Dixon's asymptotic without the classification of finite simple groups

Without using the classification of finite simple groups (CFSG), we show that the probability that two random elements of Sn$$ {S}_n $$ generate a primitive group smaller than An$$ {A}_n $$ is at most exp(−c(nlogn)1/2)$$ \exp \left(-c{\left(n\log n\right)}^{1/2}\right) $$ . As a corollary we get Dixon's asymptotic expansion 1−1/n−1/n2−4/n3−23/n4−⋯$$ 1-1/n-1/{n}^2-4/{n}^3-23/{n}^4-\cdots $$for the probability that two random elements of Sn$$ {S}_n $$ (or An$$ {A}_n $$ ) generate a subgroup containing An$$ {A}_n $$ .


introduction
We give a CFSG-free proof of the following result.
Theorem 1.Let G be the subgroup of S n generated by two random elements.The probability that G is contained in a primitive subgroup of S n smaller than A n is bounded by exp(−c(n log n) 1/2 ) for some c > 0.
This improves [EV,Theorems 1.3 and 1.6].By combining with the results of [D] we have the following corollary.(See also [O, A113869].) Corollary 2. The probability that two random elements of A n generate the group is The same asymptotic expansion is valid for the probability that two random elements of S n generate at least A n .

Satisfaction probability for unimodal words
Let F 2 = F {x, y} be the free group on two letters x, y.We write {x, y} * for the set of positive words, i.e., the submonoid generated by {x, y}.Let G = S n = Sym(Ω) for Ω = {1, . . ., n}.
SE is supported by the Royal Society.
Proof.Write w = w 1 • • • w ℓ with ℓ > 0 and w i ∈ {x ±1 , y ±1 } for each i.We may assume this expression is cyclically reduced.
We use the query model for random permutations (see [BS] or [EJ,Section A.1]).We gradually expose a random permutation π ∈ Sym(Ω) by querying values of our choice.At every stage x and y are partially defined permutations.We may query the value of any π ∈ {x ±1 , y ±1 } at any point ω ∈ Ω.If ω is already in the known domain of π, the known value is returned; this is a forced choice.Otherwise, a random value is chosen uniformly from the remaining possibilities (the complement of the known domain of π −1 ); this is a free choice.If the result of a free choice is a point in the known domain of any of x ±1 , y ±1 we say there was a coincidence.It is standard and easy to see that this process results in uniformly random permutations x and y once all values are revealed.
Begin by choosing any ω 1 ∈ Ω and exposing the trajectory For this event to occur we claim it is necessary there was some coincidence among our queries of the form ω w t = ω 1 (this is the crucial part of the argument).If ℓ(u) = 0 or ℓ(v) = 0 the argument is easy, so assume u and v have positive length.We may assume w 1 = x and w ℓ = y −1 since w is cyclically reduced.If there is no coincidence of the given form, the trajectory of ω 1 under u does not return to ω 1 , so ω 1 cannot be added to the known domain of y.Subsequently, during the negative part of the trajectory, unless there is a coincidence of the given form, ω 1 can be added to the known domains of x −1 and y −1 only.Therefore at the final step ω 1 is not in the known domain of y, so if the final step is forced then the result is not ω 1 , and if the final step is free then the result is not ω 1 by hypothesis.This proves the claim.
Since the probability that any given free choice results in ω 1 is at most 1/(n − ℓ), it follows by a union bound that Prob(ω w 1 = ω 1 ) ≤ ℓ/(n − ℓ).Conditional on the event E 1 choose a new point ω 2 outside the trajectory of ω 1 , examine the trajectory of ω 2 , and so on.In general, at iteration i, conditional on the event j<i E j where E j = {ω w j = ω j }, choose a point ω i ∈ Ω outside the union of the trajectories of ω 1 , . . ., ω i−1 and query the trajectory of ω i .In order for the event E i = {ω w i = ω i } to occur it is necessary that there be a coincidence of the form ω wt = ω i .Therefore .
Remark 4. The proof above is essentially that of [GHS + , Section 3].An error in that argument was identified in [E], but the problem does not arise for words of the special form w = uv −1 , as explained in the third paragraph of the proof.

The order of the group
Now let x, y ∈ S n be uniformly random and let G = x, y .
Proposition 5.There is a constant c > 0 such that Proof.Consider the elements of G of the form u with u ∈ {x, y} * and ℓ(u) < r (for some r).The number of such Applying the previous proposition, the probability that any two such u and u ′ are equal is bounded by 4 r (4r/n) ⌊n/4r⌋ ≤ exp(c 1 r − c 2 (n/r) log(n/r))) for some constants c 1 , c 2 > 0. Choosing r = c 3 (n log n) 1/2 for a small enough constant c 3 > 0, we obtain a bound of the required form.Failing this event, |G| ≥ 2 r − 1, so the result is proved.
A beautiful recent result of Sun and Wilmes [SW, SW2] (building on seminal work of Babai [B]) classifies primitive coherent configurations with more than exp(Cn 1/3 (log n) 7/3 ) automorphisms.A corollary is a CFSG-free determination of the uniprimitive subgroups of S n of order greater than the same bound.Much stronger bounds for the order of 2-transitive groups have been known for a long time [B2, P].Thus we know there are at most two conjugacy classes of primitive maximal subgroups M < S n apart from A n such that |M| > exp(Cn 1/3 (log n) 7/3 ), and each satisfies |M| = exp O(n 1/2 log n) .Since the number of pairs of permutations lying in a common conjugate of a maximal subgroup M is at most 1/[S n : M], Theorem 1 follows.