I am reading the paper Shuffling Cards and Stopping Times by P. Diaconis and D. Aldous (American Math. Monthly, 1986).
Context.
Let $G$ be a finite group and $Q$ be a probability distribution on $G$. We get a random walk on $G$ defined as follows: the probability of going from $x\in G$ to $gx\in G$ in one step is $Q(g)$. The $k$-fold convolution of $Q$, written $Q^{k^*}$, is a probability distribution on $G$ defined as follows: $Q^{k^*}(g)$ is the probability of landing at $g$ at the $k$-th step if we start at $e$ (identity) at the zeroth step.
At the bottom of the second page of the paper (near equation 2.2), one reads:
A fundamental result is that repeated convolutions converge to the uniform distribution $U$: $$Q^{k^*}(g)\to U(g) \text{ as } k\to \infty$$ unless $Q$ is concentrated on some coset of some subgroup.
(By "$Q$ is concentrated on $X$ we mean $Q$ vanishes outside $X$").
I am trying to see why the above is true. I know the following result:
If $\Sigma$ is the support of $Q$, that is, $\Sigma=\{x\in G:\ Q(x)>0\}$, then
$\bullet$ The chain driven by $Q$ is irreducible if and only if $\Sigma$ generates $G$.
$\bullet$ Assuming $\Sigma$ generates $G$, the random walk driven by $Q$ is aperiodic if and only if $\Sigma$ is not contained in a coset of a proper normal subgroup.
So if $Q$ is not concentrated in some coset of some proper subgroup, then the chain driven by $Q$ is ergodic, and thus we have a unique stationary distribution. It's easy to check that the uniform distribution is stationary. So the statement given in the aforementioned paper is proven in this direction.
Question.
I am unable to do the other direction.
That is, if $Q$ is assumed to be concentrated on some coset of some proper (the authors didn't put proper in their writing) subgroup of $G$, then the $k$-fold convolution does not converge to $G$.
Further, the authors mention that Poincare has given a proof using Fourier analysis (Calcul Probabilites, Gauthier-Villars, Paris, 1912). I would really like to learn this proof, but I do not know French. If you know this proof, or any Fourier analytical proof, then I am very grateful if you cna share that.
The comprehensive statement is:
Proof here (Theorem 1.3.2).