in the different components $M_{d_i}(\Bbb{C})$ and $M_{d_j}(\Bbb{C})$.
In this case, we view $x\cdot s$ as a prodcut of two elements $x$ and $s$ in $\Bbb{C}G$.
Then
\begin{multline*}
\overline{\rho_i}\left(x\right)(s)=
\overline{\rho_i}\left(\sum_{g\in G}z_g g\right)(s)
=\left(\sum_{g\in G}z_g \rho_i(g)\right)(s)\\
=\sum_{g\in G}z_g (\rho_i(g)(s))
\stackrel{(1)}{=}\left(\sum_{g\in G}z_g g\right)\cdot s
=x\cdot s
=0
\end{multline*}
It follows that $\overline{\rho_i}\left(x\right)=0$ if $x\in M_{d_j}(\Bbb{C})$ for $j\neq i$.

*For any $s\in S_i$
and $x=\sum_{g\in G}z_g g\in M_{d_i}(\Bbb{C})$ (Notice the subscript!),
\begin{multline*}
\overline{\rho_i}\left(x\right)(s)
=\overline{\rho_i}\left(\sum_{g\in G}z_g g\right)(s)
=\left(\sum_{g\in G}z_g \rho_i(g)\right)(s)\\
=\sum_{g\in G}z_g (\rho_i(g)(s))
\stackrel{(1)}{=}\left(\sum_{g\in G}z_g g\right)\cdot s
=x\cdot s.
\end{multline*}
In this case, we view the action of $x$ on $s$ as a linear transformation on $S_i$.
It follows that $\overline{\rho_i}\left(x\right)=x$.

*In addition, $\text{End}_{\Bbb{C}}{(S_i)}\cong \text{End}_{\Bbb{C}}{(L_i)}\cong M_{d_i}(\Bbb{C})$.
That is why the author says that $\overline{\rho_i}:\Bbb{C}G\to M_{d_i}(\Bbb{C})$ is a projection.
