Irredundant bases for the symmetric group

An irredundant base of a group $G$ acting faithfully on a finite set $\Gamma$ is a sequence of points in $\Gamma$ that produces a strictly descending chain of pointwise stabiliser subgroups in $G$, terminating at the trivial subgroup. Suppose that $G$ is $\operatorname{S}_n$ or $\operatorname{A}_n$ acting primitively on $\Gamma$, and that the point stabiliser is primitive in its natural action on $n$ points. We prove that the maximum size of an irredundant base of $G$ is $O\left(\sqrt{n}\right)$, and in most cases $O\left((\log n)^2\right)$. We also show that these bounds are best possible.


Introduction
Let G be a finite group that acts faithfully and transitively on a set Γ with point stabiliser H.A sequence (γ 1 , . . ., γ l ) of points of Γ is an irredundant base for the action of G on Γ if Let b(G, H) and I(G, H) denote the minimum and the maximum sizes of an irredundant base in Γ for G respectively.Recently, Gill & Liebeck showed in [7] that if G is an almost simple group of Lie type of rank r over the field F p f of characteristic p and G is acting primitively, then I(G, H) 177r 8 + Ω(f ), where Ω(f ) is the number of prime factors of f , counted with multiplicity.
Suppose now that G is the symmetric group S n or the alternating group A n .An upper bound for I(G, H) is the maximum length of a strictly descending chain of subgroups in G, known as the length, ℓ(G), of G. Define ε(G) := ℓ(G/ soc G).Cameron, Solomon, and Turull proved in [4] that where b n denotes the number of 1s in the binary representation of n.For n 2, this gives This type of upper bound is best possible for such G in general, in that for the natural action of S n or A n on n points, the maximum irredundant base size is n − 2 + ε(G).A recent paper [8] by Gill & Lodà determined the exact values of I(G, H) when H is maximal and intransitive in its natural action on n points, and in each case I(G, H) n − 3 + ε(G).
In this article, we present improved upper bounds for I(G, H) in the case where H is primitive.Note that whenever we refer to the "primitivity" of a subgroup of G, we do so with respect to the natural action of G on n points.We say that a primitive subgroup H of G is large, if there are integers m and k such that H is (S m ≀ S k ) ∩ G in product action on n = m k points or there are integers m and r such that H is S m ∩ G acting on the r-subsets of a set of size m, i.e. on n = m r points.Logarithms are taken to the base 2.
Theorem 1. Suppose G is S n or A n (n 7) and H = A n is a primitive maximal subgroup of G.
(ii) There are infinitely many such G and H for which I(G, H) √ n.
(iii) There are infinitely many such G and H for which I(G, H) > (log n) 2 /(2(log 3) 2 ) + log n/(2 log 3) and H is not large.
We also state our upper bounds for I(G, H) in terms of t := |G : H|.It is easy to show that I(G, H) b(G, H) log t.Burness, Guralnick, and Saxl showed in [3] that with finitely many (known) exceptions, in fact b(G, H) = 2, in which case it follows that I(G, H) 2 log t.
Similar O(log t) upper bounds on the maximum irredundant base size were recently shown to hold for all non-large-base primitive groups of degree t [9,10], raising the question of whether such bounds are best possible in our case.Using Theorem 1, we shall obtain better bounds in terms of t.

Corollary 2. (i) There exist constants
(ii) There is a constant c 3 ∈ R >0 and infinitely many such G and H for which (iii) There is a constant c 4 ∈ R >0 and infinitely many such G and H for which I(G, H) > c 4 (log log t) 2 and H is not large.A sequence B of points in Γ is independent if no proper subsequence B ′ satisfies G (B ′ ) = G (B) .The maximum size of an independent sequence for the action of G on Γ is denoted H(G, H).

It can be shown that b(G, H)
H(G, H) I(G, H).Another closely related property of the action is the relational complexity, denoted RC(G, H), a concept which originally arose in model theory.Cherlin, Martin, and Saracino defined RC(G, H) in [5] under the name "arity" and showed that RC(G, H) H(G, H) + 1.

Corollary 4. Suppose
The maximal subgroups of the symmetric and alternating groups were classified in [1,11].In order to prove statements (i) and (ii) of Theorem 1, we examine two families of maximal subgroups in more detail and determine lower bounds on the maximum irredundant base size, given in the next two results.
If d 2 and p = 3, 5, then If d 2 and p 7, then After laying out some preliminary results in § 2, we shall prove Theorems 5 and 6 in § 3 and § 4 respectively, before proving Theorem 1 and Corollary 2 in § 5.

The maximum irredundant base size
In this section, we collect two general lemmas.Let G be a finite group acting faithfully and transitively on a set Γ with point stabiliser H.If (γ 1 , . . ., γ l ) is an irredundant base of G, then it satisfies (1).The tail of the chain in (1) is a strictly descending chain of subgroups in G γ 1 , which is conjugate to H. Therefore, To obtain a lower bound for I(G, H), one approach is to look for a large explicit irredundant base.The following lemma says it suffices to find a long chain of subgroups in G such that every subgroup in the chain is a pointwise stabiliser of some subset in Γ.
Lemma 2.1.Let l be the largest natural number such that there are subsets ∆ 0 , ∆ 1 , . . ., Proof.Since l is maximal, we may assume that ∆ 0 = ∅ and ∆ l = Γ and that ) is a base for G and every subgroup G (∆ i ) appears in the corresponding chain of point stabilisers.Therefore, by removing all redundant points, we obtain an irredundant base of size at least l, so I(G, H) l.
Once we have an upper or lower bound for I(G, H), we can easily obtain a corresponding bound for the maximum irredundant base size of various subgroups of G.

The affine case
In this section, we prove Theorem 5.The upper bounds will follow easily from examinations of group orders.Therefore, we focus most of our efforts on the construction of an irredundant base, leading to the lower bounds.
Let p be a prime number and d be an integer such that p d 7 and let V be a d-dimensional vector space over the field F p .Let G be Sym(V ) or Alt(V ).Consider the affine group AGL(V ), the group of all invertible affine transformations of V , and let H := AGL(V ) ∩ G.
In this section, we only consider the case where p is odd.Owing to Lemma 2.2, we shall assume G = Sym(V ) and H = AGL(V ) for now.In the light of Lemma 2.1, we introduce a subgroup T of diagonal matrices and look for groups containing T that are intersections of G-conjugates of H ( § 3.1) and subgroups of T that are such intersections ( § 3.2), before finally proving Theorem 5 ( § 3.3).

Subspace stabilisers and the diagonal subgroup
Let T be the subgroup of all diagonal matrices in GL(V ) with respect to a basis b 1 , . . ., b d .Let µ be a primitive element of F p .We now find a strictly descending chain of groups from Sym(V ) to T consisting of intersections of G-conjugates of H.We treat the cases d = 1 and d 2 separately.
Then x / ∈ H and so x does not normalise H.But x normalises m µ , as m µ x = m µ −1 .Therefore, The following two lemmas concern the case d 2.An affine subspace of V is a subset of the form v + W , where v ∈ V and W is a vector subspace of V .The (affine) dimension of v + W is the linear dimension of W .For an affine transformation h = gt u with g ∈ GL(V ) and t u denoting the translation by some u ∈ V , if fix(h) is non-empty, then fix(h) is an affine subspace of V , since fix(h) = v + ker(g − id V ) for any v ∈ fix(h).Lemma 3.3.Suppose d 2, p 3, and G = Sym(V ).Let W be a proper, non-trivial subspace of V and let K < GL(V ) be the setwise stabiliser of W . Then there exists x ∈ G such that We first show that K = C H (x) and then that g, and so K C H (x). Now, let h be an element of C H (x) and write h = gt u with g ∈ GL(V ) and u ∈ V , so that h Suppose for a contradiction that there exists v ∈ W \ {0} with λv g + u / ∈ W . Then Since λ = 1, this is a contradiction and so for all v ∈ W , Hence u = 0 and v g ∈ W .Therefore, h = gt 0 stabilises W , whence h ∈ K. Then This is a contradiction as p 3, and so We now construct a long chain of subgroups of G by intersecting subspace stabilisers.
We now show that, for all (i, j) ∈ I, . ., b j .Hence g i,j ∈ K k,l for all (k, l) < (i, j) but g i,j / ∈ K i,j .Therefore, the K i,j 's, ordered lexicographically by the subscripts, are as required.
We have now found the initial segment of an irredundant base of Sym(V ).The next subsection extends this to a base.

Subgroups of the diagonal subgroup
We now show that, with certain constraints on p, every subgroup of T is an intersection of G-conjugates of T , and hence, by Lemma 3.4, an intersection of G-conjugates of H.We first prove a useful result about subgroups of the symmetric group generated by a k-cycle.Lemma 3.5.Let s ∈ S m be a cycle of length k < m and let a be a divisor of k.Suppose that (k, a) = (4, 2).Then there exists x ∈ S m such that Proof.Without loss of generality, assume s = (1 2 • • • k) and a > 1.If a = k, then take x := (1 m), so that s ∩ s x = 1, as m / ∈ supp(s i ) and m ∈ supp((s i ) x ) for all 1 i < k.Hence we may assume a < k and k = 4.We find that Then (s a ) x = s a .Hence s a = s a x s ∩ s x .To prove that equality holds, suppose s a < s ∩ s x .Then there exists b ∈ {1, . . ., a − 1} such that (s b ) x = s c for some c not divisible by a. Computing On the other hand, Comparing ( 4) and ( 5), we see that b = a−1 and k = 2a.In particular, a = 2 by the assumption that k = 4.It follows that a s c = a s a+b = a − 1, whereas a contradiction.The result follows.
This completes our preparations for the proof of Theorem 5.

Proof of Theorem 5
Recall the assumption that G is S p d or A p d (p is an odd prime and p d 7), which we identify here with Sym(V ) or Alt(V ), and H = AGL d (p) ∩ G, which we identify with AGL(V ) ∩ G.

The product action case
In this section, we prove Theorem 6.Once again, most work goes into the explicit construction of an irredundant base in order to prove the lower bounds, while the upper bounds will be obtained easily from the length of S n .
Throughout this section, let m 5 and k 2 be integers, and let G be S m k or A m k .Let M := S m ≀ S k act in product action on ∆ := {(a 1 , . . ., a k ) | a 1 , . . ., a k ∈ {1, . . ., m}} and identify M with a subgroup of S m k .Theorem 4.1 ([11]).The group M ∩ G is a maximal subgroup of G if and only if one of the following holds: The strategy to proving the lower bound in Theorem 6 is once again to find suitable twopoint stabilisers from which a long chain of subgroups can be built.
For each pair of points α, β ∈ ∆, let d(α, β) denote the Hamming distance between α and β, namely the number of coordinates that differ.
Define u ∈ S m to be (1 2 if m is even, so that u is an even permutation.Let U := u S m and note that C Sm (u) = U .The group U will play a central role in the next lemma.Lemma 4.3.Let i ∈ {2, . . ., k} and r ∈ {1, . . ., m}.Let T r be the stabiliser of r in S m and let W i be the pointwise stabiliser of 1 and i in S k .Then there exists x i,r ∈ A m k such that Proof.Without loss of generality, assume i = 2. Define x = x 2,r ∈ Sym(∆) by (a 1 , a 2 , . . ., a k ) otherwise.
The permutation x is a product of m k−2 disjoint |u|-cycles and is therefore even.
r, and w fixes 1 and 2. Therefore, for all α = (a 1 , a 2 , . . ., a k ) ∈ ∆, if a 2 = r, then and if a 2 = r, then Therefore, x and h commute.Since h is arbitrary, ∈ {r, r v −1 2 }.Then α and α h 1 are both fixed by x, and so α h 2 = α.On the other hand, 2 and a u = a, it must be the case that r v 2 = r and a uv 1 u −1 v −1 1 = a.Therefore, v 2 ∈ T r and, as a is arbitrary in {1, . . ., m − 1}, we deduce that The equality uses the fact that N S m (U ) = T r (as m 5).Through left multiplication by an element of K, we may assume If H is as in case (c) of Lemma 5.1(ii), then Using the lists of maximal subgroups in [6], one can check that ℓ(M where the second inequality follows from Stirling's approximation and the last inequality follows from the fact that log e/ log 100 < 0.218.We deduce further that log log t > log n and hence take c 8 = 1 for n > 100. Finally, log t/ log log t < n log n/ log n = n and log t/ log log t > 0.672 n log n/1.412 log n = 0.672 n/1.412.Therefore, for n > 100, we may take c 5 = 1, c 6 = 1.412/0.672< 2.11.For the values of c 1 , we use in addition the fact that, for any n 0 , if n n 0 , then (log n) 2 + (log n) + 1 = (log n) 2 1 + 1/ log n + 1/(log n) 2 < c 2 8 1 + 1/ log n 0 + 1/(log n 0 ) 2 (log log t) 2 .

Theorem 5 .
Let p be an odd prime number and d a positive integer such that p d 7 and let n

Theorem 6 .
Let m 5 and k 2 be integers and let n

For 1 j
< d, let g 1,j be the linear map that sends b j to b j + b j+1 and fixes b k for k = j.Then g 1,j stabilises b 1 , . . ., b j−1 and any sum of these subspaces, but not b 1 , . . ., b j .Hence g 1,j ∈ K 1,l for all l < j but g 1,j / ∈ K 1,j .For 2 i j d, let g i,j be the linear map that sends b j to b i−1 + b j and fixes b k for k = j.Then g i,j stabilises b 1 , . . ., b j−1 , b j , b i−1 , b j+1 , . . ., b d and any sum of these subspaces, but not b i , .

Corollary 2 2 . 5 . 3 .
now follows by combining Theorem 1 and Lemma 5.Remark Verifying all cases with 7 n 100 by enumerating primitive maximal subgroups of S n and A n in Magma [2], we may take c 5 = 1, c 6 = 4.03, c 7 = 0.70, and c 8 = 1.53 in the statement of Lemma 5.2.With these values of the constants and those in the proof of Lemma 5.2, it is straightforward to obtain the values of the constants c 2 , c 3 , c 4 given in Remark 3.
Finally, we prove an additional lemma.Lemma 5.2.Let t be the index of H inG.There exist constants c 5 , c 6 , c 7 , c 8 ∈ R >0 such that (i) c 5 log t/ log log t < n < c 6 log t/ log log t. (ii) c 7 log log t < log n < c 8 log log t.Proof.It suffices to prove that such constants exist for n sufficiently large, so we may assume n > 100.We first note that log t < log |G| n log n, from which we obtain log log t < log n + log log n < log n + (log n) Hence we may take c 7 = 1/1.412> 0.708 for n > 100.By Lemma 5.1(i), log t = log |G : H| = log |G| − log |H| > log √ n > (n log n − n log e − 1) − √ n log n + log 50 = n log n − n log e − √ n log n − log 100 > n log n − n(log e)