Let us start with a couple of easy lemmas, describing the structure of Alt(n).
Lemma 6.2.1. The groupAlt(n)is generated by the set of 3-cycles.
Proof. This is elementary. Any element of Alt(n) is the product of an even number of transpositionsτi, or equivalently a product ofτ2i−1τ2i = (a b)(c d): ifb=cthen τ2i−1τ2i = (a d b) is already a 3-cycle, and if b6=c then τ2i−1τ2i = (a c b)(b d c) is the product of two 3-cycles.
Lemma 6.2.2. For anyn≥5, any proper subgroup ofAlt(n)has index ≥n.
Proof. This is a standard result that uses the fact that Alt(n) is simple for all n ≥ 5 (see for instance [DF03, §4.6, Ex. 1]). A whole classification of maximal permutation subgroups exists, the O’Nan-Scott theorem [Sco80] already mentioned in §1.2, but we do not need such a powerful tool here.
LetG <Alt(n): in particular, Alt(n) acts by permuting the cosets of G(left cosets, say), so that there is a natural group homomorphism
ϕ: Alt(n)→Sym([Alt(n) :G]).
Since Alt(n) is simple, the normal subgroup ker(ϕ) is either{e}or Alt(n); however, there exists an element σ∈Alt(n)\G, and thenσinduces a nontrivial partition of the cosets of G, so that ker(ϕ) 6= Alt(n). Hence, ϕ is injective, and since we have (n−1)! = n−11 n!<12n! we can conclude that [Alt(n) :G]≥n.
Thanks to the previous lemmas, we can show the following result, which will prevent the arising of large alternating factors whenGitself is not giant (i.e. not equal to Sym(n) or Alt(n), see Definition 3.1.1). We also recall the notations GA, G(A)for setwise and pointwise stabilizers respectively, introduced in§3.3, and G|A for the restriction of the groupGto A(when it is possible to do so, namely whenGalready stabilizesA), introduced in Lemma 3.3.6(d).
Proposition 6.2.3. Let G≤Sym(n) be a transitive permutation subgroup, with n ≥ 5. Consider a set A ⊆ [n] with |A| = αn for some 23 ≤ α < 1, and let H ≤GA. Suppose thatH|A= Alt(A). ThenG≥Alt(n).
This is the kind of proposition that likely can be proved in several different fashions. If we were allowed to use CFSG for example, we could argue that G must be not only transitive but primitive, because an alternating group inside of it permuting more than half of the vertices prevents the formation of a nontrivial block system, and then we could use Cameron’s theorem to exclude the possibility of Gnot being a giant given that by hypothesis|G| ≥ 12(αn)!. For our purposes, however, we will need to provide a proof that does not rely on CFSG.
We remark that there is no particular reason to use 23 as a lower bound forα:
as one can readily check, we can prove the same for any constant arbitrarily close to 12, as long as we choosento be large enough.
Proof. By hypothesis we have thatH|A= Alt(A); the main idea is to prove that H( ¯A)|A= Alt(A) as well, where ¯A= [n]\A.
Consider an arbitraryx∈A. By the isomorphism theorems we first have that¯ [Alt(A) :Hx|A]≤[H :Hx], and in turn we also have that [H :Hx] = [H|A¯:Hx|A¯] following the same reasoning and using moreover the fact that Hx contains the kernel of the restriction map to ¯A. The subgroupHx|A¯cannot have more than|A|¯ cosets insideH|A¯(one can see this as an instance of the orbit-stabilizer theorem);
hence
[Alt(A) :Hx|A]≤(1−α)n < αn=|A|, (6.2.1) and by Lemma 6.2.2 we must haveHx|A= Alt(A). Now we can redefineH to be Hxacting onn\ {x}, and we can repeat the whole process with a newx0: notice thatαincreases, so that the second inequality inside (6.2.1) is still valid. Iterating the process for all points of the original ¯A, we obtain in the endH( ¯A)|A= Alt(A).
At this point it is easy to conclude. In fact,H (and thereforeG) contains all the 3-cycles (a b c) formed by elements a, b, c ∈ A, so we just have to use them
to get all the 3-cycles in [n] and we could conclude by Lemma 6.2.1. Take any x∈A: since¯ Gis transitive there exists ag∈Gthat sendsxto a given element y ∈A, and sinceα≥ 23 and n≥5 there exist two elementsr, s∈A\g( ¯A); then g(r s y)g−1 is the 3-cycle (g−1(r)g−1(s)x), which contains two elements of Aand one element of ¯A. Using
(a b c) = (b c a) = (c a b), (a c b) = (a b c)2,
(b c x) = (a c b)(a b x)(c a b),
we can then reorder elements and insert elements from other cycles as we please, and get all the 3-cycles of [n].
For any two groups H ≤ G, let us denote by L(G, H) the set of left cosets of H inside G: to prevent confusion we would rather avoid using the notation G/H for such a set, unless we are dealing with a normal subgroup H and G/H is the quotient group1. We are going to work with a class of Schreier graphs (see Definition 2.1.1) arising from the action on the cosets of a subgroup; incidentally, this was the context in which Schreier graphs were originally conceived [Sch27].
Definition 6.2.4. Let Gbe a group and let H ≤G. We define diam(G, H), the diameter of the pair(G, H), to be the maximum among the (undirected) diameters of all the Schreier graphs Sch(L(G, H), S), where S runs through all sets of gen-erators of Gand the actionη:G× L(G, H)→ L(G, H)defining the graphs is the left multiplicationη(g, g0H) =gg0H.
The diameter of a group diam(G) is then the same as diam(G,{e}), and ifH is normal inGthen diam(G/H) = diam(G, H). Of course, there is nothing special about our choice of “left”: we could as well defineR(G, H), and act on it through right multiplication.
We use here Schreier’s lemma (in a slightly altered form than in Lemma 5.2.1) so as to be able to use a chain of subgroups as a way to bound diameters. The use we make of it is identical to what happens with [Hel18, Lemma 4.7].
Lemma 6.2.5. LetGbe a finite group, letH ≤Gbe a proper nontrivial subgroup, and letS be a set of generators ofGwithe∈S=S−1. Then
diam(G)≤(2diam(G, H) + 1)diam(H) + diam(G, H)
≤4diam(G, H)diam(H).
Proof. First, if d= diam(G, H) thenS2d+1∩H generates H: this is almost the statement of Lemma 5.2.1. In that case however the subgroup ofGwas assumed to be normal, so that we could use the usual definition of diameter for the quotient group, while here we resort to Definition 6.2.4 and the set L(G, H); apart from that, the proof with generalH is identical to the proof of Lemma 5.2.1.
1The author is embarrassingly prone to get confused by the notation and assume thatH is normal wheneverG/His written on paper. May the reader be indulgent with him.
The result is now easy: ifS2d+1∩H generates H, then (S2d+1)diam(H) ⊇H and sinceSd contains by definition representatives of all the left cosets ofH inside Gwe haveSdH =G, thus concluding the proof.
The condition ofH being proper nontrivial is really only needed for the second inequality, since by definition diam({e}) = 0. For ease of notation, we can use the second inequality anyway and conventionally establish that diam({e}) = 1 (which we are going to do).