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The Journal of Symbolic Logic

Volume 81, Number 4, December 2016

ON THE RADICALS OF A GROUP THAT DOES NOT HAVE THE INDEPENDENCE PROPERTY

C ´EDRIC MILLIET

Abstract.We give an example of a pure group that does not have the independence property, whose Fitting subgroup is neither nilpotent nor definable and whose soluble radical is neither soluble nor definable.

This answers a question asked by E. Jaligot in May 2013.

The Fitting subgroup of a stable group is nilpotent and definable (F. Wagner [11]).

More generally, the Fitting subgroup of a group that satisfies the descending chain condition on centralisers is nilpotent (J. Derakhshan, F. Wagner [5]) and definable (F. Wagner [12, Corollary 2.5], see also [10] and [1]). The soluble radical of a super- stable group is soluble and definable (A. Baudish [2]). Whether this also holds for a stable group is still an open question.

Inspired by [9], we provide an example of a pure group that does not have the inde- pendence property, whose Fitting subgroup is neither nilpotent nor definable and whose soluble radical is neither soluble nor definable. The proofs require some algebra because we have decided to provide a precise computation of the Fitting subgroup and soluble radical of the group considered.

Definition1(independence property). LetMbe a structure. A formulaϕ(x, y) has the independence property inM if for alln∈, there are tuplesa1, . . . , anand (bJ)J⊂{1,...,n} of M such that

M |= ϕ(ai, bJ)

⇐⇒ i J. M does not have the independence property(or is NIP for short) if no formula has the independence property inM.

LetLbe a first order language,ManL-structure. A setX isinterpretableinM if there is a definable subsetY ⊂Mn inM and a definable equivalence relation E onX such thatX = Y/E. A family {Yi/Ei : i I} of interpretable sets in M isuniformly interpretable inM if the corresponding families{Yi : i ∈I} and {Ei :i ∈I}are uniformly definable inM.

LetLbe yet another first order language. AnL-structureN isinterpretablein M if its domain, functions, relations, and constants are interpretable sets inM.

Received April 20, 2015.

Key words and phrases. Model theory, independence property, Fitting subgroup and soluble radical, ultraproducts.

c 2016, Association for Symbolic Logic 0022-4812/16/8104-0012 DOI:10.1017/jsl.2015.56

1444

Konstanzer Online-Publikations-System (KOPS) URL: http://nbn-resolving.de/urn:nbn:de:bsz:352-0-393134

https://dx.doi.org/10.1017/jsl.2015.56

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A family ofL-structures{Ni :i ∈I}isuniformly interpretableinM if the family of domains is uniformly interpretable inM, as well as, for each symbols of the languageL, the family{si :i ∈I}of interpretations ofs inNi.

Lemma2 (D. Macpherson, K. Tent [9]). LetM be anL-structure that does not have the independence property and let{Ni :i ∈I}be a family ofL-structures that is uniformly interpretable inM. For every ultrafilterUonI, theL-structure

i∈INi/U does not have the independence property.

Corollary3. Let m andn be natural numbers and p a prime number. Let us consider the general linear groupGLm(Z/pnZ)over the finite ringZ/pnZ. LetU be an ultrafilter onNand letG be the ultraproduct

G =

nN

GLm(Z/pnZ) U. The pure groupGdoes not have the independence property.

Proof.Consider the group GLm(Zp) over the ringZpofp-adic integers, and the normal subgroups1+pnMm(Zp) for everyn1. One has the group isomorphism

GLm(Z/pnZ)GLm(Zp)

1+pnMm(Zp).

Therefore, the family of groups{GLm(Z/pnZ) :n∈N}is uniformly interpretable in the ring Mm(Qp), which is interpretable in the field Qp of p-adic numbers, hence NIP by [8]. By Lemma 2, the group G does not have the independence

property.

§1. Preliminaries on the normal structure of GLm(Z/pnZ). Given a field k, the normal subgroups of the general linear group GLm(k) are precisely the sub- groups of the centre and the subgroups containing the special linear group SLm(k) (J. Dieudonn´e [4]). In particular, the maximal normal soluble subgroup of GLm(k) is the centre, except for the two soluble groups GL2(F2) and GL2(F3). The situtation is different for the general linear group GLm(Z/pnZ) over the ringZ/pnZ, whose normal subgroups are classified by J. Brenner [3]. We follow also W. Klingenberg [7] who deals with the normal subgroups of the general linear group over a local ringR, which applies in particular toZ/pnZ.

The centre of GLm(Z/pnZ) is the subgroup of homotheties (Z/pnZ)×·1. The general congruence subgroup ofGLm(Z/pnZ)of orderis

GCm() = (Z/pnZ)×·1+pMm(Z/pnZ).

It is a normal subgroup of GLm(Z/pnZ). For every elementg of GLm(Z/pnZ), there is a maximal nsuch thatgbelongs to GCm(). We call thecongruence orderofg.

Thespecial linear subgroupof GLm(Z/pnZ) of matrices having determinant 1 is written SLm(Z/pnZ). Anelementary transvectionis an element of SLm(Z/pnZ) of the form1+reij forr Z/pnZandi = j. Atransvectionis a conjugate of an elementary transvection.

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Proposition1.1 (J. Brenner [3, Theorem 1.5]). Leta transvection of congruence order. The normal subgroup ofGLm(Z/pnZ)generated byis

GLm(Z/pnZ) = SLm(Z/pnZ)

1+pMm(Z/pnZ) .

Theorem1.2 (J. Brenner [3]). Letmp6andgan element ofGLm(Z/pnZ)of congruence order. The normal subgroup

gGLm(Z/pnZ) ofGLm(Z/pnZ)generated bygsatisfies

SLm(Z/pnZ)

1+pMm(Z/pnZ)

gGLm(Z/pnZ) (Z/pnZ)×·1+pMm(Z/pnZ).

For any real numberx, we writexfor thefloor ofx, that isxis the greatest integerksuch thatk x.

Lemma1.3. For anym2,1, andn1, the group1+pMm(Z/pnZ)is a normal nilpotent subgroup ofGLm(Z/pnZ)of nilpotency class

n−1

. Proof.For everyxin Mm(Z/pnZ), one has

(1+px)p =1+ p

k=1

(px)kCpk=1+p2y.

It follows that1+pMm(Z/pnZ) is a nilpotentp-group. Its iterated centres are Z(Hn) = (1 +pZ/pnZ)·1+pn1Mm(Z/pnZ)

Z2(Hn) = (1 +pZ/pnZ)·1+pn−2Mm(Z/pnZ) ...

Zn2(Hn) = (1 +pZ/pnZ)·1+p2Mm(Z/pnZ) Zn−1(Hn) =1+pMm(Z/pnZ),

so the nilpotency class of1+pMm(Z/pnZ) isn−1 whenn1. For every natural numberqsatisfyingn−q, one has

Zq

1+pMm(Z/pnZ)

=

1 +pZ/pnZ

·1+pnqMm(Z/pnZ), so the greatestqsuch that the aboveqth centre is a proper subgroup is the greatest qsatisfyingn−q > . As one has

n−q > ⇐⇒ n−1−q ⇐⇒ q n−1 1, this greatestqis precisely

n−1

1.

For any real numberx, we writexfor theceiling ofx, that isxis the least integerksuch thatk x.

Lemma1.4. For any1nandm3, the group1+pMm(Z/pnZ)is soluble of derived length

log2n

.

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Proof.Let us write PCm() =1+pMm(Z/pnZ) and show that SLm(Z/pnZ)PCm()

= PCm()= SLm(Z/pnZ)PCm(2).

Letα=1−pand=1−p be two elements of1+pMm(Z/pnZ). Then αα11=

1−p 1−p 1+p+· · ·+pnn 1+p +· · ·+pn n

=1+p2( ) +p3(· · ·) +· · · ,

so PCm() is included in SLm(Z/pnZ)PCm(2). Conversely, consider the two elementary transvections=1+pe12and=1+pe21. One has

−1−1 = (1+pe12)(1+pe21)(1−pe12)(1−pe21)

=1+p2e11−p2e22−p3e12+p3e21. It follows that

SLm(Z/pnZ)PCm()

contains an element that lies in PCm(2)\PCm(2+ 1). As

SLm(Z/pnZ)PCm()

is a characteristic subgroup of SLm(Z/pnZ) PCm(), it is normal in GLm(Z/pnZ). By Theorem 1.2, SLm(Z/pnZ)PCm()

contains SLm(Z/pnZ)PCm(2).

We have thus shown that for every natural numberk, thekth derived subgroup of PCm() is

PCm()(k)= SLm(Z/pnZ)PCm(2k).

The derived length of PCm() is the leastksuch that 2k n.

Lemma1.5. For natural numbersk andn k2+k, letn(k) = 1 + n

k+ 1

. Thenn(k)is the smallest natural number satisfying the equality

n n(k)

=k.

Proof.Letn=q(k+ 1) +rbe the Euclidian division ofnbyk+ 1, withq k and 0r < k+ 1. Then one has

0<k+ 1−r

1 +q 1 hence n

n(k)

=

(k+ 1)q+r 1 +q

=

k+ 1−k+ 1−r 1 +q

=k, son(k) satisfies the equality. It is the smallest such, as one has

n n(k)1

=

nn

k+1

= n

q

=

k+ 1 + r q

k+ 1.

Lemma 1.6. For natural numbersk 1 andn 2k, let dn(k) =

n 2k

. Then dn(k)is the smallest natural number satisfying the equality

log2 n dn(k)

=k. Proof.One has

n 2k n

2k

< n 2k + 1 hence

k−1k−log2

1 + 2k n

<log2 n

2nk

k,

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so thatdn(k) satisfies the equality. It is the smallest such, as

log2

nn

2k

1

=

log2

2k

2k n

n

2k

2nk

=

k−log2 2k

n n

2k 2k

n

k+ 1.

§2. Radicals ofG. We now consider G =

n∈N

GLm(Z/pnZ) U.

We call Fitting subgroup ofG and write F(G) the subgroup generated by all its normal nilpotent subgroups. By Zorn’s Lemma, any nilpotent subgroup ofG of nilpotency classkis contained in a maximal such, which might not be unique.

Lemma2.1. There is a first order formulaϕk in the language of groups such that, for any group N, the group N is nilpotent of classkif and only ifN |=ϕk.

Proof.Consider the formula

∀x1· · · ∀xk[x1,[x2,[· · ·,[xk−1, xk]· · ·]]] = 1

∧ ∃y1· · · ∃yk1[y1,[y2,[· · · ,[yk2, yk1]· · ·]]]= 1.

Theorem2.2 (Łos). Let(Mi)iN be a collection ofL-structure,U an ultrafilter on N, and M the ultraproduct

iMi

U. One hasM |= ϕ if and only if the set {i N:Mi |=ϕ}is inU.

Theorem 2.3 (Fitting subgroup of G). If the ultrafilter U is nonprincipal, for every natural numberk,G has a unique maximal normal nilpotent subgroupNk of nilpotency classk

Nk=

n∈N

(Z/pnZ)×·1+p1+n−1k+1Mm(Z/pnZ) U, hence the Fitting subgroup ofG is

F(G) = k=1

Nk. F(G)is neither nilpotent, nor definable.

Proof.Let k be a fixed natural number. By Lemma 1.3 and Lemma 1.5, the normal subgroup

(Z/pnZ)×·1+p1+nk+11Mm(Z/pnZ)

of GLm(Z/pnZ) has nilpotency classk for all but finitely manyn. AsU contains the Fr´echet filter and as being of nilpotency classk is expressible by a first order formula in the pure language of groups according to Lemma 2.1, by Łos Theorem, the ultraproduct

nN

(Z/pnZ)×·1+p1+nk+11Mm(Z/pnZ) U

is a normal nilpotent subgroup of classkofG. Reciprocally, ifgbelongs to a normal nilpotent subgroup of classk, thengG generates a normal nilpotent subgroup of

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class at mostk. By Łos Theorem, there is a setI ∈ U such that for alln ∈I, the conjugacy classgnGn generates a nilpotent normal subgroup gnGn!

of Gn of class at mostk. Letn I be fixed. By Theorem 1.2, there is a unique natural number 1 nsuch that

SLm(Z/pnZ)

1+pMm(Z/pnZ)

gnGn!

(Z/pnZ)×·1+pMm(Z/pnZ).

As1+pMm(Z/pnZ) is nilpotent of classn1

, one must havekn1

. From Lemma 1.5, it follows that n1(k) for all but finitely manyninI, so thatg belongs to the desired ultraproduct.

To show that the Fitting subgroup ofGis not definable, letgn,be the elementary transvection1+peij ofHnfor every 1 < n. By Proposition 1.1, one has

gn,Gn = SLm(Z/pnZ)

1+pMm(Z/pnZ) , hence

Zq

gn,Gn = SLm(Z/pnZ)∩Zq

1+pMm(Z/pnZ) , sogn,Gn generates a nilpotent subgroup of nilpotency class

"n

#

. For every1, let n=kn+rnwhereknandrndenote the quotient and rest of the Euclidian division ofnbyand letg denote the class moduloU of

1GLm(Z/Z),1GLm(Z/pZ), . . . ,1GLm(Z/pZ), g+1,1, g+2,1, . . . , g2,2, g2+1,2, . . . , gn,kn, . . . . As

n kn

= holds for every n 2, the normal closure gG is nilpotent of nilpotency classfor every 1. Note thatGis1-saturated by [6, Theorem 5.6].

By [10, Theorem 1.3], the Fitting subgroup ofGis not definable.

Lemma2.4. There is a first order formulaϕ in the language of groups such that, for any groupS, the groupSis soluble of derived lengthif and only ifS |=ϕ.

Proof.Consider the termt(x1, . . . , x2) defined inductively by settingt1(x1, x2) to [x1, x2] and t+1(x1, . . . , x2+1) to [t(x1, . . . , x2), t(x2+1, . . . , x2+1)]. Then consider the formula

∀x1· · · ∀x2t(x1, . . . , x2) = 1∧ ∃y1· · · ∃y21t1(y1, . . . , y21)= 1.

We callsoluble radical ofGand writeR(G) the subgroup generated by all its normal soluble subgroups.

Theorem2.5 (soluble radical ofG). If the ultrafilterUis nonprincipal, for every natural number, G has a unique maximal normal soluble subgroupS of derived length

S =

nN

(Z/pnZ)×·1+p2nMm(Z/pnZ) U, hence the soluble radical ofGis

R(G) = =1

S =F(G).

R(G)is neither soluble, nor definable.

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Proof.By Lemma 1.4, Lemma 1.6, Lemma 2.4 and Łos Theorem,Sis a normal soluble subgroup ofGof derived length. By Theorem 1.2 and Lemma 1.6,S is maximal such. Note that 1 +

"

n 2+1

#n

2

holds for everynand, so that one has

S ⊂N2 henceR(G) andF(G) coincide.

§3. Acknowledgments. Many thanks to T. Altinel for stimulating exchanges and A. Fehm for a stimulating conversation.

REFERENCES

[1]T. AltınelandP. Baginski,Definable Envelopes of Nilpotent Subgroups of Groups with Chain Conditions on Centralizers.Proceedings of the American Mathematical Society, vol. 142 (2014), no. 5, pp. 1497–1506.

[2]A. Baudisch,On superstable groups.Journal of the London Mathematical Society (2), vol. 42 (1990), no. 3, pp. 452– 464.

[3]J. Brenner,The linear homogeneous group.Annals of Mathematics (2), vol. 39 (1938), no. 2, pp. 472– 493.

[4]J. Dieudonn´e, La G´eom´etrie Des Groupes Classiques, Ergebnisse der Mathematik und ihrer Grenzgebiete (N.F.), Heft 5, Springer-Verlag, Berlin, 1955.

[5]J. DerakhshanandF. O. Wagner,Nilpotency in groups with chain conditions.Quarterly Journal of Mathematics, vol. 48 (1997), no. 192, pp. 453– 466.

[6]H. J. Keisler, The ultraproduct construction, Ultrafilters Across Mathematics, Contemporary Mathematics, vol. 530, American Mathematical Society, Providence, RI, 2010, pp. 163–179.

[7]W. Klingenberg,Linear groups over local rings.Bulletin of the American Mathematical Society, vol. 66 (1960), pp. 294–296.

[8]L. Matthews,The Independence Property in Unstable Algebraic Structures I: p-Adically Closed Fields, unpublished preprint, 1993.

[9]D. MacphersonandK. Tent,Pseudofinite groups with NIP theory and definability in finite simple groups,Groups and Model Theory, Contemporary Mathematics, vol. 576, American Mathematical Society, Providence, RI, 2012, pp. 255–267.

[10]A. O. Houcine,A remark on the definability of the Fitting subgroup and the soluble radical.

Mathematical Logic Quarterly, vol. 59 (2013), no. 1–2, pp. 62–65.

[11]F. O. Wagner,The Fitting subgroup of a stable group.Journal of Algebra, vol. 174 (1995), no. 2, pp. 599–609.

[12] ,Nilpotency in groups with the minimal condition on centralizers.Journal of Algebra, vol. 217 (1999), no. 2, pp. 448– 460.

UNIVERSIT ¨AT KONSTANZ

FACHBEREICH MATEMATIK UND STATISTIK 78457 KONSTANZ, GERMANY

and

P ˆOLE DE MATH ´EMATIQUES DE L’INSA DE LYON B ˆATIMENT L ´EONARD DE VINCI – 21

AVENUE JEAN CAPELLE

69621 VILLEURBANNE, FRANCE E-mail: cedric.milliet@insa-lyon.fr

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