We know that $\Bbb{C}S_4\cong \Bbb{C}\oplus \Bbb{C}\oplus M_{2}(\Bbb{C})\oplus M_{3}(\Bbb{C})\oplus M_{3}(\Bbb{C})$ as a $\Bbb{C}$-algebra.
I decomposed it into a direct sum of simple $\Bbb{C}S_4$-submodules. \begin{eqnarray*} \Bbb{C}S_4 &\cong& \Bbb{C}\oplus \Bbb{C}\oplus M_{2}(\Bbb{C})\oplus M_{3}(\Bbb{C})\oplus M_{3}(\Bbb{C})\\ &\cong& \Bbb{C}\oplus \Bbb{C}\oplus M_{2}(\Bbb{C})E_{11}^{(2)}\oplus M_{2}(\Bbb{C})E_{22}^{(2)}\\ &\oplus& M_{3}(\Bbb{C})E_{11}^{(3)}\oplus M_{3}(\Bbb{C})E_{22}^{(3)}\oplus M_{3}(\Bbb{C})E_{33}^{(3)}\\ &\oplus& M_{3}(\Bbb{C})E_{11}^{(3)}\oplus M_{3}(\Bbb{C})E_{22}^{(3)}\oplus M_{3}(\Bbb{C})E_{33}^{(3)}, \end{eqnarray*} where $E_{ii}^{(j)}$ is the $j\times j$ matrix with $1$ at the $(i,i)$-position and $0$ elsewhere.
I think $M_{3}(\Bbb{C})E_{11}^{(3)}\cong M_{3}(\Bbb{C})E_{22}^{(3)}\cong M_{3}(\Bbb{C})E_{33}^{(3)}$ as a simple $\Bbb{C}S_4$-module. So there is only one irreducible representation (up to equivalent) of degree $3$ and only one irreducible character of degree $3$. However, there are TWO irreducible characters of degree $3$ of $S_4$.
Where is the problem in my argument? Thanks for any help.
Thanks for @JyrkiLahtonen comments.
I thought I got where my problem is. I should not write $R=M_2(\Bbb{C})\oplus M_2(\Bbb{C}) \cong M_2(\Bbb{C})E_{11}^{(2)}\oplus M_2(\Bbb{C})E_{22}^{(2)} \oplus M_2(\Bbb{C})E_{11}^{(2)}\oplus M_2(\Bbb{C})E_{22}^{(2)}$.
This notation might cause a misunderstanding $M_2(\Bbb{C})E_{11}^{(2)}\oplus 0\oplus 0\oplus 0 \not\cong 0\oplus 0\oplus M_2(\Bbb{C})E_{11}^{(2)}\oplus 0$
and
$M_2(\Bbb{C})E_{11}^{(2)}\oplus 0\oplus 0\oplus 0 \not\cong 0\oplus M_2(\Bbb{C})E_{22}^{(2)}\oplus 0\oplus 0$
as $R$-modules.
I had better to write $$M_2(\Bbb{C})\oplus M_2(\Bbb{C}) \cong \left\{\left(\begin{pmatrix}a&0\\b&0\end{pmatrix}, 0\right)\right\} \oplus \left\{\left(\begin{pmatrix}0&a\\0&b\end{pmatrix}, 0\right)\right\} \oplus \left\{\left(0, \begin{pmatrix}a&0\\b&0\end{pmatrix}\right)\right\} \oplus \left\{\left(0, \begin{pmatrix}0&a\\0&b\end{pmatrix}\right)\right\}$$