TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 348, Number 11, November 1996

INFINITE PRODUCTS OF FINITE SIMPLE GROUPS

JAN SAXL, SAHARON SHELAH, AND SIMON THOMAS

Abstract. We classify the sequences ⟨S_{n} | n ∈ ℕ⟩ of finite simple nonabelian groups such that ∏_{n} S_{n} has uncountable cofinality.

1. Introduction

Suppose that G is a group that is not finitely generated. Then G can be expressed as the union of a chain of proper subgroups. The cofinality of G, written c(G), is defined to be the least cardinal λ such that G can be expressed as the union of a chain of λ proper subgroups. Groups of uncountable cofinality were first considered by Serre in his study of groups acting on trees.

Definition 1.1. [Se, p. 58] A group H has property (FA) if and only if whenever H acts without inversion on a tree T, then there exists a vertex t ∈ T such that h(t) = t for all h ∈ H.

In [Se], Serre characterised the groups which have property (FA).

Theorem 1.2. [Se] The group H has property (FA) if and only if the following three conditions are satisfied.

(1) H is not a nontrivial free product with amalgamation.

(2) ℤ is not a homomorphic image of H.

(3) If H is not finitely generated, then c(H) > ω.

This result led to the question of whether there exist any natural examples of uncountable groups with property (FA). Let ⟨G_{n} | n ∈ ℕ⟩ be a sequence of nontrivial finite groups. Then ∏_{n} G_{n} denotes the full direct product of the groups G_{n}, n ∈ ℕ. By Bass [Ba], if H is a profinite group and H acts without inversion on the tree T, then for every h ∈ H there exists t ∈ T such that h(t) = t. This implies that H satisfies conditions 1.2(1) and 1.2(2). In particular, we see that the profinite group ∏_{n} G_{n} has property (FA) if and only if c(∏_{n} G_{n}) > ω. The following result, which was proved by Koppelberg and Tits, provided the first examples of uncountable groups with property (FA).

Theorem 1.3. [KT] Let F be a nontrivial finite group and let G_{n} = F for all n ∈ ℕ. Then c(∏_{n} G_{n}) > ω if and only if F is perfect.

\begin{matrix}c \\ ◯ \end{matrix}1996 American Mathematical Society

Received by the editors September 21, 1995.

1991 Mathematics Subject Classification. Primary 20E15, 20A15; Secondary 03E35, 20D06, 20E08.

The research of the second author was partially supported by the U.S.-Israel Binational Science Foundation. This paper is number 584 in the cumulative list of the second author’s publications.

The research of the third author was partially supported by NSF Grants.

Suppose that F is perfect. Since |∏_{n} G_{n}| = 2^{ω}, Theorem 1.3 yields that

ω_{1} ≤ c (∏_{n} G_{n}) ≤ 2^{ω}.

This suggests the problem of trying to compute the exact value of c(∏_{n} G_{n}). (Of course, this problem is only interesting if 2^{ω} > ω_{1}.) The following result is an immediate consequence of a theorem of Koppelberg [Ko].

Theorem 1.4. If F is a nontrivial finite perfect group and G_{n} = F for all n ∈ ℕ, then c(∏_{n} G_{n}) = ω_{1}.

Proof. If ⟨g(n)⟩_{n} ∈ ∏_{n} G_{n} and π ∈ F, let X_{π}(g) = {n ∈ ℕ | g(n) = π}. Then {X_{π}(g) | π ∈ F } yields a partition of ℕ into finitely many pieces. Consider the powerset P(ℕ) as a Boolean algebra. By Koppelberg [Ko], we can express

P(ℕ) = ⋃_{α<ω_{1}}B_{α}

as the union of a chain of ω_{1} proper Boolean subalgebras. For each α < ω_{1}, define

H_{α} = {g ∈ ∏_{n} G_{n} | X_{π}(g) ∈ B_{α} for all π ∈ F }.

Then it is easily checked that H_{α} is a proper subgroup of ∏_{n} G_{n}. Clearly ∏_{n} G_{n} = ⋃_{α<ω_{1}} H_{α}, and so c(∏_{n} G_{n}) ≤ ω_{1}.

The above results suggest the following questions.

Question 1.5. For which sequences ⟨S_{n} | n ∈ ℕ⟩ of finite simple nonabelian groups do we have c(∏_{n} S_{n}) > ω?

Question 1.6. Suppose that ⟨S_{n} | n ∈ ℕ⟩ is a sequence of finite simple nonabelian groups such that c(∏_{n} S_{n}) > ω. Is it possible to compute the exact value of c(∏_{n} S_{n})?

It may be helpful to give a word of explanation concerning Question 1.6. The point is that it may be impossible to compute the exact value of c(∏_{n} S_{n}) in ZF C. For example, consider the group Sym(ℕ) of all permutations of ℕ. In [MN], Macpherson and Neumann showed that c(Sym(ℕ)) > ω. Later Sharp and Thomas [ST1] proved that it is consistent that c(Sym(ℕ)) and 2^{ω} can be any two prescribed regular uncountable cardinals subject only to the requirement that c(Sym(ℕ)) ≤ 2^{ω}. Hence it is impossible to compute the exact value of c(Sym(ℕ)) in ZF C. (The theorem of Macpherson and Neumann suggests that Sym(ℕ) is probably another natural example of an uncountable group with property (FA). In the final section of this paper, we shall confirm that this is true.)

The following result shows that there exist sequences ⟨S_{n} | n ∈ ℕ⟩ of finite simple nonabelian groups such that c(∏_{n} S_{n}) = ω.

Theorem 1.7. Let ⟨S_{n} | n ∈ ℕ⟩ be a sequence of finite simple nonabelian groups. Suppose that there exists an infinite subset I of ℕ such that the following conditions are satisfied.

(1) There exists a fixed (possibly twisted) Lie type L such that for all n ∈ I, S_{n} = L(q_{n}) for some prime power q_{n}.

(2) If n, m ∈ I and n < m, then q_{n} < q_{m}. Then c(∏_{n} S_{n}) = ω.

Here L(q_{n}) denotes the group of Lie type L over the finite field GF (q_{n}). The proof of Theorem 1.7 makes use of the following easy observation.

Lemma 1.8. Suppose that N ◁ G and that G/N is not finitely generated. Then c(G) ≤ c(G/N).

Proof of Theorem 1.7. By Lemma 1.8, we can suppose that I = ℕ. Let D be a nonprincipal ultrafilter on ℕ, and let N be the set of elements g = ⟨g(n)⟩_{n} ∈ ∏_{n} S_{n} such that {n ∈ ℕ | g(n) = 1} ∈ D. Then N is a normal subgroup of ∏_{n} S_{n}, and ∏_{n} S_{n}/N is the ultraproduct G = ∏_{n} S_{n}/D. (See Section 9.5 of Hodges [H].) By Lemma 1.8, it is enough to show that c(G) = ω.

There exists a fixed integer d such that each of the groups L(q_{n}) has a faithful d-dimensional linear representation over the field GF (q_{n}). Since the class of groups with a faithful d-dimensional linear representation is first-order axiomatisable, it follows that G has a faithful d-dimensional linear representation over some field K. (For example, see Section 6.6 of Hodges [H]. It is perhaps worth mentioning that every known proof only yields the existence of a set of axioms for this class. The problem of finding an explicit intelligible set of axioms remains open.) To simplify notation, we shall suppose that G ⩽ GL(d, K). We also suppose that K has been chosen so that G ∩ GL(d, K′) is a proper subgroup of G for every proper subfield K′ of K. By Exercise 9.5.5 of Hodges [H], |G| = 2^{ω}. It follows that |K| = 2^{ω}, and hence K has transcendence dimension 2^{ω} over its prime subfield k. Let B be a transcendence basis of K over k. Express B = ⋃_{n<ω}B_{n} as the union of a chain

of proper subsets. For each n < ω, let K_{n} be the algebraic closure of B_{n} in K. Then each K_{n} is a proper subfield of K, and K = ⋃_{n<ω}K_{n}. For each n < ω, let

G_{n} = G ∩ GL(d, K_{n}). Then G = ⋃_{n<ω}G_{n}, and each G_{n} is a proper subgroup of G.

Hence c(G) = ω.

The main result of this paper is that the converse of Theorem 1.7 is also true.

Theorem 1.9. Suppose that ⟨S_{n} | n ∈ ℕ⟩ is a sequence of finite simple nonabelian groups such that c(∏_{n} S_{n}) = ω. Then there exists an infinite subset I of ℕ such that conditions 1.7(1) and 1.7(2) are satisfied.

Now suppose that ⟨S_{n} | n ∈ ℕ⟩ is a sequence of finite simple nonabelian groups such that c(∏_{n} S_{n}) > ω. If there exists an infinite subset J of ℕ such that S_{n} = S_{m} for all n, m ∈ J, then Lemma 1.8 and Theorem 1.4 imply that c(∏_{n} S_{n}) = ω_{1}. This is the only case in which we have been able to compute the exact value of c(∏_{n} S_{n}) in ZF C.

Question 1.10. Is it consistent that there exists a sequence ⟨S_{n} | n ∈ ℕ⟩ of finite simple nonabelian groups such that c(∏_{n} S_{n}) > ω_{1}?

We hope that Question 1.10 has a positive answer, as this would lead to some very attractive problems. For example, consider the following question. (We suspect that it cannot be answered in ZF C.)

Question 1.11. Is it true that c(∏_{n} Alt(n + 5)) = c(∏_{n} P SL(n + 3, 2))?

In Section 5, we shall prove the following consistency result. Among other things, it shows that it is impossible to prove in ZF C that c(Sym(ℕ)) = c(∏_{n} S_{n}) for some sequence ⟨S_{n} | n ∈ ℕ⟩ of finite simple nonabelian groups.

Theorem 1.12. It is consistent that both of the following statements are true.

(1) c(Sym(ℕ)) = ω_{2} = 2^{ω}.

(2) c(∏_{n} G_{n}) ≤ ω_{1} for every sequence ⟨G_{n} | n ∈ ℕ⟩ of nontrivial finite groups.

The following problem is also open. (Of course, a negative answer to Question 1.10 would yield a negative answer to Question 1.13.)

Question 1.13. Is it consistent that there exists a sequence ⟨S_{n} | n ∈ ℕ⟩ of finite simple nonabelian groups such that c(∏_{n} S_{n}) > c(Sym(ℕ))?

This paper is organised as follows. In Section 2, we shall prove that if ⟨S_{n} | n ∈ ℕ⟩ is a sequence of finite alternating groups, then c(∏_{n} S_{n}) > ω. In Section 3, we shall prove Theorem 1.9 in the special case when each S_{n} is a projective special linear group. In Section 4, we shall complete the proof of Theorem 1.9. Our proof makes use of the fact that there are only finitely many sporadic finite simple groups, and thus relies on the classification of the finite simple groups. Section 5 contains the proof of Theorem 1.12. In Section 6, we shall prove that Sym(ℕ) has property (FA).

Our notation is standard, but a couple of points should be mentioned. Suppose that G is a subgroup of Sym(Ω). If each nonidentity element g ∈ G is fixedpoint-free, then G is said to act semiregularly on Ω. If G acts transitively and semiregularly, then G is said to act regularly on Ω. In this paper, permutation groups and linear groups always act on the left. Thus, for example, we have that

( 1 2 3 )( 1 3 5 7 )( 1 2 3 )^{−1} = ( 2 1 5 7 ).

We follow the usual convention of regarding each ordinal as the set of its predecessors. Thus ω = ℕ. Also if a, b are natural numbers such that a > b, then their set-theoretic difference is a ∖ b = {b, b + 1, …, a − 1}. If A is a matrix, then A^{T} denotes the transpose of A.

2. Infinite products of alternating groups

In this section, we shall prove the following special case of Theorem 1.9.

Theorem 2.1. Let ⟨S_{n} | n ∈ ℕ⟩ be a sequence of finite simple nonabelian groups. If each S_{n} is an alternating group, then c(∏_{n} S_{n}) > ω.

We shall make use of the following two results, which will be used repeatedly throughout this paper.

Proposition 2.2. [Th] Suppose that G is not finitely generated and that H is a subgroup of G. If G is finitely generated over H, then c(H) ≤ c(G).

Proof. Let c(G) = λ. Express G = ⋃_{α<λ}G_{α} as the union of a chain of λ proper

subgroups. Let H_{α} = H ∩ G_{α}. Then H = ⋃_{α<λ}H_{α}. Since G is finitely generated

over H, each H_{α} is a proper subgroup of H. Thus c(H) ≤ λ.

Proposition 2.3. Let ⟨S_{n} | n ∈ ℕ⟩ be a sequence of nontrivial finite perfect groups. Suppose that there exists a finite set F of groups such that S_{n} ∈ F for all n ∈ ℕ. Then c(∏_{n} S_{n}) > ω.

Proof. By Proposition 2.2, we can suppose that for each group S ∈ F, the set {n ∈ ℕ | S_{n} = S} is either infinite or empty. Since the class of groups of uncountable cofinality is closed under taking finite direct products, Theorem 1.4 implies that c(∏_{n} S_{n}) > ω.

We shall begin the proof of Theorem 2.1 by making a couple of easy reductions. For each m ≥ 5, let P_{m} = ∏_{n} S_{n}^{m}, where S_{n}^{m} = Alt(m) for all n ∈ ℕ. Let G_{0} = ∏_{m≥5} P_{m}. Then Lemma 1.8 implies that it is enough to prove that c(G_{0}) > ω. Let G_{1} = ∏_{m≥8} P_{m}. Then G_{0} = P_{5}×P_{6}×P_{7}×G_{1}. By Theorem 1.4, c(P_{m}) = ω_{1} for all m ≥ 5. Hence it is enough to prove that c(G_{1}) > ω. Finally let G_{2} = ∏_{m≥3} P_{2^{m}}. Then Theorem 2.1 is an immediate consequence of the following two results.

Lemma 2.4. c(G_{1}) = c(G_{2}).

Theorem 2.5. c(G_{2}) > ω.

First we shall prove Lemma 2.4. Note that Lemma 1.8 implies that c(G_{1}) ≤ c(G_{2}). Our proof that c(G_{2}) ≤ c(G_{1}) is based upon Proposition 2.2.

Let I = {⟨m, n⟩ | 8 ≤ m ∈ ℕ, n ∈ ℕ}. Then G_{1} = ∏_{⟨m,n⟩∈I}S_{n}^{m}, where S_{n}^{m} =

Alt(m). For each ⟨m, n⟩ ∈ I, let t be the integer such that 2^{t} ≤ m < 2^{t+1} and let T_{n}^{m} = Alt(2^{t}) ⩽ S_{n}^{m}. Then we can identify G_{2} with the subgroup ∏_{⟨m,n⟩∈I}T_{n}^{m} of G_{1}.

By Proposition 2.2, it is enough to prove the following result.

Lemma 2.6. G_{1} is finitely generated over G_{2}.

This is the first of the many places in this paper where we need to prove that an infinite product of groups is finitely generated over an infinite product of subgroups. A moment’s thought shows that such results require “uniform generation” results for the corresponding sequences of groups. We shall make repeated use of the following easy observation.

Proposition 2.7. Let ⟨H_{n} | n ∈ ℕ⟩ and ⟨G_{n} | n ∈ ℕ⟩ be sequences of groups such that H_{n} ⩽ G_{n} for all n ∈ ℕ. Suppose that there exists a word w(x_{1}, …, x_{s}, y_{1}, …, y_{t}) from the free group on {x_{1}, …, x_{s}, y_{1}, …, y_{t}} such that the following condition is satisfied. (2.7) For all n ∈ ℕ, there exist elements θ_{1}, …, θ_{t} ∈ G_{n} such that each ϕ ∈ G_{n} can

be expressed as ϕ = w(ψ_{1}, …, ψ_{s}, θ_{1}, …, θ_{t}) for some ψ_{1}, …, ψ_{s} ∈ H_{n}. Then there exist elements g_{1}, …, g_{t} ∈ ∏_{n} G_{n} such that

∏_{n} G_{n} = ⟨∏_{n} H_{n}, g_{1}, …, g_{t}⟩.

Lemma 2.6 is a consequence of the following “uniform generation” results for the finite alternating groups, which will also be needed in the proof of Theorem 2.5.

Lemma 2.8. Let m ≥ 3 and let θ = ( m − 2 m − 1 )( m m + 1 ) ∈ Alt(m + 1). Then for every ϕ ∈ Alt(m + 1), there exist ψ_{1}, ψ_{2}, ψ_{3} ∈ Alt(m) such that

ϕ = ψ_{1}θψ_{2}θψ_{3}.

Proof. If ϕ ∈ Alt(m), then we can take ψ_{1} = ϕ and ψ_{2} = ψ_{3} = id. So suppose that ϕ ∈ Alt(m + 1) ∖ Alt(m). Let ψ_{2} = ( m − 2 m − 1 m ). Then τ = θψ_{2}θ = ( m − 1 m − 2 m + 1 ). Since Alt(m + 1) acts 2-transitively on {1, …, m + 1}, we have the double coset decomposition

Alt(m + 1) = Alt(m) ∪ Alt(m)τAlt(m).

Thus ϕ ∈ Alt(m)τAlt(m), and so there exist ψ_{1}, ψ_{3} ∈ Alt(m) such that ϕ = ψ_{1}θψ_{2}θψ_{3}.

Let m = 4n for some n ≥ 2. A permutation π ∈ Alt(m) is said to have type 2^{2n} if π is the product of 2n disjoint transpositions. Thus π^{2} = id and π is fixed point free. The set of permutations π ∈ Alt(m) of type 2^{2n} forms a single conjugacy class in Alt(m).

Lemma 2.9 (Brenner [Br]). Let m = 4n for some n ≥ 2. Let C be the conjugacy class of Alt(m) consisting of all permutations of type 2^{2n}. Then for every ϕ ∈ Alt(m), there exist π_{1}, …, π_{4} ∈ C such that ϕ = π_{1} … π_{4}.

Lemma 2.10. Suppose that m = 8n for some n ≥ 1. Let Δ_{0} = {1, …, 4n}, Δ_{1} = {4n+1, …, 8n} and let Γ = Alt(Δ_{0})×Alt(Δ_{1}). Let θ = ∏_{i=1}^{4n}( i 2i )( 4n+i 2i−1 ). Then every ϕ ∈ Alt(m) can be expressed as a product

ϕ = ψ_{1}θψ_{2}θψ_{3}θψ_{4}θψ_{5}θψ_{6}θψ_{7}θψ_{8}θψ_{9}

for some ψ_{1}, …, ψ_{9} ∈ Γ.

Proof. By Lemma 2.9, it is enough to show that each permutation ϕ ∈ Alt(m) of type 2^{4n} can be expressed as a product ϕ = ψ_{1}θψ_{2}θψ_{3} for some ψ_{1}, ψ_{2} , ψ_{3} ∈ Γ. Let A = {ℓ ∈ Δ_{0} | ϕ(ℓ) ∈ Δ_{0}} and B = {ℓ ∈ Δ_{1} | ϕ(ℓ) ∈ Δ_{1}}. Then |A| = |B| = 2s for some 0 ≤ s ≤ 4n. Let C = {1, …, 8n}∖(A∪B). Then |C∩Δ_{0}| = |C∩Δ_{1}| = 4n−2s. Let t = 4n − 2s. Let Δ_{2} = {2i | 1 ≤ i ≤ 4n} and Δ_{3} = {2i − 1 | 1 ≤ i ≤ 4n}.

Case 1. Suppose that t ≥ 2n. Choose a subset D ⊆ C ∩ Δ_{0} of size 2n, and let E = ϕ[D]. Then there exists ψ_{1} ∈ Γ such that ψ_{1}[D] = Δ_{2} ∩ Δ_{0} and ψ_{1}[E] = Δ_{2} ∩ Δ_{1}. This implies that

ψ_{1}ϕψ_{1}^{−1} ∈ Alt(Δ_{2}) × Alt(Δ_{3}) = θ (Alt(Δ_{0}) × Alt(Δ_{1})) θ.

Thus we have that

ψ_{2} = θψ_{1}ϕψ_{1}^{−1}θ ∈ Γ,

and so

ϕ = ψ_{1}^{−1}θψ_{2}θψ_{1}

is a suitable product.

Case 2. Suppose that t < 2n. Then s > n. Choose ϕ-invariant subsets D ⊆ Δ_{0} and E ⊆ Δ_{1} such that |D| = |E| = 2n. Then there exists ψ_{1} ∈ Γ such that ψ_{1}[D] = Δ_{2} ∩ Δ_{0} and ψ_{1}[E] = Δ_{2} ∩ Δ_{1}. Arguing as in Case 1, we see that there exists ψ_{2} ∈ Γ such that ϕ = ψ_{1}^{−1}θψ_{2}θψ_{1}.

Proof of Lemma 2.6. For each 0 ≤ i ≤ 7, let

H_{i} = ∏{S_{n}^{m} | ⟨m, n⟩ ∈ I, m ≡ i (mod 8) }

and

K_{i} = ∏{T_{n}^{m} | ⟨m, n⟩ ∈ I, m ≡ i (mod 8) }.

Then G_{1} = ∏_{i=0}^{7} H_{i} and G_{2} = ∏_{i=0}^{7} K_{i}. Clearly it is enough to show that H_{i} is finitely generated over K_{i} for each 0 ≤ i ≤ 7.

First consider the case when i = 0. Let ⟨m, n⟩ ∈ I satisfy m = 8s for some s ≥ 1, and let t be the integer such that 2^{t} ≤ m < 2^{t+1}. Define Δ_{0}^{m,n} = {1, …, 4s} and Δ_{1}^{m,n} = {4s + 1, …, 8s}. Then Alt(Δ_{0}^{m,n}) ⩽ T_{n}^{m} = Alt(2^{t}). There exists an element ϕ ∈ Alt(m) = S_{n}^{m} such that ϕAlt(Δ_{0}^{m,n})ϕ^{−1} = Alt(Δ_{1}^{m,n}). Hence there exists g_{1} ∈ H_{0} such that

∏{Alt(Δ_{0}^{m,n}) × Alt(Δ_{1}^{m,n}) | ⟨m, n⟩ ∈ I, m ≡ 0 (mod 8) } ⩽ ⟨K_{0}, g_{1}⟩.

Now Lemma 2.10 implies that there exists an element g_{2} ∈ H_{0} such that H_{0} = ⟨K_{0}, g_{1}, g_{2}⟩.

Next consider the case when i = 1. For each ⟨m, n⟩ ∈ I with m ≡ 1 (mod 8), let U_{n}^{m} = Alt(m − 1) ⩽ S_{n}^{m}. By the previous paragraph, there exist elements g_{1}, g_{2} ∈ H_{1} such that

∏{U_{n}^{m} | ⟨m, n⟩ ∈ I, m ≡ 1 (mod 8) } ⩽ ⟨K_{1}, g_{1}, g_{2}⟩.

Now Lemma 2.8 implies that there exists an element g_{3} ∈ H_{1} such that H_{1} = ⟨K_{1}, g_{1}, g_{2}, g_{3}⟩. Continuing in this fashion, we can successively deal with the remaining cases.

The rest of this section is devoted to the proof of Theorem 2.5. Suppose that c(G_{2}) = ω. Express G_{2} = ⋃_{t<ω}H_{t} as the union of a chain of ω proper subgroups.

Our strategy will be to define by induction on t < ω 1. a sequence of elements f_{t} ∈ G_{2};

2. a strictly increasing sequence of integers i_{t} such that f_{t} ∈ H_{i_{t}};

3. a sequence of elements g_{t} ∈ G_{2} ∖ H_{i_{t}}. These sequences will be chosen so that there exists an element h ∈ G_{2} such that hf_{t}h^{−1} = g_{t} for all t < ω. But this implies that h ∉ ⋃_{t<ω}H_{t}, which is the desired

contradiction.

Let J = {⟨m, n⟩ | 3 ≤ m ∈ ℕ, n ∈ ℕ}. Then G_{2} = ∏_{⟨m,n⟩∈J}A_{n}^{m}, where A_{n}^{m} =

Alt(2^{m}) for all n ∈ ℕ. The elements g_{t} = ⟨g_{t}(m, n)⟩_{m,n} ∈ ∏_{m,n} A_{n}^{m}, t < ω, will be chosen so that for each ⟨m, n⟩ ∈ J, the sequence

g_{0}(m, n), g_{1}(m, n), …, g_{t}(m, n)

is a generic sequence of elements of Alt(2^{m}), in the following sense.

Definition 2.11. If 0 ≤ t ≤ m − 1, then the sequence π_{0}, …, π_{t} of elements of Alt(2^{m}) is a generic sequence if

1. the subgroup ⟨π_{0}, …, π_{t}⟩ is elementary abelian of order 2^{t+1};

2. if id ≠ ϕ ∈ ⟨π_{0}, …, π_{t}⟩, then ϕ is a permutation of type 2^{2^{m−1}}. (In other

words, ⟨π_{0}, …, π_{t}⟩ acts semiregularly on {1, …, 2^{m}}.) If m − 1 ≤ t < ω, then the sequence π_{0}, …, π_{t} of elements of Alt(2^{m}) is a generic sequence if

(a) π_{0}, …, π_{m−1} is a generic sequence;

(b) π_{ℓ} = π_{m−1} for all m − 1 ≤ ℓ ≤ t.

It is an easy exercise to show that for each t < ω, there exists a unique generic sequence π_{0}, …, π_{t} in Alt(2^{m}) up to conjugacy within Sym(2^{m}); and two such generic sequences up to conjugacy within Alt(2^{m}) if m ≥ 3 and 2 ≤ t < ω. (We shall not make use of this observation in the proof of Theorem 2.5.)

To begin the induction, choose any element f_{0} = ⟨f_{0}(m, n)⟩_{m,n} ∈ G_{2} = ∏_{m,n} A_{n}^{m} such that f_{0}(m, n) is a permutation of type 2^{2^{m−1}} for each ⟨m, n⟩ ∈ J; and let i_{0} be an integer such that f_{0} ∈ H_{i_{0}}. Let f_{0}^{G_{2}} be the conjugacy class of f_{0} in G_{2}. Then Lemma 2.9 implies that G_{2} = ⟨f_{0}^{G_{2}}⟩. Hence there exists an element h_{0} ∈ G_{2} such that g_{0} = h_{0}f_{0}h_{0}^{−1} ∉ H_{i_{0}}. Now suppose that t ≥ 0 and that we have defined

1. a sequence of elements f_{j} ∈ G_{2},

2. a strictly increasing sequence of integers i_{j} such that f_{j} ∈ H_{i_{j}},

3. a sequence of elements g_{j} ∈ G_{2} ∖ H_{i_{j}}, and

4. a sequence of elements h_{j} ∈ G_{2} for 0 ≤ j ≤ t such that the following conditions hold.

(a) f_{0}(m, n), …, f_{t}(m, n) is a generic sequence in A_{n}^{m} = Alt(2^{m}) for all m, n.

(b) If 0 ≤ j ≤ k ≤ t, then h_{k}f_{j}h_{k}^{−1} = g_{j}.

(c) If m − 1 ≤ j ≤ t and n ∈ ℕ, then h_{j}(m, n) = h_{m−1}(m, n).

First we shall define f_{t+1}.

Case 1. Suppose that m − 1 ≤ t and n ∈ ℕ. Then we define f_{t+1}(m, n) = f_{m−1}(m, n).

Case 2. Suppose that t < m − 1 and n ∈ ℕ. We shall set up some notation which will be used during the rest of this section. Let

E(m, n) = ⟨g_{0}(m, n), …, g_{t}(m, n)⟩.

Then E(m, n) is an elementary abelian group of order 2^{t+1} acting semiregularly on {1, …, 2^{m}}. Let

{1, …, 2^{m}} = Φ_{1}^{m,n} ∪ ⋯ ∪ Φ_{2^{m−t−1}}^{m,n}

be the decomposition into E(m, n)-orbits. Then E(m, n) acts regularly on Φ_{i}^{m,n} for each 1 ≤ i ≤ 2^{m−t−1}. Choose α_{i} ∈ Φ_{i}^{m,n} for each 1 ≤ i ≤ 2^{m−t−1}. Let E(m, n) = {π_{k} | 1 ≤ k ≤ 2^{t+1}}, where π_{1} = id, and define

Δ_{k}^{m,n} = {π_{k}(α_{i}) | 1 ≤ i ≤ 2^{m−t−1}}

for each 1 ≤ k ≤ 2^{t+1}. Then the diagonal subgroup

D(m, n) = Diag(Alt(Δ_{1}^{m,n}) × ⋯ × Alt(Δ_{2^{t+1}}^{m,n}))

= { ∏_{i=1}^{2^{t+1}} π_{i}ϕπ_{i} | ϕ ∈ Alt(Δ_{1}^{m,n})}

is contained in the centraliser of E(m, n) in Alt(2^{m}). Let τ(m, n) ∈ D(m, n) be any permutation of type 2^{2^{m−1}}, and define

f_{t+1}(m, n) = h_{t}(m, n)^{−1}τ(m, n)h_{t}(m, n).

This completes the definition of f_{t+1} = ⟨f_{t+1}(m, n)⟩_{m,n}.

Next we choose i_{t+1} to be an integer such that

(i) i_{t} < i_{t+1} and f_{t+1} ∈ H_{i_{t+1}}; and (ii) P^{∗} = ∏{A_{n}^{m} | 3 ≤ m ≤ t + 3, n ∈ ℕ} ⩽ H_{i_{t+1}}.

(Proposition 2.3 implies that c(P^{∗}) > ω. Hence i_{t+1} can be chosen so that (ii) also holds.)

Finally we shall define h_{t+1} and g_{t+1}.

Case 1. Suppose that m − 1 ≤ t and n ∈ ℕ. Then we define h_{t+1}(m, n) = h_{m−1}(m, n) and g_{t+1}(m, n) = g_{m−1}(m, n).

Case 2. Suppose that t < m − 1 and n ∈ ℕ. Then we choose a suitable element σ(m, n) ∈ D(m, n) and define

h_{t+1}(m, n) = σ(m, n)h_{t}(m, n)

and

g_{t+1}(m, n) = h_{t+1}(m, n)f_{t+1}(m, n)h_{t+1}(m, n)^{−1}
= σ(m, n)τ(m, n)σ(m, n)^{−1}.

Of course, a suitable choice means one such that g_{t+1} = ⟨g_{t+1}(m, n)⟩_{m,n} ∉ H_{i_{t+1}}. This completes the successor step of the induction, provided that a suitable choice exists.

Claim 2.12. There exists a choice of σ(m, n) for m > t + 1 and n ∈ ℕ such that g_{t+1} ∉ H_{i_{t+1}}.

Proof. Suppose that for every choice of the sequence

⟨σ(m, n) | t < m − 1, n ∈ ℕ⟩

we have that g_{t+1} ∈ H_{i_{t+1}}. Then we shall prove that G_{2} is finitely generated over H_{i_{t+1}}. But this means that there exists r ∈ ℕ with i_{t+1} ≤ r such that H_{r} = G_{2}, which is a contradiction.

Let J′ = {⟨m, n⟩ | t + 4 ≤ m ∈ ℕ, n ∈ ℕ} and let P′ = ∏{A_{n}^{m} | ⟨m, n⟩ ∈ J′}. Thus G_{2} = P^{∗} × P′. Note that if π(m, n) ∈ D(m, n) is any element of type 2^{2^{m−1}}, then there exists σ(m, n) ∈ D(m, n) such that σ(m, n)τ(m, n)σ(m, n)^{−1} = π(m, n). Using the fact that P^{∗} ⩽ H_{i_{t+1}}, we see that the following statement holds. (†) Suppose that π = ⟨π(m, n)⟩_{m,n} ∈ P′. If π(m, n) ∈ D(m, n) is an element of

type 2^{2^{m−1}} for all ⟨m, n⟩ ∈ J′, then π ∈ H_{i_{t+1}}. Using Lemma 2.9, we see that ∏{D(m, n) | ⟨m, n⟩ ∈ J′} ⩽ H_{i_{t+1}}. Now let θ_{1} = ⟨θ_{1}(m, n)⟩_{m,n} ∈ P′ be an element such that θ_{1}(m, n) ∈ Alt(Δ_{1}^{m,n}) is a permutation of type 2^{2^{m−t−2}} for each ⟨m, n⟩ ∈ J′. Then {ψθ_{1}(m, n)ψ^{−1} | ψ ∈ D(m, n)} is the conjugacy class in Alt(Δ_{1}^{m,n}) of all permutations of type 2^{2^{m−t−2}}. Using Lemma 2.9 again, we see that

∏{Alt(Δ_{1}^{m,n}) | ⟨m, n⟩ ∈ J′} ⩽ ⟨H_{i_{t+1}}, θ_{1}⟩.

Continuing in this fashion, we find that there exist θ_{1}, …, θ_{2^{t+1}} ∈ P′ such that

∏{Alt(Δ_{1}^{m,n}) × ⋯ × Alt(Δ_{2^{t+1}}^{m,n}) | ⟨m, n⟩ ∈ J′} ⩽ ⟨H_{i_{t+1}}, θ_{1}, …, θ_{2^{t+1}}⟩.

By repeatedly applying Lemma 2.10, we now see that there exists a finite subset F of P′ such that

P′ ⩽ ⟨H_{i_{t+1}}, θ_{1}, …, θ_{2^{t+1}}, F ⟩.

Hence G_{2} = P^{∗} × P′ is finitely generated over H_{i_{t+1}}, which is a contradiction.

Thus the induction can be carried out for all t < ω. Define the element h = ⟨h(m, n)⟩_{m,n} ∈ G_{2} by h(m, n) = h_{m−1}(m, n). Then we have that hf_{t}h^{−1} = g_{t} for all t < ω, which is a contradiction. This completes the proof of Theorem 2.5.

3. Infinite products of special linear groups

In this section, we shall prove the following result.

Theorem 3.1. Suppose that ⟨SL(d_{n}, q_{n}) | n ∈ ℕ⟩ is a sequence of finite special linear groups which satisfies the following conditions.

(1) If d_{n} = 2, then q_{n} > 3.

(2) There do not exist an infinite subset I of ℕ and an integer d such that (a) d_{n} = d for all n ∈ I; and

(b) if n, m ∈ I and n < m, then q_{n} < q_{m}. Then c(∏_{n} SL(d_{n}, q_{n})) > ω.

Using Theorem 3.1 and Lemma 1.8, we see that Theorem 1.9 is true in the special case when each S_{n} is a projective special linear group.

Our strategy in the proof of Theorem 3.1 will be the same as that in the proof of Theorem 2.1. We shall begin by defining the notion of a generic sequence of elements in SL(2^{m}, q). Let V = V (n, q) be an n-dimensional vector space over GF (q), and let B be a basis of V. Then Sym(B) denotes the group of permutation matrices with respect to the basis B. Note that for any finite field GF (q), we have that Alt(B) ⩽ SL(n, q).

Definition 3.2. If 0 ≤ t ≤ m − 1, then the sequence π_{0}, …, π_{t} of elements of SL(2^{m}, q) is a generic sequence if there exists a basis B of V (2^{m}, q) such that

1. the group ⟨π_{0}, …, π_{t}⟩ is an elementary abelian subgroup of Alt(B) of order 2^{t+1};

2. ⟨π_{0}, …, π_{t}⟩ acts semiregularly on B. If m − 1 ≤ t < ω, then the sequence π_{0}, …, π_{t} of elements of SL(2^{m}, q) is a generic sequence if

(a) π_{0}, …, π_{m−1} is a generic sequence;

(b) π_{ℓ} = π_{m−1} for all m − 1 ≤ ℓ ≤ t.

First we shall prove an analogue of Lemma 2.9. Let m = 4n for some n ≥ 1. Then C(m, q) denotes the conjugacy class in SL(m, q) consisting of all elements π such that π is represented by a permutation matrix of type 2^{2n} with respect to some basis B of V (m, q). (It is easily checked that the set of such elements forms a single conjugacy class in SL(m, q).) Note that SL(m, q) has a maximal torus T_{1} of order (q^{4n} − 1)/(q − 1). By Zsigmondy’s theorem [Zs], there exists a primitive prime divisor p > 2 of q^{4n} − 1. Let ψ ∈ T_{1} be an element of order p. Then ψ is clearly a regular element of T_{1}. (A semisimple element g ∈ SL(m, q) is regular if and only if it lies in a unique maximal torus.)

Lemma 3.3. With the above hypotheses, there exist elements π_{1}, π_{2} ∈ C(m, q) such that ψ = π_{1}π_{2}.

Proof. Regard K = GF (q^{4n}) as a 4n-dimensional vector space over GF (q). Let τ ∈ K generate a normal basis of K over GF (q); i.e. B = {τ, τ^{q}, τ^{q^{2}}, …, τ^{q^{4n−1}}} is a basis of K over GF (q). Let f ∈ Aut(K) be the Frobenius automorphism; so that f(α) = α^{q} for all α ∈ K. Let g = f^{2n}. Then g is represented by a permutation matrix of type 2^{2n} with respect to the basis B. Thus g ∈ C(m, q).

Let K^{∗} = ⟨α⟩, and let β = α^{q^{2n}−1}. Then β has order q^{2n} + 1, and g(β) = β^{−1}. Consider the primitive prime divisor p of q^{4n}−1 = (q^{2n}−1)(q^{2n}+1). Then clearly p divides q^{2n}+1. Thus there exists an element γ ∈ ⟨β⟩ of order p; and we can suppose that ψ ∈ T_{1} is the linear transformation defined by ψ(x) = γx for all x ∈ K. Since g(γ) = γ^{−1}, we see that gψg^{−1} = ψ^{−1}. Since ψ has odd order, the involutions g and gψ are conjugate in the dihedral group ⟨g, gψ⟩. The result follows.

Theorem 3.4. Suppose that m = 4n for some n ≥ 1. Then for every ϕ ∈ SL(m, q), there exist π_{1}, …, π_{10} ∈ C(m, q) such that ϕ = π_{1} … π_{10}.

Proof. Let G = SL(4n, q) and let ψ ∈ T_{1} be as above. Let τ ∈ G be an element of order q^{4n−1} − 1 and let T_{2} be the maximal torus which contains τ. (Of course, τ is a regular element of T_{2}.) Let C_{1}, C_{2} be the conjugacy classes of ψ, τ respectively. We claim that the product C_{1}C_{2} of these two classes covers all of G ∖ Z(G). Using Lemma 3.3, this implies that each element of C_{2} is a product of 3 elements of C(m, q); and hence every element of G∖Z(G) is a product of 5 elements of C(m, q). The result follows.

The proof of the claim uses character theory and follows [MSW, pp. 96–99] very closely. For any conjugacy class C_{3} of G and σ ∈ C_{3}, define

m(C_{1}, C_{2}, C_{3}) = \begin{matrix}|G| & 2 \\ |C_{G}(ψ)| & |C_{G}(τ)| & |C_{G}(σ)| \end{matrix} ∑ \frac{χ(ψ)χ(τ)χ(σ)}{χ(1)}

where the summation runs over the irreducible characters χ of G. By a well-known class formula, m(C_{1}, C_{2}, C_{3}) is equal to the number of triples (a_{1}, a_{2}, a_{3}) such that a_{i} ∈ C_{i} and a_{1}a_{2}a_{3} = 1. It therefore suffices to show that the character sum involved in the formula for m(C_{1}, C_{2}, C_{3}) is positive for any class C_{3} of non-central elements of G.

Now the values of the irreducible characters of G on semisimple elements can be calculated from the values of the Deligne-Lusztig characters. (See [Ca2, Chapter 7].) The Deligne-Lusztig characters R_{T,θ} are parametrized by pairs (T, θ). The equivalence relation of geometric conjugacy on these pairs yields a partition of the irreducible characters of G into disjoint series as follows. The geometric conjugacy classes of pairs (T, θ) can be parametrized by the conjugacy classes (s) of semisimple elements in the dual group Ĝ = P GL(4n, q). Let E(s) be the set of irreducible characters occurring as a constituent in one of the R_{T,θ} with (T, θ) corresponding to (s). Then the sets E(s), where (s) runs over the set of conjugacy classes of semisimple elements of Ĝ, form a partition of the set of irreducible characters of G. ( See [Ca2, 7.3.8 and 7.5.8].)

The R_{T,θ} span the space of class functions restricted to semisimple elements. (See [Ca2, 7.5.7].) In particular, suppose that ρ is a semisimple element of G and that χ ∈ E(s). Then if R_{T,θ}(ρ) = 0 for all pairs (T, θ) corresponding to (s), we have that χ(ρ) = 0. Now R_{T,θ} = R(s) vanishes on the regular elements of the torus T′ if the element s is not conjugate in Ĝ to an element of the dual T̂′ of T′. Hence the R_{T,θ} not vanishing on either of the classes C_{1}, C_{2} will correspond to semisimple classes (s) in Ĝ such that s ∈ T̂_{1} ∩ T̂_{2}. Let T̄_{i} be the preimage of T̂_{i} in GL(4n, q). Then T̄_{1} is cyclic of order q^{4n} − 1, and T̄_{2} is the product of two cyclic groups of orders q^{4n−1} − 1 and q − 1. Furthermore, both T̄_{1} and T̄_{2} contain Z(GL(4n, q)).

Since (|T̄_{1}| , |T̄_{2}|) = (q − 1)(4n, q − 1), it follows that T̂_{1} ∩ T̂_{2} = 1. Thus we need only consider the set E(1) of unipotent characters of G. It is well-known that if the degree of an irreducible character is divisible by the full power of a prime r dividing the order of a group, then the character vanishes on all r-singular elements of the group. Using this result, an inspection of the degrees of the unipotent characters of G in [Ca2, p. 465] shows that only two irreducible characters contribute to the character sum in the above formula; namely, the principal character and the Steinberg character St. It follows that

m(C_{1}, C_{2}, C_{3}) = \begin{matrix}|G| & 2 \\ |C_{G}(ψ)| & |C_{G}(τ)| & |C_{G}(σ)| \end{matrix} (1 + \frac{St(ψ)St(τ)St(σ)}{St(1)})

and so [Ca2, 6.4]

m(C_{1}, C_{2}, C_{3}) = \begin{matrix}|G| & 2 \\ |C_{G}(ψ)| & |C_{G}(τ)| & |C_{G}(σ)| \end{matrix} (1 − \frac{St(σ)}{St(1)}) .

This is 0 precisely when σ ∈ Z(G), as claimed.

Next we shall prove the analogue of Lemma 2.8. It is easier to state the result in terms of infinite products of groups, rather than in terms of “uniform generation”. Let ⟨SL(d_{n}, q_{n}) | n ∈ ℕ⟩ be a sequence of special linear groups. Fix some n ∈ ℕ. Let SL(d_{n}, q_{n}) act on the vector space V (d_{n}, q_{n}) in the natural manner. Extend this action to V (d_{n} + 1, q_{n}) = V (d_{n}, q_{n}) ⊕ ⟨v_{d_{n}+1}⟩ by specifying that π(v_{d_{n}+1}) = v_{d_{n}+1} for all π ∈ SL(d_{n}, q_{n}). Using this extended action, we can regard SL(d_{n}, q_{n}) as a subgroup of SL(d_{n} + 1, q_{n}).

Lemma 3.5. ∏_{n} SL(d_{n} + 1, q_{n}) is finitely generated over ∏_{n} SL(d_{n}, q_{n}).

Proof. We shall make use of the Bruhat decomposition

SL(d, q) = ⋃_{w∈W}BwB

of the special linear group, where B is a Borel subgroup and W is the Weyl group. Fix some integer n ∈ ℕ. Choose a basis {v_{i} | 1 ≤ i ≤ d_{n}} of V (d_{n}, q_{n}). We shall regard each element of SL(d_{n} + 1, q_{n}) as a matrix with respect to the basis B = {v_{i} | 1 ≤ i ≤ d_{n} + 1}. Note that we have identified SL(d_{n}, q_{n}) with the subgroup

S_{n} = \{(\begin{matrix}A & 0 \\ 0 & 1 \end{matrix}) | A ∈ SL(d_{n}, q_{n})\}

of SL(d_{n} + 1, q_{n}). Define

T_{n} = \{(\begin{matrix}1 & 0 \\ 0 & A \end{matrix}) | A ∈ SL(d_{n}, q_{n})\} .

Then there exists π ∈ SL(d_{n} + 1, q_{n}) such that πS_{n}π^{−1} = T_{n}. Hence there exists an element g_{0} ∈ ∏_{n} SL(d_{n} + 1, q_{n}) such that ∏_{n} T_{n} ⩽ G_{0} = ⟨∏_{n} S_{n}, g_{0}⟩. Let U_{n} be the subgroup of strictly upper triangular matrices in SL(d_{n} + 1, q_{n}), and let H_{n} be the subgroup of diagonal matrices. Then B_{n} = U_{n} ⋊ H_{n} is a Borel subgroup of SL(d_{n} + 1, q_{n}). We shall show that ∏_{n} B_{n} ⩽ G_{0}.

First we shall show that ∏_{n} H_{n} ⩽ G_{0}. Fix some n ∈ ℕ. Let

D = diag(λ_{1}, …, λ_{d_{n}+1}) ∈ H_{n}.

Then D_{1} = diag(λ_{1}, λ_{1}^{−1}, 1, …, 1) ∈ H_{n} ∩ S_{n}, D_{2} = diag(1, λ_{1}λ_{2}, λ_{3}, …, λ_{d_{n}+1}) ∈ H_{n} ∩ T_{n} and D = D_{1}D_{2}. The result follows.

Next we shall show that ∏_{n} U_{n} ⩽ G_{0}. Fix some n ∈ ℕ. Note that

(\begin{matrix}1 & 0 & 0 \\ 0 & A & b \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix}1 & c & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) = (\begin{matrix}1 & c & 0 \\ 0 & A & b \\ 0 & 0 & 1 \end{matrix})

for each (d_{n} − 1) × (d_{n} − 1)-matrix A. Also note that if Z ∈ U_{n} has the form

(\begin{matrix}1 & 0 & d \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})

then there exist X ∈ U_{n} ∩S_{n} and Y ∈ U_{n} ∩T_{n} such that [X, Y] = Z. (For example, this follows from Chevalley’s commutator formula [Ca1, 5.2.2].) Hence if ϕ ∈ U_{n} is arbitrary, then there exist θ, τ ∈ U_{n} ∩S_{n} and ψ, σ ∈ U_{n} ∩T_{n} such that ϕ = ψθ[τ, σ]. The result follows.

Let N_{n} be the subgroup of SL(d_{n} + 1, q_{n}) consisting of the elements which stabilise the frame {⟨v_{i}⟩ | 1 ≤ i ≤ d_{n} + 1}. Then the Weyl group of SL(d_{n} + 1, q_{n}) is W_{n} = N_{n}/B_{n} ∩ N_{n}; and W_{n} is isomorphic to Sym(d_{n} + 1) acting on the set {v_{i} | 1 ≤ i ≤ d_{n} + 1}. Note that N_{n} ∩ S_{n} corresponds to the subgroup Sym(d_{n}) of W_{n}. Let θ = ( d_{n} d_{n} + 1 ). Arguing as in the proof of Lemma 2.8, we see that for every ϕ ∈ Sym(d_{n} + 1), there exist ψ_{1}, ψ_{2}, ψ_{3} ∈ Sym(d_{n}) such that ϕ = ψ_{1}θψ_{2}θψ_{3}. Hence there exists g_{1} ∈ ∏_{n} SL(d_{n} + 1, q_{n}) such that ∏_{n} N_{n} ⩽ G_{1} = ⟨G_{0}, g_{1}⟩. It follows that G_{1} = ∏_{n} SL(d_{n} + 1, q_{n}).

Finally we shall prove the analogue of Lemma 2.10. Consider a product of the form ∏_{n} SL(8d_{n}, q_{n}). Fix some n ∈ ℕ. Let SL(8d_{n}, q_{n}) act on the vector space V_{n} = V (8d_{n}, q_{n}) in the natural manner, and let B_{n} = {v_{i} | 1 ≤ i ≤ 8d_{n}} be a basis of V_{n}. Let E_{0} = ⟨v_{i} | 1 ≤ i ≤ 4d_{n}⟩ and E_{1} = ⟨v_{i} | 4d_{n} + 1 ≤ i ≤ 8d_{n}⟩. We regard SL(E_{0}) as the subgroup of SL(8d_{n}, q_{n}) consisting of the elements π such that π[E_{0}] = E_{0} and such that π(v_{i}) = v_{i} for all 4d_{n} + 1 ≤ i ≤ 8d_{n}. We also regard SL(E_{1}) as a subgroup of SL(8d_{n}, q_{n}) in the obvious fashion. Let Γ_{n} = SL(E_{0}) × SL(E_{1}) ⩽ SL(8d_{n}, q_{n}).

Lemma 3.6. ∏_{n} SL(8d_{n}, q_{n}) is finitely generated over ∏_{n} Γ_{n}.

Proof. Once again, we shall make use of the Bruhat decomposition of the special linear group. Fix some n ∈ ℕ. We shall regard SL(8d_{n}, q_{n}) as a group of matrices with respect to the ordered basis (v_{1}, …, v_{8d_{n}}) of V_{n}. Let B_{n} = U_{n} ⋊ H_{n} be the Borel subgroup consisting of the upper triangular matrices of SL(8d_{n}, q_{n}) First we shall show that there exists a subgroup G_{0} of ∏_{n} SL(8d_{n}, q_{n}) such that

1. G_{0} is finitely generated over ∏_{n} Γ_{n}, and

2. ∏_{n} U_{n} ⩽ G_{0}.

Fix some n ∈ ℕ. Let M_{n} be the ring of all 4d_{n} × 4d_{n}-matrices over GF (q_{n}), and let

T_{n} = \{(\begin{matrix}1 & S \\ 0 & 1 \end{matrix}) | S ∈ M_{n}\} .

Then it is enough to find G_{0} such that ∏_{n} T_{n} ⩽ G_{0}. Note that for each A ∈ SL(4d_{n}, q_{n}), we have that

(\begin{matrix}A & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix}1 & S \\ 0 & 1 \end{matrix}) (\begin{matrix}A^{−1} & 0 \\ 0 & 1 \end{matrix}) = (\begin{matrix}1 & AS \\ 0 & 1 \end{matrix}) .

Regard M_{n} as an SL(4d_{n}, q_{n})-module with the natural action, S ⟼^{A} AS. Then the existence of a suitable subgroup G_{0} is an immediate consequence of the following claim.

Claim 3.7. ∏_{n} M_{n} is finitely generated as a ∏_{n} SL(4d_{n}, q_{n})-module.

Proof of Claim. We shall prove that M_{n} is “uniformly generated” as an SL(4d_{n}, q_{n})- module. The result will then follow. Fix some n ∈ ℕ. Throughout this proof, each of the matrices will be expressed in terms of 2d_{n} × 2d_{n} blocks. Let

J_{1} = (\begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}) ,  J_{2} = (\begin{matrix}0 & 0 \\ 0 & 1 \end{matrix}) ,  J_{3} = (\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}) , J_{4} = (\begin{matrix}0 & 0 \\ 1 & 0 \end{matrix}) .

If B ∈ GL(2d_{n}, q_{n}), then ( \begin{matrix}B & 0 \\ 0 & B^{−1} \end{matrix} ), ( \begin{matrix}B^{−1} & 0 \\ 0 & B \end{matrix} ) ∈ SL(4d_{n}, q_{n}); and

(\begin{matrix}B & 0 \\ 0 & B^{−1} \end{matrix}) (\begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}) = (\begin{matrix}B & 0 \\ 0 & 0 \end{matrix}) and (\begin{matrix}B^{−1} & 0 \\ 0 & B \end{matrix}) (\begin{matrix}0 & 0 \\ 1 & 0 \end{matrix}) = (\begin{matrix}0 & 0 \\ B & 0 \end{matrix}) , etc.

Thus if B_{1}, …, B_{4} ∈ GL(2d_{n}, q_{n}), then there exist C_{1}, …, C_{4} ∈ SL(4d_{n}, q_{n}) such that

(\begin{matrix}B_{1} & B_{2} \\ B_{3} & B_{4} \end{matrix}) = ∑_{i=1}^{4} C_{i}J_{i}.

Now suppose that ( \begin{matrix}S_{1} & S_{2} \\ S_{3} & S_{4} \end{matrix} ) ∈ M_{n} is arbitrary. By [Ze], each of the matrices S_{i} is the sum of two non-singular ones. Hence there exist C_{1}, …, C_{4}, D_{1}, …, D_{4} ∈ SL(4d_{n}, q_{n}) such that

(\begin{matrix}S_{1} & S_{2} \\ S_{3} & S_{4} \end{matrix}) = ∑_{i=1}^{4} C_{i}J_{i} + ∑_{i=1}^{4} D_{i}J_{i}.

Thus each M_{n} is “uniformly generated” from the generators J_{1}, …, J_{4}.

Next we shall show that there exists an element g_{0} ∈ ∏_{n} SL(8d_{n}, q_{n}) such that ∏_{n} H_{n} ⩽ G_{1} = ⟨G_{0}, g_{0}⟩; and hence ∏_{n} B_{n} ⩽ G_{1}. For each λ ∈ GF (q_{n})^{∗}, let D_{λ} = diag(λ, 1, …, 1) ∈ GL(4d_{n}, q_{n}). Define

F_{n} = \{(\begin{matrix}D_{λ} & 0 \\ 0 & D_{λ}^{−1} \end{matrix}) | λ ∈ GF (q_{n})^{∗}\} .

Since ∏_{n} Γ_{n} ⩽ G_{0}, it is enough to find an element g_{0} such that ∏_{n} F_{n} ⩽ ⟨G_{0}, g⟩. For each λ ∈ GF (q_{n})^{∗}, let E_{λ} = diag(λ, λ^{−1}, 1, …, 1) ∈ GL(4d_{n}, q_{n}). Define

K_{n} = \{(\begin{matrix}E_{λ} & 0 \\ 0 & 1 \end{matrix}) | λ ∈ GF (q_{n})^{∗}\} .

Then ∏_{n} K_{n} ⩽ ∏_{n} Γ_{n} ⩽ G_{0}. Also there exists an element π ∈ N_{n} such that πK_{n}π^{−1} = F_{n}. The existence of a suitable element g_{0} follows easily.

Finally we shall show that there exists an element g_{1} ∈ ∏_{n} SL(8d_{n}, q_{n}) such that ∏_{n} N_{n} ⩽ G_{2} = ⟨G_{1}, g_{1}⟩. Fix some n ∈ ℕ. Let E_{0}^{n} = {v_{i} | 1 ≤ i ≤ 4d_{n}} and E_{1}^{n} = {v_{i} | 4d_{n}+1 ≤ i ≤ 8d_{n}}, so that B_{n} = E_{0}^{n}∪E_{1}^{n}. Then the groups of permutation matrices Alt(E_{0}^{n}), Alt(E_{1}^{n}) are subgroups of Γ_{n}. Lemma 2.10 implies that there exists an element g_{1} ∈ ∏_{n} Alt(B_{n}) such that ∏_{n} Alt(B_{n}) = ⟨∏ (Alt(E_{0}^{n}) × Alt(E_{1}^{n})) , g_{1}⟩. Now let π = ⟨π(n)⟩_{n} ∈ ∏_{n} N_{n} be an arbitrary element. Let X be the subset of ℕ consisting of those n such that π(n) corresponds to an odd permutation of B_{n}. Then there exists an element ψ_{X} = ⟨ψ_{X}(n)⟩_{n} ∈ ∏_{n} (N_{n} ∩ Γ_{n}) such that

1. ψ_{X}(n) corresponds to the odd permutation ( v_{1} v_{2} ) if n ∈ X, and

2. ψ_{X}(n) = 1 if n ∉ X. Since ∏_{n} H_{n}, ∏_{n} Alt(B_{n}) ⩽ ⟨G_{1}, g_{1}⟩, it follows that πψ_{X} ∈ ⟨G_{1}, g_{1}⟩, and hence π ∈ ⟨G_{1}, g_{1}⟩. Thus ∏_{n} N_{n} ⩽ ⟨G_{1}, g_{1}⟩.

Now we are ready to begin the proof of Theorem 3.1. Since the proof is very similar to that of Theorem 2.1, we shall just sketch the main points. Suppose that ⟨SL(d_{n}, q_{n}) | n ∈ ℕ⟩ is a sequence of finite special linear groups which satisfies the hypotheses of Theorem 3.1. By Proposition 2.3, we can suppose that {d_{n} | n ∈ ℕ} is an infinite subset of ℕ. Arguing as in the proof of Lemma 2.4, we can reduce to the case when each d_{n} has the form 2^{m_{n}} for some m_{n} ≥ 2. (Since the sequence satisfies condition 3.1(2), there exists a finite set F of groups such that if d_{n} ≤ 3, then SL(d_{n}, q_{n}) ∈ F. By Propositions 2.2 and 2.3, we can safely ignore these factors.) Let G = ∏_{n} SL(2^{m_{n}}, q_{n}) and suppose c(G) = ω. Express G = ⋃_{t<ω}H_{t} as

the union of a chain of ω proper subgroups. Now suppose that t ≥ 0 and that we have defined

1. a sequence of elements f_{j} = ⟨f_{j}(n)⟩_{n} ∈ G,

2. a strictly increasing sequence of integers i_{j} such that f_{j} ∈ H_{i_{j}},

3. a sequence of elements g_{j} = ⟨g_{j}(n)⟩_{n} ∈ G ∖ H_{i_{j}}, and

4. a sequence of elements h_{j} = ⟨h_{j}(n)⟩_{n} ∈ G for 0 ≤ j ≤ t such that the following conditions hold.

(a) f_{0}(n), …, f_{t}(n) is a generic sequence in SL(2^{m_{n}}, q_{n}) for each n ∈ ℕ.

(b) If 0 ≤ j ≤ k ≤ t, then h_{k}f_{j}h_{k}^{−1} = g_{j}.

(c) If m_{n} − 1 ≤ j ≤ t, then h_{j}(n) = h_{m_{n}−1}(n). We must show that it is possible to continue the induction. There is no difficulty in defining f_{t+1} and i_{t+1}. The problem is to show that there exist suitable elements h_{t+1} and g_{t+1} = h_{t+1}f_{t+1}h_{t+1}^{−1} such that g_{t+1} ∉ H_{i_{t+1}}. As in the proof of Theorem 2.1, we shall show that if no such elements exist, then G is finitely generated over H_{i_{t+1}}; which is a contradiction. So suppose that no such elements exist. Let P^{∗} = ∏{SL(2^{m_{n}}, q_{n}) | m_{n} ≤ t+3} and P′ = ∏{SL(2^{m_{n}}, q_{n}) | m_{n} ≥ t+4}, so that G = P^{∗}×P′. Since ⟨SL(2^{m_{n}}, q_{n} | n ∈ ℕ⟩ satisfies condition 3.1(2), either c(P^{∗}) > ω or P^{∗} is finite. Thus we can suppose that i_{t+1} was chosen so that P^{∗} ⩽ H_{i_{t+1}}. Fix some n ∈ ℕ such that m_{n} ≥ t + 4. Let B_{n} be a basis of V (2^{m_{n}}, q_{n}) chosen so that the group E(n) = ⟨g_{0}(n), …, g_{t}(n)⟩ is an elementary abelian subgroup of Alt(B_{n}) of order 2^{t+1}, which acts semiregularly on B_{n}. Let {Φ_{i}^{n} | 1 ≤ i ≤ 2^{m_{n}−t−1}} be the set of orbits of E(n) on B_{n}. For each 1 ≤ i ≤ 2^{m_{n}−t−1}, choose v_{i}^{1} ∈ Φ_{i}^{n}. Let E(n) = {ψ_{k} | 1 ≤ k ≤ 2^{t+1}}. For each 1 ≤ k ≤ 2^{t+1} and 1 ≤ i ≤ 2^{m_{n}−t−1}, define v_{i}^{k} = ψ_{k}(v_{i}^{1}). For each 1 ≤ k ≤ 2^{t+1}, let V_{k}^{n} = ⟨v_{i}^{k} | 1 ≤ i ≤ 2^{m_{n}−t−1}⟩. Then V (2^{m_{n}}, q_{n}) = V_{1}^{n} ⊕ ⋯ ⊕ V_{2^{t+1}}^{n}, and the diagonal subgroup

D_{n} = Diag (SL(V_{1}^{n}) × ⋯ × SL(V_{2^{t+1}}^{n}))

is contained in the centraliser of ⟨g_{0}(n), …, g_{t}(n)⟩ in SL(2^{m_{n}}, q_{n}). Since each candidate π for g_{t+1} satisfies π ∈ H_{i_{t+1}}, we find that the following statement holds.

(†) Suppose that π ∈ P′. If π(n) ∈ D_{n}∩C(2^{m_{n}}, q_{n}) for all n such that m_{n} ≥ t+4,

then π ∈ H_{i_{t+1}}. Using Theorem 3.4, this implies that ∏{D_{n} | m_{n} ≥ t + 4} ⩽ H_{i_{t+1}}. Arguing as in the proof of Theorem 2.1, we see that there exists a subgroup Γ_{0} of G such that

1. Γ_{0} is finitely generated over H_{i_{t+1}}, and

2. ∏{SL(V_{1}^{n}) × ⋯ × SL(V_{2^{t+1}}^{n}) | m − n ≥ t + 4} ⩽ Γ_{0}. By repeatedly applying Lemma 3.6, we next see that there exists a subgroup Γ_{1} of G such that

1. Γ_{1} is finitely generated over Γ_{0}, and

2. P′ = ∏{SL(2^{m_{n}}, q_{n}) | m_{n} ≥ t + 4} ⩽ Γ_{1}. But this means that Γ_{1} = G, and so G is finitely generated over H_{i_{t+1}}. This contradiction shows that the induction can be carried out for all t < ω. But this yields an element h ∈ G such that hf_{t}h^{−1} = g_{t} for all t < ω, which is impossible. Thus c(G) > ω. This completes the proof of Theorem 3.1.

4. The proof of Theorem 1.9

In this section, we shall complete the proof of Theorem 1.9. Most of our work will go into proving the special cases of Theorem 1.9 in which each S_{n} is a classical group of a fixed kind. We shall deal successively with the symplectic groups, the unitary groups and the orthogonal groups over finite fields. The general result will then follow easily. (Clear accounts of the classical groups can be found in [Ca1] and [Ta].)

4.1. Symplectic groups. Suppose that ⟨Sp(2d_{n}, q_{n}) | n ∈ ℕ⟩ is a sequence of finite symplectic groups such that d_{n} ≥ 2 for each n ∈ ℕ. Fix some n ∈ ℕ. Then there exists a basis e f = (e_{i} | 1 ≤ i ≤ d_{n}) (f_{i} | 1 ≤ i ≤ d_{n}) of the corresponding symplectic space such that (e_{i}, f_{j}) = δ_{ij} and (e_{i}, e_{j}) = (f_{i}, f_{j}) = 0 for all 1 ≤ i, j ≤ d_{n}. (Such a basis is called a normal basis.) We shall consider Sp(2d_{n}, q_{n}) as a group of matrices with respect to the ordered basis e f. Let E_{d_{n}} = ⟨e_{1}, …, e_{d_{n}}⟩ and F_{d_{n}} = ⟨f_{1}, …, f_{d_{n}}⟩. Then the setwise stabiliser of the subspaces E_{d_{n}} and F_{d_{n}} in Sp(2d_{n}, q_{n}) contains the subgroup

G_{n} = \{(\begin{matrix}A & 0 \\ 0 & (A^{−1})^{T} \end{matrix} ) | A ∈ SL(d_{n}, q_{n})\} .

Theorem 4.1. Suppose that d_{n} ≥ 3 for all n ∈ ℕ. Then ∏_{n} Sp(2d_{n}, q_{n}) is finitely generated over the subgroup ∏_{n} G_{n}.

Corollary 4.2. Suppose that ⟨S_{n} | n ∈ ℕ⟩ is a sequence of finite simple symplectic groups such that there does not exist an infinite subset I of ℕ for which conditions 1.7(1) and 1.7(2) are satisfied. Then c(∏_{n} S_{n}) > ω.

Proof of Corollary 4.2. For each n ∈ ℕ, let S_{n} = P Sp(2d_{n}, q_{n}). Put J = {n ∈ ℕ | d_{n} < 3}, so that ∏_{n∈ℕ} S_{n} = (∏_{n∈J} S_{n}) × (∏_{n∉J} S_{n}). By assumption, there exists a finite set F of groups such that S_{n} ∈ F for all n ∈ J. By Proposition 2.3, either c(∏_{n∈J} S_{n}) > ω or ∏_{n∈J} S_{n} is finite. Hence if ∏_{n∉J} S_{n} is finite, then the result follows from Proposition 2.2. So we can suppose that ∏_{n∉J} S_{n} is infinite; and it is enough to prove that c(∏_{n∉J} S_{n}) > ω. To simplify notation, we shall suppose that J = ∅. Let ∏_{n} G_{n} be the subgroup of ∏_{n} Sp(2d_{n}, q_{n}) defined above.

By Theorem 3.1, c(∏_{n} G_{n}) > ω. So using Theorem 4.1 and Proposition 2.2, we see that c(∏_{n} Sp(2d_{n}, q_{n})) > ω. Hence c(∏_{n} P Sp(2d_{n}, q_{n})) > ω.

We shall approach Theorem 4.1 via the Bruhat decomposition

Sp(2d, q) = ⋃_{w∈W}BwB

of the symplectic group, where B is a Borel subgroup and W is the Weyl group. Fix some n ∈ ℕ. Let e f = (e_{i} | 1 ≤ i ≤ d_{n}) (f_{i} | 1 ≤ i ≤ d_{n}) be our distinguished normal basis. For each 1 ≤ i ≤ d_{n}, let E_{i} = ⟨e_{1}, … e_{i}⟩. Then the stabiliser B_{n} of the flag of totally isotropic subspaces

E_{1} ⩽ E_{2} ⩽ ⋯ ⩽ E_{d_{n}}

is a Borel subgroup of Sp(2d_{n}, q_{n}). Let N_{n} be the subgroup of Sp(2d_{n}, q_{n}) which stabilises the symplectic frame {⟨e_{i}⟩, ⟨f_{i}⟩ | 1 ≤ i ≤ d_{n}}. Then the Weyl group of Sp(2d_{n}, q_{n}) is N_{n}/B_{n} ∩ N_{n}. Let H_{n} = B_{n} ∩ N_{n}. Then H_{n} consists of the matrices of the form

(\begin{matrix}D & 0 \\ 0 & D^{−1} \end{matrix})

where D ∈ GL(d_{n}, q_{n}) is a diagonal matrix. Let UT_{n} be the subgroup of strictly upper triangular matrices in SL(d_{n}, q_{n}), and define

U_{n} = \{(\begin{matrix}P & P S \\ 0 & (P^{−1})^{T} \end{matrix} ) | P ∈ UT_{n}, S^{T} = S\} .

Then B_{n} = U_{n} ⋊ H_{n}.

First we shall show that there exists a subgroup Γ_{0} of ∏_{n} Sp(2d_{n}, q_{n}) such that 1. Γ_{0} is finitely generated over ∏_{n} G_{n}, and

2. ∏_{n} U_{n} ⩽ Γ_{0}. Note that for each A ∈ SL(d_{n}, q_{n}), we have

(\begin{matrix}A & 0 \\ 0 & (A^{−1})^{T} \end{matrix} ) (\begin{matrix}1 & S \\ 0 & 1 \end{matrix}) (\begin{matrix}A^{−1} & 0 \\ 0 & A^{T} \end{matrix} ) = (\begin{matrix}1 & ASA^{T} \\ 0 & 1 \end{matrix} ) .

Let M_{n} be the left SL(d_{n}, q_{n})-module of symmetric d_{n} × d_{n}-matrices, with the action

S ⟼^{A} ASA^{T} .

Then it is enough to prove that ∏_{n} M_{n} is finitely generated as a ∏_{n} SL(d_{n}, q_{n})- module. We shall consider M_{n} in three different cases, and show that in each case M_{n} is “uniformly generated” as a SL(d_{n}, q_{n})-module. The result will then follow. Let p = char(GF (q_{n})).

Case 1. Suppose that p > 3. Since p is odd, every S ∈ M_{n} is congruent to a diagonal matrix. This easily implies that there exists A ∈ SL(d_{n}, q_{n}) such that ASA^{T} is a diagonal matrix. Thus we need only consider diagonal matrices D = diag(λ_{1}, …, λ_{d_{n}}) ∈ M_{n}. Let D_{1} = diag(1, 0, …, 0) and D_{2} = diag(0, 1, …, 1), so that 1 = D_{1} + D_{2}. If α = (α_{1}, …, α_{d_{n}}) ∈ (GF (q_{n})^{∗})^{d_{n}}, let

R_{α} = diag(α_{1}, 1, …, 1, α_{1}^{−1})

and

S_{α} = diag((α_{2} … α_{d_{n}})^{−1}, α_{2}, …, α_{d_{n}}).

Then R_{α}D_{1}R_{α}^{T} = diag(α_{1}^{2}, 0, …, 0) and S_{α}D_{2}S_{α}^{T} = diag(0, α_{2}^{2}, …, α_{d_{n}}^{2}). Since p > 3, for each λ ∈ GF (q_{n}), there exist β_{1}, …, β_{4} ∈ GF (q_{n})^{∗} such that λ = ∑_{i=1}^{4} β_{i}^{2}. (For example, see Chapter 4 [Sm].) Thus we can “uniformly generate” each diagonal matrix D ∈ M_{n} from the generators D_{1} and D_{2}.

Case 2. Suppose that p = 3. Once again, we need only consider diagonal matrices D ∈ M_{n}. Let D_{1} = diag(1, 1, 0, …, 0) and D_{2} = diag(δ, 1, 1, …, 1), where δ = 1 if d_{n} is even and δ = 0 if d_{n} is odd. For each subset X of {1, …, d_{n}}, let D_{1}^{X} = diag(χ_{X}(1), 0, …, 0) and D_{2}^{X} = diag(0, χ_{X}(2), …, χ_{X}(d_{n})), where χ_{X} is the characteristic function of X. It is easy to check that if

S ∈ \{(\begin{matrix}0 & 0 \\ 0 & 0 \end{matrix}) , (\begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}) , (\begin{matrix}0 & 0 \\ 0 & 1 \end{matrix}) , (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix})\}

then there exist A_{1}, …, A_{6} ∈ SL(2, 3) such that S = ∑_{i=1}^{6} A_{i}A_{i}^{T}. Hence there exist B_{i}, C_{i} ∈ SL(d_{n}, 3) for 1 ≤ i ≤ 6 such that ∑_{i=1}^{6} B_{i}D_{1}B_{i}^{T} = D_{1}^{X} and ∑_{i=1}^{6} C_{i}D_{2}C_{i}^{T} = D_{2}^{X}. Now it is easy to complete the proof of this case. Let D = diag(λ_{1}, …, λ_{d_{n}}) ∈ M_{n}. Then for each 1 ≤ i ≤ d_{n}, there exist α_{i}, β_{i} ∈ GF (q_{n}) such that λ_{i} = α_{i}^{2} + β_{i}^{2}. (It is well-known that if 𝔽 is any finite field, then every element of 𝔽 is a sum of two squares. Unfortunately it is often not possible to express an element as the sum of two nonzero squares.) Let X = {i | α_{i} ≠ 0} and Y = {i | β_{i} ≠ 0}. Then there exist diagonal matrices R_{j} ∈ SL(d_{n}, q_{n}) for 1 ≤ j ≤ 4 such that R_{1}D_{1}^{X}R_{1}^{T} = diag(α_{1}^{2}, 0, …, 0), R_{2}D_{2}^{X}R_{2}^{T} = diag(0, α_{2}^{2}, …, α_{d_{n}}^{2}), R_{3}D_{1}^{Y} R_{3}^{T} = diag(β_{1}^{2}, 0, …, 0) and R_{4}D_{2}^{Y} R_{4}^{T} = diag(0, β_{2}^{2}, …, β_{d_{n}}^{2}). Thus we can “uniformly generate” each diagonal matrix from the generators D_{1} and D_{2}.

Case 3. Suppose that p = 2. If S = (s_{ij}) ∈ M_{n}, then S is said to be alternating if s_{ii} = 0 for all 1 ≤ i ≤ d_{n}. If S is not alternating, then S is congruent to a diagonal matrix. Clearly for each S ∈ M_{n}, there exist B_{k} ∈ M_{n} for 1 ≤ k ≤ 3 such that S = ∑_{k=1}^{3} B_{k} and none of the B_{k} are alternating. Thus we need only consider diagonal matrices D ∈ M_{n}. It is easily checked that if C is a diagonal 3 × 3-matrix over GF (2), then there exist A_{i} ∈ SL(3, 2) for 1 ≤ i ≤ 4 such that C = ∑_{i=1}^{4} A_{i}A_{i}^{T}. It is now easy to adapt the argument of Case 2. (In fact, the argument is even simpler in this case, as every element in GF (q_{n}) is a square.) This completes the proof of the existence of the subgroup Γ_{0} of ∏_{n} Sp(2d_{n}, q_{n}).

Next we shall show that there exists a subgroup Γ_{1} of ∏_{n} Sp(2d_{n}, q_{n}) such that 1. Γ_{1} is finitely generated over Γ_{0}, and

2. ∏_{n} H_{n} ⩽ Γ_{1}. For each n ∈ ℕ, let LT_{n} be the subgroup of strictly lower triangular matrices in SL(d_{n}, q_{n}), and define

V_{n} = \{(\begin{matrix}Q & 0 \\ SQ & (Q^{−1})^{T} \end{matrix} ) | Q ∈ LT_{n}, S^{T} = S\} .

Then there exists an element π ∈ Sp(2d_{n}, q_{n}) such that πU_{n}π^{−1} = V_{n}. Hence there exists g_{1} ∈ ∏_{n} Sp(2d_{n}, q_{n}) such that ∏_{n} V_{n} ⩽ Γ_{1} = ⟨Γ_{0}, g_{1}⟩. We shall prove that ∏_{n} H_{n} ⩽ Γ_{1}. For each λ ∈ GF (q_{n})^{∗}, let D_{λ} = diag(λ, 1, …, 1) ∈ GL(d_{n}, q_{n}). Define

F_{n} = \{(\begin{matrix}D_{λ} & 0 \\ 0 & D_{λ}^{−1} \end{matrix}) | λ ∈ GF (q_{n})^{∗}\} .

Since ∏_{n} G_{n} ⩽ Γ_{1}, it suffices to show that ∏_{n} F_{n} ⩽ Γ_{1}. For each t ∈ GF (q_{n}), let S(t) be the symmetric d_{n} × d_{n}-matrix with t in the upper left position and 0 elsewhere. Define

X(t) = (\begin{matrix}1 & S(t) \\ 0 & 1 \end{matrix}) ∈ U_{n} and T (t) = (\begin{matrix}1 & 0 \\ S(t) & 1 \end{matrix}) ∈ V_{n}.

Then it is easily checked that for each λ ∈ GF (q_{n})^{∗},

(\begin{matrix}D_{λ} & 0 \\ 0 & D_{λ}^{−1} \end{matrix}) = X(λ)Y (−λ^{−1})X(λ)X(−1)Y (1)X(−1).

(This is essentially [Ca1, 6.4.4].) Hence we have that ∏_{n} F_{n} ⩽ Γ_{1}, as required.

Finally we shall show that there exists a subgroup Γ_{2} of ∏_{n} Sp(2d_{n}, q_{n}) such that

1. Γ_{2} is finitely generated over Γ_{1}; and

2. ∏_{n} N_{n} ⩽ Γ_{2}. This implies that Γ_{2} = ∏_{n} Sp(2d_{n}, q_{n}), and hence completes the proof of Theorem 4.1. Again fix some n ∈ ℕ. Then W_{n} is generated by the images of the elements {w_{i} | 1 ≤ i ≤ d_{n}} of N_{n} defined as follows.

(a) If 1 ≤ i < d_{n}, then w_{i} is the permutation matrix corresponding to the permutation ( e_{i} e_{i+1} )( f_{i} f_{i+1} ).

(b) w_{d_{n}}(e_{d_{n}}) = −f_{d_{n}}, w_{d_{n}}(f_{d_{n}}) = e_{d_{n}} and w_{d_{n}} fixes the remaining elements of e f. (Thus w_{d_{n}} corresponds to the odd permutation ( e_{d_{n}} f_{d_{n}} ).)

It follows that W_{n} is isomorphic to ℤ_{2}^{d_{n}} ⋊ Sym(d_{n}), where ℤ_{2}^{d_{n}} is the natural permutation module for Sym(d_{n}). Let t = ⌊d_{n}/2⌋ and let v ∈ ℤ_{2}^{d_{n}} be a vector of weight t. Let E_{n} be the submodule of ℤ_{2}^{d_{n}} consisting of the vectors of even weight. Then for every u ∈ E_{n}, there exist π, ϕ ∈ Sym(d_{n}) such that u = π(v) + ϕ(v).

Let W_{n}^{+} be the subgroup of W_{n} consisting of the even permutations of the set {e_{i}, f_{i} | 1 ≤ i ≤ d_{n}}. Then W_{n}^{+} can be regarded as a subgroup of Sp(2d_{n}, q_{n}). Also notice that W_{n}^{+} corresponds to the subgroup E_{n} ⋊ Sym(d_{n}) of ℤ_{2}^{d_{n}} ⋊ Sym(d_{n}). So the argument of the previous paragraph shows that there exists g_{2} ∈ ∏_{n} Sp(2d_{n}, q_{n}) such that ∏_{n} W_{n}^{+} ⩽ ⟨Γ_{1}, g_{2}⟩. We shall show that ∏_{n} N_{n} ⩽ ⟨Γ_{1}, g_{2}⟩. Once again fix some n ∈ ℕ. Consider the element w = w_{d_{n}}w_{d_{n}−1}w_{d_{n}}. Since w corresponds to an even permutation of {e_{i}, f_{i} | 1 ≤ i ≤ d_{n}}, it follows that w ∈ ⟨W_{n}^{+}, B_{n}⟩. Using the standard properties of groups with BN-pairs [Ca1, Section 8.2], we have that

(B_{n}w_{d_{n}}B_{n}) (B_{n}wB_{n}) = B_{n}w_{d_{n}}wB_{n} ∪ B_{n}wB_{n}.

In particular, w ∈ (B_{n}w_{d_{n}}B_{n}) (B_{n}wB_{n}). Hence there exist elements b_{1}, b_{2}, b_{3} ∈ B_{n} such that w = b_{1}w_{d_{n}}b_{2}wb_{3}, and so w_{d_{n}} = b_{1}^{−1}wb_{3}^{−1}w^{−1}b_{2}^{−1}. Obviously if we let 1 = c_{1} = c_{2} = c_{3} ∈ B_{n}, then 1 = c_{1}^{−1}wc_{3}^{−1}w^{−1}c_{2}^{−1}. Hence for each subset X of ℕ, ψ_{X} = ⟨ψ_{X}(n)⟩_{n} ∈ ⟨Γ_{1}, g_{2}⟩, where ψ_{X}(n) = w_{d_{n}} if n ∈ X and ψ_{X}(n) = 1 if n ∉ X. It follows easily that ∏_{n} N_{n} ⩽ ⟨Γ_{1}, g_{2}⟩. (Cf. the final paragraph of the proof of Lemma 3.6.)

4.2. Unitary groups. In this subsection, we consider products ∏_{n} SU(d_{n}, q_{n}) of finite special unitary groups. First consider the case when d_{n} is even. Then the corresponding unitary space has a normal basis e f. Arguing as in Subsection 4.1, we obtain the following result.

Theorem 4.3. Suppose that ⟨SU(2d_{n}, q_{n}) | n ∈ ℕ⟩ is a sequence of special unitary groups which satisfies the following conditions.

(1) If d_{n} = 1, then q_{n} > 3.

(2) There do not exist an infinite subset I of ℕ and an integer d such that (a) d_{n} = d for all n ∈ I; and

(b) if n, m ∈ I and n < m, then q_{n} < q_{m}. Then c(∏_{n} SU(2d_{n}, q_{n})) > ω.

Next consider a product of the form ∏_{n} SU(2d_{n} + 1, q_{n}), where d_{n} ≥ 2 for all n ∈ ℕ. Fix some n ∈ ℕ. Then there exists a basis

e f (w) = (e_{i} | 1 ≤ i ≤ d_{n}) (f_{i} | 1 ≤ i ≤ d_{n}) (w)

of the corresponding unitary space such that (e_{i}, f_{j}) = δ_{ij} and (e_{i}, e_{j}) = (f_{i}, f_{j}) = 0 for all 1 ≤ i, j ≤ d_{n}; and (w, w) = 1 and (w, e_{i}) = (w, f_{i}) = 0 for all 1 ≤ i ≤ d_{n}. Then we can regard SU(2d_{n}, q_{n}) as the subgroup of SU(2d_{n} + 1, q_{n}) consisting of the elements π such that π(w) = w.

Theorem 4.4. Suppose that d_{n} ≥ 2 for all n ∈ ℕ. Then ∏_{n} SU(2d_{n} + 1, q_{n}) is finitely generated over ∏_{n} SU(2d_{n}, q_{n}).

Proof. We shall make use of the Bruhat decomposition

SU(2d + 1, q) = ⋃_{w∈W}BwB

of the special unitary group, where B is a Borel subgroup and W is the Weyl group. Fix some n ∈ ℕ. For each 1 ≤ i ≤ d_{n}, let E_{i} = ⟨e_{1}, …, e_{i}⟩. Then the stabiliser B_{n} of the flag of totally isotropic subspaces E_{1} ⩽ E_{2} ⩽ ⋯ ⩽ E_{d_{n}} is a Borel subgroup of SU(2d_{n} + 1, q_{n}). Let N_{n} be the subgroup of SU(2d_{n} + 1, q_{n}) which stabilises the polar frame {⟨e_{i}⟩, ⟨f_{i}⟩ | 1 ≤ i ≤ d_{n}}. Then the Weyl group of SU(2d_{n} +1, q_{n}) is W_{n} = N_{n}/B_{n} ∩N_{n}. Note that N_{n} ∩SU(2d_{n}, q_{n}) already contains representatives of each element of W_{n}. Thus it suffices to prove that there exists an element g such that ∏_{n} B_{n} ⩽ ⟨∏_{n} SU(2d_{n}, q_{n}), g⟩. We shall regard SU(2d_{n} +1, q_{n}) as a group of matrices with respect to the ordered basis (e_{1}, …, e_{d_{n}}, w, f_{d_{n}}, …, f_{1}). Thus B_{n} is the subgroup of the upper triangular matrices which are contained in SU(2d_{n} + 1, q_{n}). Let U_{n} be the subgroup of B_{n} consisting of the strictly upper triangular matrices. Let H_{n} = B_{n} ∩N_{n}. Then H_{n} consists of the diagonal matrices of the form

(\begin{matrix}D & 0 & 0 \\ 0 & λ^{−1} & 0 \\ 0 & 0 & D^{∗} \end{matrix})

where 1. D = diag(λ_{1}, …, λ_{d_{n}}) ∈ GL(d_{n}, q_{n}^{2});

2. D^{∗} = diag(λ̄_{d_{n}}^{−1}, …, λ̄_{1}^{−1}); and

3. λ = det(DD^{∗}). Here σ ⟼ σ̄ is the automorphism of GF (q_{n}^{2}) of order 2. (Notice that for all diagonal matrices D ∈ GL(d_{n}, q_{n}^{2}), we have that λλ̄ = 1 and hence (λ^{−1}w, λ^{−1}w) = (w, w).) We have that B_{n} = U_{n} ⋊ H_{n}.

First we shall show that there exists an element g ∈ ∏_{n} SU(2d_{n} + 1, q_{n}) such that ∏_{n} U_{n} ⩽ ⟨∏_{n} SU(2d_{n}, q_{n}), g⟩. Later we shall see that also

∏_{n} H_{n} ⩽ ⟨∏_{n} SU(2d_{n}, q_{n}), g⟩;

and so g satisfies our requirements. Once more, fix some n ∈ ℕ. Let E_{d_{n}}^{+} = ⟨E_{d_{n}}, w⟩, and let Γ_{n} be the setwise stabiliser of E_{d_{n}}^{+} in SU(2d_{n}+1, q_{n}). Let ρ : Γ_{n} → GL(E_{d_{n}}^{+}) be the restriction map. We shall regard GL(E_{d_{n}}^{+}) as a group of matrices with respect to the ordered basis (e_{1}, …, e_{d_{n}}, w). Note that for each A ∈ SL(d_{n}, q_{n}^{2}), we have that

(\begin{matrix}A & 0 \\ 0 & 1 \end{matrix}) ∈ ρ [SU(2d_{n}, q_{n}) ∩ Γ_{n}] ;

and that for each x ∈ GF (q_{n}^{2})^{d_{n}}, we have that

(\begin{matrix}A & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix}1 & x \\ 0 & 1 \end{matrix}) (\begin{matrix}A^{−1} & 0 \\ 0 & 1 \end{matrix}) = (\begin{matrix}1 & Ax \\ 0 & 1 \end{matrix}) .

Now choose θ ∈ U_{n} such that ρ(θ) = (\begin{matrix}1 & x_{0} \\ 0 & 1 \end{matrix}), where x_{0} ∈ GF (q_{n}^{2})^{d_{n}} is any

nonzero vector. Let ϕ ∈ U_{n} be an arbitrary element. Then ρ(ϕ) = (\begin{matrix}B & y \\ 0 & 1 \end{matrix}) for

some y ∈ GF (q_{n}^{2})^{d_{n}} and some strictly upper triangular matrix B ∈ GL(d_{n}, q_{n}^{2}). Clearly there exist A_{1}, A_{2} ∈ SL(d_{n}, q_{n}^{2}) such that

(\begin{matrix}1 & −y \\ 0 & 1 \end{matrix}) = (\begin{matrix}1 & A_{1}x_{0} + A_{2}x_{0} \\ 0 & 1 \end{matrix}) = (\begin{matrix}1 & A_{1}x_{0} \\ 0 & 1 \end{matrix}) (\begin{matrix}1 & A_{2}x_{0} \\ 0 & 1 \end{matrix}) .

Hence there exist ψ_{1}, ψ_{2} ∈ SU(2d_{n}, q_{n}) ∩ Γ_{n} such that ρ(ψ_{1}θψ_{1}^{−1}ψ_{2}θψ_{2}^{−1}ϕ) =

(\begin{matrix}B & 0 \\ 0 & 1 \end{matrix}) and hence ψ_{1}θψ_{1}^{−1}ψ_{2}θψ_{2}^{−1}ϕ ∈ SU(2d_{n}, q_{n}). Thus we can “uniformly

generate” U_{n} using the element θ. It follows that there exists an element g ∈ ∏_{n} SU(2d_{n} + 1, q_{n}) such that ∏_{n} U_{n} ⩽ ⟨∏_{n} SU(2d_{n}, q_{n}), g⟩.

Let ψ ∈ N_{n} ∩ SU(2d_{n}, q_{n}) correspond to the permutation ( e_{1} f_{1} ) … ( e_{d_{n}} f_{d_{n}} ). Then V_{n} = ψU_{n}ψ^{−1} is the unipotent subgroup of strictly lower triangular matrices of SU(2d_{n} + 1, q_{n}); and also ∏_{n} V_{n} ⩽ ⟨∏_{n} SU(2d_{n}, q_{n}), g⟩.

We can regard SU(3, q_{n}) as the subgroup of SU(2d_{n} + 1, q_{n}) consisting of those elements π such that π(e_{i}) = e_{i} and π(f_{i}) = f_{i} for all 1 ≤ i ≤ d_{n} − 1. Now let h ∈ ∏_{n} H_{n} be an arbitrary element. Then there exists g ∈ ∏_{n} (H_{n} ∩ SU(2d_{n}, q_{n})) such that hg ∈ ∏_{n} (H_{n} ∩ SU(3, q_{n})). Consequently, in order to show that ∏_{n} H_{n} is contained in ⟨∏_{n} SU(2d_{n}, q_{n}), g⟩, it is enough to show that ∏ (H_{n} ∩ SU(3, q_{n})) ⩽ ⟨∏_{n} SU(2d_{n}, q_{n}), g⟩. To accomplish this, we shall use a slightly modified form of [Ca1, pp. 239–242]. For the rest of this proof, we shall write the elements of SU(3, q_{n}) as 3 × 3-matrices with respect to the ordered basis (e_{d_{n}}, w, f_{d_{n}}). Fix an element ϵ ∈ GF (q_{n}^{2}) such that ϵϵ̄ = −1. Suppose that λ, t ∈ GF (q_{n}^{2}) satisfy λ^{−1} + λ̄^{−1} = tt̄. Then the matrices

A_{1} = (\begin{matrix}1 & ϵ^{−1}λt & λ \\ 0 & 1 & ϵλ̄t̄ \\ 0 & 0 & 1 \end{matrix}) and A_{2} = (\begin{matrix}1 & ϵ^{−1}λ̄t & λ \\ 0 & 1 & ϵλt̄ \\ 0 & 0 & 1 \end{matrix})

are elements of U_{n} ∩ SU(3, q_{n}), and the matrix

B = (\begin{matrix}1 & 0 & 0 \\ −ϵt̄ & 1 & 0 \\ λ̄^{−1} & −ϵ^{−1}t & 1 \end{matrix})

is an element of V_{n} ∩ SU(3, q_{n}). The product of these matrices is

A_{1}BA_{2} = (\begin{matrix}0 & 0 & λ \\ 0 & −λ^{−1}λ̄ & 0 \\ λ̄^{−1} & 0 & 0 \end{matrix}) .

Let L_{n} be the subset of GF (q_{n}^{2})^{∗} consisting of those elements λ such that there exists t ∈ GF (q_{n}^{2}) such that λ^{−1} + λ̄^{−1} = tt̄. By [Ca1, 13.7.3], each λ ∈ GF (q_{n}^{2})^{∗} can be expressed as λ = λ_{1}λ̄_{2}^{−1} for some λ_{1}, λ_{2} ∈ L_{n}. Hence we can “uniformily generate” each element of H_{n} ∩ SU(3, q_{n}) via the equation

(\begin{matrix}λ & 0 & 0 \\ 0 & λ^{−1}λ̄ & 0 \\ 0 & 0 & λ̄^{−1} \end{matrix}) = (\begin{matrix}0 & 0 & λ_{1} \\ 0 & −λ_{1}^{−1}λ̄_{1} & 0 \\ λ̄_{1}^{−1} & 0 & 0 \end{matrix}) (\begin{matrix}0 & 0 & λ_{2} \\ 0 & −λ_{2}^{−1}λ̄_{2} & 0 \\ λ̄_{2}^{−1} & 0 & 0 \end{matrix}) .

We can now easily obtain the following result.

Corollary 4.5. Suppose that ⟨S_{n} | n ∈ ℕ⟩ is a sequence of finite simple unitary groups such that there does not exist an infinite subset I of ℕ for which conditions 1.7(1) and 1.7(2) are satisfied. Then c(∏_{n} S_{n}) > ω.

4.3. Orthogonal groups. In this subsection, we shall consider products of finite orthogonal groups. First consider the case when each group has the form Ω^{+}(2d, q). Then the corresponding orthogonal space has a normal basis e f. Arguing as in Subsection 4.1, we obtain the following result.

Theorem 4.6. Suppose that ⟨Ω^{+}(2d_{n}, q_{n}) | n ∈ ℕ⟩ is a sequence of orthogonal groups which satisfies the following conditions.

(1) d_{n} ≥ 3 for each n ∈ ℕ.

(2) There does not exist an infinite subset I of ℕ and an integer d such that (a) d_{n} = d for all n ∈ I; and

(b) if n, m ∈ I and n < m, then q_{n} < q_{m}. Then c(∏_{n} Ω^{+}(2d_{n}, q_{n})) > ω.

Now we shall consider products of the form ∏_{n} Ω(2d_{n} + 1, q_{n}), where d_{n} ≥ 2 for each n ∈ ℕ. Fix some n ∈ ℕ. Let Q be the quadratic form on the corresponding orthogonal space, and let (u, v) = Q(u+v)−Q(u)−Q(v) be the associated bilinear map. We can suppose that there exists a basis

e f (w) = (e_{i} | 1 ≤ i ≤ d_{n}) (f_{i} | 1 ≤ i ≤ d_{n}) (w)

of the orthogonal space such that

(e_{i}, f_{j}) = δ_{ij} and (e_{i}, e_{j}) = (f_{i}, f_{j}) = Q(e_{i}) = Q(f_{i}) = 0

for all 1 ≤ i, j ≤ d_{n}; and

Q(w) = 1 and (w, e_{i}) = (w, f_{i}) = 0

for all 1 ≤ i ≤ d_{n}. Then we can regard Ω^{+}(2d_{n}, q_{n}) as the subgroup of Ω(2d_{n}+1, q_{n}) consisting of the elements π such that π stabilises the subspace ⟨e_{i}, f_{i} | 1 ≤ i ≤ d_{n}⟩ setwise and π(w) = w. Clearly this situation is very similar to that which we considered in Subsection 4.2. The main difference is that the Weyl group gets larger in the passage from Ω^{+}(2d_{n}, q_{n}) to Ω(2d_{n} + 1, q_{n}). The Weyl group W_{n} of Ω(2d_{n} + 1, q_{n}) is ℤ_{2}^{d_{n}} ⋊ Sym(d_{n}), acting on the set {e_{i}, f_{i} | 1 ≤ i ≤ d_{n}} with blocks of imprimitivity {e_{i}, f_{i} | 1 ≤ i ≤ d_{n}}. The Weyl group of Ω^{+}(2d_{n}, q_{n}) is the subgroup W_{n}^{+} of W_{n} consisting of the even permutations of {e_{i}, f_{i} | 1 ≤ i ≤ d_{n}}. But this point has already been dealt with during our treatment of the symplectic groups in Subsection 4.1. Hence we can easily obtain the following result.

Theorem 4.7. Suppose that d_{n} ≥ 2 for all n ∈ ℕ. Then ∏_{n} Ω(2d_{n} + 1, q_{n}) is finitely generated over ∏_{n} Ω^{+}(2d_{n}, q_{n}).

Finally we shall consider products of the form ∏_{n} Ω^{−}(2d_{n} +2, q_{n}), where d_{n} ≥ 3 for each n ∈ ℕ. In this case, there exists a basis e f (w, z) of the corresponding orthogonal space such that

(e_{i}, f_{j}) = δ_{ij} and (e_{i}, e_{j}) = (f_{i}, f_{j}) = Q(e_{i}) = Q(f_{i}) = 0

for all 1 ≤ i, j ≤ d_{n}; and

(w, e_{i}) = (w, f_{i}) = (z, e_{i}) = (z, f_{i}) = 0

for all 1 ≤ i ≤ d_{n}; and the subspace ⟨w, z⟩ does not contain any singular vectors. So we can regard Ω^{+}(2d_{n}, q_{n}) as the subgroup of Ω^{−}(2d_{n} + 2, q_{n}) consisting of the elements π such that π stabilises the subspace ⟨e_{i}, f_{i} | 1 ≤ i ≤ d_{n}⟩ setwise, π(w) = w and π(z) = z.

Theorem 4.8. Suppose that d_{n} ≥ 3 for each n ∈ ℕ. Then ∏_{n} Ω^{−}(2d_{n} + 2, q_{n}) is finitely generated over ∏_{n} Ω^{+}(2d_{n}, q_{n}).

Proof. As before, we shall make use of the Bruhat decomposition

Ω^{−}(2d + 2, q) = ⋃_{w∈W}BwB,

where B is a Borel subgroup and W is the Weyl group. Fix some n ∈ ℕ. For each 1 ≤ i ≤ d_{n}, let E_{i} = ⟨e_{1}, …, e_{i}⟩. Then the stabiliser B_{n} of the flag of totally singular subspaces

E_{1} ⩽ E_{2} ⩽ ⋯ ⩽ E_{d_{n}}

is a Borel subgroup of Ω^{−}(2d_{n} + 2, q_{n}). Let N_{n} be the subgroup of Ω^{−}(2d_{n} + 2, q_{n}) which stabilises the polar frame {⟨e_{i}⟩, ⟨f_{i}⟩ | 1 ≤ i ≤ d_{n}}. Then the Weyl group of Ω^{−}(2d_{n} + 2, q_{n}) is W_{n} = N_{n}/B_{n} ∩ N_{n}. Once again, W_{n} is ℤ_{2}^{d_{n}} ⋊ Sym(d_{n}) acting on the set {e_{i}, f_{i} | 1 ≤ i ≤ d_{n}} with blocks of imprimitivity {e_{i}, f_{i}} for 1 ≤ i ≤ d_{n}. As before, the main point is to show that there exists a subgroup G of ∏_{n} Ω^{−}(2d_{n} + 2, q_{n}) such that

1. G is finitely generated over ∏_{n} Ω^{+}(2d_{n}, q_{n}), and

2. ∏_{n} B_{n} ⩽ G.

Let U_{n} be the subgroup of unipotent elements of B_{n} and let H_{n} = B_{n} ∩ N_{n}, so that B_{n} = U_{n} ⋊ H_{n}. First we shall show that there exists an element g_{0} ∈ ∏_{n} Ω^{−}(2d_{n} + 2, q_{n}) such that ∏_{n} U_{n} ⩽ G_{0} = ⟨∏_{n} Ω^{+}(2d_{n}, q_{n}), g_{0}⟩. Note that if π ∈ U_{n}, then there exist vectors x, y ∈ E_{d_{n}} such that π(w) = w + x and π(z) = z + y. Let E_{d_{n}}^{+} = ⟨E_{d_{n}}, w, z⟩ and let Γ_{n} be the setwise stabiliser of E_{d_{n}}^{+} in Ω^{−}(2d_{n}+2, q_{n}). Let ρ : Γ_{n} → GL(E_{d_{n}}^{+}) be the restriction map. We regard GL(E_{d_{n}}^{+}) as a group of matrices with respect to the ordered basis (e_{1}, …, e_{d_{n}}, w, z). So for each A ∈ SL(d_{n}, q_{n}), we have that

(\begin{matrix}A & 0 \\ 0 & 1 \end{matrix}) ∈ ρ [Ω^{+}(2d_{n}, q_{n}) ∩ Γ_{n}] .

Arguing as in the proof of Theorem 4.4, we see that the existence of a suitable element g_{0} ∈ ∏_{n} Ω^{−}(2d_{n} +2, q_{n}) is a consequence of the following easy observation.

Claim 4.9. Suppose that d ≥ 3. Let S = SL(d, q) and V = V (d, q). Let S act on V × V via the action A(x, y) = (Ax, Ay). Suppose that a, b ∈ V are linearly independent. Then for all (x, y) ∈ V × V, there exist A, B ∈ S such that (x, y) = A(a, b) + B(a, b).

Finally we shall show that there exists an element g_{1} ∈ ∏_{n} Ω^{−}(2d_{n} + 2, q_{n}) such that ∏_{n} H_{n} ⩽ G_{1} = ⟨G_{0}, g_{1}⟩. We shall regard Ω^{−}(4, q_{n}) as the subgroup of Ω^{−}(2d_{n} + 2, q_{n}) consisting of the elements π such that π stabilises the subspace ⟨e_{d_{n}}, f_{d_{n}}, w, z⟩ setwise and such that π(e_{i}) = e_{i} and π(f_{i}) = f_{i} for all 1 ≤ i ≤ d_{n} − 1. Since ∏_{n} Ω^{+}(2d_{n}, q_{n}) ⩽ G_{0}, it is enough to find an element g_{1} such that ∏_{n} (H_{n} ∩ Ω^{−}(4, q_{n})) ⩽ ⟨G_{0}, g_{1}⟩. We shall make use of the fact that

Ω^{−}(4, q_{n}) ≃ SL(2, q_{n}^{2})/{±1}.

(For example, see [Ta, 12.42].) Let p_{n} = char(GF (q_{n})). Then U_{n} ∩ Ω^{−}(4, q_{n}) is a group of order q_{n}^{2}, and hence is a Sylow p_{n}-subgroup of Ω^{−}(4, q_{n}). It is easily checked that SL(2, q_{n}^{2}) is “uniformly generated” by the two subgroups UT (2, q_{n}^{2}) and LT (2, q_{n}^{2}), consisting of the strictly upper triangular matrices and strictly lower triangular matrices of SL(2, q_{n}^{2}). (See [Ca1, 6.4.4].) Since UT (2, q_{n}^{2}) and LT (2, q_{n}^{2}) are Sylow p_{n}-subgroups of SL(2, q_{n}^{2}), it follows that there exists an element g_{1} ∈ ∏_{n} Ω^{−}(2d_{n} + 2, q_{n}) such that ∏_{n} Ω^{−}(4, q_{n}) ⩽ ⟨G_{0}, g_{1}⟩.

Corollary 4.10. Suppose that ⟨S_{n} | n ∈ ℕ⟩ is a sequence of finite simple orthogonal groups such that there does not exist an infinite subset I of ℕ for which conditions 1.7(1) and 1.7(2) are satisfied. Then c(∏_{n} S_{n}) > ω.

4.4. Conclusion. We can now complete the proof of Theorem 1.9. Suppose that ⟨S_{n} | n ∈ ℕ⟩ is sequence of finite simple nonabelian groups such that there does not exist an infinite subset I of ℕ for which conditions 1.7(1) and 1.7(2) are satisfied. Let G = ∏_{n} S_{n}. Let

• C_{0} be the set of 26 sporadic finite simple groups,

• C_{1} be the set of finite simple alternating groups,

• C_{2} be the set of finite simple projective special linear groups,

• C_{3} be the set of finite simple symplectic groups,

• C_{4} be the set of finite simple unitary groups,

• C_{5} be the set of finite simple orthogonal groups, and

• C_{6} be the set of finite simple groups of Lie types E_{6}, E_{7}, E_{8}, F_{4}, G_{2}, ^{2}E_{6},

2B_{2}, ^{2}G_{2}, ^{2}F_{4} and ^{3}D_{4}. By the classification of the finite simple groups, each finite simple nonabelian group lies in one of the above sets. Some groups lie in more than one of these sets. For example, Alt(8) ≃ P SL(4, 2). For the rest of this argument, we shall suppose that we have slightly modified the above sets so that they yield a partition of the finite simple nonabelian groups.

For each 0 ≤ i ≤ 6, let J_{i} = {n ∈ ℕ | S_{n} ∈ C_{i}} and let P_{i} = ∏_{n∈J_{i}} S_{n}. Then G = ∏_{i=0}^{6} P_{i}. Using Proposition 2.2, it is enough to show that for each 0 ≤ i ≤ 6, either c(P_{i}) > ω or P_{i} is finite. If 1 ≤ i ≤ 5, this has been proved in Sections 2, 3 and 4. And if i = 0, this is an immediate consequence of Proposition 2.3. Finally consider P_{6}. Our hypothesis on ⟨S_{n} | n ∈ ℕ⟩ implies that there exists a finite set of simple groups F ⊆ C_{6} such that S_{n} ∈ F for all n ∈ J_{6}. So the result once again follows from Proposition 2.3. This completes the proof of Theorem 1.9.

5. A consistency result

In this section, we shall prove Theorem 1.12. Our notation follows that of Kunen [Ku]. Thus if ℙ is a notion of forcing and p, q ∈ ℙ, then q ≤ p means that q is a strengthening of p. If V is the ground model, then we denote the generic extension by V^{ℙ} when we do not want to specify a particular generic filter G ⊆ ℙ.

Definition 5.1. A notion of forcing ℙ is said to have the Laver property if the following holds. Suppose that

1. ⟨A_{n} | n ∈ ℕ⟩ is a sequence of finite sets;

2. f : ℕ → ℕ is a function such that f(n) ≥ 1 for all n ∈ ℕ and f(n) → ∞ as n → ∞;

3. p ∈ ℙ, g̃ is a ℙ-name and p ⊩ g̃ ∈ ∏_{n} A_{n}. Then there exist q ≤ p and a sequence ⟨B_{n} | n ∈ ℕ⟩ such that

(a) B_{n} ⊆ A_{n} and |B_{n}| ≤ f(n);

(b) q ⊩ g̃ ∈ ∏_{n} B_{n}.

Theorem 1.12 is an immediate consequence of the following two results.

Theorem 5.2. Suppose that V ⊨ CH, and that ⟨ℙ_{α}, ℚ̃_{α} | α < ω_{2}⟩ is a countable support iteration of proper notions of forcing such that for all α < ω_{2}

(1) ⊩_{α} ℚ̃_{α} has the cardinality of the continuum; and

(2) ⊩_{α} ℚ̃_{α} has the Laver property. Then in V^{ℙ_{ω_{2}}} , c (∏_{n} G_{n}) ≤ ω_{1} for every sequence ⟨G_{n} | n ∈ ℕ⟩ of nontrivial finite groups.

Theorem 5.3. Suppose that V ⊨ CH. Then there exists a countable support iteration ⟨ℙ_{α}, ℚ̃_{α} | α < ω_{2}⟩ of proper notions of forcing such that

(a) ⟨ℙ_{α}, ℚ̃_{α} | α < ω_{2}⟩ satisfies conditions 5.2(1) and 5.2(2); and

(b) V^{ℙ_{ω_{2}}} ⊨ c(Sym(ℕ)) = ω_{2} = 2^{ω}.

First we shall prove Theorem 5.2.

Definition 5.4. Let ⟨G_{n} | n ∈ ℕ⟩ be a sequence of nontrivial finite groups.

1. A cover is a function c : ℕ → [ ⋃_{n∈ℕ}G_{n}]^{<ω} such that for all n ∈ ℕ

(a) ∅ ≠ c(n) ⊆ G_{n};

(b) the identity element 1_{G_{n}} ∈ c(n);

(c) if a ∈ c(n), then a^{−1} ∈ c(n).

2. If g = ⟨g(n)⟩_{n} ∈ ∏_{n} G_{n}, then c covers g if g(n) ∈ c(n) for all n ∈ ℕ.

3. If c is a cover and f : ℕ → ℕ, then c is an f-cover if |c(n)| ≤ f(n) for all n ∈ ℕ.

4. If c_{1} and c_{2} are covers, then the cover c_{1} ∗ c_{2} is defined by

(c_{1} ∗ c_{2})(n) = {ab, (ab)^{−1} | a ∈ c_{1}(n), b ∈ c_{2}(n)}.

Lemma 5.5. If c_{1} is an f_{1}-cover and c_{2} is an f_{2}-cover, then c_{1}∗c_{2} is a 2f_{1}f_{2}-cover.

Proof. Obvious.

It is perhaps worth mentioning that ∗ is generally not an associative operation on the set of covers of ∏_{n} G_{n}.

Definition 5.6. If C is a set of covers of ∏_{n} G_{n}, then its closure cℓ(C) is the least set of covers satisfying

1. C ⊆ cℓ(C); and

2. if d_{1}, d_{2} ∈ cℓ(C) then d_{1} ∗ d_{2} ∈ cℓ(C).

Lemma 5.7. Suppose that C is a set of covers of ∏_{n} G_{n}. Then

{g ∈ ∏_{n} G_{n} | There exists d ∈ cℓ(C) such that g is covered by d}

is a subgroup of ∏_{n} G_{n}.

Proof. Easy.

From now on, let f : ℕ → ℕ be the function defined by f(n) = 2^{n+2} for all n ∈ ℕ.

Lemma 5.8. Suppose that ⟨G_{n} | n ∈ ℕ⟩ is a sequence of finite groups such that |G_{n}| ≥ 2^{(n+2)^{2}} for all n ∈ ℕ. If C is a countable set of f-covers of ∏_{n} G_{n}, then

{g ∈ ∏_{n} G_{n} | There exists d ∈ cℓ(C) such that g is covered by d}

is a proper subgroup of ∏_{n} G_{n}.

Proof. Suppose that d ∈ cℓ(C) is an m-fold ∗-product of c_{1}, …, c_{m} ∈ C in some order. (Remember that ∗ is not an associative operation.) Then Lemma 5.5 implies that d is a 2^{m−1}f^{m}-cover. So we can enumerate cℓ(C) = {d_{n} | n ∈ ℕ} in such a way that d_{n} is a ϕ_{n}-cover for all n ∈ ℕ, where ϕ_{n} = 2^{n}f^{n+1}. In particular,

|d_{n}(n)| ≤ 2^{n}f(n)^{n+1} = 2^{n^{2}+4n+2} < |G_{n}| .

Hence there exists g = ⟨g(n)⟩_{n} ∈ ∏_{n} G_{n} such that g(n) ∈ G_{n} ∖ d_{n}(n) for all n ∈ ℕ. Clearly g is not covered by any element d ∈ cℓ(C).

Proof of Theorem 5.2. Suppose that V ⊨ CH and that ⟨ℙ_{α}, ℚ̃_{α} | α < ω_{2}⟩ is a countable support iteration of proper notions of forcing such that for all α < ω_{2}

1. ⊩_{α} ℚ̃_{α} has the cardinality of the continuum; and

2. ⊩_{α} ℚ̃_{α} has the Laver property.

From now on, we shall work inside V^{ℙ_{ω_{2}}} . Let ⟨G_{n} | n ∈ ℕ⟩ be a sequence of nontrivial finite groups. First suppose that there exists an infinite subset I of ℕ and a finite group G such that G_{n} = G for all n ∈ I. By Lemma 1.8 and Theorems 1.3 and 1.4, c (∏_{n} G_{n}) ≤ c (∏_{n∈I} G_{n}) ≤ ω_{1}. Hence we can assume that no such subset I of ℕ exists. Then there exists an infinite subset J = {j_{n} | n ∈ ℕ} of ℕ such that |G_{j_{n}}| ≥ 2^{(n+2)^{2}} for all n ∈ ℕ. By Lemma 1.8, c (∏_{n} G_{n}) ≤ c (∏_{n∈J} G_{n}). To simplify notation, we shall suppose that |G_{n}| ≥ 2^{(n+2)^{2}} for all n ∈ ℕ.

Since the sequence ⟨G_{n} | n ∈ ℕ⟩ of finite groups can be coded by a real number, there exists α < ω_{2} such that ⟨G_{n} | n ∈ ℕ⟩ ∈ V^{ℙ_{α}}. By Shelah (III 4.1 of [Sh-b]), V^{ℙ_{α}} ⊨ CH. Let {c_{β} | β < ω_{1}} be an enumeration of the f-covers c ∈ V^{ℙ_{α}} of ∏_{n} G_{n}. For each γ < ω_{1}, let C_{γ} = {c_{β} | β < γ} and define

H_{γ} = {g ∈ ∏_{n} G_{n} | There exists d ∈ cℓ(C_{γ}) such that g is covered by d}.

By Lemma 5.8, H_{γ} is a proper subgroup of ∏_{n} G_{n} for all γ < ω_{1}. Thus it suffices to show that ∏_{n} G_{n} = ⋃_{γ<ω_{1}}H_{γ}.

Let g ∈ ∏_{n} G_{n} be any element. By Shelah ([Sh-b, VI, Section 2] and [Sh-326, Appendix]), the Laver property is preserved by countable support iterations of proper notions of forcing. This implies that there exists a sequence ⟨B_{n} | n ∈ ℕ⟩ ∈ V^{ℙ_{α}} such that

(a) B_{n} ⊆ G_{n} and |B_{n}| ≤ 2^{n}; and

(b) g(n) ∈ B_{n} for all n ∈ ℕ. Define the function c by

c(n) = B_{n} ∪ {a^{−1} | a ∈ B_{n}} ∪ {1_{G_{n}}}

for all n ∈ ℕ. Then c ∈ V^{ℙ_{α}} is an f-cover of ∏_{n} G_{n}, and so c = c_{β} for some β < ω_{1}.

Hence g ∈ ⋃_{γ<ω_{1}}H_{γ}.

The rest of this section will be devoted to the proof of Theorem 5.3. Each of the notions of forcing which we shall use in our iteration will satisfy Axiom A. It is well-known that if ℙ satisfies Axiom A, then ℙ is proper. (For example, see [J, p.101]. )

Definition 5.9. A notion of forcing ℙ satisfies Axiom A if there is a collection {≤_{n}| n ∈ ω} of partial orderings of ℙ which satisfies the following conditions.

(1) p ≤_{0} q iff p ≤ q.

(2) If p ≤_{n+1} q, then p ≤_{n} q.

(3) If ⟨p_{n} | n ∈ ω⟩ is a sequence such that p_{n+1} ≤_{n} p_{n} for all n ∈ ω, then there exists q ∈ ℙ such that q ≤_{n} p_{n} for all n ∈ ω.

(4) For each p ∈ ℙ, n ∈ ω and an ordinal name α̃, there exist a countable set B and a condition q ∈ ℙ such that q ≤_{n} p and q ⊩ α̃ ∈ B.

Definition 5.10. Fix a partition {I_{n} | n ∈ ℕ} of ℕ into infinitely many finite subsets such that the following conditions hold.

1. |I_{n}| ≥ 2 for all n ∈ ℕ.

2. For each t ≥ 2, there exist infinitely many n ∈ ℕ such that |I_{n}| = t.

3. If n < m, then max(I_{n}) < min(I_{m}). (Thus each I_{n} consists of a finite set of

consecutive integers.) The notion of forcing 𝔹 consists of all functions p such that

(a) there exists a subset J of ℕ such that dom p = ⋃_{n∈J}I_{n};

(b) if n ∈ J, then p ↾ I_{n} ∈ Sym(I_{n});

(c) if t ≥ 2, then there exist infinitely many n ∈ ℕ ∖ J such that |I_{n}| = t. If p, q ∈ 𝔹, then we define q ≤ p if and only if q ⊇ p.

Lemma 5.11. 𝔹 satisfies Axiom A and has the Laver property.

Proof. For each p ∈ 𝔹 and t ≥ 2, let

S^{t}(p) = {m ∈ ℕ | m ∉ dom p and |I_{m}| = t},

and for each n ≥ 1, let S_{n}^{t}(p) be the set of the first n elements of S^{t}(p). If t ≤ 1 or n = 0, let S_{n}^{t}(p) = ∅. For each n ∈ ω, define a partial ordering ≤_{n} on 𝔹 by setting q ≤_{n} p if and only if

1. q ⊇ p, and

2. for each t ≤ n, S_{n}^{t}(p) ⊆ 𝔹 ∖ dom p. Then it is easily checked that the partial orderings {≤_{n}| n ∈ ω} satisfy clauses (1)–(4) of Definition 5.9. It is also easy to verify that 𝔹 has the Laver property.

It follows that 𝔹 satisfies conditions 5.2(1) and 5.2(2). (It is also easily seen that 𝔹 is ^{ω}ω-bounding. However, we shall not need this fact in the proof of Theorem 5.3.) After first introducing some group theoretic notation, we shall explain the relevance of 𝔹 to the problem of computing c(Sym(ℕ)).

Definition 5.12. Suppose that {a_{n} | n ∈ ℕ} is the increasing enumeration of the infinite subset A of ℕ. If π ∈ Sym(ℕ), then π^{A} ∈ Sym(A) is defined by π^{A}(a_{n}) = a_{π(n)} for all n ∈ ℕ. If Γ is a subgroup of Sym(ℕ), then Γ^{A} = {π^{A} | π ∈ Γ}.

Definition 5.13. If g : ℕ → ℕ is a strictly increasing function, then

P_{g} = ∏_{n} Sym(g(n) ∖ g(n − 1)).

(Here we use the convention that g(−1) = 0.)

Definition 5.14. 1. If f, g : ℕ → ℕ, then f ≤^{∗} g iff there exists n_{0} ∈ ℕ such

that f(n) ≤ g(n) for all n ≥ n_{0}.

2. If g : ℕ → ℕ is a strictly increasing function, then

S_{g} = ⟨π ∈ Sym(ℕ | π, π^{−1} ≤^{∗} g⟩.

𝔹 was designed so that the following density condition would hold.

Lemma 5.15. Suppose that g : ℕ → ℕ is a strictly increasing function and that p ∈ 𝔹. Then there exists an infinite subset A of ℕ ∖ dom p such that p ∪ π ∈ 𝔹 for all π ∈ P_{g}^{A}.

Proof. Let dom p = ⋃_{n∈J}I_{n}. Then it is easy to find a suitable set A of the form

⋃_{n∈K}I_{n}, where K is an appropriately chosen subset of ℕ ∖ J.

Arguing as in Section 2 of [ST2], we can now easily obtain the following result.

Lemma 5.16. Let V ⊨ CH and let ⟨ℙ_{α}, ℚ̃_{α} | α < ω_{2}⟩ be a countable support iteration of proper notions of forcing such that for each α < ω_{2}, ℙ_{α} ⊩ |ℚ̃_{α}| = 2^{ω}. Suppose that S ⊆ {α < ω_{2} | cf(α) = ω_{1}} is a stationary subset of ω_{2}, and that ℚ̃_{α} = 𝔹̃ for all α ∈ S. (Here 𝔹̃ is the notion of forcing 𝔹 in the generic extension V^{ℙ_{α}}.) Then the following statements are equivalent in V^{ℙ_{ω_{2}}} .

(1) c(Sym(ℕ)) = ω_{1}.

(2) It is possible to express Sym(ℕ) = ⋃_{i<ω_{1}}G_{i} as the union of a chain of proper

subgroups such that for each strictly increasing function g : ℕ → ℕ, there exists i < ω_{1} with S_{g} ⩽ G_{i}.

Definition 5.17. Laver forcing 𝕃 consists of the set of all trees T ⊆ ^{<ω}ω with the following property. There exists an integer k such that

1. if n < k, then |T ∩^{n}ω| = 1;

2. if n ≥ k and η ∈ T ∩ ^{n}ω, then there exist infinitely many i ∈ ω such that

η ⟨i⟩ ∈ T. If S, T ∈ 𝕃, then S ≤ T iff S ⊆ T.

The following result is well-known.

Lemma 5.18. (1) Suppose that V ⊨ ZF C. Then there exists a function g ∈

ℕℕ ∩ V^{𝕃} such that f ≤^{∗} g for all f ∈ ^{ℕ}ℕ ∩ V.

(2) 𝕃 satisfies Axiom A and has the Laver property.

It is now easy to complete the proof of Theorem 5.3. Let V ⊨ CH. Define a countable support iteration ⟨ℙ_{α}, ℚ̃_{α} | α < ω_{2}⟩ of proper notions of forcing with the Laver property inductively as follows. If cf(α) = ω_{1}, let ℚ̃_{α} = 𝔹̃. Otherwise, let ℚ̃_{α} = 𝕃̃. From now on, we work inside V^{ℙ_{ω_{2}}} . Clearly 2^{ω} = ω_{2}. Suppose that c(Sym(ℕ)) = ω_{1}. By Lemma 5.16, we can express Sym(ℕ) = ⋃_{i<ω_{1}}G_{i} as the

union of a chain of proper subgroups such that for each strictly increasing function g : ℕ → ℕ, there exists i < ω_{1} with S_{g} ⩽ G_{i}. Lemma 5.18 implies that there exists a sequence ⟨g_{α} : ℕ → ℕ | α < ω_{2}⟩ of strictly increasing functions such that

1. if α < β < ω_{2}, then g_{α} ≤^{∗} g_{β}; and

2. for all f : ℕ → ℕ, there exists α < ω_{2} such that f ≤^{∗} g_{α}.

There exist i < ω_{1} and a cofinal subset C of ω_{2} such that S_{g_{α}} ⩽ G_{i} for all α ∈ C. But this means that G_{i} = Sym(ℕ), which is a contradiction. Hence c(Sym(ℕ)) = ω_{2}.

6. Sym(ℕ) has property (FA)

In this section, we shall prove that Sym(ℕ) has property (FA). By Macpherson and Neumann [MN], c(Sym(ℕ)) > ω. Also, since every proper normal subgroup of Sym(ℕ) is countable, ℤ is not a homomorphic image of Sym(ℕ). Thus it is enough to prove the following result.

Theorem 6.1. Sym(ℕ) is not a nontrivial free product with amalgamation.

Suppose that Sym(ℕ) is a nontrivial free product with amalgamation. Then there exists a tree T such that

1. Sym(ℕ) acts without inversion on T; and

2. there exists π ∈ Sym(ℕ) such that π(t) ≠ t for all t ∈ T.

(See Theorem 7 in [Se].) Thus it suffices to prove that whenever Sym(ℕ) acts without inversion on a tree T, then for every π ∈ Sym(ℕ) there exists a vertex t ∈ T such that π(t) = t. (This also yields a second proof that ℤ is not a homomorphic image of Sym(ℕ).) We shall make use of the following theorems of Serre.

Theorem 6.2. [Se, Theorem 16] SL(3, ℤ) has property (FA).

Theorem 6.3. [Se, Proposition 27] Let G = ⟨g_{1}, …, g_{n}⟩ be a finitely generated nilpotent group acting without inversion on the tree T. Suppose that for each 1 ≤ i ≤ n, there exists t_{i} ∈ T such that g_{i}(t_{i}) = t_{i}. Then there exists t ∈ T such that g(t) = t for all g ∈ G.

For the rest of this section, let Sym(ℕ) act without inversion on the tree T.

Lemma 6.4. If π ∈ Sym(ℕ) contains no infinite cycles, then there exists t ∈ T such that π(t) = t.

Proof. There exists a sequence ⟨G_{n} | n ∈ ℕ⟩ of nontrivial finite cyclic groups such that π ∈ ∏_{n} G_{n} ⩽ Sym(ℕ). By Bass [Ba], whenever the profinite group ∏_{n} G_{n} acts without inversion on a tree T, then for every g ∈ ∏_{n} G_{n} there exists t ∈ T such that g(t) = t.

Lemma 6.5. Suppose that π ∈ Sym(ℕ) contains infinitely many infinite cycles and no nontrivial finite cycles. Then there exists t ∈ T such that π(t) = t.

Proof. Consider the left regular action of SL(3, ℤ) on itself. Let h ∈ SL(3, ℤ) be an element of infinite order, and let H = ⟨h⟩. Then H has infinite index in SL(3, ℤ). Hence in the left regular action on SL(3, ℤ), h contains infinitely many infinite cycles and no finite cycles. Consequently if Ω = {n ∈ ℕ | π(n) ≠ n}, then there exists a subgroup G of Sym(ℕ) such that

1. π ∈ G ⩽ Sym(Ω) ⩽ Sym(ℕ); and

2. the permutation group (G, Ω) is isomorphic to the left regular action of

SL(3, ℤ) on itself. By Theorem 6.2, there exists t ∈ T such that g(t) = t for all g ∈ G.

Lemma 6.6. Suppose that π ∈ Sym(ℕ) contains finitely many infinite cycles and no nontrivial finite cycles, and that π fixes infinitely many n ∈ ℕ. Then there exists t ∈ T such that π(t) = t.

Proof. There exist ϕ_{1}, ϕ_{2} ∈ Sym(ℕ) such that the following conditions are satisfied.

1. ϕ_{1} and ϕ_{2} both contain infinitely many infinite cycles and no nontrivial finite cycles.

2. [ϕ_{1}, ϕ_{2}] = 1.

3. π = ϕ_{1}ϕ_{2}. By Lemma 6.5 and Theorem 6.3, there exists t ∈ T such that g(t) = t for all g ∈ G = ⟨ϕ_{1}, ϕ_{2}⟩.

Proof of Theorem 6.1. Let π ∈ Sym(ℕ) be any element. We shall show that π fixes a vertex of T. Express π = ϕψ as a product of disjoint permutations such that ψ has no infinite cycles and ϕ has no nontrivial finite cycles. By Lemma 6.4 and Theorem 6.3, it is enough to show that ϕ fixes a vertex of T. Suppose that ϕ ≠ 1. By Lemma 6.5, we can assume that ϕ contains only finitely many infinite cycles. Let θ = ϕ^{2}. Then θ contains ℓ infinite cycles for some 2 ≤ ℓ ∈ ℕ. Hence there exist τ_{1}, τ_{2} ∈ Sym(ℕ) such that the following conditions are satisfied.

1. τ_{1} and τ_{2} both contain finitely many infinite cycles and no nontrivial finite cycles.

2. τ_{1} and τ_{2} are disjoint permutations.

3. θ = τ_{1}τ_{2}. By Lemma 6.6 and Theorem 6.3, θ = ϕ^{2} fixes a vertex of T. By 6.3.4 of [Se], ϕ also fixes a vertex of T.

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Department of Pure Mathematics and Mathematical Statistics, Cambridge University, 16 Mill Lane, Cambridge CB2 1SB, England

   E-mail address: j.saxl@pmms.cam.ac.uk

Department of Mathematics, The Hebrew University, Jerusalem, Israel Current address: Department of Mathematics, Rutgers University, New Brunswick, New Jersey

08903

   E-mail address: shelah@sunset.ma.huji.ac.il

Department of Mathematics, Bilkent University, Ankara, Turkey

Current address: Department of Mathematics, Rutgers University, New Brunswick, New Jersey 08903

   E-mail address: sthomas@math.rutgers.edu