Let $\mathbb C[S_3]$ be the group algebra of $S_3$. We have an isomorphism of $\mathbb C$-algebras
$$\mathbb C[S_3]\cong \mathbb C \times \mathbb C \times M_2(\mathbb C).$$
The existence of such an isomorphism is not difficult to show using Maschke's theorem. Moreover, one can show using character theory, as done, e.g., in this post, that this decomposition corresponds precisely to the (iso classes of) irreducible representations of $S_3$. However, how does one construct an explicit isomorphism between $\mathbb C[S_3]$ and its decomposition, as $\mathbb C$-algebras?
I do know that one copy of $\mathbb C$ corresponds to the trivial representation, one copy corresponds to the sign representation, and the matrix algebra $M_2(\mathbb C)$ corresponds to the "standard" representation. I also know that elements in $\mathbb C[S_3]$ have the form $\sum_{1 \leq i \leq 6}\lambda_i \sigma_i$ for some $\lambda_i \in \mathbb C$ and $\sigma_i \in S_3$. I also think this decomposition above into irreducible factors is somehow related to primitive idempotents? However, I'm not sure how to start constructing an explicit isomorphism.
Best Answer
The answer by Lukas Heger of course settles all finite groups at once. Still, for $S_3$ it is very possible to give an explicit hands-on description of this map.
Such an isomorphism is essentially nothing more than writing out the decomposition of $\mathbb{C}[S_3]$ into primary components explicitly. This is not too difficult to do: For notation, let $\tau$ be any nontrivial 3-cycle (so that $A_3=\{1,\tau, \tau^2\}$) and $\sigma$ be any transposition.
The subspace of fixed points of $\mathbb{C}[S_3]$ is just $T$, the line of all vectors with constant coefficients, i.e. all multiples of the vector $\sum_{g\in S_3}g$.
Similarly, the subspace $S$ corresponding to the sign representation can be easily determined to be all multiples of the vector $$1+\tau+\tau^2-\sigma(1+\tau+\tau^2),$$ i.e. all those vectors that have equal coefficients on any coset of $A_3$ but overall coordinate sum of $0$. (This can for instance be determined from the observation that both the sign and the trivial representation are fixed by $A_3$ - hence both components must lie in the subspace of $A_3$-fixed points, which are just the multiples of $1+\tau+\tau^2$.)
Thus, the last remaining component is the four dimensional invariant complement of $T\oplus S$, which is just $$V=\left \{ x\in \mathbb{C}[S_3]:\sum_{g\in A_3}x_g g=\sum_{g\in \sigma A_3}x_g g=0 \right \}. $$
These three subspaces $T,S$ and $V$ are now exactly your desired $\mathbb{C}, \mathbb{C}$ and $M_2(\mathbb{C})$. The isomorphism from $\mathbb{C}[S_3]$ can then be read off from the projection onto these subspaces (i.e. primary idempotents). These can be cleanly written in terms of $u=\frac{1}{3}(1+\tau+\tau^2)$: The idempotent in $T$ is $e_T=\frac{u+\sigma u}{2}$, in $S$ we have $e_S=\frac{u-\sigma u}{2}$ and for $V$ we have the complementary $$e_V=1-e_T-e_S=1-u.$$ Altogether this explicitly determines the algebra isomorphism $$\mathbb{C}[S_3]\to T\times S\times V, x\mapsto (e_Tx,e_Sx,e_Vx).$$
The further identifications $T\simeq S\simeq \mathbb{C}$ are of course canonical, while $V\simeq M_2(\mathbb{C})$ is determined only up to isomorphism, which is essentially a choice on how to split $V$ further into two complementary $S_3$-invariant subspaces. I will leave it to you to figure out a specific such isomorphism.