This exercise asks us to prove that if is a probability measure on the finite group , then if and only if . Here is the convolution product, is the uniform probability on defined by for all and is defined by .
The `if’ direction is easy: just note that and .
The `only if’ direction is a bit harder, and has a nice solution using the Fourier setup. The Fourier transform of at the representation is
If is unitary, then for all (where is the transposed complex conjugate of the matrix ). Hence
Hence, taking the Fourier transform of at a non-trivial unitary irreducible representation , we get
But if is a complex matrix such that then . So we’ve shown that for all unitary irreducible representations . Moreover, since is a probability measure,
So the Fourier transform of agrees with the Fourier transform of on all unitary irreducible representations. Any representation is equivalent to a direct sum of such representations, so by the Fourier Inversion Theorem, .