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
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 =
n∈N
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.
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+pn−1Mm(Z/pnZ)
Z2(Hn) = (1 +pZ/pnZ)·1+pn−2Mm(Z/pnZ) ...
Zn−2(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+pn−qMm(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
.
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 αα−1−1=
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,
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· · · ∃yk−1[y1,[y2,[· · · ,[yk−2, yk−1]· · ·]]]= 1.
Theorem2.2 (Łos). Let(Mi)i∈N 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+1−1Mm(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
n∈N
(Z/pnZ)×·1+p1+nk+1−1Mm(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
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 classn−1
, one must havekn−1
. From Lemma 1.5, it follows that n−1(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 thatGisℵ1-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· · · ∃y2−1t−1(y1, . . . , y2−1)= 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 =
n∈N
(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.
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.
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