rudiger gobel and saharon shelah- philip hall's problem on non-abelian splitters

8/3/2019 Rudiger Gobel and Saharon Shelah- Philip Hall's Problem On Non-Abelian Splitters

1/11

738

re

vision:2000-08-28

modified:2000-08-29

Philip Halls Problem On Non-Abelian

Splitters

Rudiger Gobel and Saharon Shelah

Abstract

Philip Hall raised around 1965 the following question which is stated in theKourovka Notebook [11, p. 88]: Is there a non-trivial group which is isomorphicwith every proper extension of itself by itself? We will decompose the probleminto two parts: We want to find non-commutative splitters, that are groups

G = 1 with Ext(G,G) = 1. The class of splitters fortunately is quite largeso that extra properties can be added to G. We can consider groups G withthe following properties: There is a complete group L with cartesian product

L = G, Hom (L, S) = 0 (S the infinite symmetric group acting on ) andEnd(L,L) = InnL {0}. We will show that these properties ensure that G is asplitter and hence obviously a Hall-group in the above sense. Then we will apply

a recent result from our joint paper [8] which also shows that such groups exist,in fact there is a class of Hall-groups which is not a set.

1 Introduction

Philip Hall investigated in his lectures at Cambridge around 1965 the following classof groups, which are characterized by our first

Definition 1.1 We will say that a group G is a Hall-group if any extension H of Gis isomorphic to G provided G normal in H and H/G = G.

John Lennox communicated in Kourovkas Notebook Halls question concerningthe existence of these Hall-groups. We want to show the existence of Hall-groups using

1991 Mathematics Subject Classification: Primary 13C05, 18E40, 18G05, 20K20, 20K35, 20K40;Secondary: 13D30, 18G25, 20K25, 20K30, 13C10Key words and phrases: self-splitting non-commutative groups, infinite simple groupsThe authors are supported by the project No. G 0545-173, 06/97 of the German-Israeli Foundationfor Scientific Research & Development.[GbSh:738] in Shelahs list of publications.

1


2/11

738

re

vision:2000-08-28

modified:2000-08-29

some terminology which turned out to be particular important in module theory (and

abelian groups) recently, see [9, 10].

Definition 1.2 A group G is a weak splitter if any extension of G by G splits. Inhomological terms this is to say that Ext(G, G) = 1, or equivalently any short exactsequence

1 G

H

G 1

gives raise to a splitting map : G H such that = idG. Here maps are actingon the right. Hence H = ker Im = G Im and if Im is also normal in H,then we say that G is a splitter. Hence G H with H/G = G implies H = G N forsome normal subgroup N H which is isomorphic to G.

Recall that G = N U is the semidirect product of N and U, where N is normalin G, and if also U is normal, then we write N U for the direct product.

In case of abelian groups classical splitters are free abelian groups as well as torsion-free cotorsion groups, which are well-known for a long time, see Fuchs [4]. Othersplitters were constructed only recently, see [9, 10]. They were also fundamental forsolving the flat cover conjecture for modules. Here we will study non-commutativesplitters, which however are obtained quite differently. Such groups will be based onthe following ad hoc

Definition 1.3We will say that a group L = 1 is rigid if the following holds:

(i) L is complete, i.e. L has trivial center zL = 1 and no outer automorphismsAut L = Inn L.

(ii) Hom(L, S) = 0.

(iii) End L = Inn L {0}.

Here S is the full symmetric group acting on a countable set = {0, 1, 2, . . . }.Aut L and Inn L denote the automorphism group and the group of inner automorphismsof L, respectively. Moreover Hom (A, B) is the set of homomorphisms from the group

A to the group B, where 0 is the zero-homomorphism mapping any element to 1; inparticular End A = Hom (A, A) is the near endomorphism ring ofA. By A we denotethe cartesian power over the cardinal of the group A. Using these natural definitionswe have a possibility to find Hall-group by means of our

Main Theorem 1.4 If L is a rigid group and G = L, then the following holds

(a) Aut G = Inn G S.

2


3/11

738

re

vision:2000-08-28

modified:2000-08-29

(b) G is a splitter.

(c) G G = G.

Here we note that many groups with (a) in the Main Theorem 1.4 are constructedin [6, 5, 3, 2] - in fact for arbitrary groups in place of S, but (b), (c) are also crucial.We have an immediate

Corollary 1.5 If L is a rigid group, then L is a Hall-group.

In view of Theorem 1.4 it remains show the existence of a rigid group. In Section3 where we will provide the following result from [8] and sketch the essentials of its

(lengthy) proof in a few lines. We also note that condition (i) of Definition 1.3 doesnot speak about the group L but only of its infinite cartesian power. It would bedesirable for convenience and for esthetic reasons to have a group theoretic conditionimmediately on L. This is established in Proposition 2.1 and leads to our main theoremfrom [8] mentioned as Theorem 3.1 in the last section. Here are the parts needed forapplication in the proof of our Main Theorem 1.4.

Theorem 1.6 For any infinite cardinals < with regular, = and = + >20 there is a group H with the following properties.

(i) H is a simple group of cardinality .

(ii) There are an element h H such that any element of H is a product of at most4 conjugates of h.

(iii) H is rigid.

2 Proof of Main Theorem 1.4

Proof. Condition (c) is obvious because G G = L L = L = G.In order to show (a) let S act naturally on and write G = n Len, hence

x =n

xnen with xn L (2.1)

denotes a general element of G. Also let [x] = {n : xn = 1} denote the support ofx. Moreover

GA = {x G : [x] A} G for any A .

3


4/11

738

re

vision:2000-08-28

modified:2000-08-29

If S, then induces an automorphism of G (also denoted by ) given by

: G G (x =n

xnen x =n

xnen).

Hence S Aut G and ifg G, then let

g : G G (x xg = g1xg)

denote the inner automorphism, conjugation by g. Hence

G = Inn G = {g : g G}

is normal in Aut G and visibly G

S = 1. The semidirect product

G S Aut G

is a subgroup of Aut G and we claim that the two groups are equal. Let Aut G bea given automorphism and n, m . Then we define a homomorphism

nm : L L (x xen xen (xen)m),

where clearly (xen)m is the mth coordinate ofxen =

i(xen)iei as follows fromour convention (2.1). Using the canonical embedding

n : L Len G (x xen)

and the canonical projection

m : G L (x =n

xnen xm),

we can also say thatnm = nm End L.

From Definition 1.3 (iii) we have mn Inn L {0} and if mn = 0 then there istmn L such that mn = tmn. Let

Y = {(n, m) : tmn = mn},

hencemn = 0 (n, m) \ Y.

Now we define An = {m : (n, m) Y} and claim that

the An (n ) are pair-wise disjoint. (2.2)

4


5/11

738

re

vision:2000-08-28

modified:2000-08-29

If m An1 An2 and n1 = n2 then (n1, m), (n2, m) Y. Take any xi L and

let yi = xieni (i = 1, 2). Hence [y1, y2] = 1 in G and [Ln1m, L n2m] = 1 follows fromLnim = Lenim L. If Lnim = L, then [L, L] = 1 and L would be abelian, whichcontradicts zL = 1 = L. If Ln1m = L then n1m = 0 by Definition 1.3 (iii), whichcontradicts (n1, m) Y and (2.2) is shown.

Next we observe:

If x =i

xiei G and xn = 1 then xm = 1 for all m An. (2.3)

Note that x G\{n}, G = G{n} G\{n} and xm G\{n}m. From

[G{n}, G\{n}] = 1

follows that G\{n}m and G{n}m commute, and m An implies that

Gnm = Lnm = L.

Hence G\{n}m = 1 from zL = 1 and in particular xm = 1.Now we want to show that

|An| = 1 for all n . (2.4)

If An = and 1 = x L, then xenm = 1 for all m , hence xen = 1 and1 = xen ker contradicts Aut G. If |An| > 1, then choose n1 = n2 An anddefine C = G\{n1,n2} and D = G{n1,n2}. Hence G = C D, D

= L L and considerthe canonical projection D : G D.

From (2.3) follows D C = 0, and nD maps L into D. This map is anisomorphism as follows from n1, n2 An, hence Len = Len1 Len2 . The right handside has the outer automorphism switching coordinates, while Aut L = L by Definition1.3(i). This is impossible, and |An| = 1 follows.

In the next step we show that

n

An = (2.5)

Otherwise there is 1 = y G with [y]

n An = . Let x G be the preimage of y

under the automorphism , hence x = y. We want to show that x =n

xnen = 1,

which is a contradiction. By (2.4) we can write An = {m}, hence

1 = ym = xm = xnnm = xntnm

5


6/11

738

re

vision:2000-08-28

modified:2000-08-29

using m / [y] and (2.3) for all n . We get xn = 1 and x = 1 follows. From (2.4)

and (2.5) follows that

= : (n m) if An = {m}

is a permutation S, and it is easy to see that

1 = ()1 (2.6)

We view Aut G as described at the beginning. If we replace by = 1then = id is obvious and

acts component-wise, inducing conjugations tn for eachn . If t =

ntnen G then

= t, hence = t G S and (a) is shown.

It remains to show (b). Consider

1 Gid

H

M 1 with M = G

and let : M H be a map of representatives for in M, that is x = x forall x M. If x M, then (x) is an inner automorphism of H which induces anautomorphism = (x) G of GH. From (a) we have = yxx for some yx Gand x S.

If we replace by : M H with x = y1x (x), then is again a coset

representation for and if we call the new map again, then yx = 1 for all x M.

Recall that (x) G = x for all x M,

and consider the map : G S (x x).

It is easy to check that is a homomorphism, hence Hom(G, S) = 0 by Definition1.3 (ii), which sends every x to the identity in S. This is to say that x cHG forall x H. Let C = cHG H denote the centralizer of G in H. From zG = 1 followsthat G C = 1 moreover by the above H is generated by the normal subgroups Gand C, hence H = G C. By the exact sequence above C : C M = G is an

isomorphism and we arrive at H = G G hence G is a splitter. P

Proposition 2.1 Let be a cardinal and K and L are groups with the following prop-erties

(a) |K| < |L|.

(b) L is simple, moreover

6


7/11

738

re

vision:2000-08-28

modified:2000-08-29

(c) g L, m such that any x L is product of at most m conjugates of g

Then Hom(L, K) = 0.

We derive a

Corollary 2.2 If L is a simple group > 20 such that End L = Inn L {0} andcondition (c) of Proposition 2.1 holds, then L is rigid and L is a Hall-group.

Proof of the corollary: We can apply Proposition 2.1 for K = S, hence

Hom(L, S) = 0.

As L is simple, it must be complete, so L is rigid. By Corollary 1.5 it follows that L

is a Hall-group.

Proof of Proposition 2.1: Let g L and m be as in (c). IfG = L and isany homomorphism from Hom (G, L), then consider any x =

i


8/11

738

re

vision:2000-08-28

modified:2000-08-29

3 A result from [8]

It remains to find simple groups L of cardinality |L| > 20 such that End L = Inn L{0}with the extra property that there are g L, m and any element of L is productof at most m conjugates of g.

Relatives of these groups are constructed in [7] in references given there and forinstance in [2, 3, 5, 6]. However they are not good enough for our purpose in this paper.When working on [8] we noticed the connection to the Hall problem and extended theconstruction in order to incorporate its use above.

Theorem 3.1 LetA be a family of suitable groups and < be infinite cardinals suchthat is regular uncountable and = . Then we can find a group H of cardinality = + such that the following holds.

(i) H is simple. Moreover, if 1 = g H, then any element of H is a product of atmost four conjugates of g.

(ii) Any A A is a subgroup of H and if A is not empty, then H[A] = H, wereH[A] is the subgroup of H generated by all subgroups of H isomorphic to A.

(iii) Any monomorphism : A H for some A A is induced by some h H, thatis there is some h H such that = h A.

(iv) If A H is an isomorphic copy of some A A, then the centralizer cHA = 1is trivial.

(v) Any monomorphism H H is an inner automorphism.

For our application we can assume A = . Otherwise the following definition isneeded.

Definition 3.2 LetA be any group with trivial center and view A Aut(A) as innerautomorphisms of A. Then A is called suitable if the following conditions hold:

(i) A = 1 is a finite group.

(ii) If A Aut(A) and A = A then A = A.

(iii) Aut(A) is complete.

8


9/11

738

re

vision:2000-08-28

modified:2000-08-29

Note that Aut (A) has trivial center because A has trivial center. Hence the last

condition only requires that Aut (A) has no outer automorphisms. It also follows fromthis that any automorphism of A extends to an inner automorphism of Aut (A). Agroup A is complete ifA has trivial center zA and any automorphism is inner. We alsorecall the easy observation from [7] which is a consequence of the classification of finitesimple groups:

All finite simple groups are suitable.Also note that there are many well-known examples of suitable groups which are

not simple. Just apply Wielands theorem on automorphism towers of finite groupswith trivial center, see [13].

The proof of the Theorem 3.1 is a transfinite induction building the group H (which

has cardinal = +

the successor cardinal of) as a union of a chain of subgroups Hof cardinality . The inductive steps are separated by four disjoint stationary subsetSi (0 i 3) of, where ordinals in S0S1S2 are limit ordinals of cofinality whileordinals in S3 have cofinality . Passing from H to H+1 now depends on the positionof . If does not belong to one of these stationary subsets, then H+1 = H Zis just a free product of H with a (new) infinite cyclic group Z. If belongs toone of the first three stationary subsets, then HNN-extensions are used as in [12]. Incase S0 we must deal with the conjugacy problem for (i) of the theorem. Here itis enough to ensure that all elements of infinite order are conjugate and this is whatHNN is designed for. Similarly we can deal with (ii) and (iii) by free products withamalgamated subgroup using S1 and S2 respectively. An enumeration ofelements with repetitions ensures that nobody is forgotten. Condition (iv), which isnot needed her, is pure group theory, a book keeping proof by transfinite induction.The more complicated demand is Condition (v) of the theorem which is a strengtheningfor complete. As there are obviously more possible monomorphisms to deal with thenelements in the group, a combinatorial principle is needed, a Black Box must be appliedwhich still allows to deal with the possible injections one after another while runs withrepetitions through S3 (a set of only elements) while enumerating partial injectivemaps on the set H. The basic tool is that the group H is build in such a way thatthere are many elements h H with large abelian centralizer cH(h) of cardinality .These centralizer can be arranged to come from a rigid family of abelian groups, this

is to say from a theorem shown more then a decade ago for abelian groups in Corner,Gobel [1, p.465]:

Theorem 3.3 For each subset X of the set (the cardinal) there is an 1-freeabelian group GX of cardinal such that the following holds.

Hom(GX , GY) =

Z : if X Y0 : if X Y

9


10/11

738

re

vision:2000-08-28

modified:2000-08-29

The proof of the theorem on abelian groups uses an earlier Black Box from Shelah, see

also [1] for more details.An abelian group is 1-free if all its countable subgroups are free abelian. Hence

many centralizers are algebraically very different. And as centralizers must be mappedunder monomorphisms into centralizer, an idea often used for characterizing certain(automorphism) groups, it seems perhaps convincing that such a covering with a rigidsystem of abelian centralizer almost forces monomorphism to be well-behaving, to beconjugation. The details however are a bit lengthy checking done in [8].

References

[1] A. L. S. Corner, R. Gobel, Prescribing endomorphism algebras - a unified treat-ment, Proc. London Math. Soc. (3) 50, 447 479 (1985).

[2] M. Dugas, R. Gobel, On locally finite pgroups and a problem of Philip Hall,Journal of Algebra 159 (1993), 115 138

[3] M. Dugas, R. Gobel, Torsionfree nilpotent groups and Emodules, Archiv derMath. 54 (1990), 340 351

[4] L. Fuchs, Infinite abelian groups - Volume 1,2 Academic Press, New York 1970,

1973.

[5] R. Gobel, A. Paras, Outer automorphism groups of countable metabelian groups,1998, pp. 309317 Proceedings of the Dublin Conference on Abelian Groups,Birkhauser Verlag, Basel 1999

[6] R. Gobel, A. Paras, Outer automorphism groups of metabelian groups, Journal ofPure and Appl. Algebra (2000)

[7] R. Gobel, J. L. Rodrguez and S. Shelah, Large localizations of finite simple groups,submitted to Crelle Journal 1999

[8] R. Gobel, S. Shelah, Localizations of groups, in submitted for publication

[9] R. Gobel, S. Shelah, Cotorsion theories and splitters, to appear in Trans. Amer.Math. Soc. (2000)

[10] R. Gobel, S. Shelah, Almost free splitters, Colloquium Math. 81 (1999) 193 221.

[11] The Kourovka Notebook, Unsolved problems in group theory, Novosibirsk 1999

10


11/11

738

re

vision:2000-08-28

modified:2000-08-29

[12] R. C. Lyndon and P. E. Schupp, Combinatorial Group Theory, Springer Ergeb-

nisberichte 89, BerlinHeidelbergNew York 1977.

[13] D. Robinson, A Course in the Theory of Groups, Graduate Texts in Math. vol 80,BerlinHeidelbergNew York 1996.

[14] P. Schultz, Self-splitting groups, Preprint series of the University of Western Aus-tralia at Perth (1980)

[15] P. E. Schupp, A characterization of inner automorphisms , Proc. Amer. Math. Soc.101 (1987), 226228.

Rudiger GobelFachbereich 6, Mathematik und InformatikUniversitat Essen, 45117 Essen, Germanyemail: [email protected]

andSaharon ShelahDepartment of MathematicsHebrew University, Jerusalem, Israeland Rutgers University, Newbrunswick, NJ, U.S.Ae-mail: [email protected]

11

rudiger gobel and saharon shelah- philip hall's problem on non-abelian splitters

Documents