Random Walks on Groups that are neither aperiodic nor irreducible

Question

This question is about random walks on finite groups that are periodic but not irreducible. It follows on from this MO question.

Let $G$ be a finite group and $\Sigma\subset G$. Suppose $\Sigma$ generates a proper subgroup $H\leq G$. Suppose further that $\Sigma\subset Nh$, a coset of $N\rhd H$, a proper normal subgroup.

In this case $H/N\cong C_d$, where $d$ is the period of the random walk (equal, e.g. to the order of the coset $Nh$ in $H/N$).

So we have $H\leq G$, and the random walk, essentially, bounces around $H/N\cong C_d$.

Suppose further that there exists disjoint subsets $S_0,S_1,\dots,S_{d-1}\subset G$ of size $|G|/d$ such that:

• $\displaystyle \bigcup_{i=0}^{d-1}S_{i}=G$
• $\Sigma^{(k)}\subset Ng^k\subset S_{k}$, where $k$ is understood$\mod d$
• again for $k,\,\ell$ understood$\mod d$, for any $k$, $\ell$, $\displaystyle \Sigma^{(k)}\subset S_{k+\ell}(S_\ell)^{-1}$

My question is whether the cyclic nature of $H/N$ is reflected in these $S_k$.

Can there exist an element $x\in S_k$ such that there exists elements in $y\in S_m$ (but not $Nh^m$) and $z\in S_n$ (but not in $Nh^n$) such that $$yz=x,$$ but $m+n\neq d$.

I am fairly sure that there is a counterexample for groups $N\rhd H\not\rhd G$. I was trying to construct one using direct products but had to give up. Perhaps a counterexample might exist for $S_3\times S_4$ or $S_4\times S_4$.

Answer 1

The following follows from the linked question.

Let $G=S_3\times C_2$ and let $H\leq G$ be $H=\{(e,0),((12),0),(e,1),((12),1)\}$ and $N\rhd H$ be given as $N=\{(e,0),((12),0)\}$.

Let $\Sigma=\{(e,1),((12),1)\}$ and note this generates $H$ and $\Sigma=N(e,1)$ and the walk jumps between $N$ and $N(e,1)$.

Now define $S_0=\{(e,0),((12),0),((13),1),((23),1),((123),1),((132),1)\}$ and $S_1=G\backslash S_0$. This is a partition of unity and also the powers of $N(e,1)^k$ lie in $S_{k}$ (understood $\mod 2$)... and we have the inclusions of $\Sigma^{(k)}\subset S_{k+\ell}(S_\ell)^{-1}$.

Now let $y=z=((132),0)$ both in $S_1$. Note $$x=yz=((123),0)\in S_1,$$

and $1+1\neq 1\mod 2$, so we have a counterexample