Abelian sections of the symmetric groups with respect to their index

We show the existence of an absolute constant $\alpha>0$ such that, for every $k \geq 3$, $G:=\mathop{\mathrm{Sym}}(k)$, and for every $H \leqslant G$ of index at least $3$, one has $|H/[H,H]| \leq |G:H|^{\alpha/ \log \log |G:H|}$. This inequality is the best possible for the symmetric groups, and we conjecture that it is the best possible for every family of arbitrarily large finite groups.


Introduction
Abelian quotients of permutation groups attracted attention for the first time in [4], where the authors show that an abelian section of Sym(k) has order at most 3 k/3 , for every k ≥ 3. Better bounds hold for primitive groups [1], and for transitive groups [5]. In these notes, a different aspect concerning the subgroups of the symmetric groups is revealed: as the index increases, the abelian quotients grow as slowly as possible. This bound is sharp in a number of situations, which make the proof by induction somewhat challenging. For example, equality is satisfied for infinitely many k, by elementary abelian groups of order p ⌊k/p⌋ having all of their orbits of cardinality p. More is said in Section 5, where we show some evidences towards the fact that Theorem 1.1 is the best possible for every family of arbitrarily large finite groups.

Preliminaries
Unless explicitly stated otherwise, all the logarithms are to base 2, and exp(x) := 2 x . We will often use inequalities for the factorial function: to avoid useless calculations, we recall three estimates, which provide increasing accuracy.
Proof. The right side of (i) is obvious and the left side is equivalent to √ k ≤ (k!) 1/k . These are the geometric means of {1, k} and {1, 2, ..., k} respectively. Since the product j(k − j) is maximum where j is close to k/2, then the first mean is at most the second, as desired. For (ii), let us notice that To obtain the left side, we use j+1 j ≤ e 1/j , so that To obtain the right side, we use j j+1 ≤ e −1/(j+1) , so that Finally, (iii) is the sharp estimate of Robbins [8].
As it is easy to see, taking logarithms in Lemma 2.1 (i), we see that 1 2 k log k ≤ log(k!) ≤ k log k, and that 1 2 log k ≤ log log(k!) ≤ 2 log k for all k ≥ 3.
Lemma 2.2. Let G be an arbitrary group, N ⊳ G, and G/N ∼ = Q. Then To prove Theorem 1.1, we seek for an absolute constant C > 0 such that, for every sufficiently large k and every H Sym(k) of index at least 3, one has H| .
By f (n) ≪ g(n) and g(n) ≫ f (n) we mean the same thing, namely, that there is C > 0 such that f (n) ≤ C · g(n) for all n ≥ C. Moreover, we will use frequently the fact that the function x/ log x is an increasing function when x ≥ 3. Remark 2.3 (Small subgroups). Choose a large k, and let G := Sym(k). We notice that (2.1) is true for subgroups H such that log |H| ≤ 1 4 k log k. In fact, via Lemma 2.1 (i), we have log |G : H| ≥ log(k!) − 1 4 k log k ≥ 1 4 k log k. From the main theorem of [4] we have |H/H ′ | ≤ 3 k/3 , and so From the main theorem of [6], a primitive subgroup of G which does not contain Alt(k) has size at most 4 k . Since Alt(k) is perfect, primitive groups do not affect the proof of Theorem 1.1.

Transitive subgroups
In short, looking inside Sym(k), we use two sharp inequalities with respect to k: one for the index of a maximal transitive group, and one for the abelianization of a transitive group, which is provided by [5]. The structure of the groups we are left with is trapped by some smaller symmetric or alternating group, and this allows to argue that their abelian quotients are small.
Proof. We will prove the same inequality with the natural logarithm in place of log. Taking the logarithm in Lemma 2.1 (ii), we obtain . Using three times these inequalities, we can write It rests to prove that this is at least 1 3 k ln b, and arranging the terms we see that this is equivalent to Now we observe that, for large enough k we have because 2 3 k − 1 /(k/2) converges to 4/3, while ln(k)/ ln(k/2) converges to 1. Finally, for every 2 ≤ b ≤ k/2, the right side of (3.1) is at least b/ ln b, because x/ ln x is an increasing function.
In reality, the proof of Theorem 1.1 requires a slightly better result than Theorem 3.2 itself. Given a finite group R and a prime p ≥ 2, let a p (R) be the number of the abelian composition factors of R of order p. We define Informally, this is the logarithm of the "abelian portion" of R. We also introduce some more notation about wreath products. Let W := R b ⋊ Sym(b). We denote by ρ Sym(b) : W → Sym(b) the projection over the top group, and for every j = 1, ..., b, we denote by b j=1 R j the base subgroup. For every j we have This allows to consider the projections ρ j : is true under our hypothesis. Since ρ(G) Sym(k) is transitive, putting back Theorem 3.2, and noting that 2/ √ π < 2, we obtain the claimed inequality.
Proof of Theorem 1.1 for transitive H. We will always suppose that k is larger than any constant. We have already settled primitive groups at the end of Section 2, so let H G be transitive but not primitive, and contained in a maximal transitive group W := Sym(a) b ⋊ Sym(b) as in Lemma 3.1. In particular, we choose W in such a way that a is the smallest possible (equivalently, the blocks of imprimitivity have minimal size). Then log |G : H| ≥ log |G : For such a fixed b, we take a closer look at H W . Let us recall the notation we introduced just before the statement of Proposition 3.3. First, the projection over the top group ρ Sym(b) (H) Sym(b) is transitive (otherwise H Sym(k) itself would be not transitive). This implies that the projections ρ j of N H (Sym(a)) in Sym(a) are all isomorphic for every 1 ≤ j ≤ b. Let us denote by H proj Sym(a) one of these projections. Since the blocks of imprimitivity have minimal size by the construction of W , we get that H proj Sym(a) is primitive. If H proj does not contain Alt(a), then from the main theorem of [6] we have |H proj | ≤ 4 a . By the imprimitive embedding theorem we have H (H proj ) b ⋊ Sym(b), and so |H| ≤ 4 ab · b!. Using also Lemma 2.1 (i), and (3.2), it follows that As we have seen in Remark 2.3, this implies that H is too much small. We are left with the cases where either H proj = Alt(a) or H proj = Sym(a). If H proj = Alt(a), then from Proposition 3.3 (notice that a(R) = 0 in this case) we have Using again (3.2), we obtain log |G : H| log log |G : H| .
If H proj = Sym(a), then from Proposition 3.3 (notice that a(R) = 1 in this case) we have Using a last time (3.2), we obtain log |H/H ′ | ≪ log |G:H| log log |G:H| as before, and the proof of the transitive case is complete.

Intransitive subgroups
Let H be contained in Sym(a) × Sym(b) for some a + b = k and 1 ≤ b ≤ a ≤ k − 1. Consider the projection ρ : H → Sym(b). The factorized subgroup Ker(ρ) × ρ(H) Sym(a) × Sym(b) has the same size of H, and not smaller abelianization from Lemma 2.2. Thus, we can suppose H = A × B for some A Sym(a) and B Sym(b). We can also suppose that a, as k, is larger than any constant. We have for all a large enough. Set x := a!/|A|. We notice that, for all a ≥ 3, We can assume x ≥ a. Now we compute the following limit.
Lemma 4.2. Let X, Y, K be positive integers larger than 2. If then log x log log x + log y log log y ≤ log(xy · K) log log(xy · K) for every 3 ≤ x ≤ X and 3 ≤ y ≤ Y .
Proof. We will argue replacing x, y, X, Y, K with their logarithms (to the base 2). Fix K ≥ log 3, and set We will prove that f (x, y) is non-increasing in x and y. To do this, we can replace log 2 with ln in the definition of f . When considered in (1, +∞) × (1, +∞), f (x, y) is an analytic function. Computing the partial derivative with respect to x, we obtain ∂f ∂x Since the expression of f is symmetric with respect to x and y, we have ∂f ∂y ≤ 0 as before, and the proof follows.
From the previous lemma, it is enough to check (4.2) when x = a! and y = b!. The next inequality is really about the inverse function of the gamma function, and concludes the proof of Theorem 1.1. Proof. Taking the natural logarithm in Lemma 2.1 (iii), we obtain k(ln k − 1) + ln(2πk) 2 ≤ ln(k!) ≤ k(ln k − 1) + ln(2πk) 2 + 1 12k .
Indeed, the inequality in the middle is true for sufficiently large a and b, because comparing the leading terms in the asymptotic expansions of both sides we obtain .