## Brownian motion and nonabelian finite groups

Based on the previous post, I can’t help make the following observations.

Let $G$ be a non-abelian finite group, and $G'$ its derived group.  Let $Pr(G)$ be the probability that two elements of $G$ commute: $Pr(G):=N_c(G)/N_t(G)$,  where $N_c(G)$ is the number of ordered pairs $(x,y)\in G\times G$ such that $xy=yx$,   and $N_t(G)$  is the total number of ordered pairs in $G\times G$.

Rusin has shown that $Pr(G) \leq \frac{1}{4}+\frac{3}{4}\frac{1}{|G'|}$.  Then the maximum value of $Pr(G)$ is $5/8$, and is attained for $G=H_8$ and $G=D_4$, two groups of order 8. Moreover, when $|G'|<8$ we have $Pr(G)>11/32$.

Now to another world where these very numbers also appear.   For any planar lattice $L$, denote by $a_{n,L}$ the number of self-avoiding nearest-neighbour paths, and call $\mu_L:=\inf_{n\geq 1}(a_{n,L})^{1/n}$ the connectivity constant.

Then a famous conjecture of Nienhuis is that for any lattice $L$ one has $a_n(L)=(\mu_L)^nn^{11/32+o(1)}$ as $n\rightarrow +\infty$. This rate has been proven conditionally by Lawler-Schramm-Werner.

Also, these authors had previously shown the following result: let $B$ and $B'$ be two independent Brownian motions with $|B_0-B_0'|=1$, then there exists a constant $c$ such that  the probability that these two random paths do not intersect is governed by the exponent $5/8$, namely one has $t^{-5/8}/c\leq P[ B(0,t) \cap B'(0,t)=\emptyset ]\leq c t^{-5/8}$. (Their results are vastly more general, but boil down to this for two families made by one Brownian each).

So, in both cases of non-abelian finite groups and of self-avoiding random walks we have these two rather noticeable numbers 5/8 and 11/32 appearing:   is there something deep going on, or is this just an accident?