Artin-Wedderburn isomorphism of $\mathbb{C}[S_3]$

The map is obviously an homomorphism and the spaces have the same dimension (which is $6$), so it's enough to show that it is injective.

Let \[ \sum_{g \in S_3} a_g g \] be any linear combination such that the image is $0$. Since the coefficient of the first component is $0$, then \[ \sum_{g \in S_3} a_g = 0 \] and since the coefficient of the second component is $0$ then \[ \sum_{g \in S_3} sgn(g) a_g = 0 \] so, by taking the sum and the difference, we immediately have $a_1 + a_{(123)} + a_{(132)} = 0$ and $a_{(12)} + a_{(13)} + a_{(23)}= 0$.

Now, look at the corresponding matrices (see attachment). The coefficient of the entry $(1,1)$ is $a_{1} - a_{(132)} + a_{(23)} - a_{(13)}$, and the coefficient of the entry $(2,2)$ is $a_{1} - a_{(123)} - a_{(23)} + a_{(13)}$. Taking the sum, we have $2 a_1 - a_{(132)}- a_{(123)} = 0$, but since $a_1 + a_{(123)} + a_{(132)} = 0$ we get $a_1 = 0$ and so $a_{(123)} = -a_{(132)}$. Taking the difference instead, we get $a_{(23)} = a_{(13)}$. Looking at the entries $(1,2)$ and $(2,1)$, we have $-a_{(123)} + a_{(132)} + a_{(12)} - a_{(23)} = 0$ and $a_{(123)} - a_{(132)} + a_{(12)} - a_{(13)} = 0$, and now taking the sum, since $a_{(13)} = a_{(23)}$, we have $a_{(12)} = a_{(13)}$ as well and so they are all $0$, since the sum is $0$. Taking now the difference, we get $a_{(123)} = a_{(132)}$, but they are opposite and so they are both $0$ as well.

In conclusion, all the coefficients are $0$, so the map is injective, so it is an isomorphism.