# Probability of a group being finite

Suppose $$F_m := F[x_1, … , x_m]$$ is a free group on $$m$$ generators $$x_1, … , x_m$$ and lets define Cayley ball $$B_m^n := \{e, x_1, x_1^{-1}, … , x_m, x_m^{-1}\}^n$$ as the set of all elements with Cayley length $$n$$ or less.

Suppose $$R_1, … , R_l$$ are $$l$$ random elements chosen uniformly from $$B_m^n$$. Then we can define a random group as $$G(m, l, n) := \frac{F_m}{\langle \langle \{R_1, … , R_l\} \rangle \rangle}$$.

Now will suppose that $$m$$ is fixed and $$l = l(n)$$ depends on $$n$$. We say that the random group $$G(m, l, n)$$ belongs to a class of groups $$\mathfrak{U}$$ almost surely iff $$\lim_{n \to \infty} P(G(m, l, n) \in \mathfrak{U}) = 1$$.

A following theorem was proved by Ollivier:

If $$\lim_{n \to \infty} \frac{\ln(l(n))}{n(\ln(2m - 1))} > \frac{1}{2}$$ then $$G(m, l, n)$$ is almost surely finite.

If $$\lim_{n \to \infty} \frac{\ln(l(n))}{n(\ln(2m - 1))} < \frac{1}{2}$$ then $$G(m, l, n)$$ is almost surely infinite

My question is:

Is there some sort of exact expression for limit probability $$\lim_{n \to \infty} P(G(m, l(n), n) \text{ is finite})$$ for arbitrary non-decreasing $$l: \mathbb{N} \to \mathbb{N}$$?

I, personally think, that it is very likely to be of the form

$$\lim_{n \to \infty} P(G(m, l(n), n) \text{ is finite}) = \lim_{n \to \infty} a^{-b^{\ln(2m - 1)n - 2\ln(l(n))}}$$

for some positive real numbers $$a(m)$$ and $$b(m)$$.

However, I do not know that for sure. That is just intuition mostly based on analogies with a somewhat similar "phase transition theorem" from a completely different field, that states:

Suppose $$G(n, p(n))$$ is an Erdos-Renyi random graph with $$n$$ vertices and edge probability $$p(n)$$. Then $$\lim_{n \to \infty} P(G(n, p(n)) \text{ is connected}) = \lim_{n \to \infty} e^{-e^{\ln(n) - np(n)}}$$

• This seems more suited to MathOverflow than here, especially given that it's not been answered in over a year. Jul 6, 2021 at 21:09