aa r X i v : . [ m a t h . G R ] J a n J ´ONSSON GROUPS OF VARIOUS CARDINALITIES
SAMUEL M. CORSON
Abstract.
A group G is J´onsson if ∣ H ∣ < ∣ G ∣ whenever H is a proper sub-group of G . Using an embedding theorem of Obraztsov it is shown that thereexists a J´onsson group G of infinite cardinality κ if and only if there existsa J´onsson algebra of cardinality κ . Thus the question as to which cardinalsadmit a J´onsson group is wholly reduced to the well-studied question of whichcardinals are not J´onsson. As a consequence there exist J´onsson groups of arbi-trarily large cardinality. Another consequence is that the infinitary edge-orbitconjecture of Babai is true. Introduction
A group G is J´onsson (named for Bjarni J´onsson) if every proper subgroup of G has cardinality strictly less than that of G . Finite groups are J´onsson and countablyinfinite examples include the quasi-cyclic groups Z ( p ∞ ) and the Tarski monstersconstructed by Ol’shanskii ([8], [9, Theorem 28.1]). A J´onsson group of cardinality ℵ was constructed by Shelah in [10] and further examples of J´onsson groups ofcardinality ℵ with striking properties were later obtained by Obraztsov ([6], [9,Corollary 35.4]).There are known to exist J´onsson groups of cardinality ℵ , ℵ , . . . (see [5, The-orem E]) but there do not appear to be any of cardinality ℵ ω or higher in theliterature which are constructed only from the standard ZFC axioms. It is knownthat if there is a J´onsson group of cardinality κ then there exists one of cardinaity κ + [5], where κ + denotes the successor cardinal to κ (the smallest cardinal which isstrictly greater than κ ). If 2 λ = λ + then there exists a J´onsson group of cardinality λ + [10], but failure of this equality at every cardinal λ > algebra is an ordered pair ( A, F ) where A is a set and F is a collection of finitaryoperations on A . An algebra ( A, F ) is J´onsson if F is at most countable and anyproper subalgebra of ( A, F ) , that is- any proper subset of A which is closed underapplications of the operations in F is of cardinality strictly less than ∣ A ∣ . The mainresult is the following. Theorem 1.
For κ an infinite cardinal, there exists a J´onsson algebra of cardinality κ if and only if there exists a simple, torsion-free J´onsson group of cardinality κ .Naturally a J´onsson group is a J´onsson algebra, so we are only concerned withproducing a J´onsson group provided the existence of a J´onsson algebra. FromTheorem 1 one obtains many new examples of J´onsson groups. Mathematics Subject Classification.
Key words and phrases.
J´onsson group, J´onsson algebra.This work was supported by the Heilbronn Institute for Mathematical Research, Bristol, UK..
Corollary 1.
There is a J´onsson group of cardinality κ when(1) κ is the successor of a regular cardinal [14];(2) κ = λ + , where λ is singular and not a limit of weakly inaccessible cardinals[11]; or(3) κ = ℶ + ω [12].For example, there exist J´onsson groups of cardinality ℵ , ℵ , ℵ ω + , ℵ ω + , and ℵ ω + . Since the successor of a cardinal is regular, we obtain from (1) that foreach infinite cardinal κ there exists a J´onsson group of cardinality κ ++ . Thus thereare J´onsson groups of arbitrarily large (regular) cardinality and this in itself has aconsequence in combinatorics which we mention.The infinitary edge-orbit conjecture of L. Babai [1, Conjecture 5.22] is the fol-lowing: For each cardinal κ there exists a group G such that whenever G ≃ Aut ( Γ ) forsome graph Γ the action of Aut ( Γ ) on the edges of Γ has at least κ orbits. It was shown by Simon Thomas that this conjecture is true provided there existJ´onsson groups of arbitrarily large regular cardinality [13] and so the following isimmediate.
Corollary 2.
The infinitary edge-orbit conjecture is true.For another application, one can combine Theorem 1 with [5, Theorem G] to seethat for any infinite cardinal κ there exists a topological group H of cardinality κ +++ which is not discrete, with H simple and J´onsson and every proper subgroup of H is discrete. Keisler and Rowbottom have shown that there are J´onsson algebrasof every infinite cardinality in G¨odel’s constructible universe L [2, Corollary 9.1])and so in the constructible universe there exist J´onsson groups of every nonzerocardinality.We caution the reader that there are some results about J´onsson algebras whichdo not carry over into the area of J´onsson groups. Recall that an algebra is locallyfinite if each finitely generated subalgebra is finite. It is known that there exists aJ´onsson algebra of cardinality κ if and only if there exists a locally finite J´onssonalgebra of cardinality κ [2, Theorem 3.10]. On the other hand if κ is an uncountableregular cardinal then there cannot exist a locally finite J´onsson group of cardinality κ [4, Theorem 2.6]. 2. The proof
The proof of Theorem 1 will make use of a classical result regarding J´onssonalgebras as well as a group embedding theorem. The following is obtained fromsome results of Los (see [2, Theorems 3.3, 3.4]).
Lemma 2.
There is a J´onsson algebra on an infinite set X if and only if thereexists a binary operation j ∶ X × X → X such that ( X, { j }) is J´onsson.We prepare to state a group embedding theorem by recalling some definitions.For X a set we let P( X ) denote the powerset of X and P fin ( X ) denote the collectionof finite subsets of X . Definitions 3.
Given a collection of groups { G i } i ∈ I we call the set ⋃ i ∈ I G i , where { } = G i ∩ G j whenever i ≠ j , the free amalgam of the groups { G i } i ∈ I and willdenote it by Ω . We will let Ω ⊆ Ω denote the subset Ω ∖ { } . An embedding of ´ONSSON GROUPS OF VARIOUS CARDINALITIES 3 Ω to a group G is an injective function E ∶ Ω → G where each restriction E ↾ G i isa homomorphism. An embedding defines a homomorphism from the free product ∗ i ∈ I G i to G . Definition 4.
Let { G i } i ∈ I be a collection of groups, Ω be the free amalgam, andΩ = Ω ∖ { } as above. A function f ∶ P( Ω ) ∖ {∅} → P( Ω ) is generating provided(a) if X ⊆ G i for some i ∈ I then f ( X ) = ⟨ X ⟩ ∖ { } ;(b) if X ⊆ Ω is finite and X /⊆ G i for all i ∈ I then f ( X ) = Y where Y is an arbitrarycountable subset of Ω such that X ⊆ Y and if Z ⊆ Y is finite nonempty then f ( Z ) ⊆ Y ;(c) if X ⊆ Ω is infinite and X /⊆ G i for all i ∈ I then f ( X ) = ⋃ Z ∈P fin ( X ) f ( Z ) .It is clear that if f is generating then f ( f ( X )) = f ( X ) . The following is aspecialization of a beautiful embedding theorem of Obraztsov (see [7, Theorem A]and let H in that statement be the trivial group). Proposition 5.
Suppose that { G i } i ∈ I is a collection of groups and that f is agenerating function on this collection. Then there is an embedding E ∶ Ω → G of the free amalgam which induces a homomorphism φ ∶ ∗ i ∈ I G i → G satisfying thefollowing properties:(1) G = ⟨⋃ i ∈ I φ ( G i )⟩ (i.e. φ is surjective);(2) if h ∈ G is not conjugate in G to an element in one of the groups φ ( G i ) then h is of infinite order;(3) each subgroup M of G is either cyclic, or conjugate in G to a subgroup ofone of the φ ( G i ) , or conjugate in G to a subgroup of form ⟨ C ⟩ where C ⊆ Ωsatisfies f ( C ) = C . Proof of Theorem 1.
Suppose that there is a J´onsson algebra of cardinality κ . Sinceit is known that there exist J´onsson groups of cardinality ℵ , we may assume that κ is uncountable. Let { G α } α < κ be a collection of groups where each G α is infinitecyclic and generated by z α . Let Ω be the free amalgam of the groups { G α } α < κ and Ω = Ω ∖ { } . Let τ ∶ Ω → { z α } α < κ be given by g ↦ z α where g ∈ G α .As there is a J´onsson algebra of cardinality κ , we have by Lemma 2 a binaryoperation j ∶ { z α } α < κ × { z α } α < κ → { z α } α < κ such that ({ z α } α < κ , { j }) is a J´onssonalgebra. For each subset Y ⊆ { z α } α < κ we let j ( Y ) ⊆ { z α } α < κ denote the subalgebragenerated by Y under the operation j . Let π ∶ { z α } α < κ → κ be given by z α ↦ α .We claim that the function f ∶ P( Ω ) ∖ {∅} → P( Ω ) defined by f ( X ) = { ⟨ X ⟩ ∖ { } if X ⊆ G α for some α ∈ κ, (⋃ α ∈ π ( j ( τ ( X ))) G α ) ∖ { } otherwiseis generating. Certainly condition (a) of Definition 4 holds. To check condition(b) we let X ⊆ Ω be finite with X /⊆ G α for all α < κ . The set τ ( X ) ⊆ { z α } α < κ isfinite, and so the set j ( τ ( X )) generated by τ ( X ) under j will be countable. Then π ( j ( τ ( X ))) is countable. Thus the set f ( X ) = (⋃ α ∈ π ( j ( τ ( X ))) G α )∖{ } is countableas a countable union of countable sets. Certainly f ( X ) ⊇ X in this case, since X ⊆ (⋃ α ∈ π ( τ ( X )) G α ) ∖ { } ⊆ (⋃ α ∈ π ( j ( τ ( X ))) G α ) ∖ { } = f ( X ) .Moreover given a finite nonempty Z ⊆ f ( X ) we either have Z ⊆ G α for some α < κ ,in which case α ∈ π ( j ( τ ( X ))) and SAMUEL M. CORSON f ( Z ) = ⟨ Z ⟩ ∖ { } ⊆ G α ∖ { } ⊆ f ( X ) or else we have f ( Z ) = (⋃ α ∈ π ( j ( τ ( Z ))) G α ) ∖ { } ⊆ (⋃ α ∈ π ( j ( τ ( X ))) G α ) ∖ { } = f ( X ) since τ ( Z ) ⊆ j ( τ ( X )) . Thus condition (b) holds. For condition (c) we let X ⊆ Ωbe infinite such that X /⊆ G α for each α ∈ κ . Given g ∈ f ( X ) we have g ∈ G α ∖ { } for some α ∈ π ( j ( τ ( X ))) . Thus z α ∈ j ( τ ( X )) and we may select Z ⊆ τ ( X ) which is finite such that z α ∈ j ( Z ) , and we may assume without loss of generalitythat Z has at least two elements (since τ ( X ) has at least two elements). Selecta finite Z ⊆ X such that τ ( Z ) = Z and it is clear that g ∈ f ( Z ) . Therefore f ( X ) ⊆ ⋃ Z ∈P fin ( X ) f ( Z ) and the inclusion f ( X ) ⊇ ⋃ Z ∈P fin ( X ) f ( Z ) has alreadybeen established, so condition (c) holds as well.We may therefore apply Proposition 5 to produce a simple group G into whichthe free amalgam Ω embeds and satisfying the listed properties of Proposition 5.Let φ ∶ ∗ α < κ G α → G be the induced homomorphism. To simplify notation we willidentify each group G α with the isomorphic image φ ( G α ) and thus consider G α asa subgroup of G .Notice that G is torsion-free by Proposition 5 (2), since each of the subgroups G α is torsion-free. Also, ∣ G ∣ = κ since Ω injects into G and G = ⟨⋃ α < κ G α ⟩ with ∣ G α ∣ = ℵ for each α < κ .It remains to see that G is J´onsson. Letting M be a subgroup of G with ∣ M ∣ = κ ,we know by Proposition 5 (3) that up to conjugation M is of form ⟨ C ⟩ where C ⊆ Ωsatisfies f ( C ) = C . Thus by conjugating if necessary we may assume without lossof generality that M = ⟨ C ⟩ for such a C . Since ∣ M ∣ = κ it is clear that ∣ C ∣ = κ .As κ is uncountable we know C is not a subset of any of the G α , so C = f ( C ) =(⋃ α ∈ π ( j ( τ ( C ))) G α ) ∖ { } . Since the function τ ∶ Ω → { z α } α < κ is countable-to-one wesee that ∣ τ ( C )∣ = κ , and as j is as in Lemma 2 we get j ( τ ( C ))) = { z α } α < κ . Thus π ( j ( τ ( C ))) = κ and C = f ( C ) = (⋃ α ∈ π ( j ( τ ( C ))) G α ) ∖ { } = Ωfrom which we determine that M ≥ ⟨⋃ α ∈ π ( j ( τ ( C ))) G α ⟩ = G and we are done. (cid:3) References [1] L. Babai,
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