Saharon Shelah- The Height of the Automorphism Tower of a Group

8/3/2019 Saharon Shelah- The Height of the Automorphism Tower of a Group

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THE HEIGHT OF THE AUTOMORPHISMTOWER OF A GROUP

SH810

Saharon Shelah

Institute of MathematicsThe Hebrew University of Jerusalem

Einstein Institute of Mathematics

Edmond J. Safra Campus, Givat RamJerusalem 91904, Israel

Department of MathematicsHill Center-Busch Campus

Rutgers, The State University of New Jersey110 Frelinghuysen Road

Piscataway, NJ 08854-8019 USA

Abstract. For a group G with trivial center there is a natural embedding ofG intoits automorphism group, so we can look at the latter as an extension of the group.

So an increasing continuous sequence of groups, the automorphism tower, is defined,the height is the ordinal where this becomes fixed, arriving to a complete group. Weshow that for many such there is such a group of cardinality which is of height> 2, so proving that the upper bound essentially cannot be improved.

I would like to thank Alice Leonhardt for the beautiful typing.Research partially supported by NSF Grant No. NSF-DMS 0100794Paper Number 810First done Fall 2000First Typed - 02/Sept/18Latest Revision - 07/Dec/15

Typeset by AMS-TEX

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2 SAHARON SHELAH

0 Introduction

For a group G with trivial center there is a natural embedding of G into its

automorphism group Aut(G) where g G is mapped to the inner automorphismx gxg1 which is defined and is not the identity for g = eG as G has a trivialcenter, so we can view Aut(G) as a group extending G. Also the extension Aut(G) isa group with trivial center, so we can continue defining G increasing with forevery ordinal ; let G be when we stop, i.e., the first such that G

= G

(or = but see below) hence > G = G, (see Definition 0.2).How large can G be?

Weilandt [Wel39] proves that for finite G, G is finite. Thomas [Th85] celebratedwork proves for infinite G that G (2

|G|)+, in fact as noted by Felgner andThomas G < (2

|G|)+. Thomas shows also that +. Later he ([Th98]) showed

that if = + for some . An important lemmathere which we shall use (see 0.6 below) is that if G is the automorphism group ofa structure of cardinality , H G, |H| then G,H, the normalizer length ofH in G (see Definition 0.3(2)), is < . Concerning groups with center Hamkinsshow that G < the first strongly inaccessible cardinal > |G|. On the subject see

the forthcoming book of Thomas.

We shall show, e.g.

0.1 Theorem. If is strong limit singular of uncountable cofinality then > 2.

It would have been nice if the lower bound for , + would (consistently) be the

correct one for all simultaneously, but Theorem 0.1 shows that this is not so.Note that Theorem 0.1 shows that provably in ZFC, in general the upper bound(2)+ cannot be improved. See Conclusion 3.12 for proof of the theorem, quotingresults from pcf theory. We thank Simon Thomas, the referee, Itay Kaplan and

Daniel Herden for many valuable complaints detecting serious problems in earlierversions.The program, described in a simplified way, is that for each so called -parameter

p which includes a partial order I = Ip, we define a group Gp and a two elementsubgroup Hp such that nor

Gp

(Hp) : rkp reflects rkp = rk


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THE HEIGHT OF THE AUTOMORPHISM TOWER OF A GROUP SH810 3

We use an inverse system s = J, pu, u,v : u J v of -parameters whereu,v maps Ipv to Ipu ; however, in general the u,vs do not preserve order (but dopreserve it in some weak global sense) where J is an 1-directed partial order. Now

for each u J, we can define the group Gpu ; and we can take inverse limit in twoways.

Way 1: The inverse limit ps (with u,s for u J ofs) is a -parameter and so thegroup Gps is well defined.

Way 2: The inverse system Gpu , u,v : u J v of groups, where u,v is the(partial) homomorphism from Gpv to Gpu induced by u,v, has an inverse limitGs.Now

(A) concerning Gps, we normally have good control over rkps hence on thenormalizer length of Hps inside Gps

(B) Gs is (more exactly can be represented good enough as) inverse limit ofgroups of cardinality hence is isomorphic to Aut(A) for some structureA of cardinality

(C) in the good case Gps = Gs so we are done (by 0.6).

In 3 we work to get the main result.There are obvious possible improvement of the results here, say trying to prove (see Definition 0.5) for every . But more importantly, a natural conjecture,at least for me was = because all the results so far on have a parallel for

(though not inversely). In particular it seemed reasonable that for = 0 thelower bound was right, i.e., = 1. See more in Kaplan-Shelah [KpSh 882].

0.2 Definition. 1) For a group G with trivial center, define the group G withtrivial center for an ordinal , increasing continuous with such that G = Gand G is the group of automorphisms of G identifying g G withthe inner automorphisms it defines. We may stipulate G = {eG}.[We know that G is a group with trivial center increasing continuous with and for some < (2|G|+0)+ we have > G = G.]2) The automorphism tower height of the group G is G =

atwG = Min{ : G

=

G

}; clearly G G

= G

, atw stands for automorphismtower.3) Let =

atw be the least ordinal such that G < for every group G of

cardinality ; we call it the group tower ordinal of .

Now we define normalizer (group theorist write NG(H), but probably for othersnorG(H) will be clearer, at least this is so for the author).


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4 SAHARON SHELAH

0.3 Definition. 1) Let H be a subgroup of G.We define norG(H), a subgroup of G, by induction on the ordinal , increasing

continuous with . We may add nor1G (H) = {eG}.

Case 1: = 0.nor0G(H) = H.

Case 2: = + 1.

norG(H) = norG(norG(H)), see below.

Case 3: a limit ordinal

norG(H) = {norG(H) : < }

where

norG(H) = {g G :g normalizes H, i.e. gH g1 = H, equivalently

(x H)[gxg1 H & g1xg H]}.

2) Let G,H = nlgG,H, the normalizer length of H in G, be Min{ : nor

G(H) =

nor+1G (H)}; so G,H nor

G(H) = nor

G(H); nlg stands for normalizer

length.3) Let =

nlg be the least ordinal such that >

G,H whenever G = Aut(A)

for some structure A on and H G is a subgroup satisfying |H| .

4) = nlf is the least ordinal such that >

nlgG,H wherever G = Aut(A), A a

structure of cardinality , H a subgroup ofG of cardinality and norG (H) ={norG(H) : an ordinal} = G.

0.4 Definition. We say that G is a -automorphism group if G is the automor-phism group of some structure of cardinality .

0.5 Definition. Let = () be the first ordinal such that there is no sentence L+, satisfying:

(a) < is a linear order

(b) for every < there is a model M of such that (|M|,


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0.6 Lemma. .

0.7 Question: 1) Is it consistent that for some , < ? Is this provable in ZFC?

Is the negation consistent?2) Similarly for the inequalities <

, (and <

< ).

0.8 Observation. For every 0 we have atw nlg

nlf .

Proof. By 0.6 and checking the definitions of nlg , nlf . In fact we mostly work on

proving that in 0.1, nlf > 2.

Notation: For a group G and A G let AG be the subgroup of G generated byA.

A more detailed explanation of the proof:We would like to derive the desired group from a partial order I representing the

ordinal desired as G,H in some way and the tower of normalizers of an appropriatesubgroup of this length. It seems natural to say that if t I represent the ordinal then the s


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6 SAHARON SHELAH

and try to show that the normalizer tower of the subgroup HI = {e, heG


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1 Constructing groups from partialorders and long normalizer sequences

Discussion: Our aim is, for a partial order I, to define a group G = GI and asubgroup H = HI such that the normalizer length ofH inside G reflects the depthof the well founded part ofI. Eventually we would like to use I of large depth suchthat |HI| and the normalizer length of H inside GI is > 2, even equal to thedepth of I.

For clarity we first define an approximation, in particular, H appears only in2. How do we define the group G = GI from the partial order I? For eacht I we would like to have an element associated with it (it is g(,)) suchthat it will enter norG(H) exactly for = rkI(t) + 1. We intend that amongthe generators of the group commuting is the normal case, and we need witnesses

that g(,) / nor+1G (H) wherever < rkI(t), > 0. It is natural thatif rkI(t1) = and t1


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8 SAHARON SHELAH

1.2 Definition. 1) s is a -p.o.w.i.s. (partial order weak inverse system) when:

(a) s = (J, I, ) so J = Js = J[s], I = Is, = s

(b) J is a directed partial order of cardinality (c) I = Iu : u J = Isu : u J

(d) Iu = Is

u is a partial order of cardinality

(e) = u,v : u J v

(f) u,v is a partial mapping from Iv into Iu (no preservation of order is re-quired!)

(g) if u J v J w then u,w = u,v v,w.

2) s is a p.o.w.i.s. means -p.o.w.i.s. for some .3) For u J let Xu = X

s

u be the set of x such that for some n < :

(a) x = (t, ) = (tx, x)

(b) x is a function from {0, . . . , n 1} to {0, 1}

(c) t = t : n = tx : n where t I

s

u is


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1.3 Definition. Assume s is a -p.o.w.i.s. and u Js.1) Let Gu = G

s

u be the group generated by {gx : x Xs

u} freely except the equationsin u =

s

u where u consists of

(a) g1x = gx, that is gx has order 2, for each x Xu

(b) gy1gy2 = gy2gy1 when y1, y2 Xu and n(y1) = n(y2)

(c) gxgy1g1x = gy2 when

u,sx,y1,y2

, see below.

1A) Let x,y = ux,y =

u,sx,y mean that x,y1,y2 for some y1, y2 such that y

{y1, y2}, see below.1B) Let x,y1,y2 =

ux,y1,y2

= u,sx,y1,y2 mean that:

(a) x, y1, y2 Xu

(b) n(x) < n(y1) = n(y2)(c) y1 n(x) = y2 n(x)

(d) ty1 = ty2

(e) for < n(y1) we have: y1() = y2() iff = n(x) x = y1 n(x).

2) Let G


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10 SAHARON SHELAH

2) If x Xsu then {(y1, y2) : u,sx,y1,y2

holds} is a permutation of order two of{y Xsu : n(y) > n(x)}.3) Moreover, the permutation in part (2) maps each {y Xsu : n(y) = k} onto itself

when k (n(x), ) and it maps u{yXsu:n(y)>k} onto itself when n(x) k < .4) u,sx,y1,y2 iff

u,sx,y2,y1

.5) For x, y Xsu, in the group G

s

u the elements gx, gy commute except when x =y (x = y n(x) y = x n(y)). In this case, if n(x) < n(y) there is y = y such

that x,y,y and y() = y

() = n(x).

Proof. Straight (details on (2),(3) see the proof of 1.6). 1.5

We first sort out how elements in Gsu and various subgroups can be (uniquely)represented as products of the generators.

1.6 Claim. Assume that s is a -p.o.w.i.s., u Js and n(y) then x < y.

1) Any member of Gu is equal to a product of the form gx1 . . . gxm(x Xu) wherex n(x), ty = tx+1 , [i n(y) > n(x) and y n(x) = y n(x) = x hence byDefinition 1.3(1B), x,y,y ,x,y1,y1 ,x,y2,y2 hence

k,mx maps y, y1, y2 to y, y1, y2

respectively, so the desired conclusion is trivial. If (y n(x) = x) (y n(y) = y)or (y n(x) = x) (y n(y) = y) we can also get the result. So we can assumey n(x) = x and y n(y) = y and as above y n(x) = x for = 1, 2. So by

Definition 1.3(1B) as x,y,y we have ty = ty

, y(i) = y

(i) i < n(y) i = n(x)

and as x,y,y we have ty = ty , y(i) = y

(i) i < n(y) i = n(x) for = 1, 2

and as y,y1,y2 we have ty = ty (n(y) + 1 ), y1 n(y) = y2 n(y) = y, ty1 = ty2

and y1(i) = y2(i) i < n(y1) i = n(y).

Hence ty

= ty (n(y) + 1), ty

1 = ty

2 , y

1 n(y) = y

2 n(y) = y

, and

y

1(i) = y

2(i) i < n(y1) i = n(y) recalling y

1(i) = 1 y

1(i) = 0. So wehave finished proving clause (i).

Clause (ii) of1 follows from clause (i).As for clause (iii) note that for x1 = x2 X such that n(x1) = k = n(x2) and

y X we have k,mx1 (y) = y y n(x1) = x1 y n(x2) = y n(x1) =

x1 = x2 k,mx2 (y) = y, so k,mx1

, k,mx2 commute follows, hence by (ii) it follows

that k,mx1 , k,mx2

commute as required.

Lastly, clause (iv) follows from k,mx is a permutation of order two of X.

We prove this revised formulation of the uniqueness, the one on Gu,X byinduction on m k.

Note that (recalling assumption of 1.6)

() if x X, y X and x


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We can prove the induction step.

Now we define a mapping from {gx : x X} to Aut(G) by

x k,mx . Now above describes G and by 1 the mapping maps

to equations which are satisfied by Aut(G), hence there is ahomomorphism from G into Aut(G).

Hence by 1.8 the twisted product G = G G is well defined.

Let be the following mapping from {gx : x X} to G: if x X

then (gx) := (gx, eG) G G and ifx X then

(gx) := (eG, gx) G G.

Now easily every equation from is mapped by to an equation satisfiedin G (if it is from then we use the definition ofG = Gu,X,

if it is from \, then we check by cases according to the clauses ofDefinition 1.3(1), if it is clause (a) the equation has the form g2x = e, x X

and use G |= g2x = e. If the equation is from clause (b) then it has theform gxgy = gygx where x, y X

and use G is abelian.

Lastly, if the equation is from clause (c) then the equation has the form gxgy1g1x =

gy2 where x X, y1, y2 X and x,y1,y2 holds; then we use (e) of

1.8(2).

So as G is generated by {gx : x X} freely except the equationsfrom it follows that can be (uniquely) extended to a homomorphism fromG into G. Let us return to the statment in 5. So assume x

1 1, G a group and J a setwith:

(a) ft is an automorphism of G of order n for t J (i.e. fnt = idG)

(b) ft, fs Aut(G) commute for any s, t J.

Then there are K and gt : t J such that

() K is a group

() G is a normal subgroup of K

() K is generated by G {gt : t J}

() if a G and t J then g1t agt = ft(a)

() if


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3) If H1, H2 are subgroups of G1, G2 respectively, and g1 H1 (g1) maps H2onto itself and : H1 Aut(H2) is (x) = (x) H2 then {(h1, h2) : h1 H1, h2 H2} is a subgroup of G1 G2 and is in fact H1 H2; we denote

by

[H1, H2].4) If the pairs (Ha1 , H

a2 ) and (H

b1, H

b2) are as in part (3) and H

c1 := H

a1 H

b1 , H

c2 :=

Ha2 Hb2 then the pair (H

c1 , H

c2) is as in part (3) and (H

a1 [Ha1 ,Ha2 ]H

a2 )(H

b1 [Hb1,Hb2 ]

Hb2) = (Hc1 [Hc1 ,Hc2 ] H

c2).

Proof. Known and straight. 1.8

1.9 Claim. Lets be a-p.o.w.i.s., u Js andIu = Is

u be non-trivial, see Definition1.1(6).

1) If 0 < then the normalizer of G


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18 SAHARON SHELAH

Let y1 = y and by 1.5(1),(5) and Definition 1.3(1A) there is y2 such that u,sx,y1,y2

hence Gu |= gxgyg1x = gy2 and t

y = ty1 = ty2 , so rk2u(y2) rkIu(ty2n(y2)

) =

rkIu(ty1n(y1)

) < hence y2 X


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Note that

()6 xm() = y n(xm()) and y X


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()13 g1gyg = gyk+1 .

[Why? We can prove by induction on = 1, . . . , k + 1 that(gx1 . . . gx1)

1gy(gx1 . . . gx1) = gy , by the definition of the ys, i.e., by

()8 and they are well defined by ()9 + ()10 + ()12.]()14 g

1gyg = gm()+1.[Why? By ()12 and ()13.]

()15 g1gyg / G


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2 Correcting the group

The Gsus from 1 have long towers of normalizers but the base, G


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(f) for z Zu let n(z) = {k : k < k} ifz = (tk : k,

k) : k < k Z1uand n(z) is already defined ifz Z0u in clause (b).

2.2 Observation. In Definition 2.1:1) For u Js, Ku is well defined and Gu, Lu are subgroups of Ku (after theidentification).2) For I Isu let L

s

u,I be the subgroup of Ls

u generated by {hgG


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2.3 Definition. 1) IfI is a partial order then kI is the set of t = t : < k wheret I.2) If t kI then tpqf(t, , I) = {(, 1, 2) : = 0 and I |= t1 < t2 or = 1 and

t1 = t2 or = 2 and I |= t1 > t2 and = 3 if none of the previous cases}.2A) Let Sk = {tpqf(t, , I) : t kI and I is a partial order}.3) We say t kI realizes p Sk when p = tpqf(t, , I).4) If k1 < k2 and p2 Sk2 then p1 := p2 k1 is the unique p1 Sk1 such that ifp2 = tpqf(t, , I) then p1 = tpqf(t k1, , I).

Remark. Below each member of 0k, 1k, 2k will be a description of an element of

Gsu,As

u , Ks

u respectively from a k-tuple of members of Is

u. Of course, a member ofZsu is a description of a generator of K

s

u.

2.4 Definition. 1) For k < let 0k = {0k,p : p S

k} where for p Sk we let

0k,p be the set of sequences of the form (j, j) : j < j() such that:

(a) for each j for some n = n(j , j) we have j = j,i : i n(j , j) is asequence of numbers < k of length n + 1 such that p = tpqf(t, , I) tj,i : i n(j , j) is


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24 SAHARON SHELAH

3) For k < let 1k = {1k,p : p S

k} where for p Sk we let 1k,p be

the set of = (j, j) : j < j() 0k,p such that: for every s and u J

s

if t k(Isu) realizes p then there is no 0k,p with supp(

) supp() and

satisfying gu,st, G


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(2j , 2j ) = (

1(j), 1(j)) (so 2 is a permutation of 1, compare

1.6(7))

(b) for 1, 2 1k,p we have

() gu,st,1

G


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26 SAHARON SHELAH

3) For u J[s] v let u,v be 0u,v

1u,v.

4) For u Js and z Zu let u,z be the following permutation of Du = Dsu whereDu is from Definition 2.1(3)(a).

For each (v, g) Du we define u,z((v, g)) as follows:

Case 1: z Dom(0v,u) Z0u and v,u(z) X

s

v , i.e., v,u(tz) : n(z) is


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As Lu is commutative, h and (g, h) commute in Ku iff in Lu

(hu(g))(heGuG


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28 SAHARON SHELAH

fifth by the definition of norGu(), the sixth by 1.9(1), the seventh by()1.] 2.10

2.11 Observation. Let s be a p.o.w.i.s.1) For u Js and x Zsu we have: u,x is a well defined function and is apermutation of Dsu.2) Ifu J[s] v then D

s

u Ds

v.3) Ifu J[s] v and y Z

s

v and x = u,v(y) then u,x = v,y Du.

Proof. Straight.

2.12 Definition. Let s be a -p.o.w.i.s.1) Let Sk = {q : q is a function with domain Sk and for p Sk, q(p) 2k,p}, on

2k,p, see Definition 2.4(4) above.

2) We say that q Sk is disjoint when supp(q(p)) : p Sk is a sequence ofpairwise disjoint sets. We say that q is reduced when q(p) is p-reduced for everyp Sk.3) Let Z2u = Z

2,su be {Z

2,ku : k < }, where Z

2,ku = Z

2,k,su is the set of pairs (t, q)

where t k(Isu) and q Sk.

4) For z = (t, q) Z2u let u,z = s

u,z be the following permutation of Du: if

v J[s] u and (v, g) {v} Kv then s

u,z((v, g)) = (v, gg) where g = gv,s

v,u(t),q(p)where p = tpqf(v,u(t), , I

s

v), and, of course, v,u(t : < k) = v,u(t) : < k.If v,u(t) is not well-defined set g

= 1 trivially again.5) For (t, q) Z2u let gt,q = g

ut,q

= gu,st,q

= gt,q(p) where p = tpqf(t, , Iu). Let

gvt,q

= gv,st,q

= gvv,u(t),q

when v J[s] u and v,u(t) is well-defined.

2.13 Remark. We can add {su,z : z Z2,su } to the generators of F

s

u defined in 2.15below.

2.14 Observation. In Definition 2.12(4), su,z is a well defined permutation of Ds

u.

Proof. Easy.


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2.15 Definition. Let s be a p.o.w.i.s.1) Let Fu = F

s

u be the subgroup of the group of permutations of Ds

u generated by{u,z : z Z

s

u}.

2) For a p.o.w.i.s. s let Ms be the following model:set of elements: {(u, g) : u Js and g Ksu} {(1, u , f ) : u J

s and f Fsu}.

relations: PMs1,u , a unary relation, is {(u, g) : g Ku} for u Js,

PMs2,u , a unary relation is {(1, u , f ) : f Fu} for u Js

RMsu,v,h, a binary relation, is {((v, g), (1, u , f )) : f Fu, g Kv and f((v, h)) =(v, g)} for u Js and v J[s] u and h Kv.

2.16 Observation. If s is a -p.o.w.i.s. and v J[s] u and f Fu then f maps

{v} Kv = PMs1,v onto itself.

Remark. If Fsu and v J[s] u then ({v} Kv) comes directly from Ks

v , butthe relation between the ({v} Kv) : v J[s] u are less clear.

2.17 Claim. Lets be a p.o.w.i.s.1) is an automorphism of Ms iff:

(a) is a function with domain Ms

(b) for every u Js we have:

() Du Fsu

() letting fu = Du we have (1, u , f ) PMs2,u ((1, u , f ))

= (1, u , f uf) where fuf is the product in Fu.

2) If fu Fu for u Js and fu fv for u J[s] v then there is one and only one

automorphism of Ms such that u Js fu .

Proof. First assume that f = fu : u Js is as in part (2). We define f, a

function with domain Ms by:

1 (a) if a = (u, g) PMs1,u and u J

s then f(a) = fu(a)

(b) if a = (1, u , f ) PMs2,u then f(a) = (1, u , f uf).


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So

2 (a) f is a well defined function

(b) f is one to one(c) f is onto Ms

(d) f maps PMs1,u onto P

Ms1,u and P

Ms2,u onto P

Ms2,u for u J

s

(e) also f = f1u : u Js satisfies the condition of part (2) and

f is the inverse off

(f) f maps RMsu,v,h onto itself.

[Why? The only non-trivial one is clause (f) and in it by clause (e) it is enough

to prove that f maps RMsu,v,h into R

Msu,v,h. So assume v J[s] u, h Kv and

((v, g), (1, u , f )) RMsu,v,h hence f Fu, g Kv and f((v, h)) = (v, g). So f((v, g)) =fv((v, g)) and f(1, u , f ) = (1, u , f uf) and we would like to show that (fv((v, g)), (1, u , f uf)) RMsu,v,h.

This means that (fuf)((v, h)) = fv((v, g)). We know that f((v, h)) = (v, g) hence(fuf)((v, h)) = fu(f((v, h))) = fu((v, g)) so we have to show that fu((v, g)) =fv((v, g)). But v J[s] u hence (by the assumption on f) we have fv fu hencefu((v, g)) = fv((v, g)) so we are done.]

So we have shown that

3 if f = fu : u Js is as in part (2) then f is an automorphism of Ms.

Next

4 if Aut(Ms) and Du is the identity for each u Js then = idMs.

[Why? By the PMs2,u s, RMsu,v,hs and F

s

u being a group of permutations of Du.]

5 the mapping Du : u Js is a homomorphism from Aut(Ms) into{f : f as above} with coordinatewise product, with kernel { Aut(Ms) : Du = idDu for every u J

s}.

[Why? Easy. Observe that Du Fu for every u Js.]

6 the mapping above is onto.

[Why? Easy by 3.Given Aut(Ms), let fu = Du. Clearly fu Fu and u J[s] v fu fv

so f = fu : u Js is as above so by 3 we know f is an automorphism of Ms


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and f1 is an automorphism of Ms which is the identity on each Du hence by

4 is idMs . So = f, is as required.]

7

the mapping above is one to one.

[Why? Easy by 4.]Together both parts should be clear. 2.17

2.18 Definition. 1) We say that q1, q2 Sk are S-equivalent where S Sk

when p S q1(p)E2k,pq2(p).

2) Omitting S means S = Sk.

2.19 Claim. Lets be a nice -p.o.w.i.s. (or just Dom(u,v) = Iv for all u J[s]).

1) If u Js and f Fsu then for some k and t = t : < k k(Isu) and q S

k

we have:

() f = u,(t,q) (so if v J[s] u then f ({v} Ks

v) is moving by multiplicationby gv

v,u(t),q, e.g. g Kv f((v, g)) = (v, gvv,u(t),qg).

2) {u,(t,q) : (t, q) Z2u} is a group of permutations of D

s

u which includes Fs

u.

3) For every q Sk there is a reduced q Sk which is equivalent to it (seeDefinition 2.12(2)).

Proof. 2),3) Straight.1) We use freely Definition 2.12. Recall that Fsu is the group of permutations of D

su

generated by {u,z : z Zs

u}. Hence it is enough to prove that f Fs

u satisfies theconclusion of the claim in the following cases.

Case 0: f is the identity.It is enough to let k = 0 so t = ,Sk is a singleton {} and q() is the empty

sequence 2k of length 1, i.e. we use in Definition 2.12(3) the case k = 0and in Definition 2.4(1) the case j() = 0.

Case 1: f = u,z where z Z0u.

So z = (tz

, z

). We set k = n(z) + 1, t = tz

k

(Is

u) and define q as follows:(a) if p Sk describes a decreasing sequence then

q(p) = (0, 1, 2, . . . , k 1, z) 2k

as sequence of length 1

(b) if not, then q(p) = as in Case 0.


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Case 2: f = u,z where z Z1u.

Also clear.

Case 3: f = f1f2 (product in Fs

u) where f1, f2 Fs

u satisfy the conclusion of theclaim.

Just combine the definitions. Here we make use ofs being a nice -p.o.w.i.s. and2.9(3) to avoid those cases where it is impossible to choose t Dom v,u, meaningthat f = u,(t,q) always acts trivially on {v} K

s

v while f1, f2 may not be trivialthemselves.

Case 4: f = f1 where f Fsu satisfies the conclusion of the claim.Easy, too. 2.19

2.20 Remark. If q Sk and q1, q2 Sk and v J[s] u, t k(Iu) and q =tpqf(

s

v,u(t), , Iv) and q1(q), q2(q) are not E2k,q-equivalent, then g

vt,q1

= gvt,q2

.

Proof. This is by Claim 2.6(3C).


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3 The main result

We can prove that every -p.o.w.i.s has a limit, but for our application it is more

transparent to consider -p.o.w.i.s s which is the -p.o.w.i.s. t + its limit.

3.1 Definition. We say that s is the limit of t as witnessed by v when (both arep.o.w.i.s. and)

(a) Jt Js and Js = Jt {v}, v / Jt and u Js u J[s] v

(b) Isu = It

u and s

u,v = t

u,v when u J[s] v


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3.4 Main Claim. Ksv is an almost -automorphism group (see below) when:

(a) s, t are both p.o.w.i.s.

(b) s is an existential limit of t as witnessed by v(c) Jt is 1-directed, linear (i.e., for every u, v Jt we have

u J[t] v or v J[t] u) and unbounded

(d) t is a -p.o.w.i.s. (so |Jt| and |Itu| for u Jt)

(e) t is nice (see Definition 1.2(7)).

3.5 Definition. G is an almost -automorphism group when: there is a -automorphismgroup G+ and a normal subgroup G of G+ of cardinality such that G is iso-morphic to G+/G, i.e., there is a homomorphism from G+ onto G with kernel

G.

Before proving 3.4 we explain: why will being almost -automorphism group helpus in proving our intended result?

Recalling Definition 0.3 and Observation 0.8:

3.6 Claim. For any ordinal , if there is an almost -automorphism group G witha subgroup H of cardinality such that G,H = [such that nor

G(H) = G

( < )(nor

G

(H) = G)] then there is a-automorphism group G with a subgroupH of cardinality such that G,H = [such that nor

G(H

) = G ( 0, < ()

< and = . Then wecan find Fi : i < , S , D , J satisfying the conditions from 3.8 with = (andmore).

Proof. By 3.11 and [Sh:g]. 3.10

3.11 Claim. Assume

(a) = i : i < is an increasing sequence of regular cardinals withlimit

(b) = tcf(i


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2) We can choose Fi j 2

.

Proof. By 3.8 and Claim 3.10. 3.12

3.13 Remark. 1) If = 0

do we have atw

nlg

nlf

> +

? But if = 0 then quite easily yes.2) In 3.12 we can weaken is strong limit. E.g. if has uncountable cofinalityand < ||cf() < , then nlf >

cf(); see more in [Sh:E12, 18].3) We elsewhere will weaken the assumption in 3.7, 3.8 but deduce only that nlgis large.


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REFERENCES.

[JShT 654] Winfried Just, Saharon Shelah, and Simon Thomas. The automorphismtower problem revisited. Advances in Mathematics, 148:243265, 1999.math.LO/0003120.

[KpSh 882] Itay Kaplan and Saharon Shelah. The automorphism tower of a center-less group without choice. Archive for Mathematical Logic, 48:799815,2009. math.LO/0606216.

[Sh:E12] Saharon Shelah. Analytical Guide and Corrections to [Sh:g].math.LO/9906022.

[Sh:c] Saharon Shelah. Classification theory and the number of nonisomorphicmodels, volume 92 of Studies in Logic and the Foundations of Math-

ematics. North-Holland Publishing Co., Amsterdam, xxxiv+705 pp,1990.

[Sh:g] Saharon Shelah. Cardinal Arithmetic, volume 29 of Oxford LogicGuides. Oxford University Press, 1994.

[Th85] Simon Thomas. The automorphism tower problem. Proceedings of theAmerican Mathematical Society, 95:166168, 1985.

[Th98] Simon Thomas. The automorphism tower problem II. Israel Journalof Mathematics, 103:93109, 1998.

[Wel39] H. Wielandt. Eine Verallgemeinerung der invarianten Untergruppen.Math. Z., 45:209244, 1939.

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Saharon Shelah- The Height of the Automorphism Tower of a Group

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