More about existentially closed locally finite CA-groups

Andreas Ecker

March 13, 2000

Abstract In [3] we gave a proof, that e.c. locally finite solvable CA-groups are the semidirect product of a e.c. abelian π-group by the direct product of Pruefer groups of type q^{∞} for q ∈ π′. But this is only a necessary condition. We give also sufficient conditions for a group of this kind to be an e.c. locally finite CA-group. We also show that there are many non-isomorphic e.c. groups in this class, even if π is fixed.

Introduction

This article is a subsequent paper of [3]. We therefore refer to the first paper for notation and supplementary results. A group G is called a CA-group if the centralizer of every non-identity element in G is abelian. We use the following notation.

Notation. The class of solvable locally finite CA-groups will be denoted with C.

In [3] we showed, that a solvable existentially closed (e.c.) locally finite CA-group (i.e. an e.c. C-group) is the semidirect product of an e.c. abelian π-group by ⨁_{q∈π′} Z_{q^{∞}}. (Here π is a set of primes, and π′ its complement in the set of all primes.) That poses the question, if the converse is true, i.e. are all CA-groups of this type e.c. in C? This is not true, as we show in this paper. Moreover we give sufficient conditions for such a group to be e.c. in C.

Another question is, how many non-isomorphic countable e.c. groups in C there are. Of course, as π is an arbitrary set of primes, there are 2^{ℵ_{0}} non-isomorphic groups, which is the maximal number. But what can be said, if we fix π? For example, if π is the set of all primes, then a existentially closed group G in C is isomorphic to ⨁_{p∈π} Z_{p^{∞}}^{(ω)}. We show that in all other cases there are 2^{ℵ_{0}} non-isomorphic groups for fixed π.

1 Cyclotomic Polynomials over ℤ_{p}

We showed in [3] that the e.c. groups in C are isomorphic to A ⋊ Z, where there is a set of primes π, so that A is e.c. in the class of locally finite abelian π-groups and Z is a divsible locally cyclic π′-group. In the countable case we have thus

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

Definition 1.1. We call a C-group G of type π, if π is a non-empty set of primes, the Fitting subgroup F (G) of G is a π-group, and the complement of F (G) in G is a π′-group.

In [3] we used group rings in the following way. Let G = A⋊Z be a CA-group, A abelian and Z locally cyclic. Then A can be considered as a module over the group ring ℤZ. If A is a p-group, we can use instead ℤ_{p}Z, where ℤ_{p} is the ring of p-adic integers. We will see, that this ring is more useful to describe all possible actions. Let ℤ_{p} be the ring of p-adic integers, ℚ_{p} be the field of p-adic numbers. For our examinations we consider the special case, that Z is finite and cyclic. Let z be a generating element of Z and k the order of z. Then ℤ_{p}Z is isomorphic to a factor ring of the polynomial ring ℤ_{p}[x]:

ℤ_{p}Z ≅ ℤ_{p}[x]/(x^{k} − 1).

Therefore we want to examine some properties of ℤ_{p}[x]. Let F_{k}(x) the k-th cyclotomic polynomial. Then, according to the Lemma of Hensel ([5]), F_{k}(x) is reducible over 𝔽_{p} (the field with p elements), if and only if it is reducible over ℤ_{p}, and the decompositions are compatible.

We need a proposition. It is a special case of a Theorem from [5] (p. 63).

Proposition 1.2. Let g, h ∈ ℤ_{p}[x] polynomials with no common divisors, so that g und h have no common divisors modulo p. Then there exist u, v ∈ ℤ_{p}[x], so that ug + vh = 1.

Proof. Because ℚ_{p}[x] is a principal ideal ring, there are polynomials u′, v′ over the field of p-adic numbers ℚ_{p}, so that u′g + v′h = 1. (Because of a theorem of Gauss u′ and v′ have also no common divisors over ℚ_{p}[x].) The polynomials can be chosen, so that the degree of v′ (repectively u′) smaller is than that of g (respectively h). Now let m be the smallest natural number, so that u := p^{m}u′ and v := p^{m}v′ are in ℤ_{p}[x]. Then ug + vh = p^{m} in ℤ_{p}. We consider this equation modulo p. If m > 0, then ug + vh ≡ 0 (mod p). But this is not possible, because u ≡ 0 (mod p) and v ≡ 0 (mod p) would be a contradiction to the choice of m. As the degree of v (respectively u) is smaller than that of g (respectively h), g and h must have a common divisor modulo p. That is a contradiction to the assumption. Thus m = 0.

We use the previous proposition to show another result.

Proposition 1.3. Let F_{k}(x) = g_{1}(x)·. . .·g_{n}(x) be the decomposition of the k-th cyclotomic polynomial in its irreducible factors over ℤ_{p}. Let h_{i}(x) := ∏_{j≠i} g_{j}(x) for i = 1, …, n. (If n = 1 set h_{1}(x) = 1.) Then there are polynomials u_{i}(x) ∈ ℤ_{p}[x] (i = 1, …, n), so that u_{1}(x)h_{1}(x) + … + u_{n}(x)h_{n}(x) = 1.

Proof. We prove with induction over m, that there exist polynomials v_{1}^{(m)}, …, v_{m}^{(m)} ∈ ℤ_{p}[x] for m ≤ n, so that

v_{1}^{(m)}h_{1} + … + v_{m}^{(m)}h_{m} = ∏_{j=m+1}^{n} g_{j}.

m = 1: Let v_{1}^{(1)} := 1

m ↦ m + 1: ∏_{j=1}^{m} g_{j} and g_{m+1} have no common divisor in ℤ_{p}[x] and 𝔽_{p}. According to Proposition 1.2 there exist s_{1}, s_{2} ∈ ℤ_{p}[x], so that

s_{1} · g_{m+1} + s_{2} · ∏_{j=1}^{m} g_{j} = 1.  (1)

We multiply (1) with ∏_{j=m+2}^{n} g_{j}, and replace

∏_{j=m+1}^{n} g_{j}  by  v_{1}^{(m)} · h_{1} + … + v_{m}^{(m)} · h_{m}.

We get

∏_{j=m+2}^{n} g_{j} = s_{1} · g_{m+1} · ∏_{j=m+2}^{n} g_{j} + s_{2} · ∏_{j=1}^{m} g_{j} · ∏_{j=m+2}^{n} g_{j}
= s_{1} · ∏_{j=m+1}^{n} g_{j} + s_{2} · h_{m+1}
= s_{1} · v_{1}^{(m)} · h_{1} + … + s_{1} · v_{m}^{(m)} · h_{m} + s_{2} · h_{m+1}

Let v_{i}^{(m+1)} := s_{1} · v_{i}^{(m)} for i ≤ m and v_{m+1}^{(m+1)} := s_{2}.

2 Construction of CA-groups

We turn back to our problem. We show, that fixed-point-free actions of order k can be described with help of the k-th cyclotomic polynomial.

Proposition 2.1. Let A ⋊ Z a CA-group, where A is an abelian p-group, k not divisible by p and Z cyclic of order k. Furthermore let a ∈ A and z a generating element of Z. Then there is a divisor g of the k-th cyclotomic polynomial F_{k} in ℤ_{p}[x], so that g(z) ∗ a = 0.

Proof. It is sufficient to show, that F_{k}(z)∗a = 0, when F_{k} is the k-th cyclotomic polynomial. As k is the order of z, we have (z^{k} − 1) ∗ a = 0. Moreover F_{k}(x) is a divisor of x^{k} − 1. Thus there is a polynomial H(x) ∈ ℤ_{p}[x], so that x^{k} − 1 = F_{k}(x) · H(x). We show, that H(z) acts injective on A. For that purpose it is sufficient to show, that if f(x) is an irreducible divisor of H(x), then f(z) acts injective on A. Thus let f(x) be an irreducible divisor of H(x), and 0 ≠ b ∈ A. Then f is a divisor of some l-th cyclotomic polynomial F_{l}, where l is a divisor of k. If f(z) ∗ b = 0, then (z^{l} − 1) ∗ b = 0. Thus b is fixed under the action of z^{l}. That is not possible, because z^{l} ≠ 1 acts fixed-point-free.

There is a converse to the Proposition. The following Theorem generalizes Theorem 3.1 from [3]. We will use the notation C_{k} for the cyclic group ℤ/(k · ℤ).

Theorem 2.2. Let k be a natural number and p a prime not dividing k. Let f(x) = x^{n} +λ_{n−1} ·x^{n−1} +…+λ_{1} ·x+λ_{0} be a divisor of the k-th cyclotomic polynomial over ℤ_{p} and A an abelian p-group. Then the group A^{(n)} ⋊ C_{k} with the action (here let z be a generating element of C_{k})

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

is a CA-group.

Proof. The proof is exactly the same as the proof of Theorem 3.1 in [3]. All we have to change is the argument 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(φ)(a_{1}, …, a_{n}) & = & (0, …, 0). \end{aligned}

The polynomials f and x^{l} − 1 have no common divisor in ℤ_{p}[x] and modulo p. Therefore there are (see Proposition 1.2) polynomials u(x), v(x) ∈ ℤ_{p}[x], so that

u(x) · f(x) + v(x) · (x^{l} − 1) = 1.

Therefore

(a_{1}, …, a_{n}) = (u(φ) · f(φ) + v(φ) · (φ^{l} − id))(a_{1}, …, a_{n})
= u(φ)(f(φ)(a_{1}, …, a_{n})) + v(φ)((φ^{l} − id)(a_{1}, …, a_{n}))
= (0, …, 0).

3 A Decomposition

Now let G = A⋊Z be a locally finite solvable CA-group. We decompose the k-th cyclotomic polynomial F_{k}(x) = g_{1}(x) · . . . · g_{n}(x) in its irreducible factors over ℤ_{p}. We define for i = 1, …, n,

A_{i} = A_{i}(z) := {a ∈ A; g_{i}(z) ∗ a = 0}.

Theorem 3.1. Let A ⋊ C_{k} be a locally finite CA-group, A an abelian p-group and z a generating element of C_{k}. Let F_{k}(x) = g_{1}(x) · . . . · g_{n}(x) be the decomposition of the k-th cyclotomic polynomial into its irreducible factors over ℤ_{p}. Define for i = 1, …, n: A_{i} := {a ∈ A; g_{i}(z) ∗ a = 0}.

Then the following holds. The subgroups A_{i} (i ≤ n) are invariant under the action of C_{k}, and A = A_{1} ⊕ … ⊕ A_{n}.

Proof. 1. The A_{i} are subgroups and invariant under the action.

That is obvious.

2. The A_{i} are disjoint.

Suppose a ∈ A_{i} ∩ ⟨⋃_{j≠i} A_{j}⟩. With 1.3 we get

\begin{aligned}a & = & 1 ∗ a & = & (u_{1}(z)h_{1}(z) + … + u_{n}(z)h_{n}(z)) ∗ a \\ & & & = & u_{1}(z)h_{1}(z) ∗ a + … + u_{n}(z)h_{n}(z) ∗ a. \end{aligned}

Because a ∈ A_{i}, we have h_{j}(z) ∗ a = 0 for j ≠ i. Hence a = u_{i}(z)h_{i}(z) ∗ a. On the other hand a ∈ ⟨⋃_{j≠i} A_{j}⟩, i.e. there are b_{j} ∈ A_{j} for j ≤ n with b_{i} = 0, so that a = b_{1} + … + b_{n}. We get

a = u_{i}(z)h_{i}(z) ∗ a = u_{i}(z)h_{i}(z) ∗ b_{1} + … + u_{i}(z)h_{i}(z) ∗ b_{n} = 0,

because for all j ≤ n we have u_{i}(z)h_{i}(z) ∗ b_{j} = 0.

3. A is generated by the A_{i}. Let a ∈ A. With 1.3 we get

a = 1 ∗ a = (u_{1}(z)h_{1}(z) + … + u_{n}(z)h_{n}(z)) ∗ a
= u_{1}(z)h_{1}(z) ∗ a + … + u_{n}(z)h_{n}(z) ∗ a
= h_{1}(z) ∗ (u_{1}(z) ∗ a) + … + h_{n}(z) ∗ (u_{n}(z) ∗ a)

and h_{i}(z) ∗ (u_{i}(z) ∗ a) ∈ A_{i}.

We want to decompose the blocks A_{i} into smaller ones. The A_{i} are modules over the ring R_{i} := ℤ_{p}[x]/I_{i} (let I_{i} be the ideal generated by g_{i}). The action of x on A_{i} corresponds to that of z. Remark, that R_{i} is isomorphic to the ring ℤ_{p} with an adjointed primitive k-th root of unit. Therefore all the R_{i} are isomorphic. R_{i} is a principal ideal domain, moreover the ideals in R_{i} are of the type (p^{l}) for some l ∈ ℕ. In what follows we will consider A_{i} as R_{i}-module. The A_{i} (for i = 1, …, n) are divisible R_{i}-modules, if and only if they are divisible as groups. This is because all elements not divisible by p are invertible in R_{i}.

Proposition 3.2. We assume the same conditions as in 3.1 and X a divisible R_{i}-submodule of A_{i}. Then X is a direct summand in A_{i}.

Proof. We remarked above, that R_{i} is a principal ideal domain. Divisible submodules over principal ideal domains are direct summands.

In the following proposition we need only “maximal divisibility”. We follow in the proof largely that for divisible submodules (see for example [4], Theorem 21.2).

Proposition 3.3. Prerequisites as in 3.1 and A = A_{1} ⊕ … ⊕ A_{n}. Let p^{m} be the exponent of A_{i} and C_{p^{m}}^{(s)} ≅ X ≤ A_{i}, so that X is a submodule of A_{i}. Then X is a direct summand (as a R_{i}-module).

Proof. Let Y be a maximal R_{i}-submodule with X ∩ Y = {0}. Suppose there is z ∈ A_{i} − (X + Y). Because of the maximality of Y we get X ∩ ⟨Y ∪ {z}⟩_{R_{i}} ≠ {0}. Thus I := {r ∈ R_{i}; rz ∈ X + Y } is a non-trivial ideal in R_{i}. Hence I is generated by p^{k} for some k ∈ ℕ. Therefore there are x ∈ X, y ∈ Y with x = y + p^{k}z. We show, that x is divisible by p^{k}, that means, there is x_{1} ∈ X, so that p^{k}x_{1} = x. We have p^{m−k}x = p^{m−k}y + p^{m}z = p^{m−k}y. Because X ∩ Y = {0}, we get p^{m−k}x = 0. If p^{t} is the order of x, then x is divisible by p^{m−t}. As m − k ≥ t, x is also divisible by p^{m−(m−k)} = p^{k}. Thus we can write p^{k}x_{1} = y + p^{k}z, respectively y = p^{k}(x_{1} − z). Now consider Y′ := ⟨Y ∪ {x_{1} − z}⟩_{R_{i}}. We claim, that X ∩ Y′ = {0}. To prove this, let x_{2} ∈ X, y_{2} ∈ Y and r ∈ R_{i}, so that x_{2} = y_{2} + r(x_{1} − z). Then rz ∈ X + Y, and hence r ∈ I. Therefore there is s ∈ R_{i}, so that r(x_{1} − z) = sp^{k}(x_{1} − z) = sy ∈ Y. Thus x_{2} = y_{2} + sy ∈ Y, which means x_{2} = 0. We found a contradiction, as Y ⫋ Y′, but we did choose Y maximal.

Remark. One can choose Y in the proof so, that Y contains a certain invariant subgroup, that is disjoint to X.

Definition 3.4. Let M be a R-module. We call M (directly) indecomposable, if there are no proper submodules A, B ≤ M, so that M = A ⊕ B.

Proposition 3.5. Let the conditions be as in 3.1, and R_{i} as defined above. Furthermore let A_{i} be finite. Then we can decompose A_{i} in a direct sum of indecomposable R_{i}-modules.

Proof. Let a be an element of maximal order in A_{i} and U the span of a as R_{i}-module (i.e. U = {r ∗ a; r ∈ R_{i}}). We claim, that U is indecomposable. Suppose U = X_{1} ⊕ X_{2}. Then J_{i} := {r ∈ R_{i}; r ∗ a ∈ X_{i}} for i = 1, 2 would be ideals in R_{i}, that are not contained in each other. But this is not possible in R_{i}. U is of the same type as X in 3.3. (Let m be the degree of g_{i}, then U is as group isomorphic to ⟨{a}⟩ ⊕ … ⊕ ⟨{a^{z^{m−1}}}⟩.) Thus U is a direct summand. Let A_{i} = U ⊕ Y, then we can find analogously a direct summand in Y. Inductively we get the desired decomposition.

In the proof of our main theorem (Theorem 4.1) we will have to extend a decomposition as in 3.5 to a decomposition in a group containing A ⋊ C_{k}. We will use the following proposition.

Proposition 3.6. Let A ⋊ C_{k} ≤ B ⋊ C_{k} be finite CA-groups, so that A ≤ B ≅ C_{p^{m}}^{(s)} are abelian p-groups. Let z be a generating element of C_{k}, so that g_{i}(z) ∗ b = 0 for all b ∈ B. Let A = U_{1} ⊕ … ⊕ U_{l} be a decomposition of A in non-trivial indecomposable R_{i}-modules. Then there are indecomposable R_{i}-modules V_{1}, …, V_{l} and a R_{i}-module V, so that U_{j} ≤ V_{j} (for j = 1, …, l) and B = V_{1} ⊕ … ⊕ V_{l} ⊕ V.

Proof. For j ∈ {1, …, l} let v_{j} ∈ B, so that v_{j} is of order p^{m}, and U_{j} ⊆ ⟨{v_{j}}⟩_{R_{i}}. (Such v_{j} exist, because if x is of maximal order in U_{j}, then U_{j} is generated by x. Let v_{j} be a suitable root of x.) Let V_{j} := ⟨{v_{j}}⟩_{R_{i}}. We claim, that ⟨V_{j}; j = 1, …, l⟩ = V_{1} ⊕ … ⊕ V_{l}. To prove this, suppose z_{1} + … + z_{l} = 0, so that z_{j} ∈ V_{j} for j = 1, …, l. Choose k minimal, so that p^{k}z_{j} ∈ U_{j} for all j = 1, …, l. Then p^{k}z_{j} = 0 (for j = 1, …, l), as A = U_{1} ⊕ … ⊕ U_{l}. Suppose k ≥ 1. Because z_{j} is in the span of v_{j}, there is a polynomial h(x) ∈ ℤ_{p}[x] of smaller degree than g_{i}, so that z_{j} = h(z) ∗ v_{j}. As p^{k} ∗ z_{j} = 0, the polynomial h(x) is divided by p^{m−k}. Thus h(x) = p^{m−k} · h′(x) for some polynomial h′(x) ∈ ℤ_{p}[x]. On the other hand we have p^{m−1}v_{j} ∈ U_{j}, and therefore

p^{k−1}z_{j} = p^{k−1}p^{m−k} · h′(z) ∗ v_{j} = h′(z) ∗ p^{m−1}v_{j} ∈ U_{j}.

That is a contradiction to the minimality of k. Thus k = 0 and z_{j} = 0 for j = 1, …, l. It follows with 3.3, that V_{1} ⊕ … ⊕ V_{l} is a direct summand. Hence there is some V, so that B = V_{1} ⊕ … ⊕ V_{l} ⊕ V.

We now have a decomposition of locally finite, abelian-by-finite CA-groups.

The Fitting subgroup A can be decomposed in the p-Sylow-groups A_{p} and these in blocks A_{1}, …, A_{n}. The blocks can be decomposed in direct summands, because invariant subgroups, which are maximal divisible in A_{i}, are direct summands in A_{i}.

Proposition 3.7. Let G = A ⋊ ⟨{z}⟩ be a finite solvable CA-group, A abelian, and k be the order of z. Then G is embeddable in a group B^{ϕ(k)} ⋊ ⟨{z}⟩, where the action of z is according to 2.2 and f = F_{k}, the k-th cyclotomic polynomial.

Proof. It is clear, that it suffices to consider the case, that A is a p-group. Let p^{s} be the exponent of A. According to 3.1 let F_{k}(x) = g_{1}(x) · . . . · g_{n}(x) and A = A_{1} ⊕ … ⊕ A_{n}. Let m := \frac{ϕ(k)}{n} (the degree of the polynomials g_{i}(x)). If we decompose A_{i} according to 3.5 in indecomposable R_{i}-modules, we get

A_{i} = C_{p^{i_{1}}} ^{m} ⊕ ⋯ ⊕ C_{p^{i_{l}}} ^{m}.

We write D_{j} := C_{p^{i_{j}}} ^{m}. We can embed D_{j} in E_{j} := C_{p^{s}}^{m} as an abelian group. Choose a ∈ D_{j} with order p^{i_{j}}. Let

b_{0} ∈ E_{j}  so that p^{s−i_{j}} ∗ b_{0} = a,
b_{1} ∈ E_{j}  so that p^{s−i_{j}} ∗ b_{1} = z ∗ a,
…
b_{m−1} ∈ E_{j} so that p^{s−i_{j}} ∗ b_{m−1} = z^{m−1} ∗ a.

It follows, that E_{j} is (as a group) isomorphic to

⟨{b_{0}}⟩ ⊕ ⋯ ⊕ ⟨{b_{m−1}}⟩.

Therefore one can define an action of z on E_{j} according to 2.2 with f = g_{i}. This action extends the action of z on D_{j}. If we do this for all j we get an embedding

A_{i} ⋊ ⟨{z}⟩ → (C_{p^{s}}^{m})^{r_{i}} ⋊ ⟨{z}⟩ for some r_{i} ∈ ℕ.

Let r := max_{1≤i≤n}{r_{i}} and B′_{i} := (C_{p^{s}}^{m})^{r}, then we get an embedding

A_{i} ⋊ ⟨{z}⟩ → B′_{i} ⋊ ⟨{z}⟩.

Now let B := C_{p^{s}}^{r}. When we decompose the group B^{ϕ(k)} ⋊⟨{z}⟩ (with the action according to 2.2 and f = F_{k}) according to 3.1, we get

B^{ϕ(k)} = B_{1} ⊕ ⋯ ⊕ B_{n},

where the B_{i} are R_{i}-modules. It is easy to see, that B_{i} ≅ B′_{i} as R_{i}-modules. Hence there is an embedding

A ⋊ ⟨{z}⟩ → B^{ϕ(k)} ⋊ ⟨{z}⟩.

4 A Characterization of the E.C. Groups in C

It is now possible to characterize e.c. C-groups. We already know, that an e.c. C-group is isomorphic to a group of kind

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

for a set of primes π and infinite cardinals κ_{p} for p ∈ π. The question is, which groups of this kind are existentially closed.

Theorem 4.1. For a locally finite CA-group G = A ⋊ Z with

A ≅ ⨁_{p∈π} Z_{p^{∞}}^{(κ_{p})} and Z ≅ ⨁_{q∈π′} Z_{q^{∞}}

are equivalent:

(i) G is existentially closed in the class of C-groups.

(ii) Every finite solvable CA-group of type π is embeddable in G.

(iii) For all p ∈ π, l ∈ ω and every π′-number k the group (C_{p}^{ϕ(k)})^{l} ⋊ C_{k} can be embedded in G. The action of C_{k} on a block C_{p}^{ϕ(k)} is according to Theorem 2.2 with f = F_{k}, the k-th cyclotomic polynomial.

(iv) For all p ∈ π and for every π′-number k the p-subgroup A^{(p)} of A can be decomposed in the following way. Let f(x) = g_{1}(x) · . . . · g_{n}(x) the decomposition of the k-th cyclotomic polynomial in its irreducible factors over ℤ_{p}. Furthermore let z be an element of order k in Z and A_{i} := {a ∈ A^{(p)}; g_{i}(z) ∗ a = 0}. Then for every i = 1, …, n : A_{i} ≠ {0}.

Proof. (i) ⇒ (iii). Let G be an e.c. C-group, p ∈ π, l ∈ ω, and k a π′-number. The Proof of the Amalgamation Theorem 5.1 in [3] shows, that there is a locally finite solvable CA-group H containing G and (C_{p}^{ϕ(k)})^{l} ⋊ C_{k}. The operations in this group can be described through an existential sentence, which is fullfilled in H. Thus the sentence must also to be true in G. That means, (C_{p}^{ϕ(k)})^{l} ⋊ C_{k} is embeddable in G.

(ii) ⇒ (iii). This is trivial.

(iii) ⇒ (ii). Let G′ = A′ ⋊ ⟨{z}⟩ be a finite solvable CA-group of type π. According to 3.7, G′ is embeddable in a group B^{ϕ(k)} ⋊ ⟨{z}⟩. As B is a finite abelian group, we have B = C_{p_{1}^{m_{1}}} ⊕ ⋯ ⊕ C_{p_{k}^{m_{k}}}. As (C_{p_{1}} ⊕ ⋯ ⊕ C_{p_{k}})^{ϕ(k)} ⋊ ⟨{z}⟩ is embeddable in G, and A is divisible, G′ must also be embeddable in G.

(iii) ⇒ (iv). U ⋊ C_{k} = (C_{p}^{ϕ(k)})^{l} ⋊ C_{k} is decomposable in (U_{1} ⊕ … ⊕ U_{n}) ⋊ C_{k}, where U_{i} := \{a ∈ U; g_{i}(z) ∗ a = 0\}. Every U_{i} has p^{\frac{ϕ(k)}{n} ·l} − 1 elements of order p (and thus is non-trivial) and is a subgroup of A_{i}.

(iv) ⇒ (i): Let ∃x̄ ψ(ḡ, x̄) be an existential sentence with parameters ḡ from G, that is true in a group H ≥ G. Let K be the finite subgroup of H, that is generated by ḡ and the fullfilling elements for x̄. Let U := G∩K. Let Z′ := C_{H}(Z). As Z is maximal as a locally cyclic π-group, it follows Z′ = Z. We may assume, that K ∩ Z ≠ {1}, because otherwise choose 1 ≠ z ∈ Z, and consider ⟨K ∪ {z}⟩ instead of K. Let A′ := F(H) be the Fitting subgroup of H (thus H = A′ ⋊ Z), D := F(K) and Y′ := K ∩ Z (thus K = D ⋊ Y′). Furthermore let C := F(U) and Y = U ∩ Z (then Y = Y′, as Y′ = K ∩ Z = K ∩ (G ∩ Z) = (K ∩ G) ∩ Z = U ∩ Z = Y). We get the following diagram.

[figure: 1; U = C ⋊ Y; G = A ⋊ Z; K = D ⋊ Y; H = A′ ⋊ Z]

Let k be the order of Y. Let p ∈ π, and C^{(p)} the p-subgroup of C, D^{(p)} that of D. Then there are according to 3.1 decompositions C^{(p)} = C_{1}^{(p)} ⊕ … ⊕ C_{n}^{(p)} and D^{(p)} = D_{1}^{(p)} ⊕ … ⊕ D_{n}^{(p)}, so that C_{i}^{(p)} ≤ D_{i}^{(p)} and C_{i}^{(p)} < A_{i} for i = 1, …, n. All A_{i} are divisible. (For a ∈ A_{i} let c ∈ A, so that p·c = a. Then c′ := u_{i}(z)h_{i}(z)∗c ∈ A_{i} and p·c′ = a.) Now let i ∈ {1, …, n}. We have to find a subgroup in A_{i}, that is isomorphic to D_{i}^{(p)}. The isomorphism shall fix C_{i}^{(p)}. We can assume, that D_{i}^{(p)} ≅ C_{p^{m}}^{(s)} for some m, s ∈ ω. That is possible, because H is embeddable in a divisible C-group. With 3.5 we get a decomposition C_{i}^{(p)} = U_{1}⊕…⊕U_{l} in indecomposable R_{i}-modules. According to 3.6 there are indecomposable R_{i}-modules V_{1}, …, V_{l} and a R_{i}-module V, so that U_{j} ≤ V_{j} (for j = 1, …, l) and D_{i}^{(p)} = V_{1} ⊕ … ⊕ V_{l} ⊕ V. Now let v_{j} be a generating element of V_{j} (j = 1, …, l) as a R_{i}-module. Let u_{j} = p^{n_{j}}v_{j}, with n_{j} minimal, so that u_{j} ∈ U_{j} and w_{j} ∈ A_{i} with u_{j} = p^{n_{j}}w_{j}. The mapping

V_{1} ⊕ … ⊕ V_{l} → ⟨{w_{1}}⟩ ⊕ … ⊕ ⟨{w_{l}}⟩
defined by  v_{j} ↦ w_{j}  (for j = 1, …, l)

is a R_{i}-isomorphism. Because of (iv) there are infinitely many elements of order p in A_{i}. In particular there is a finite R_{i}-submodule of exponent p in A_{i}, which is disjoint to ⟨{w_{1}}⟩ ⊕ … ⊕ ⟨{w_{l}}⟩ and has cardinality ≥ |V |. This submodule is embeddable in a finite submodule X of exponent p^{|V |} in A_{i}, which is disjoint to ⟨{w_{1}}⟩⊕…⊕⟨{w_{l}}⟩. Obviously V is isomorphic to a submodule W of this module. Let W be a submodule of A_{i}, that is isomorphic to V and disjoint to ⟨{w_{1}}⟩ ⊕ … ⊕ ⟨{w_{l}}⟩, then one gets an embedding

V_{1} ⊕ … ⊕ V_{l} ⊕ V → ⟨{w_{1}}⟩ ⊕ … ⊕ ⟨{w_{l}}⟩ ⊕ W ↪ A_{i}

The mapping fixes C_{i}^{(p)}. One can do the same for all i = 1, …, n and analogously for all prime numbers p ∈ π. Altogether one gets an embedding of K in G. That means, that the sentence ∃x̄ ψ(ḡ, x̄) is fullfilled in G. Hence G is existentially closed.

We get a second charcterization of e.c. C-groups.

Corollary 4.2. Let G be a C-group. Then the following is equivalent:

(i) G is a e.c. C-group.

(ii) G is divisible, and there is π ⊆ ℙ, so that every finite solvable CA-group of type π is embeddable in G.

Proof. (i) ⇒ (ii): The divisibility folllows from Theorem 6.1 from [3]. It follows from Theorem 4.1, that every finite solvable CA-group of type π is embeddable.

(ii) ⇒ (i): Let A := [G, G]. Then there is a locally cyclic subgroup Z ≤ G, so that G = A ⋊ Z. It is clear, that G has to be of type π. For every p ∈ π there must be infinitely many elements of order p in A. That is because for every π′-number k the group (C_{p})^{ϕ(k)}⋊C_{k} is embeddable in G. As G and therefore also A are divisible, it follows A ≅ ⨁_{p∈π} Z_{p^{∞}}^{(κ_{p})}. It is also easy to see, that Z ≅ ⨁_{q∈π′} Z_{q^{∞}}. The rest follows with 4.1.

We now have determined the structur of existentially closed C-groups quite clearly. The question, that poses itself, is how many non-isomorphic e.c. C-groups of type π exist for fixed π.

5 A Tree

Our next aim will be to determine, how many non-isomorphic countable e.c. C-groups of type π exist. For that purpose we will construct a tree. The idea is to continue the decompositon according to 3.1. Let G = A ⋊ Z be an e.c. C-group of type π, and let k_{1}, k_{2} be π′-numbers, so that k_{1} divides k_{2}. Let z_{i} ∈ Z with order k_{i} for i = 1, 2. Then the decomposition of A with z_{2} is a refinement of the decomposition with z_{1}. If we refine the decomposition step by step, we receive a tree. We also can describe this tree with cyclotomic polynomials. We prefer this representation, because it is independent of the chosen e.c. C-group.

Definition 5.1. Let π be a non-empty set of primes, and p ∈ π. Let {q_{n}; n ∈ ω} be an enumeration of π′, so that every prime occurs infinitely many times. Let k_{0} := 1 and k_{n} := ∏_{i<n} q_{i} for n ≥ 1.

We define L_{n} as the set of the normed irreducible divisors of the k_{n}-th cyclotomic polynomial over ℤ_{p} (respectively ℚ_{p}). We define on T := ⋃_{n∈ω} L_{n} the following relation.

g ≤ h ⇔ g ∈ L_{n}, h ∈ L_{m}, so that n ≤ m,
and h(x) divides g(x^{\frac{km}{kn}}).

Let T = T_{π}^{(p)} := ⟨T, ≤⟩.

Proposition 5.2. T is a tree, and L_{n} is the n-th level of T .

Proof. We show first, that T is partially ordered.

“≤” is reflexive. That is obvious.

“≤” is anti-symmetric. Let g ≤ h and h ≤ g, g ∈ L_{n}, h ∈ L_{m}. Then n = m. That means, g(x) divides h(x), and h(x) divides g(x). As g and h are irreducible and normed, it follows g = h.

“≤” is transitive. Let g_{1} ≤ g_{2} and g_{2} ≤ g_{3} and g_{i} ∈ L_{n_{i}} for i = 1, 2, 3. Let r_{1} := \frac{k_{n_{2}}}{k_{n_{1}}} , r_{2} := \frac{k_{n_{3}}}{k_{n_{2}}} , and s := \frac{k_{n_{3}}}{k_{n_{1}}} . Then there are polynomials h_{1}, h_{2} ∈ ℤ_{p}[x], so that

g_{1}(x^{r_{1}}) = g_{2}(x) · h_{1}(x) and g_{2}(x^{r_{2}}) = g_{3}(x) · h_{2}(x).

It follows

g_{1}(x^{s}) = g_{1}((x^{r_{2}})^{r_{1}}) = g_{2}(x^{r_{2}}) · h_{1}(x^{r_{2}}) = g_{3}(x) · h_{2}(x) · h_{1}(x^{r_{2}}).

Hence g_{3}(x) is a divisor of g_{1}(x^{s}) and thus g_{1} ≤ g_{3}.

Now we prove, that a element h ∈ L_{m} has exactly one predecessor in every level L_{n} with n < m.

There is g ∈ L_{n}, so that g ≤ h. Let K be the splitting field of h over ℚ_{p} and a ∈ K a zero of h. Then the multiplicative order of a is k_{m}. Let r := \frac{k_{m}}{k_{n}}. The element b := a^{r} has therefore multiplicative order k_{n}, and is hence a zero of the k_{n}-th cyclotomic polynomial. Thus there is g ∈ L_{n} with g(b) = 0. That means, a is a zero of g(x^{r}). As a is also a zero of h, it follows, that h(x) is a divisor of g(x^{r}), because h is irreducible.

Let g_{1}, g_{2} ∈ L_{n} with g_{1}, g_{2} ≤ h. Then g_{1} = g_{2}. Let again K be the splitting field of h over ℚ_{p} and a ∈ K a zero of h. Let again r := \frac{k_{m}}{k_{n}}. Then h(x) is a divisor of g_{1}(x^{r}) and g_{2}(x^{r}). Thus a is a zero of g_{1}(x^{r}) and g_{2}(x^{r}). The element b := a^{r} is hence a zero of g_{1}(x) and g_{2}(x). If g_{1} ≠ g_{2}, then they can have no common zero. Hence g_{1} = g_{2}.

Thus the set of elements below an element h is well-ordered. The level S_{0} consists only of the polynomial x − 1. This is the minimal element of the tree.

We can consider three interpretations of the tree T . As an order on the irreducible divisors of the k_{n}-th cyclotomic polynomials, as a decomposition of the Sylow-p-group of an e.c. C-group of type π, or as an order on the k_{n}-th roots of unity in ℚ̃_{p}.

Let us now consider the relationship to the decomposition of the p-subgroup of an e.c. C-group of type π. Let G be an e.c. C-group of type π and A^{(p)} the p-Sylow-group of G. Let z_{n} ∈ G with order k_{n} and z_{n+1} ∈ G with z_{n+1}^{q_{n}} = z_{n}. Furthermore let g ∈ L_{n}, h ∈ L_{n+1}. Then

V := {a ∈ A^{(p)}; h(z_{n+1}) ∗ a = 0} ⊆ U := {a ∈ A^{(p)}; g(z_{n}) ∗ a = 0} ⇔ g ≤ h.

That means, we can choose {z_{n}; n ∈ ω} ⊆ G, so that z_{n} has order k_{n} and z_{n+1}^{q_{n}} = z_{n} for all n ∈ ω, and identify g ∈ L_{n} with U := {a ∈ A^{(p)}; g(z_{n}) ∗ a = 0}. The ordering of T is then realized by the U in A^{(p)}. Remark, that because of 4.1 none of the U is trivial.

We turn to the interpretation as an order on the roots of unity. Let n ≤ m, g an irreducible divisor of the k_{n}-th and h an irreducible divisor of the k_{m}-th cyclotomic polynomial. Let r := \frac{k_{m}}{k_{n}} and ℚ̃_{p} the algebraic closure of ℚ_{p}. Then

\begin{aligned}g ≤ h & ⇔ & ∀z ∈ ℚ̃_{p} : h(z) = 0 → g(z^{r}) = 0 \\ & ⇔ & ∃z ∈ ℚ̃_{p} : h(z) = 0 ∧ g(z^{r}) = 0. \end{aligned}

Let E = E(π′) := {z ∈ ℚ̃_{p}; There is a π′-number k with z^{k} = 1} the set of π′-roots of unity. As a multiplicative group E is isomorphic zu ⨁_{q∈π′} Z_{q^{∞}}. On E we can define an equivalence relation R. For z_{1}, z_{2} ∈ E let

z_{1} ∼ z_{2} ⇔ There is a π′-number k and an irreducible divisor g

of the k-th cyclotomic polynomial with g(z_{1}) = g(z_{2}) = 0.

The equivalence classes of this relation can be mapped one-to-one on the set of irreducible divisors of the k-th cyclotomic polynomial for π′-numbers k. An order on E/R can be defined in the following way.

[z_{1}] ≤ [z_{2}] ⇔ z_{1} has order r, z_{2} has order s,
r divides s, and z_{2}^{\frac{s}{r}} ∈ [z_{1}].

The order of the tree is a restriction of this order on the subset

{[z]; The order of z is k_{n} for some n ∈ ℕ}.

Now let K be the closure of ℚ_{p} under the π′-roots of unity, i.e. K = ℚ_{p}(E). We consider E as a subgroup of the multiplicative group of the field K. We define A := Aut(E). We consider also a subgroup of A. We consider first the set of field automorpisms of K, that fix ℚ_{p}, i.e. the Galois group G(K : ℚ_{p}). Then we define U := {ϕ ↿_{E} ; ϕ ∈ G(K : ℚ_{p})} as a subgroup of A. The equivalence classes of the relation R are exactly the orbits of E unter U. The structure of A is now easy to determine. As E ≅ ⨁_{q∈π′} Z_{q^{∞}} and Aut(Z_{q^{∞}}) = ℤ_{q}^{∗} (the group of unities of q-adic integers), it follows

A ≅ ∏_{q∈π′} ℤ_{q}^{∗}.

(Here let ∏ be the cartesian product.) Hence A is abelian. One gets a homomorphism

\begin{aligned}f : & A/U & → & Aut(T ) \\ & ϕ · U & ↦ & ϕ′ : [z] ↦ [ϕ(z)]. \end{aligned}

As A is transitive on the sets of primitive k-th roots of unity, Aut(T ) is also transitive on the single levels. In other words, we proved the following proposition.

Proposition 5.3. T is homogeneous, i.e. given two elements g_{1}, g_{2} of a level, then g_{1} can be mapped on g_{2} by an isomorphism of the tree.

We will need a number-theoretic proposition.

Proposition 5.4. Let π be a set of primes, so that π′ := ℙ − π is finite, and p ∈ π. Let

k := \{ \begin{matrix}∏_{q∈π′} q & if 2 ∉ π, \\ 2 · ∏_{q∈π′} q & if 2 ∈ π. \end{matrix}

Let k′ be a multiple of k and s the smallest natural number, so that k′ divides p^{s} − 1, and let r be the largest π′-number, that divides p^{s} − 1. Then q · r is the largest π′-number, that divides p^{q·s} − 1. Moreover q · s is the smallest number t, so that q · r is a divisor of p^{t} − 1.

Proof. Let x ∈ ℕ, so that p^{s} = 1 + xr. It follows,

p^{qs} = (1 + xr)^{q} = ∑_{i=0}^{q} \binom{q}{i}x^{i}r^{i},

respectively

p^{qs} − 1 = ∑_{i=1}^{q} \binom{q}{i}x^{i}r^{i}.

It follows directly, that qr divides qxr, r^{2}, …, r^{i}, …, and thus also ∑_{i=1}^{q} \binom{q}{i}x^{i}r^{i}. That means, qr divides p^{qs} − 1. Now suppose, that there is q′ ∈ π′, so that q′qr is a divisor of p^{qs} − 1. Suppose q > 2, then q′qr is a divisor of \frac{q(q−1)}{2}r^{2} and of r^{i} for i > 2. If q = 2, then according to our assumption 4 is also a divisor of r. It follows, that q′qr is a divisor of r^{i} for i ≥ 2. Hence in both cases q′qr must be a divisor of qxr, which means, that q′ is a divisor of x. That is a contradiction, because x is obviously a π-number. Now let t be minimal, so that qr is a divisor of p^{t} − 1. According to the assumption s is the smallest number, so that r is a divisor of p^{s} − 1, and hence s < t. Moreover s is a divisor of t, and t a divisor of qs. That is only possible for t = qs.

Lemma 5.5. The tree T has finitely many branches, if π′ is finite, and 2^{ℵ_{0}} branches, if π′ is infinite.

Proof. To get a survey of the branches of T , it is most convenient, to examine the width of the tree. We make the following definitions.

l_{n} :=|L_{n}|, the number of irreducible divisors of F_{k_{n}} over ℤ_{p},
α_{n} := min{l ∈ ℕ; k_{n} divides p^{l} − 1}.

The number l_{n} can be calculated with the degree of an irreducible divisor. Every irreducible divisor of the k_{n}-th cyclotomic polynomial over ℚ_{p} has degree α_{n}. That is because the field GF(p^{α_{n}}) with p^{α_{n}} elements is the smallest field of characteristic p, that contains an element of multiplicative order k_{n} (and therefore the splitting field of the k_{n}-th cyclotomic polynomial over 𝔽_{p}). Now let ϕ be the Euler function, then the number of irreducible divisors is l_{n} := \frac{ϕ(k_{n})}{α_{n}}. We want to examine, in which cases l_{n+1} > l_{n}. It is clear, that α_{n} ≤ α_{n+1}. Moreover α_{n} is a divisor of α_{n+1}. That is because the splitting field GF(p^{α_{n+1}}) of the k_{n+1}-th cyclotomic polynomial is an extension of GF(p^{α_{n}}).

We have to consider several cases.

Case 1. q_{n} divides k_{n}.

Then ϕ(k_{n+1}) = q_{n} · ϕ(k_{n}). Moreover α_{n+1} is a divisor of q_{n} · α_{n}. (That is analogous to Proposition 5.4.)

Subcase 1.a. α_{n+1} = α_{n}

Then k_{n+1} is already a divisor of p^{α_{n}} − 1. It follows l_{n+1} > l_{n}.

Subcase 1.b. α_{n+1} = q_{n} · α_{n}

Then l_{n+1} = \frac{ϕ(k_{n+1})}{α_{n+1}} = \frac{q_{n}·ϕ(k_{n})}{q_{n}·α_{n}} = \frac{ϕ(k_{n})}{α_{n}} = l_{n}.

Case 2. q_{n} does not divide k_{n}. Then ϕ(k_{n+1}) = (q_{n} − 1) · ϕ(k_{n}).

Subcase 2.a. α_{n} even and q_{n} > 2.

k_{n+1} = q_{n}·k_{n} divides p^{\frac{(qn−1)·αn}{2}} −1, because (p^{\frac{αn}{2}} )^{q_{n}−1} ≡ 1 (mod q_{n}). Thus α_{n+1} is a divisor of \frac{(q_{n}−1)·α_{n}}{2} . We get l_{n+1} = \frac{ϕ(k_{n+1})}{α_{n+1}} ≥ \frac{(q_{n}−1)·ϕ(k_{n})}{\frac{(qn−1)·αn}{2}} = \frac{2·ϕ(k_{n})}{α_{n}} > l_{n}.

Subcase 2.b. α_{n} odd or q_{n} = 2.

From l_{n+1} = l_{n} it follows α_{n+1} = (q_{n} − 1) · α_{n}. Thus α_{n+1} is even for q_{n} > 2.

Else l_{n+1} > l_{n}.

Now let π′ finite. Then for sufficient large n only Case 1 can occur. We show, that Subcase 1.a can happen only finitely often. Define k′ := k, s and r like in Proposition 5.4. Let m be the smallest number, so that r is a divisor of k_{m}. It follows inductively from Proposition 5.4, that α_{n} = \frac{k_{n}}{r} · s for n ≥ m. Hence l_{n} = l_{m} for n ≥ m.

For infinite π′ Case 2 occurs infinitely often. It suffices to consider this case for q_{n} > 2. If l_{n+1} = l_{n} and q_{n} > 2 in Subcase 2.b, then α_{m} is even for m > n. Hence we get for sufficient large n always l_{n+1} > l_{n}, if Case 2 occurs.

6 The Number of Non-Isomorphic E.C. C-Groups

Theorem 6.1. If π′ is infinite, then there are 2^{ℵ_{0}} non-isomorphic countable e.c. C-groups of type π.

Proof. Let p ∈ π, and T be tree defined in 5.1. Let S = ⟨S, ≤⟩ be a subtree of T without blind alleys and with the property, that for all a, b, c ∈ T holds, that if a ≤ b ≤ c and a, c ∈ S, then b ∈ S. (In other words, every branch of S shall be a branch of T .) Then we can construct a model-theoretic type p^{S}(x) in the following way. Let n ∈ ω and g_{i_{1}}, …, g_{i_{m}} the irreducible divisors of the k_{n}-th cyclotomic polynomial, that lie in S. Let g(x), h_{l}(x) ∈ ℤ[x], so that g(x) ≡ ∏_{j=1}^{m} g_{i_{j}}(x) (mod p) and h_{l}(x) ≡ ∏_{j≠l} g_{i_{j}}(x) (mod p) for l = 1, …, m. Let ϕ_{n}^{S}(x) be the formula

ϕ_{n}^{S}(x) :≏ x^{p} = 1 ∧ ∃y : y^{k_{n}} = 1 ∧ g(y) ∗ x = 1 ∧ ⋀_{j=1}^{m} h_{j}(y) ∗ x ≠ 1.

For g(y) = y^{m} + a_{m−1}y^{m−1} + … + a_{0} the term g(y) ∗ x shall be an abbreviation for

(y^{m})^{−1} · x · y^{m} · ((y^{m−1})^{−1} · x · y^{m−1})^{a_{m−1}} · . . . · x^{a_{0}}.

Set

p^{S}(x) := {ϕ_{n}^{S}(x); n ∈ ω}.

We have to show, that p^{S}(x) is consistent. Let n < m. From G ⊧ ϕ_{m}^{S}(a) for some a ∈ G it follows, that G ⊧ ϕ_{n}^{S}(a). Let G be a C-group and a ∈ G, so that G ⊨ ϕ_{m}^{S}(a). Let z a fullfilling element for y in ϕ_{m}^{S}(a). Then G ⊨ ϕ_{n}^{S}(a), as z^{k_{m}/k_{n}} is a fullfilling element for y, because S is a subtree. The formula ϕ_{n}^{S}(x) is realized in every e.c. C-group. Let z ∈ Z with order k_{n} and A^{(p)} = A_{1} ⊕ … ⊕ A_{l} the decomposition of the p-Sylow-group according to 3.1. Choose 0 ≠ a_{i} ∈ A_{i} with order p and set a := a_{i_{1}} + … + a_{i_{m}}. Then G ⊨ ϕ_{n}^{S}(a), as z (inserted for y) fullfills the sentence.

Now let S_{1} = ⟨S_{1}, ≤⟩ and S_{2} = ⟨S_{2}, ≤⟩ be subtrees, so that there is n ∈ ω with |S_{1}∩L_{n}| ≠ |S_{2} ∩ L_{n}|. Then p^{S_{1}}(x) and p^{S_{2}}(x) cannot be realized by the same element. Suppose G = A ⋊ Z is an e.c. C-group of type π, and a ∈ G realizes p^{S_{1}}(x) and p^{S_{2}}(x). Then there is z_{i} ∈ Z, so that z_{i} is a fullfilling element for ϕ_{n}^{S_{i}}[a] (i = 1, 2). As z_{1} and z_{2} have the same order, there is a group automorphism of Z, which maps z_{1} on z_{2}. The induced tree automorphism of T then maps S_{1} ∩ L_{n} on S_{2} ∩ L_{n}, a contradiction.

Let σ := {n ∈ ω; a_{n} < a_{n+1}}, where a_{n} is defined as in Lemma 5.5. For τ ⊆ σ we define a subtree S_{τ} recursively as follows. Suppose, the m-th level L_{τ,m} of the tree is constructed for m ≤ n. If n ∈ τ, then the n + 1-th level L_{τ,n+1} consists of all elements, which are in T successors of an element of L_{τ,n}. If n ∉ τ choose for every element in L_{τ,n} exactly one successor. For any two of these subtrees there is obviously always a n ∈ ω, so that the n-th level has distinct cardinality. As π′ is infinite, σ is also infinite. It follows, that there are 2^{ℵ_{0}} different types (over the empty set) in an e.c. C-group of type π. For each of these types there is a countable e.c. C-group of type π, in which the type is realized. On the other side each countable e.c. C-group can of course only realize countably many types.

If π′ is finite, one can also show, that there are 2^{ℵ_{0}} non-isomorphic countable e.c. Cgroups of type π. We have to introduce some kind of dimension. Let G = A ⋊ Z be a C-group of type π. Let (z_{n})_{n∈ω} be a sequence of elements in Z, so that z_{n} has order k_{n}, and z_{m}^{\frac{km}{kn}} = z_{n} holds for n < m let. The tree T constructed as in 5.1 has only finitely many branches. Thus one can decompose the p-subgroup A^{(p)} of A in the following manner. Let Z_{i} := (g_{n}^{(i)})_{n∈ω} be a branch in T . Set A^{(i)} := {a ∈ A^{(p)}; ∀n ∈ ω : g_{n}^{(i)}(z_{n}) ∗ a = 0}. One gets a decomposition

A^{(p)} = A^{(1)} ⊕ … ⊕ A^{(m)}.

We now have to consider, how “large” these blocks are as ℤ_{p}[Z]-modules.

To this aim we consider U_{i} := {a ∈ A^{(i)}; p · a = 0}. Let K = 𝔽_{p}(E) the closure of 𝔽_{p} under the k-th cyclotomic polynomials for all π′-numbers k. We claim, that U_{i} is a K-vector space. It is clear, that U_{i} is a module over the group ring R := 𝔽_{p}Z. We can define the ideal

I_{i} := {r ∈ R; ∀u ∈ U_{i} : r ∗ u = 0}

in R. I_{i} is a maximal ideal in R. To show that, let s ∈ R − I_{i}. As Z is locally cyclic, one can find n ∈ ω, z ∈ Z and λ_{l} for l ≤ n, so that s = λ_{n}z^{n} + … + λ_{1}z + λ_{0}. Here z can be chosen, so that z = z_{m} for some m ∈ ω. Let f(x) be the polynomial of smallest degree in 𝔽_{p}[x], so that f(z) ∗ u = 0 for all u ∈ U_{i}. According to the assumption f is irreducible and therefore has no common divisor with λ_{n}x^{n} + … + λ_{1}x + λ_{0}. Thus there exist polynomials u, v ∈ 𝔽_{p}[x], so that 1 = u(x)f(x) + v(x)(λ_{n}x^{n} + … + λ_{1}x + λ_{0}). It follows, that 1 lies in the ideal generated by I_{i} and s. Hence F_{i} := R/I_{i} is a field and U_{i} in natural manner a F_{i}-vector space. Moreover F_{i} is isomorphic to the field K we described above.

The dimension of U_{i} over K can be finite or infinite in e.c. C-groups, as one can easily verify by Theorem 4.1. As a conclusion we get the following theorem.

Theorem 6.2. Let π ≠ ∅ be a set of primes, which does not contain all primes. Then there are 2^{ℵ_{0}} non-isomorphic countable e.c. C-groups of type π.

Proof. For infinite π′ the claim was proven in 6.1.

Now let π′ be finite. Then π is infinite. Like above one gets for each p ∈ π a decomposition of the p-subgroup in finitely many direct summands. We can attach to each summand a dimension, which can be finite or countable. For any σ ⊆ π there exists a e.c. C-group G_{σ} of type π, so that for p ∈ σ the dimension of each direct summand in the decomposition is 1, and for p ∈ π − σ the dimension is ω. The so constructed 2^{ℵ_{0}} groups are obviously pairwise non-isomorphic.

7 An Example

We want to make an example for the decomposition. Suppose you want to know, how many finite solvable CA-groups of order 147 = 3 · 7^{2} there are. Of course there are the abelian groups C_{3} ⊕ C_{49} and C_{3} ⊕ C_{7} ⊕ C_{7}. Now let us turn to the non-abelian groups. It is clear, that the Fitting subgroup must contain the elements of order 7. Thus |F(G)| = 49. A complement Z to F(G) is isomorphic to C_{3}. We consider the cyclotomic polynomial F_{3}(x) = x^{2} + x + 1 over 𝔽_{7}. We get F_{3}(x) = (x + 5) · (x + 3) (mod 7). Set g_{1}(x) := x + 5 and g_{2}(x) := x + 3. Let g′_{i} be the corresponding divisors of F_{3} in ℤ_{p}[x] (i.e. g′_{i}(x) ≡ g_{i}(x) (mod 7)). We get two cases.

Case 1. F(G) ≅ C_{7} ⊕ C_{7}.

Choose a generating element z ∈ Z and decompose F(G) according to 3.1.

Subcase 1. A_{1} ≅ A_{2} ≅ C_{7}.

Subcase 2. A_{1} = F(G) or A_{2} = F(G). These cases are isomorphic (choose z^{2} instead of z).

Case 2. F(G) ≅ C_{49}.

Then A_{1} = F(G) or A_{2} = F(G).

Thus we get three non-abelian groups.

Now let π = {7}. Let us take a look how T_{π}^{(7)} is constructed. We will take here the irreducible divisors in 𝔽_{7}[x]. Suppose q_{0} = 3, q_{1} = q_{2} = q_{3} = 2 in the enumeration of the primes of π′. We have

F_{3}(x) = x^{2} + x + 1 ≡ (x + 5) · (x + 3) (mod 7),
F_{6}(x) = x^{2} − x + 1 ≡ (x + 4) · (x + 2) (mod 7),
F_{12}(x) = x^{4} − x^{2} + 1 ≡ (x^{2} + 4) · (x^{2} + 2) (mod 7),
F_{24}(x) = x^{8} − x^{4} + 1 ≡ (x^{2} + x + 4) · (x^{2} + 2x + 2)
·(x^{2} + 5x + 2) · (x^{2} + 6x + 4) (mod 7).

The tree looks as follows.

[figure: x − 1; x + 5; x + 3; x + 4; x + 2; x^{2} + 4; x^{2} + 2; x^{2} + 2x + 2; x^{2} + 5x + 2; x^{2} + x + 4; x^{2} + 6x + 4]

We can now examine, how many non-abelian CA-groups exist, that are isomorphic to (C_{7} ⊕ C_{7}) ⋊ C_{24}. We decompose C_{7} ⊕ C_{7} = A_{1} ⊕ A_{2} ⊕ A_{3} ⊕ A_{4} according to the irreducible divisors of F_{24} and a generating element z ∈ C_{24}.

Case 1. A_{1} ≅ C_{7} ⊕ C_{7}.

Isomorphic are the cases A_{j} ≅ C_{7} ⊕ C_{7} for j = 2, 3, 4.

Case 2. A_{1} ≅ A_{2} ≅ C_{7}.

Isomorphic is the case A_{3} ≅ A_{4} ≅ C_{7}.

Case 3. A_{1} ≅ A_{3} ≅ C_{7}.

Isomorphic is the case A_{2} ≅ A_{4} ≅ C_{7}.

Case 4. A_{1} ≅ A_{4} ≅ C_{7}.

Isomorphic is the case A_{2} ≅ A_{3} ≅ C_{7}.

Thus there are four non-isomorphic non-abelian groups of this type.

8 Some Remarks

There are still some questions left about the model theory of e.c. C-groups. For example one could ask, if two e.c. C-groups of type π are elementarily equivalent (i.e. they fullfill he same first-order sentences). If π is the set of all primes, then this is obviously true. But there are cases, where this is not true, as we were able to show in [2]. Of course two e.c. C-groups of type π must fullfill the same existential sentences, as any finite solvable CA-group of type π is embeddable.

Another question concerns the stability of the e.c. C-groups (in the sense of the stability theory of Morley and Shelah). If π′ is infinite, then the decomposition like in Chapter 3 supplies a descending chain of definable subgroups. According to the definable chain conditions in stable groups (see for example [1], p. 92), these groups cannot be superstable. In the case, that π is finite, we found something like a dimension to count the number of models. For the number of e.c. C-groups of type π in the cardinality ℵ_{α} one gets |α + ω|^{ℵ_{0}}. This could be seen as a hint, that these groups are superstable or even ω-stable. However, if we take the first-order theory T of an e.c. C-group of type π, there are of course models of T, in which the Fitting subgroup, as well as any complement to the Fitting subgroup contain elements of infinite order. Thus there are much more models.

References

[1] J. T. Baldwin: Fundamentals of Stability Theory. Perspectives in Mathematical Logic, Springer Verlag, Berlin, 1988.

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

[3] A. Ecker: Existentially closed locally finite CA-groups. Preprint.

[4] L. Fuchs: Infinite Abelian Groups. Academic Press, New York, 1970.

[5] K. Hensel: Theorie der algebraischen Zahlen. Teubner, Leipzig-Berlin, 1908.