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, .

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I’m adding this comment to record a simpler way to do the exercise, suggested by Jon Barrett.

From we have

for all . Putting gives

.

The Cauchy–Schwarz inequality applied to the dot-product of the all ones vector and the vector with entries for shows that

So we have equality in the Cauchy–Schwarz inequality, and so all the must be equal.