Existentially closed locally finite CA-groups

Andreas Ecker

March 9, 2000

We show, that existentially closed locally finite CA-groups are either isomorphic to SL_{2}(𝔽̃_{2}) (where 𝔽̃_{2} is the algebraic closure of the field with two elements), or to the semidirect group of an existentially closed abelian π-group by the direct sum of Pruefer groups of type q^{∞} for q ∈ π′. For this purpose we prove an Amalgamation Theorem.

1 Introduction

Definition 1.1. We call a group G a CA-group, if the centralizer of every non-identity element in G is abelian.

These groups are sometimes also called commutativity-transitive groups or CT-groups, because the relation, that two elements of a CA-group G commute, is transitive on G − {1} (and therefore an equivalence relation). Examples for CA-groups are abelian groups, free groups, free solvable groups and the groups SL_{2}(K) for fields of characteristic 2.

L. Weisner showed in [6] that finite CA-groups are simple or abelian-by-cyclic. However, as Y.-F. Wu pointed out, there is a error in the proof. For a corrected proof see [7] or [2]. Moreover the non-abelian, solvable CA-groups are Frobenius groups, where the Frobenius kernel is an abelian Hall π-subgroup and the Frobenius complement is cyclic. M. Suzuki [5] showed that there is no finite simple CA-group of odd order (this was a step towards the theorem of Feit-Thompson). Finally it was shown independently by M. Suzuki and G.E. Wall (see [1]) that the finite simple CA-groups of even order are isomorphic to SL_{2}(2^{n}) for some n > 1. The finite CA-groups are therefore completely classified.

On the other hand a classification of all CA-groups seems nearly impossible. For example free groups and more generally free products of CA-groups are again CA-groups. From the model-theoretic point of view it would be interesting to learn more about the existentially closed CA-groups. But this problem seems also quite hard, since there is no global amalgamation theorem for CA-groups. Therefore we will restrict ourselves in this paper to the smaller class of locally finite CA-groups. In this class we can use the classification of the finite CA-groups to get a classification of the existentially closed (e.c) groups. For readers not familiar with the term existentially closed we will give a short introduction in Chapter 2.

Most results in this paper are from the author’s doctoral thesis [2]. After finishing the thesis the author took notice of a paper from Y.-F. Wu [7] where some related results are proven. The latter article also deals with polycyclic and torsion-free solvable CA-groups.

In the following chapter we will select some facts about existentially closed structures and CA-groups. Most of the results in this chapter are wellknown. In Chapter 3 we will construct (metabelian) CA-groups using cyclotomic polynomials. The greatest part of the paper deals with the proof of the Amalgamation Theorem in Chapter 5. This theorem is crucial for the main result in Chapter 6. We close with an outlook to some further results from [2], which shall be published in a subsequent paper.

2 Preliminaries

We start with a model-theoretic definition.

Definition 2.1. Let L be a first-order language. We say a structure 𝔐 is existentially closed (e.c.) in K, if K is a class of L-structures, 𝔐 ∈ K, and if for every quantifier-free formula Φ(x̄, ȳ) and for every tuple ā in 𝔐, if there is a structure 𝔑 ∈ K, so that 𝔐 ⊆ 𝔑 and 𝔑 ⊨ ∃ȳ Φ(ā, ȳ), then already 𝔐 ⊨ ∃ȳ Φ(ā, ȳ).

Existentially closed structures have a rich structure and are therefore often easier to classify. For example, the e.c. fields are exactly the algebraically closed fields. In inductive classes (i.e. classes that are closed under unions of chains and under isomorphisms) every structure can be embedded in an e.c. structure. (For more model-theoretic details see [3].) It is easy to see, that the class of locally finite CA-groups is inductive.

We want to gather some facts about CA-groups. We already mentioned the following.

Fact 1. Let G be a finite CA-group. Then G is abelian-by-cyclic or isomorphic to a group SL_{2}(2^{n}) for some n.

The next result is also wellknown and quite easy to prove, because every nilpotent subgroup of a CA-group is abelian.

Fact 2. The Fitting subgroup Fit(G) of a CA-group G is abelian and the unique maximal normal abelian subgroup.

We now turn to the solvable (metabelian) case.

Fact 3. Let G be a finite metabelian CA-group and A the Fitting subgroup of G. If A ≨ G, then G is a Frobenius group and A is the Frobenius kernel. A has a cyclic complement C in G and |A| ≡ 1 (mod |C|).

It follows, that A is a Hall π-subgroup. Furthermore all complements of A are conjugate and maximal abelian subgroups. C acts fixed-point-free on A via conjugation. Remark, that we use a stronger notion than sometimes in the literature.

Definition 2.2. We call a automorphism group K ≤ Aut(H) fixed-point-free, if all k ∈ K, k ≠ id leave only the unit element of H fixed.

On the other hand, if H is an abelian group and K ≤ Aut(H) fixed-point-free and a CA-group, then H ⋊ K is a CA-group (see Propositon 3.1).

Fact 4. Let G be a finite CA-group and N a normal subgroup of G. Then G/N is a CAgroup.

For a proof see [7] or [2]. Remark. Obviously this result does not generalize for arbitrary CA-groups. For example, free groups are CA-groups and any group is isomorphic to a factor group of a free group. However the statement is still true for locally finite CA-groups.

3 Construction of CA-Groups

We will need some propositions about the construction of CA-groups. A special case of the following proposition (for N abelian) can be found in [7] (Lemma 7).

Proposition 3.1. Let G be a group and N ◁ G so that

(i) N and G are CA-groups and

(ii) ∀x ∈ N − {1} : C_{G}(x) ⊆ N.

Then G is a CA-group.

Proof. Let g ∈ G − {1}. We have to show that C_{G}(x) is abelian.

Case 1. g ∈ N.

Then C_{G}(g) = C_{N}(g) because of condition (ii), and the claim follows because of (i).

Case 2. g ∈ G − N.

Let h, k ∈ C_{G}(g). As G/N is a CA-group we have hk = kh · n for some n ∈ N. But then kh, khn ∈ C_{G}(g), and thus n ∈ C_{G}(g). Because of (ii) this is only possible for n = 1.

We want to use the proposition above for the construction of CA-groups. Let us first make some remarks. If G = A⋊⟨{z}⟩ is a group and A abelian, then A is a ℤ[z]-module in a natural way (let z act via conjugation on A). Therefore, if we want to construct a CA-group with Fitting subgroup A and a cyclic complement of order q, it suffices to define a fixed point free action of order q on A. Instead of writing a^{z} we will write z ∗ a and consider A as an additive ℤ[z]-module.

We first need a proposition about polynomials. Let 𝔽_{q} be the field with q elements.

Proposition 3.2. Let f(x), g(x) ∈ ℤ[x] be normed polynomials, so that the greatest common divisor of the polynomials is 1. For any prime number q let

P_{q} : ℤ[x] → 𝔽_{q}[x]

be the canonical projection. If for all prime numbers q the greatest common divisor of P_{q}(f(x)) and P_{q}(g(x)) is 1, then there exist polynomials r(x), s(x) ∈ ℤ[x] so that

r(x) · f(x) + s(x) · g(x) = 1.

Proof. As ℚ[x] ⊇ ℤ[x] is a Euclidean domain, there are polynomials R, S ∈ ℚ[x], so that

R · f + S · g = 1

and deg (R) < deg (g), deg (S) < deg (f). (After a theorem of Gauss gcd (f, g) = 1 in ℚ[x].) Then there exists n ∈ ℕ so that n · R, n · S ∈ ℤ[x]. Thus

(n · R) · f + (n · S) · g = n.

Now let n be the smallest natural number greater than 0, so that there exist r, s ∈ ℤ[x] with

r · f + s · g = n

and deg (r) < deg (g), deg (s) < deg (f). Suppose there is a prime number q dividing n. Then

P_{q}(r) · P_{q}(f) + P_{q}(s) · P_{q}(g) = 0.

But this is a contradiction, as deg (P_{q}(r)) < deg (P_{q}(g)) and deg (P_{q}(s)) < deg (P_{q}(f)). [If P_{q}(r) = P_{q}(s) = 0, then p divides all coefficients of r and s, which gives a direct contradiction to the minimality of n.]

Let C_{k} be the cyclic group of order k.

Theorem 3.3. Let k be a natural number. Let F_{k}(x) = x^{n} + λ_{n−1} · x^{n−1} + ⋯ + λ_{1} · x + λ_{0} the kth cyclotomic polynomial over ℤ and A an abelian group that has no elements of some order dividing k. Then the group A^{(n)} ⋊ C_{k} with the action (z a generating element of C_{k})

z ∗ (a, 0, …, 0) := (0, a, 0, …, 0)
z^{2} ∗ (a, 0, …, 0) := (0, 0, a, 0, …, 0)
⋮
z^{n−1} ∗ (a, 0, …, 0) := (0, …, 0, a)
z^{n} ∗ (a, 0, …, 0) := (−λ_{0} · a, −λ_{1} · a, …, −λ_{n−1} · a)
= (z^{n} − f(z)) ∗ (a, 0, …, 0)

is a CA-group.

Proof. The action of z is described over the following function.

\begin{aligned}φ : & A^{n} & → & A^{n} \\ & (a_{1}, …, a_{n}) & ↦ & (0, a_{1}, …, a_{n−1}) − (λ_{0}a_{n}, …, λ_{n−1}a_{n}). \end{aligned}

This function is obviously an endomorphism of A^{n}. We show, that it is injective. Suppose φ((a_{1}, …, a_{n})) = (0, …, 0). Then a_{n} = 0, as λ_{0} is invertible in ℤ. Thus

(0, …, 0) = φ((a_{1}, …, a_{n})) = (0, a_{1}, …, a_{n−1})

and therefore (a_{1}, …, a_{n}) = (0, …, 0). To prove that φ is surjective, let (b_{1}, …, b_{n}) ∈ A^{n}. We want to show, that (b_{1}, …, b_{n}) is in the image of φ. For that purpose we have to show, that the equational system

−λ_{0}a_{n} = b_{1}
a_{1} − λ_{1}a_{n} = b_{2}
…
a_{n−1} − λ_{n−1}a_{n} = b_{n}

is solvable. As λ_{0} is invertible, we get immediately

a_{n}  = −λ_{0}^{−1}b_{1}
a_{1} = b_{2} + λ_{1}a_{n} = b_{2} − λ_{1}λ_{0}^{−1}b_{1}
…
a_{n−1} = b_{n} + λ_{n−1}a_{n} = b_{n} − λ_{n−1}λ_{0}^{−1}b_{1}.

Thus φ is an isomorphism.

We defined φ, so that φ^{n}(a, 0, …, 0) = (φ^{n} − F_{k}(φ))(a, 0, …, 0). As a direct conclusion we have F_{k}(φ)(a, 0, …, 0) = (0, …, 0). Thus

\begin{aligned}φ^{k}(a, 0, …, 0) & = & ((φ^{k} − id) + id)(a, 0, …, 0) \\ & = & (φ^{k} − id)(a, 0, …, 0) + (a, 0, …, 0). \end{aligned}

As F_{k}(x) is a divisor of x^{k} − 1, we get (φ^{k} − id)(a, 0, …, 0) = (0, …, 0). Analogously

φ^{k}(φ^{l}(a, 0, …, 0)) = φ^{l}(φ^{k}(a, 0, …, 0)) = φ^{l}(a, 0, …, 0),

for l ∈ ω. Therefore φ^{k}(a_{1}, …, a_{n}) = (a_{1}, …, a_{n}) for all a_{1}, …, a_{n} ∈ A. Thus k is the order of the action.

To show that the action is fixed-point-free, suppose φ^{l}(a_{1}, …, a_{n}) = (a_{1}, …, a_{n}) for (a_{1}, …, a_{n}) ∈ A^{n} and some l < k. Then

\begin{aligned}(φ^{l} − id)(a_{1}, …, a_{n}) & = & (0, …, 0) and \\ F_{k}(φ)(a_{1}, …, a_{n}) & = & (0, …, 0). \end{aligned}

The polynomials F_{k}(x) and x^{l} − 1 have no common divisor in ℤ[x] and modulo p for all prime numbers p not dividing k. Therefore there are (compare the proof of Proposition 3.2) polynomials u(x), v(x) ∈ ℤ[x] and m ∈ ℕ so that

u(x) · F_{k}(x) + v(x) · (x^{l} − 1) = m

and every prime that divides m is also a divisor of k. Therefore

\begin{aligned}(m · a_{1}, …, m · a_{n}) & = & (u(φ) · F_{k}(φ) + v(φ) · (φ^{l} − 1))(a_{1}, …, a_{n}) \\ & = & (0, …, 0). \end{aligned}

As A cannot have any elements of order m, it follows a_{1} = ⋯ = a_{n} = 0. Thus there are no fixed points apart from 0.

Remark. The action described in 3.3 is in some way the most general action of order k in CA-groups. If G = A⋊⟨{z}⟩ is a CA-group, A = Fit(G) and z has order k, then F_{k}(z)∗a = 0 for all a ∈ A.

4 Locally finite CA-Groups

We now turn to locally finite CA-groups.

Proposition 4.1 ([7],[2]). Let G be a locally finite CA-group.

(i) If G is not solvable, then G is isomorphic to SL_{2}(K), where K is a locally finite field of characteristic 2. Thus G is countable or finite.

(ii) If G is solvable, then G is abelian or metabelian, and the Fitting subgroup A of G has a locally cyclic complement Z. A is a π-group, and Z a π′-group for some set π of primes.

Proof. (i) The class of groups isomorphic to SL_{2}(K), where K is a perfect field of characteristic 2, is inductive (a proof for that fact is given in [2]). Let G be a countable non-solvable locally finite CA-group. Then G is the union of a chain of finite non-solvable CA-groups. It follows, that G is isomorphic to SL_{2}(K) for some locally finite field of characteristic 2. Now a non-solvable locally finite CA-group of cardinality ℵ_{1} is the union of a chain of countable groups, and thus must also be isomorphic to a group SL_{2}(K′). But this is impossible, because there is no locally finite field of cardinality ℵ_{1}. It follows, that there is no uncountable non-solvable locally finite CA-group.

(ii) A finitely generated subgroup of G is a finite solvable CA-group and therefore metabelian. It follows, that G has to be metabelian. Now let A be the Fitting subgroup of G. As G is a CA-group, A is the unique maximal normal abelian subgroup. If A ≠ G, then choose g ∈ G − A and 1 ≠ a ∈ A. Let Z be the centralizer of g in G. Now let h ∈ G and H := ⟨{a, g, h}⟩. Then H is a finite metabelian CA-group, a ∈ Fit(H) = A ∩ H, and Z ∩ H is a complement of Fit(H) in H. Thus h = b · z for some b ∈ A ∩ H and z ∈ Z ∩ H. It follows h ∈ A · Z and thus G = A · Z.

Let us now take a look at the existentially closed groups in the class of the locally finite solvable CA-groups. From Proposition 4.1 it follows, that we have two cases to observe. A non-solvable e.c. locally finite CA-group G is isomorphic to a group SL_{2}(K), where K is a locally finite field of characteristic 2. It is not hard to guess, that K must be algebraically closed (because the existentially closed fields are exactly the algebraically closed). Then K has to be the algebraic closure of the field with 2 elements, since K must be locally finite. We will return to these arguments in Chapter 6.

For the solvable case we will use the fact, that the existentially closed abelian groups are divisible abelian groups, which contain infinitely many elements of order p for every prime number p. Accordingly the existentially closed abelian π-groups are divisible abelian π-groups which contain infinitely many elements of order p for every prime number p ∈ π. There is up to isomorphy only one countable group of this type, namely

⨁_{p∈π} Z_{p^{∞}}^{(ω)}.

We want to show that the solvable countable e.c. groups in the class of the locally finite CA-groups are of the form

(⨁_{p∈π} Z_{p^{∞}}^{(ω)}) ⋊ (⨁_{q∈π′} Z_{q^{∞}}),

where π is a non-empty set of prime numbers and π′ is the complement of π. Now let G = A ⋊ Z be a solvable existentially closed locally finite CA-group, so that A := Fit(G) is a π-group and Z is a π′-group. Then we have to show, that

(i) the Fitting subgroup A is existentially closed in the class of abelian π-groups,

(ii) and the complement Z is the largest possible locally cyclic π′-group.

We start with (i). Let L be the first-order theory of group theory. Furthermore let Φ(x̄, ȳ) be a quantifier-free formula in L and ā a tuple of elements of A, so that there is an abelian πgroup B ⊇ A in which ∃ȳ Φ(ā, ȳ) holds. Let b̄ be a tuple of elements of B satisfying Φ(ā, ȳ) (i.e. B ⊧ Φ(ā, b̄)). Now let U be the finite subgroup of B generated by ā, b̄. Then the sentence ∃ȳ Φ(ā, ȳ) is also true in U and more generally in every group containing U. Now, if we can show, that there is a locally finite CA-group H containing G, so that U ≤ Fit (H), then the sentence ∃ȳ Φ(ā, ȳ) is true in H. As G is e.c., the sentence is also true in G and therefore in A. (We may need to change Φ here a little bid, but the argument stays valid.)

For the proof of (ii) we have to show, that G = A ⋊ Z can be embedded in a locally finite CA-group B ⋊ ⨁_{q∈π′} Z_{q^{∞}}.

The missing two arguments will be shown in an Amalgamation Theorem in the next section.

5 The Amalgamation Theorem

Theorem 5.1 (Amalgamation Theorem). Let G be a locally finite solvable CA-group, A the Fitting subgroup of G, and Z a complement of A in G. Let π be a set of primes, so that A is a π-group and Z is a π′-group. Let U and V be finite abelian π-groups, so that there are embeddings φ_{1} : V → U and ψ_{1} : V → A. Then there exists a locally finite CA-group H = B ⋊ Y with the following properties. There are embeddings φ_{2} : U → B and ψ_{2} : G → H so that φ_{2} ∘ φ_{1} = ψ_{2} ∘ ψ_{1} and ψ_{2} maps Z into Y ≅ ⨁_{q∈π′} ℤ_{q^{∞}}.

Proof. Let us first give some comments on how to prove this theorem. As it would be very hard to define the action of Y on B (whatever B might be then), we will define the action in finite steps. The following (commuting) diagram shall give an idea.

V ⟶ A ⟶ A ⋊ ⟨{z_{1}}⟩ ⟶ … ⟶ A ⋊ ⟨{z_{n}}⟩ ⟶ … G
↓ ↓ ↓ ↓ ↓
U ⟶ B_{0} ⟶ B_{1} ⋊ ⟨{y_{1}}⟩ ⟶ … ⟶ B_{n} ⋊ ⟨{y_{n}}⟩ ⟶ … H

(All the arrows in this diagram stand for embeddings.) For easier notation we will write A and B_{n} additively as ℤ[z_{n}]- respectively ℤ[y_{n}]-modules for n ∈ ω. The first step is an amalgamation of A und U with amalgamated subgroup V. That is easy; we take the (outer) direct product of U and A and factorize modulo the subgroup {(φ_{1}(v), ψ_{1}(−v)); v ∈ V }. We will call this group B_{0}. From now on we will treat A as a subgroup of B_{0}. Let {q_{n}; n ∈ ω} be an enumeration of π′, so that every q ∈ π′ occurs infinitely many times. We define z_{0} = y_{0} = 1 ∈ ⨁_{q∈π′} ℤ_{q^{∞}} and η_{0} = ζ_{0} = 1 ∈ ℕ. Let ζ_{n+1} the biggest number dividing η_{n+1} := ∏_{i=0}^{n} q_{i}, so that there is an element of order ζ_{n+1} in Z. Let z_{n+1} ∈ Z be an element of order ζ_{n+1}, so that for ξ := \frac{ζ_{n+1}}{ζ_{n}} the equation z_{n+1}^{ξ} = z_{n} holds. Let y_{n+1} ∈ ⨁_{q∈π′} ℤ_{q^{∞}} an element of order η_{n+1}, so that y_{n+1}^{q_{n}} = y_{n}.

We now have to tell how B_{n} is constructed. Let ϕ be the Euler function (ϕ(n) is the number of primitive roots modulo n) and K_{n} := B_{0}^{ϕ(η_{n})}. Then K_{n} ⋊ ⟨{y_{n}}⟩ is a CA-group with the action described in Theorem 3.3. To construct B_{n} out of K_{n} we have to consider the action of z_{n} on A. That means we will take B_{n} as some factor group of K_{n}, so that we have an embedding A ⋊ ⟨{z_{n}}⟩ → B_{n} ⋊ ⟨{y_{n}}⟩ in a natural way.

We set γ_{n} := \frac{η_{n}}{ζ_{n}} and k := ϕ(η_{n}), and define

\begin{matrix}N_{n} := \{(z_{n} ∗ a_{1}, z_{n} ∗ a_{2}, …, z_{n} ∗ a_{k}) − y_{n}^{γ_{n}} ∗ (a_{1}, a_{2}, …, a_{k}); \\ a_{1}, a_{2}, …, a_{k} ∈ A\}. \end{matrix}

N_{n} is a subgroup of A^{k}. On A^{k} we have actions of y_{n} and z_{n}, if we define

z_{n} ∗ (a_{1}, …, a_{k}) := (z_{n} ∗ a_{1}, …, z_{n} ∗ a_{k}).

The actions of y_{n} and z_{n} commute, because

y_{n} ∗ (z_{n} ∗ (a_{1}, …, a_{k})) = y_{n} ∗ (z_{n} ∗ a_{1}, …, z_{n} ∗ a_{k})
= (0, z_{n} ∗ a_{1}, …, z_{n} ∗ a_{k−1})
−(λ_{0}(z_{n} ∗ a_{1}), …, λ_{k−1}(z_{n} ∗ a_{k}))
= z_{n} ∗ ((0, a_{1}, …, a_{k−1}) − (λ_{0}a_{1}, …, λ_{k−1}a_{k}))
= z_{n} ∗ (y_{n} ∗ (a_{1}, …, a_{k})).

(Here let F_{η_{n}}(x) = x^{k} + λ_{k−1}x^{k−1} + ⋯ + λ_{0} be the η_{n}th cyclotomic polynomial.) Thus we have

N_{n} = (z_{n} − y_{n}^{γ_{n}}) ∗ A^{k}.

It is easy to see, that N_{n} is a normal subgroup of K_{n} ⋊ ⟨{y_{n}}⟩, i.e. N_{n} is invariant under the action of y_{n}, because

y_{n} ∗ ((z_{n} − y_{n}^{γ_{n}}) ∗ (a_{1}, …, a_{k})) = (z_{n} − y_{n}^{γ_{n}}) ∗ (y_{n} ∗ (a_{1}, …, a_{k})).

Let B_{n} := K_{n}/N_{n}. As B_{n} ⋊ ⟨{y_{n}}⟩ is a factor group of a locally finite CA-group, it is itself a CA-group (see Fact 4).

We get the following (incomplete) diagram:

… → A ⋊ ⟨{z_{n}}⟩ → A ⋊ ⟨{z_{n+1}}⟩ → …
?  ?
… ? B_{n} ⋊ ⟨{y_{n}}⟩ ? B_{n+1} ⋊ ⟨{y_{n+1}}⟩ ? …
↑  ↑
… ? K_{n} ⋊ ⟨{y_{n}}⟩ ? K_{n+1} ⋊ ⟨{y_{n+1}}⟩ ? …

We have to show that we can replace the questionmarks in the diagram with embeddings. These embeddings shall make the diagram commutative. That means, we have to define mappings

I. K_{n} ⋊ ⟨{y_{n}}⟩ → K_{n+1} ⋊ ⟨{y_{n+1}}⟩

II. A ⋊ ⟨{z_{n}}⟩ → B_{n} ⋊ ⟨{y_{n}}⟩

III. B_{n} ⋊ ⟨{y_{n}}⟩ → B_{n+1} ⋊ ⟨{y_{n+1}}⟩

and show, that these mappings are

a) homomorphisms and

b) injective.

I. We start with the embedding K_{n} ⋊ ⟨{y_{n}}⟩ → K_{n+1} ⋊ ⟨{y_{n+1}}⟩. Here K_{n} is a ℤ[y_{n}]-module and K_{n+1} is a ℤ[y_{n+1}]-module and therefore (as y_{n+1}^{q_{n}} = y_{n}) also a ℤ[y_{n}]-module. Thus it suffices to show, that there is an injective ℤ[y_{n}]- module homomorphism from K_{n} to K_{n+1}. Let k := ϕ(η_{n}) and l := ϕ(η_{n+1}). We define

θ_{n} : B_{0}^{k}  → B_{0}^{l}
(b_{1}, b_{2}, …, b_{k}) ↦  (b_{1}, 0, …, 0)
+ y_{n+1}^{q_{n}} ∗ (b_{2}, 0, …, 0)
+ …
+ y_{n+1}^{(k−1)q_{n}} ∗ (b_{k}, 0, …, 0).

a) We have to show that this function is a homomorphism of ℤ[y_{n}]-modules. The function θ_{n} shall have the following property:

y_{n}^{r} ∗ (b, 0, …, 0) ↦ y_{n+1}^{r·q_{n}} ∗ (b, 0, …, 0)  for all r ∈ ℕ.  (1)

We defined θ_{n} over this condition for r < k. We have to show, that condition (1) is also satisfied for r ≥ k. Let F_{η_{n}}(x) be the η_{n}-th cyclotomic polynomial. The action of y_{n} on B_{0}^{k} was so defined, that

F_{η_{n}}(y_{n}) ∗ (b_{1}, …, b_{k}) = (0, …, 0) for all (b_{1}, …, b_{k}) ∈ K_{n}.

Therefore, if x^{r} = P_{1}(x) · F_{η_{n}}(x) + P_{2}(x) with deg(P_{2}) < k, then

y_{n}^{r} ∗ (b_{1}, …, b_{k}) = P_{2}(y_{n}) ∗ (b_{1}, …, b_{k}).

On the other hand we have x^{q_{n}·r} = P_{1}(x^{q_{n}})·F_{η_{n}}(x^{q_{n}})+P_{2}(x^{q_{n}}). Now let F_{η_{n+1}} be the η_{n+1}-th cyclotomic polynomial. Then F_{η_{n+1}}(x) is a divisor of F_{η_{n}}(x^{q_{n}}) (more strictly: F_{η_{n}}(x^{q_{n}}) = F_{η_{n+1}}(x)·F_{η_{n}}(x), if l = k·(q_{n}−1), and F_{η_{n}}(x^{q_{n}}) = F_{η_{n+1}}(x), if l = k · q_{n}). As

F_{η_{n+1}}(y_{n+1}) ∗ (b_{1}, …, b_{l}) = (0, …, 0) for all (b_{1}, …, b_{l}) ∈ K_{n+1}

we have therefore

y_{n+1}^{r·q_{n}} ∗ (b, 0, …, 0) = P_{2}(y_{n+1}^{q_{n}}) ∗ (b, 0, …, 0) for all b ∈ B.

Thus

θ_{n}(y_{n}^{r} ∗ (b, 0, …, 0)) = θ_{n}(P_{2}(y_{n}) ∗ (b, 0, …, 0))
= P_{2}(y_{n+1}^{q_{n}}) ∗ (b, 0, …, 0)
= y_{n+1}^{r·q_{n}} ∗ (b, 0, …, 0).

It follows easily, that θ_{n} is a homomorphism.

b) To show the injectivity of θ_{n} we have two cases to observe. Case 1. q_{n} divides η_{n}. Then l := ϕ(η_{n+1}) = ϕ(η_{n} · q_{n}) = q_{n} · ϕ(η_{n}) = q_{n} · k. Therefore

θ_{n}((b_{1}, b_{2}, …, b_{k})) = (\underbrace{b_{1}, 0, …, 0}_{q_{n}}, b_{2}, …, \underbrace{b_{k}, 0, …, 0}_{q_{n}}),

and it is obvious, that this is an embedding. We remark for later purposes that K_{n+1} can equivivalently be written as

\{ (\begin{matrix}b_{11} & … & b_{1k} \\ ⋮ & & ⋮ \\ b_{q_{n}1} & … & b_{q_{n}k} \end{matrix}); b_{ij} ∈ B_{0}, 1 ≤ i ≤ q_{n}, 1 ≤ j ≤ k\},

where θ_{n} maps (b_{1}, b_{2}, …, b_{k}) to

(\begin{matrix}b_{1} & … & b_{k} \\ 0 & … & 0 \\ ⋮ & & ⋮ \\ 0 & … & 0 \end{matrix}) .

[Here

\begin{aligned}c_{n+1} ∗ (\begin{matrix}b_{11} & … & b_{1k} \\ b_{21} & … & b_{2k} \\ ⋮ & & ⋮ \\ b_{q_{n}1} & … & b_{q_{n}k} \end{matrix}) & = & (\begin{matrix}0 & … & b_{q_{n}(k−1)} \\ b_{11} & … & b_{1k} \\ ⋮ & & ⋮ \\ b_{(q_{n}−1)1} & … & b_{(q_{n}−1)k} \end{matrix}) \\ & & − (\begin{matrix}λ_{0}b_{q_{n}k} & … & λ_{k−1}b_{q_{n}k} \\ 0 & … & 0 \\ ⋮ & & ⋮ \\ 0 & … & 0 \end{matrix}) , \end{aligned}

where F_{η_{n}}(x) = x^{k} +λ_{k−1}x^{k−1} +⋯+λ_{0} is the η_{n}th cyclotomic polynomial. This holds because then F_{η_{n+1}}(x) = x^{q_{n}k} + λ_{k−1}x^{q_{n}(k−1)} + ⋯ + λ_{0}.] Case 2. q_{n} does not divide η_{n}.

Then l = (q_{n} − 1) · k. Let c := y_{n+1}^{η_{n}} and t := q_{n} − 1. We construct a ℤ[y_{n+1}]- module K′_{n+1} that is isomorphic to K_{n+1}. Here the elements of K′_{n+1} are written as (t × k)-matrices over B_{0}. Let again F_{η_{n}}(x) = x^{k} + λ_{k−1}x^{k−1} + ⋯ + λ_{0} be the η_{n}th cyclotomic polynomial. We define fixed point free actions of y_{n} = y_{n+1}^{q_{n}} and c on K′_{n+1}:

y_{n} ∗ (\begin{matrix}b_{11} & … & b_{1k} \\ b_{21} & … & b_{2k} \\ ⋮ & & ⋮ \\ b_{t1} & … & b_{tk} \end{matrix}) = (\begin{matrix}0 & b_{11} & … & b_{1(k−1)} \\ 0 & b_{21} & … & b_{2(k−1)} \\ ⋮ & ⋮ & & ⋮ \\ 0 & b_{t1} & … & b_{t(k−1)} \end{matrix}) − (\begin{matrix}λ_{0}b_{1k} & … & λ_{k−1}b_{1k} \\ λ_{0}b_{2k} & … & λ_{k−1}b_{2k} \\ ⋮ & & ⋮ \\ λ_{0}b_{tk} & … & λ_{k−1}b_{tk} \end{matrix})

and

c ∗ (\begin{matrix}b_{11} & … & b_{1k} \\ b_{21} & … & b_{2k} \\ ⋮ & & ⋮ \\ b_{t1} & … & b_{tk} \end{matrix}) = (\begin{matrix}0 & … & 0 \\ b_{11} & … & b_{1k} \\ ⋮ & & ⋮ \\ b_{(t−1)1} & … & b_{(t−1)k} \end{matrix}) − (\begin{matrix}b_{t1} & … & b_{tk} \\ b_{t1} & … & b_{tk} \\ ⋮ & & ⋮ \\ b_{t1} & … & b_{tk} \end{matrix})

We remark that the definition is essentially the same as in 3.3. y_{n} acts on the columns and c acts on the rows (here the q_{n}th cyclotomic polynomial is x^{q_{n}−1} + ⋯ + x + 1). As q_{n} does not divide η_{n}, there are natural numbers 0 < r < η_{n}, 0 < s < q_{n} so that r · q_{n} + s · η_{n} = 1. Then y_{n+1} = y_{n}^{r} · c^{s} and we can define

y_{n+1} ∗ (\begin{matrix}b_{11} & … & b_{1t} \\ ⋮ & & ⋮ \\ b_{k1} & … & b_{kt} \end{matrix}) := (y_{n}^{r} · c^{s}) ∗ (\begin{matrix}b_{11} & … & b_{1t} \\ ⋮ & & ⋮ \\ b_{k1} & … & b_{kt} \end{matrix}) .

This action is welldefined and of order η_{n+1} because the actions of y_{n} and c commute. We can now define an isomorphism σ between K_{n+1} and K′_{n+1} over the condition

y_{n+1}^{r} ∗ (b, 0, …, 0) ↦ y_{n+1}^{r} ∗ (\begin{matrix}b & 0 & … & 0 \\ 0 & 0 & … & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & … & 0 \end{matrix}) for 0 ≤ r ≤ l − 1

Then σ ∘ θ_{n} maps (b_{1}, …, b_{k}) to the matrix

(\begin{matrix}b_{1} & b_{2} & … & b_{k} \\ 0 & 0 & … & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & … & 0 \end{matrix}) ,

which is obviously injective. Thus θ_{n} must be injective.

II. Now we show how A ⋊ ⟨{z_{n}}⟩ can be embedded in B_{n} ⋊ {y_{n}}.

\begin{matrix}A ⋊ ⟨\{z_{n}\}⟩ & → & B_{n} ⋊ \{y_{n}\} \\ (a, z_{n}^{r}) & ↦ & ((a, 0, …, 0) + N_{n}, y_{n}^{r·γ_{n}}). \end{matrix}

Let d_{n} := y_{n}^{γ_{n}}.

a) This function is a homomorphism. That can be concluded from

(z_{n}^{r} − d_{n}^{r}) ∗ (a, 0, …, 0) ∈ N_{n} for r ∈ ℕ,

which follows from

(z_{n}^{r} − d_{n}^{r}) = (z_{n} − d_{n}) · (z_{n}^{r−1} + z_{n}^{r−2}d_{n} + ⋯ + z_{n}d_{n}^{r−2} + d_{n}^{r−1}).

b) The function is injective. Suppose (a, 0, …, 0) ∈ N_{n}. Then (a, 0, …, 0) = (z_{n} − d_{n}) ∗ (b_{1}, …, b_{k}) for some b_{1}, …, b_{k} ∈ A. Let F_{ζ_{n}}(x) = x^{v} + μ_{v−1}x^{v−1} + ⋯ + μ_{0} be the ζ_{n}th cyclotomic polynomial. Then

F_{ζ_{n}}(z_{n}) = z_{n}^{v} + μ_{v−1}z_{n}^{v−1} + ⋯ + μ_{0}
= (z_{n} − d_{n}) · ((z_{n}^{v−1} + μ_{v−1}z_{n}^{v−2} + ⋯ + μ_{1})
+d_{n} · (z_{n}^{v−2} + μ_{v−1}z_{n}^{v−3} + ⋯ + μ_{2})
+ …
+d_{n}^{v−1})
+F_{ζ_{n}}(d_{n}).

As

F_{ζ_{n}}(z_{n}) ∗ (a_{1}, …, a_{k}) = F_{ζ_{n}}(d_{n}) ∗ (a_{1}, …, a_{k}) = (0, …, 0)

for all a_{1}, …, a_{k} ∈ A, and (a, 0, …, 0) = (z_{n} − d_{n}) ∗ (b_{1}, …, b_{k}), we have

(0, …, 0) = ((z_{n}^{v−1} + μ_{v−1}z_{n}^{v−2} + ⋯ + μ_{1})
+d_{n} · (z_{n}^{v−2} + μ_{v−1}z_{n}^{v−3} + ⋯ + μ_{2})
+ …
+d_{n}^{v−1}) ∗ (a, 0, …, 0).

The claim follows now with the same argument that we used to show the injectivity of θ_{n}. That means, because

\begin{matrix}⟨\{d_{n}^{r} ∗ (b, 0, …, 0); 0 ≤ r ≤ v − 1, b ∈ B_{0}\}⟩ \\ = ⨁_{r=0}^{v−1}⟨\{d_{n}^{r} ∗ (b, 0, …, 0); b ∈ B_{0}\}⟩, \end{matrix}

it follows a = 0.

III. It is clear, how the embedding B_{n} ⋊ ⟨{y_{n}}⟩ → B_{n+1} ⋊ ⟨{y_{n+1}}⟩ must be defined to make the diagram commutative.

\begin{matrix}B_{n} ⋊ ⟨\{y_{n}\}⟩ & → & B_{n+1} ⋊ ⟨\{y_{n+1}\}⟩ \\ ((b_{1}, …, b_{k}) + N_{n}, y_{n}^{r}) & ↦ & (θ_{n}(b_{1}, …, b_{k}) + N_{n+1}, y_{n+1}^{r·q_{n}}). \end{matrix}

This is well-defined, because if (b_{1}, …, b_{k}) ∈ N_{n}, then θ_{n}(b_{1}, …, b_{k}) ∈ N_{n+1}.

a) It is also clear, that the mapping is a homomorphism, because θ_{n} is a homomorphism and that the diagram, we have to complete, is commutative.

b) It remains to show, that this mapping is injective. Let d_{n} := y_{n}^{γ_{n}} and d_{n+1} := y_{n+1}^{γ_{n+1}}. We have to show, that if

θ_{n}(b_{1}, …, b_{k}) ∈ N_{n+1} = (z_{n+1} − d_{n+1}) ∗ A^{l},

then

(b_{1}, …, b_{k}) ∈ N_{n} = (z_{n} − d_{n}) ∗ A^{k}.

We have to observe several cases. Case 1. z_{n} = z_{n+1} and d_{n} = y_{n}^{γ_{n}} = y_{n+1}^{γ_{n+1}} = d_{n+1}. The representation of the elements of K_{n+1} as matrices shows,that there is a y_{n}- and z_{n}-linear mapping Π_{n} : K_{n+1} → K_{n}, so that Π_{n} ∘ θ_{n} is the identity on K_{n}. Thus if

\begin{aligned}θ_{n}(b_{1}, …, b_{k}) & = & (z_{n+1} − d_{n+1}) ∗ (a_{1}, …, a_{l}) \\ & = & (z_{n} − d_{n}) ∗ (a_{1}, …, a_{l}), \end{aligned}

then

(b_{1}, …, b_{k}) = Π_{n}(θ_{n}(b_{1}, …, b_{k}))
= Π_{n}((z_{n} − d_{n}) ∗ (a_{1}, …, a_{l}))
= (z_{n} − d_{n}) ∗ Π_{n}(a_{1}, …, a_{l}) ∈ N_{n}.

Case 2. z_{n} = z_{n+1}^{q_{n}} and γ_{n} = γ_{n+1}. (Then d_{n} = d_{n+1}^{q_{n}}.) We first make the remark, that there is ℤ[y_{n}]-module L_{n}, that is isomorphic to K_{n}, so that L_{n} can be written as

L_{n} = \{ (\begin{matrix}b_{11} & … & b_{1v} \\ ⋮ & & ⋮ \\ b_{u1} & … & b_{uv} \end{matrix}); b_{ij} ∈ B_{0}\},

where v = ϕ(ζ_{n}) and u = \frac{ϕ(η_{n})}{ϕ(ζ_{n})} and

d_{n} ∗ (\begin{matrix}b_{11} & … & b_{1v} \\ ⋮ & & ⋮ \\ b_{u1} & … & b_{uv} \end{matrix}) = (\begin{matrix}0 & b_{11} & … & b_{1(v−1)} \\ 0 & ⋮ & & ⋮ \\ 0 & b_{u1} & … & b_{u(v−1)} \end{matrix}) − (\begin{matrix}μ_{0}b_{1v} & … & μ_{v−1}b_{1v} \\ ⋮ & & ⋮ \\ μ_{0}b_{uv} & … & μ_{v−1}b_{uv} \end{matrix}) .

(Here let F_{ζ_{n}}(x) = x^{v} +μ_{v−1}x^{v−1} +⋯+μ_{0} again be the ζ_{n}th cyclotomic polynomial.) This can be done recursively in the way we introduced matrices in I.b. We remark, that we do not say much about the action of y_{n} on L_{n}, but this is not important here. We will identify K_{n} with L_{n} and have therefore an embedding θ_{n} : L_{n} → L_{n+1}. We need to show the following claim. Claim A.

\begin{aligned}N_{n} & = & \{(z_{n} − d_{n}) ∗ (\begin{matrix}a_{11} & … & a_{1v} \\ ⋮ & & ⋮ \\ a_{u1} & … & a_{uv} \end{matrix}); a_{ij} ∈ A\} \\ & =^{!} & \{(z_{n} − d_{n}) ∗ (\begin{matrix}a_{11} & … & a_{1(v−1)} & 0 \\ ⋮ & & ⋮ & ⋮ \\ a_{u1} & … & a_{u(v−1)} & 0 \end{matrix}); a_{ij} ∈ A\}. \end{aligned}

Proof of Claim A. Obviously it is sufficient to show, that for all b_{1}, …, b_{u} ∈ A the element

(z_{n} − d_{n}) ∗ (\begin{matrix}0 & … & 0 & b_{1} \\ ⋮ & & ⋮ & ⋮ \\ 0 & … & 0 & b_{u} \end{matrix})

can be written as

(z_{n} − d_{n}) ∗ (\begin{matrix}a_{11} & … & a_{1(v−1)} & 0 \\ ⋮ & & ⋮ & ⋮ \\ a_{u1} & … & a_{u(v−1)} & 0 \end{matrix})

with a_{ij} ∈ A (1 ≤ i ≤ u, 1 ≤ j ≤ v − 1). Set

a_{w1}  := μ_{0}z_{n}^{−1} ∗ b_{w}
a_{w2}  := (μ_{1}z_{n}^{−1} + μ_{0}z_{n}^{−2}) ∗ b_{w}
⋮
a_{w(v−1)} := (μ_{v−2}z_{n}^{−1} ∗ ⋯ + μ_{0}z_{n}^{−v+1}) ∗ b_{w}

for 1 ≤ w ≤ u. Easy computation shows

(z_{n} − d_{n}) ∗ (\begin{matrix}0 & … & 0 & b_{1} \\ ⋮ & & ⋮ & ⋮ \\ 0 & … & 0 & b_{u} \end{matrix}) = (z_{n} − d_{n}) ∗ (\begin{matrix}a_{11} & … & a_{1(v−1)} & 0 \\ ⋮ & & ⋮ & ⋮ \\ a_{u1} & … & a_{u(v−1)} & 0 \end{matrix}) .

[Remark, that

\begin{aligned}a_{w(v−1)} & := & (μ_{v−2}z_{n}^{−1} ∗ ⋯ + μ_{0}z_{n}^{−v+1}) ∗ b_{w} \\ & = & −(z_{n} + μ_{v−1}) ∗ b_{w}, \end{aligned}

because

\begin{matrix}(z_{n} + μ_{v−1} + μ_{v−2}z_{n}^{−1} ∗ ⋯ + μ_{0}z_{n}^{−v+1}) ∗ b_{w} \\ = z_{n}^{1−v} ∗ (F_{ζ_{n}}(z_{n}) ∗ b_{w}) = 0.] \end{matrix}

We proceed in Case 2 of the proof. We have to distinguish two subcases.

Subcase 2.1. q_{n} divides ζ_{n}.

Here

θ_{n} : (\begin{matrix}a_{11} & … & a_{1v} \\ ⋮ & & ⋮ \\ a_{u1} & … & a_{uv} \end{matrix}) ↦ (\begin{matrix}a_{11} & 0 & … & 0 & a_{12} & … & a_{1v} & 0 & … & 0 \\ ⋮ & & & & ⋮ & & ⋮ & & & ⋮ \\ a_{u1} & 0 & … & 0 & a_{u2} & … & a_{uv} & 0 & … & 0 \end{matrix}) .

Now suppose

θ_{n}( (\begin{matrix}a_{11} & … & a_{1v} \\ ⋮ & & ⋮ \\ a_{u1} & … & a_{uv} \end{matrix}) ) = (z_{n+1} − d_{n+1}) ∗ (\begin{matrix}b_{11} & … & b_{1(v−1)} & 0 \\ ⋮ & & ⋮ & ⋮ \\ b_{u1} & … & b_{u(v−1)} & 0 \end{matrix}) .

As the action we have to consider, is only acting on the rows, it suffices to treat the rows seperately. For 1 ≤ w ≤ u we have therefore

\begin{matrix}(\overbrace{a_{w1}, 0, …, 0}^{q_{n}}, a_{w2}, …, \overbrace{a_{wv}, 0, …, 0}^{q_{n}}) = \\ (z_{n+1} ∗ b_{w1}, z_{n+1} ∗ b_{w2} − b_{w1}, …, −b_{w(v−1)}). \end{matrix}

Omitting the index w and and using the following replacments

\begin{aligned}β_{ij} & := & b_{i+(j−1)·q_{n}} \\ q & := & q_{n} \\ z & := & z_{n+1} \end{aligned}

we can write this in matrices:

(\begin{matrix}a_{1} & … & a_{v} \\ 0 & … & 0 \\ ⋮ & & ⋮ \\ 0 & … & 0 \end{matrix}) = (\begin{matrix}z ∗ β_{11} & … & z ∗ β_{1v} − β_{q(v−1)} \\ z ∗ β_{21} − β_{11} & … & ⋮ \\ ⋮ & & ⋮ \\ z ∗ β_{q1} − β_{(q−1)1} & … & −β_{(q−1)v} \end{matrix})  (2)

Set

x_{1}  := z^{−q+1} ∗ β_{11}  = z_{n+1}^{−q_{n}+1} ∗ β_{11}
x_{2}  := z^{−q+1} ∗ β_{12}  = z_{n+1}^{−q_{n}+1} ∗ β_{12}
⋮
x_{v−1} := z^{−q+1} ∗ β_{1(v−1)} = z_{n+1}^{−q_{n}+1} ∗ β_{1(v−1)}.

We claim, that

\begin{aligned}(a_{1}, …, a_{v}) & = & (z_{n} − d_{n}) ∗ (x_{1}, …, x_{v−1}, 0) \\ & = & (z_{n} ∗ x_{1}, z_{n} ∗ x_{2} − x_{1}, …, −x_{v−1}), \end{aligned}

and therefore

(\begin{matrix}a_{11} & … & a_{1v} \\ ⋮ & & ⋮ \\ a_{u1} & … & a_{uv} \end{matrix}) ∈ N_{n}.

Using (2) we get for 1 ≤ r < v:

0 = z ∗ β_{2r} − β_{1r}
0 = z ∗ β_{3r} − β_{2r}
…
0 = z ∗ β_{qr} − β_{(q−1)r}

Thus

β_{qr} = z^{−1} ∗ β_{(q−1)r}
= z^{−2} ∗ β_{(q−2)r}
…
= z^{−q+1} ∗ β_{1r} = z_{n+1}^{−q+1} ∗ β_{1r}.

and β_{1v} = 0. We get

a_{1}  = z ∗ β_{11} = (z_{n+1}^{q_{n}} · z_{n+1}^{−q_{n}+1}) ∗ β_{11} = z_{n} ∗ x_{1}
a_{2}  = z ∗ β_{12} − β_{q1} = z_{n} ∗ x_{2} − z^{−q+1} ∗ β_{11} = z_{n} ∗ x_{2} − x_{1}
…
a_{v−1} = z_{n} ∗ x_{v−1} − x_{v−2}
a_{v}  = z ∗ β_{1v} − β_{q(v−1)} = 0 − z^{−q+1} ∗ β_{1(v−1)} = −x_{v−1}.

That is just what we had to show.

Subcase 2.2 q_{n} does not divide ζ_{n}. Analogously as in Subcase 2.1 we can treat the rows separately. We can write these rows again as matrices like in I. Case 2. and get

(\begin{matrix}a_{1} & … & a_{v} \\ 0 & … & 0 \\ ⋮ & & ⋮ \\ 0 & … & 0 \end{matrix}) = (z_{n+1} − d_{n+1}) ∗ (\begin{matrix}b_{11} & … & b_{1v} \\ b_{21} & … & b_{2v} \\ ⋮ & & ⋮ \\ b_{(q−1)1} & … & b_{(q−1)v} \end{matrix})  (3)

Set q := q_{n} and u := q − 1. We make the following claim. Let b_{ij} ∈ A for 1 ≤ i ≤ u, 1 ≤ j ≤ v. Then there are α_{ij}, β_{ij} ∈ A for 1 ≤ i ≤ u, 1 ≤ j ≤ v, so that

\begin{matrix}(z_{n+1} − d_{n+1}) ∗ (\begin{matrix}b_{11} & … & b_{1v} \\ ⋮ & & ⋮ \\ b_{u1} & … & b_{uv} \end{matrix}) = (z_{n} − d_{n})∗ & (\begin{matrix}α_{11} & … & α_{1v} \\ ⋮ & & ⋮ \\ α_{u1} & … & α_{uv} \end{matrix}) \\ +(z_{n+1}^{ζ_{n}} − d_{n+1}^{ζ_{n}})∗ & (\begin{matrix}β_{11} & … & β_{1v} \\ ⋮ & & ⋮ \\ β_{u1} & … & β_{uv} \end{matrix}) \end{matrix}

To prove this, we take a look at the polynomials x^{q} − 1 and x^{ζ_{n}} − 1. They have greatest common divisor x − 1 in ℤ[x] and modulo p (i.e. in 𝔽_{p}[x]) for all prime numbers p, because q does not divide ζ_{n}. Using proposition 3.2 there are polynomials P_{1}(x), P_{2}(x) ∈ ℤ[x] so that

x − 1 = (x^{q} − 1) · P_{1}(x) + (x^{ζ_{n}} − 1) · P_{2}(x).

Replacing x by z_{n+1} · d_{n+1}^{−1} we get

\begin{aligned}z_{n+1} · d_{n+1}^{−1} − 1 & = & ((z_{n+1} · d_{n+1}^{−1})^{q} − 1) · P_{1}(z_{n+1} · d_{n+1}^{−1}) \\ & & +((z_{n+1} · d_{n+1}^{−1})^{ζ_{n}} − 1) · P_{2}(z_{n+1} · d_{n+1}^{−1}), \end{aligned}

respectively

\begin{aligned}z_{n+1} − d_{n+1} & = & (z_{n} − d_{n}) · d_{n+1}^{1−q} · P_{1}(z_{n+1} · d_{n+1}^{−1}) \\ & & +(z_{n+1}^{ζ_{n}} − d_{n+1}^{ζ_{n}}) · d_{n+1}^{1−ζ_{n}} · P_{2}(z_{n+1} · d_{n+1}^{−1}). \end{aligned}

The claim follows directly. With Claim A equation 3 turns to (u := q − 1)

\begin{aligned}(\begin{matrix}a_{1} & … & a_{v} \\ 0 & … & 0 \\ ⋮ & & ⋮ \\ 0 & … & 0 \end{matrix}) & = & (z_{n} − d_{n}) ∗ (\begin{matrix}α_{11} & … & α_{1(v−1)} & 0 \\ α_{21} & … & α_{2(v−1)} & 0 \\ ⋮ & & ⋮ & ⋮ \\ α_{u1} & … & α_{u(v−1)} & 0 \end{matrix}) \\ & & +(z_{n+1}^{ζ_{n}} − d_{n+1}^{ζ_{n}}) ∗ (\begin{matrix}β_{11} & … & β_{1v} \\ ⋮ & & ⋮ \\ β_{(u−1)1} & … & β_{(u−1)v} \\ 0 & … & 0 \end{matrix}) \end{aligned} (4)

Let x_{i} := z_{n}^{ζ_{n}^{u−1}} ∗ α_{ui} + ⋯ + z_{n}^{ζ_{n}} ∗ α_{2i} + α_{1i} = ∑_{j=1}^{u} z_{n+1}^{ζ_{n}^{j−1}} ∗ α_{j1}. Then

\begin{aligned}(a_{1}, …, a_{v}) & = & (z_{n} − d_{n}) ∗ (x_{1}, …, x_{v−1}, 0) \\ & = & (z_{n} ∗ x_{1}, z_{n} ∗ x_{2} − x_{1}, …, z_{n} ∗ x_{v−1} − x_{v−2}, −x_{v−1}). \end{aligned}

To show that, let us consider the equations from (4). We get

β_{(u−1)1} = z_{n} ∗ α_{u1}
β_{(u−2)1} − z_{n+1}^{ζ_{n}} ∗ β_{(u−1)1} = z_{n} ∗ α_{(u−1)1}
⋮
β_{11} − z_{n+1}^{ζ_{n}} ∗ β_{21} = z_{n} ∗ α_{21}

and

β_{(u−1)i} = z_{n} ∗ α_{ui} − α_{u(i−1)}
β_{(u−2)i} − z_{n+1}^{ζ_{n}} ∗ β_{(u−1)i} = z_{n} ∗ α_{(u−1)i} − α_{(u−1)(i−1)}
⋮
β_{1i} − z_{n+1}^{ζ_{n}} ∗ β_{2i} = z_{n} ∗ α_{2i} − α_{2(i−1)}

for 1 < i < v, and

β_{(u−1)v} = −α_{u(v−1)}
β_{(u−2)v} − z_{n+1}^{ζ_{n}} ∗ β_{(u−1)v} = −α_{(u−1)(v−1)}
⋮
β_{1v} − z_{n+1}^{ζ_{n}} ∗ β_{2v} = −α_{2(v−1)}.

We get

β_{11} = z_{n} ∗ ∑_{j=2}^{u} z_{n+1}^{ζ_{n}^{j−2}} ∗ α_{j1},
β_{1i} = z_{n} ∗ ∑_{j=2}^{u} z_{n+1}^{ζ_{n}^{j−2}} ∗ α_{ji} − ∑_{j=2}^{u} z_{n+1}^{ζ_{n}^{j−2}} ∗ α_{j(i−1)}  (1 < i < v),
β_{1v} = − ∑_{j=2}^{u} z_{n+1}^{ζ_{n}^{j−2}} ∗ α_{j(v−1)}.

Moreover we get from (4)

a_{1} = z_{n} ∗ α_{11} + z_{n+1}^{ζ_{n}} ∗ β_{11}
= z_{n} ∗ ∑_{j=1}^{u} z_{n+1}^{ζ_{n}^{j−1}} ∗ α_{j1}
= z_{n} ∗ x_{1},

and similarly

a_{i} = z_{n} ∗ α_{1i} − α_{1(i−1)} + z_{n+1}^{ζ_{n}} ∗ β_{1i}
= z_{n} ∗ ∑_{j=1}^{u} z_{n+1}^{ζ_{n}^{j−1}} ∗ α_{j1} − ∑_{j=1}^{u} z_{n+1}^{ζ_{n}^{j−1}} ∗ α_{j(i−1)}
= z_{n} ∗ x_{i} − x_{i−1},

for 1 < i < v, and

a_{v} = −α_{1(v−1)} + z_{n+1}^{ζ_{n}} ∗ β_{1v}
= − ∑_{j=1}^{u} z_{n+1}^{ζ_{n}^{j−1}} ∗ α_{j(v−1)}
= −x_{v−1}.

If we take B as the direct limit of (B_{n})_{n∈ℕ}, Y as the direct limit of (⟨{y_{n}}⟩)_{n∈ℕ} and H as the direct limit of (B_{n} ⋊ ⟨{y_{n}}⟩)_{n∈ℕ}, then H = B ⋊ Y has the desired properties. That completes our proof.

6 The Main Result

We state now the main theorem of this paper.

Theorem 6.1. A countable existentially closed locally finite CA-group is isomorphic to

SL_{2}(𝔽̃_{2})

(where 𝔽̃_{2} is the algebraic closure of the field with 2 elements) or isomorphic to

(⨁_{p∈π} ℤ_{p^{∞}}^{(ω)}) ⋊ (⨁_{q∈π′} ℤ_{q^{∞}})

for some set π ≠ ∅ of primes.

An uncountable existentially closed locally finite CA-group is isomorphic to

A ⋊ (⨁_{q∈π′} ℤ_{q^{∞}}),

where A is an existentially closed abelian π-group for some set π ≠ ∅ of primes.

Proof. Let first G be a non-solvable existentially closed locally finite CA-group. In Proposition 4.1 we proved, that G is isomorphic to SL_{2}(K), where K is a locally finite field of characteristic 2. It is easy to see (compare [4]), that the field operations are definable in a group SL_{2}(K), when K is a perfect field of characteristic 2. Therefore K has to be algebraically closed. As K is also locally finite, K ≅ 𝔽̃_{2}.

We already made some remarks on the solvable case in Chapter 4. Let again Φ(x̄, ȳ) be a quantifier-free formula in the first-order language L of groups, and ā a tuple of elements of A, so that there is an abelian π-group B ⊇ A in which ∃ȳ Φ(ā, ȳ) holds. Let b̄ be a satisfying tuple of elements in B and set U := ⟨ā, b̄⟩. The Amalgamation Theorem 5.1 shows, that there is a locally finite CA-group H ≥ G, so that U ≤ Fit(H). Let 1 ≠ c ∈ Fit(G). Then

H ⊧ ∃ȳ (Φ(x̄, ȳ) ∧ ⋀_{i}[y_{i}, c] = 1).

As G is e.c., this sentence is also true in G, and hence A ⊧ ∃ȳ Φ(ā, ȳ). Like we saw in Theorem 5.1, the sentence ∃x (x^{n} = 1 ∧ ⋀_{i<n} x^{i} ≠ 1) can be fullfilled in a locally finite CA-group H ≥ G for every π′-number n, and is therefore also true in G.

7 Further Remarks

In the Main Theorem we gave necessary conditions for a group to be an e.c. locally finite CA-group. The first question one would ask is, if these conditions are sufficient. They are not, as we showed in [2]. In [2] we also gave sufficient conditions for a group to be an e.c. locally finite CA-group.

Another question concerns the number of non-isomorpic e.c. groups. As π can be an arbitrary subset of primes, there are of course 2^{ℵ_{0}} non-isomorphic countable e.c. locally finite CA-groups. But what can be said, if π is fixed? For example if π is the set of all primes, then there is only one countable e.c. group (the group is abelian). In [2] we were able to show, that in all other cases there are 2^{ℵ_{0}} non-isomorphic countable e.c. solvable locally finite CA-groups of type π.

Both results shall be presented in a subsequent paper.

References

[1] R. Brauer, M. Suzuki, G. E. Wall: A characterisation of the one-dimensional unimodular projective groups over finite fields. Illinois Journal of Mathematics, 2, (1958), 718–745.

[2] A. Ecker: Zur Modelltheorie der CA-Gruppen. Dissertation, University of Tübingen, 1999.

[3] W. Hodges: Model Theory, vol. 42 of Encyclopedia of mathematics and its applications. Cambridge University Press, 1993.

[4] P. Kissel: Beiträge zur Modelltheorie der linearen und der symplektischen Gruppen. Dissertation, University of Tübingen, 1991.

[5] M. Suzuki: The nonexistence of a certain type of simple group of odd order. Proceedings of the American Mathematical Society, 8, (1957), 686–695.

[6] L. Weisner: Groups in which the normaliser of every element except identity is abelian. Bulletin of the American Mathematical Society, 31, (1925), 413–416.

[7] Y.-F. Wu: Groups in which commutativity is a transitive relation. Journal of Algebra, 207(1), (1998), 165–181.