[math] \newcommand{\mathds}{\mathbb}[/math]

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Find an alternative, more conceptual proof for the equality

[[math]] S_3^+=S_3 [[/math]]

by considering the following morphism, called universal coaction map,

[[math]] \Phi:\mathbb C^3\to\mathbb C^3\otimes C(S_3^+)\quad,\quad e_i\to\sum_je_j\otimes u_{ji} [[/math]]

then by applying the Fourier transform over the group [math]\mathbb Z_3[/math] on the [math]\mathbb C^3[/math] part, and then observing that the coefficients of [math]u[/math], in Fourier transform, must clearly commute.