Given a random walk on finite group $W=(G,S,P)$ the separation distance is defined as $s_k=|G|max_{g \in G}(\frac{1}{|G|}-Q^k(g))$.
Now we have a lemma that says if $W=(G,S,P)$ is a directed random walk. then there exists constants $C, \alpha >0$ such that $$s_k \leq C*e^{\alpha k}$$ for all $k>0$. Here a directed random walk means that $P(s)=P(-s) \forall e \neq s \in S$
Ok here is the proof that I saw:
so let $d=d_S$ be a diameter of the group $G$(I assume by diameter they mean the diameter of the Cayley graph generated by $S$). By definition of a separation distance we have $s_1=s_2=...=s_{d-1}=1$, $s_d<1$. Then from monotonicity and submultiplicativity we get $$s_{md+i} \leq s_d^m, m \geq0, 0\leq i \leq d-1$$.(Here I don't understand, because by submultiplicativity(which we assume to be true for the separation distance), we get $s_{md+i} \leq s_d^m*s_i$?) And therefore, $s_k \leq s_d^{\lfloor\frac{k}{d}\rfloor}$. How do we prove the submultiplicativity property?
OK, lets see and example of a walk $(G=Z_6,S=\{e,1,-1\},P)$, we have $P(e)=0.5,P(1)=P(-1)=0.25$. So now, $d=3$ as the Cayley graph is a cycle of length 6. $s_1=s_2=0.1\bar{6}, s_3=0.1\bar{6}-0.25^3$. So see? $s_1 \neq 1$