I thought I had a bit of a breakthrough. So, consider the algebra of a functions on the dual (quantum) group \widehat{S_3}. Consider the projection:

\displaystyle p_0=\frac12\delta^e+\frac12\delta^{(12)}\in F(\widehat{S_3}).

Define u\in M_p(\widehat{S_3}) by:

u(\delta^\sigma)=\langle\text{sign}(\sigma)1,1\rangle=\text{sign}(\sigma).

Note

\displaystyle T_u(p_0)=\frac12\delta^e-\frac12 \delta^{(12)}:=p_1.

Note p_1=\mathbf{1}_{\widehat{S_3}}-p_0=\delta^0-p_0 so \{p_0,p_1\} is a partition of unity.

I know that p_0 corresponds to a quasi-subgroup but not a quantum subgroup because \{e,(12)\} is not normal.

This was supposed to say that the result I proved a few days ago that (in context), that p_0 corresponded to a quasi-subgroup, was as far as we could go.

For H\leq G, note

\displaystyle p_H=\frac{1}{|H|}\sum_{h\in H}\delta^h,

is a projection, in fact a group like projection, in F(\widehat{G}).

Alas note:

\displaystyle T_u(p_{\langle(123)\rangle})=p_{\langle (123)\rangle}

That is the group like projection associated to \langle (123)\rangle is subharmonic. This should imply that nearby there exists a projection q such that u^{\star k}(q)=0 for all k\in\mathbb{N}… also q_{\langle (123)\rangle}:=\mathbf{1}_{\widehat{S_3}}-p_{\langle(123)\rangle} is subharmonic.

This really should be enough and I should be looking perhaps at the standard representation, or the permutation representation, or S_3\leq S_4… but I want to find the projection…

Indeed u(q_{(123)})=0…and u^{\star 2k}(q_{\langle (123)\rangle})=0.

The punchline… the result of Fagnola and Pellicer holds when the random walk is is irreducible. This walk is not… back to the drawing board.

I have constructed the following example. The question will be does it have periodicity.

Where \rho:S_n\rightarrow \text{GL}(\mathbb{C}^3) is the permutation representation, \rho(\sigma)e_i=e_{\sigma_i}, and \xi=(1/\sqrt{2},-1/\sqrt{2},0), u\in M_p(G) is given by:

u(\sigma)=\langle\rho(\sigma)\xi,\xi\rangle.

This has u(\delta^e)=1 (duh), u(\delta^{(12)})=-1, and otherwise u(\sigma)=-\frac12 \text{sign}(\sigma).

The p_0,\,p_1 above is still a cyclic partition of unity… but is the walk irreducible?

The easiest way might be to look for a subharmonic p. This is way easier… with \alpha_\sigma=1 it is easy to construct non-trivial subharmonics… not with this u. It is straightforward to show there are no non-trivial subharmonics and so u is irreducible, periodic, but p_0 is not a quantum subgroup.

It also means, in conjunction with work I’ve done already, that I have my result:

Definition Let G be a finite quantum group. A state \nu\in M_p(G) is concentrated on a cyclic coset of a proper quasi-subgroup if there exists a pair of projections, p_0\neq p_1, such that \nu(p_1)=1, p_0 is a group-like projection, T_\nu(p_1)=p_0 and there exists d\in\mathbb{N} (d>1) such that T_\nu^d(p_1)=p_1.

(Finally) The Ergodic Theorem for Random Walks on Finite Quantum Groups

A random walk on a finite quantum group is ergodic if and only if the driving probability is not concentrated on a proper quasi-subgroup, nor on a cyclic coset of a proper quasi-subgroup.