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Nilpotent groups, o-minimal Euler characteristic, and linear algebraic groups

We establish a surprising correspondence between groups definable in o-minimal structures and linear algebraic groups, in the nilpotent case. It turns out that in the o-minimal context, like for finite groups, nilpotency is equivalent to the normalizer property or to uniqueness of Sylow subgroups. As a consequence, we show algebraic decompositions of o-minimal nilpotent groups, and we prove that a nilpotent Lie group is definable in an o-minimal expansion of the reals if and only if it is a linear algebraic group.


Introduction
Groups that are definable in o-minimal structures have been studied by many authors in the past thirty years, often in analogy with Lie groups.
For compact groups, by a conjecture of Pillay in [16], now fully proved, every definable group G has a canonical quotient G/G 00 that, endowed with the logic topology, is a compact Lie group [2] with same dimension [9], same homotopy invariants [1], and same first order theory [10].
Strong connections have been found also for groups that are not compact. For instance, every connected abelian real Lie group is the direct product of its maximal torus T by a torsion-free closed subgroup. Similarly, by [3], every o-minimal definably connected abelian group G is the direct product of a maximal abstract torus T (Definition 2.8) and the maximal torsion-free definable subgroup N (G) (Fact 2.9). Therefore every abelian o-minimal group is elementarily equivalent to a linear algebraic group of the same dimension. This is not the case, in general, for solvable groups, as shown by Hrushovski, Peterzil and Pillay in [10]. They give an example of a solvable o-minimal group that is not elementarily equivalent to any definable real Lie group. In this paper we study the intermediate class of nilpotent groups, showing a surprising similarity with the linear algebraic setting, even for finite groups. In Section 2 we prove the following: Theorem 1.1. Let G be a nilpotent group definable in an o-minimal structure. Then (1) G has maximal abstractly compact subgroups K, and where N (G) is the maximal normal definable torsion-free subgroup of G.
(2) If G is definably connected then its center Z(G) is definably connected and contains every abstractly compact subgroup of G.
As a consequence of decomposition (1) above, in Section 4 we show that linear algebraic groups are the only nilpotent Lie groups that can be defined in an ominimal expansion of the real field:  A main tool is the o-minimal Euler characteristic E, an invariant under definable bijections that has been used by Strzebonski in [17] to develop a theory of definable p-groups and definable p-Sylow subgroups, extending classical notions and results for finite groups. In Section 2 and 3 it is used to show the following equivalent characterizations to nilpotency, well-known for finite groups: (1) Assume E(G) = 0. Then the following are equivalent: (a) G is nilpotent.
(b) G has exactly one p-Sylow subgroup for each prime p dividing E(G).
(c) All p-Sylow subgroups of G are normal.
(2) Suppose E(G) = 0 and G = G 0 . Then the following are equivalent: (3) Let G be definably connected. Then the following are equivalent: Finally, Section 4 contains a digression on definable abelian torsion-free groups G, for which a decomposition in 1-dimensional definable subgroups is proved, when dim Aut(G) > 0. This is related to the problem of characterizing definable groups that are elementarily equivalent to a linear algebraic group of the same dimension. Throughout the paper groups are definable with parameters in an o-minimal structure M. We assume M satisfies the definable choice property (that is, each definable equivalence relation on a definable set has a definable set of representatives) so that, whenever H < G are definable groups, the quotient G/H is a definable set, even if H is not normal.

Nilpotency and Euler characteristic
If P is a cell decomposition of a definable set X, the o-minimal Euler characteristic E(X) is defined as the number of even-dimensional cells in P minus the number of odd-dimensional cells in P, and it does not depend on P (see [7], Chapter 4). As points are 0-dimensional cells, it follows that for finite sets cardinality and Euler characteristic coincide. Moreover, since for every definable sets A, B we have that E(A × B) = E(A)E(B), the following holds: (1) If p is a prime dividing E(G), then G contains an element of order p. In particular, if E(G) = 0 then G has elements of each prime order, and Therefore, definable torsion-free groups are definably connected. (2) Each p-subgroup is contained in a p-Sylow, and p-Sylows are all conjugate. Given a definable group G, we denote by N (G) the maximal normal definable torsion-free subgroup of G (that exists by Proposition 2.1 in [5]).
We first consider the case where E(G) = 0. If G is infinite and definably connected, then either G is torsion-free or G has elements of each finite order. So if E(G) = 0 then G is not definably connected and G 0 = N (G) is torsion-free.
Lemma 2.4. Let G be a definable group such that |E(G)| = p a , for some p prime. Then any p-Sylow subgroup H of G has order p a , H is definably isomorphic to G/G 0 and G = G 0 ⋊ H.
Let H be a p-Sylow subgroup of G. By Fact 2.3(3) we know that E(G/H) = 0 mod p, thus E(G/H) = ±1. So E(H) = |H| = |E(G)| = p a . Moreover, G 0 and H have trivial intersection, as G 0 is torsion-free and H is finite. Therefore G = G 0 ⋊H, as wanted.
Remark 2.5. The semidirect product may be not direct. E.g., But when G is nilpotent, much more can be said: Proof. If G is finite, then N (G) = {e}, and (2) and (3) are well-known. So let G be infinite with dim G = n > 0 and |E(G)| = m = p a1 1 · · · p a k k . We will prove the three statements by induction on n + m.
Suppose, by a contradiction, that Z = Z(G) is finite of cardinality r. Then G/Z is a nilpotent group of dimension n and Euler characteristic m/r < m. By induction, G/Z = F ′ × N ′ , where F ′ is the direct product of its unique p-Sylow subgroups. Let now F be the pull-back in G of F ′ . This is a finite nilpotent group so it is the direct product of its unique p-Sylow subgroups and G = N (G) × F . However this implies that the infinite center of N (G) is included in the center of G that was assumed to be finite, contradiction. So Z(G) is infinite and (1) holds. Now assume Z(G) 0 = G 0 = N (G). If k = 1 and |E(G)| = p a , then by Lemma 2.4 we know that G = G 0 ⋊ G/G 0 . But as Z(G) 0 = G 0 , the product is direct and G has exactly one p-Sylow subgroup.
Suppose k > 1. As G/G 0 is a finite nilpotent group, it is the direct product of its (unique) p i -Sylow subgroups H 1 , . . . , H k . Let K 1 < G be the pull-back of the product of the first k − 1 factors, and K 2 be the pull-back of H k . By induction (2) and (3) hold.
Remark 2.7. In the proposition above, G nilpotent is an essential assumption for all three conditions. For conditions (2) and (3), we have already noticed this in Remark 2.5. For condition (1), it is enough to consider a definable centerless torsion-free group, such as R ⋊ R >0 .
We can now show the first part of Theorem 1.3: As finite p-groups are nilpotent and we are assuming N (G) is nilpotent, it follows that G is nilpotent as well.
We now consider the case where E(G) = 0. It is well-known that G may have no maximal definably compact subgroup (for instance, see Example 5.3 in [17]). However, by Theorem 1.5 in [3], if G is definably connected then G always has maximal abstractly compact subgroups, all conjugate (if definable). We will show that when G is nilpotent (definably connected or not), then maximal abstractly compact subgroups of G are a direct complement of N (G).
Definition 2.8. Let G be a definable group and let P be a property. We say that a subgroup H < G is abstractly P if H is a section of a definable subgroup with property P in a definable quotient of G.
That is, there is a definable normal subgroup N of G and a definable subgroup H ′ of G/N with property P , whose pull-back in G is N ⋊ H. In particular, H is abstractly isomorphic to a definable group H ′ with property P . We call H an abstract torus when H ′ is a definable torus (that is, abelian, definably connected and definably compact). (2) Any maximal abstract torus T of G is contained in A, and Proof. By induction on n = dim G. If n = 1, then by Fact 2.3(4)(5), G is a 0-group and there is nothing to prove. Suppose n > 1. If G is abelian, see Fact 2.9. So let G be non-abelian and set Z = Z(G). Note that both Z and G/Z are infinite, because G is nilpotent and definably connected. Suppose first G/Z is definably compact. Then N (G) ⊆ Z, G = N (G) × T (Fact 2.9) and G has a unique 0-Sylow A that is, moreover, central. So let G/Z be not definably compact. By induction, G/Z has a unique 0-Sylow A 1 , and G/Z = N 1 × T 1 , where N 1 = N (G/Z) is definable torsion-free, and is a maximal abstract torus of G/Z. Note that A 1 is the image of any 0-Sylow A of G. By Proposition 2.6 in [3], N (A) is central in G, therefore A 1 is definably compact and T 1 = A 1 .
Let H be the pull-back of A 1 in G. As A 1 is normal, H is normal as well. The quotient G/H = N 1 is torsion-free, so H contains all 0-subgroups of G. By induction (since dim N 1 > 0, as G/Z is not definably compact), H has a unique 0-Sylow, so G has a unique 0-Sylow A = N (A) × T .
Note that since A is the only 0-Sylow of G, it contains all k-torsion elements G[k] of G, for each k ∈ N. Each G[k] is a finite normal subgroup of G, therefore central. Let S be the union of all G[k]. That is, S is the torsion subgroup of G. We claim that A is the definable subgroup generated by S. If not, let S = K A. Note that, by minimality, K is definably connected. Since A is a 0-group, it follows that E(A/K) = 0. So by Fact 2.3(1)(5), the abelian group A/K contains a 0-subgroup. But this is impossible, because K contains all torsion elements of A and it is a direct factor of A (as both A and K are abelian and divisible), so A/K is torsion-free. Therefore S = A. Since S is central, A is central as well. Finally, notice that every maximal abstract torus T of G contains S, therefore T ⊂ A.
Remark 2.11. The nilpotency assumption in Proposition 2.10 cannot be extended to solvability, not even for linear groups. For instance, the group G = R 2 ⋊ SO 2 (R), where SO 2 (R) acts on R 2 by matrix multiplication, is a centerless solvable linear group with several 0-Sylows.
We now show the second part of Theorem 1.3: Proof of Theorem 1.3 (2). Let G be a definably connected group with E(G) = 0. (a) ⇒ (b) If G is nilpotent, then by Proposition 2.10, G has exactly one 0-Sylow. (b) ⇒ (c) Obvious. (c) ⇒ (a) By Theorem 1.5 in [3], G = P H where P is a union of conjugates of a 0-Sylow A and H is definable torsion-free. Since A is normal in G by assumption, then P = A and G is solvable. Whenever G is solvable and definably connected, then G/N (G) is definably compact and therefore abelian by [13]. As we are assuming N (G) nilpotent, then G is nilpotent as well.
We conclude the section with the proof of Theorem 1.1: Proof of Theorem 1.1. Let G be a nilpotent definable group.
(1) We want to show that G has maximal abstractly compact subgroups K, and any such K is a direct complement of N (G). If E(G) = 0, then K = F from Proposition 2.6. If G = G 0 , then K = T in Proposition 2.10. If E(G) = 0 and G = G 0 , then K = F · T , where F is a finite normal subgroup of G such that G = F · G 0 [8, Theo 6.10], and T is a maximal abstract torus of G 0 .
(2) If G is torsion-free, there is nothing to prove. Set N = N (G) G. As G is definably connected, then E(G) = 0. By Proposition 2.10, for every maximal abstract torus T of G. Therefore Z(G) is definably connected and contains every abstractly compact subgroup of G.

Nilpotency and normalizers
It is well-known that a finite group G is nilpotent if and only if G has the normalizer property (also called normalizers grow). That is, every proper subgroup H of G is contained properly in its normalizer N G (H) = {g ∈ G : H g = H}.
For infinite groups one implication still holds: every nilpotent group has the normalizer property. However, there are infinite groups with this property that are not even solvable. We show below that for groups definable in o-minimal structures nilpotency is equivalent to the normalizer property, even when restricted to definable subgroups, assuming N (G) is nilpotent:

. Let G be a definably connected group such that N (G) is nilpotent. Then G is nilpotent if and only if H N G (H), for every proper definable H < G.
Proof. Assume H N G (H) for every proper definable H < G. We will show that G is nilpotent. If G is not solvable, let R be the solvable radical of G. Then the quotient of G/R by its finite center is a centerless semisimple groupḠ.
SupposeḠ is definably compact and let H be the normalizer of a maximal definable torus T ofḠ. We claim that H is self-normalizing. Suppose g ∈Ḡ normalizes H. Then T g is a maximal definable torus of H. Therefore T g = T x for some x ∈ H, and g ∈ H as well. Now the pull-back of H in G is a proper definable subgroup equal to its normalizer, contradiction.
IfḠ is not definably compact, then by [3],Ḡ =KH, whereK is definably compact andH is torsion-free. By [12], G is elementarily equivalent to a connected centerless semisimple Lie group, for which maximal compact subgroups are selfnormalizing subgroups. Therefore the pre-image ofK in G is a proper definable subgroup equal to its normalizer, contradiction.
Hence G must be solvable. If G is not torsion-free, let A be a 0-Sylow of G. Then G = N (G) · A by Fact 2.9. Let H = N G (A). If H = G, then A is normal in G. By Theorem 1.3(2), then G is nilpotent, and we are done. Assume that H is a proper subgroup of G. By Theorem 1.3 (2), this is equivalent to say that G is not nilpotent. We claim that N G (H) = H. Since A is normal in H, then by Theorem 1.3(2), H is nilpotent. Let now g ∈ G be such that H g = H. As H is nilpotent, by Proposition 2.10, A is the only 0-Sylow of G and A g = A. Therefore g ∈ N G (A) = H. So H is a proper definable subgroup of G equal to its normalizer, contradiction.
Thus we have shown that every time G is not nilpotent, there is a definable subgroup H < G such that N G (H) = H.

Nilpotent groups and linear algebraic groups
Connected solvable Lie groups that are definable in an o-minimal expansion of the reals are completely characterized in [4]. Some of them, for instance the group in [18] pg. 327, are not Lie isomorphic to any linear algebraic group. However, if we restrict to nilpotent groups, the only definable Lie groups are linear algebraic: Proof of Theorem 1.2. Clearly linear algebraic groups over the reals are definable in the real field. Conversely, let G be a nilpotent real Lie group definable in an o-minimal structure.
First assume G is connected. By Proposition 2.10, G has a closed simplyconnected normal subgroup N = N (G) and a central connected compact subgroup T such that G = N × T . By Theorem 4.5 in [4], N is a triangular group, so is isomorphic to a closed connected subgroup of U T m (R), the group of unipotent upper triangular matrices, for some m ∈ N. All such groups are algebraic, as the exponential map is polynomial for nilpotent Lie algebras. If dim T = k, then the subgroup T is Lie isomorphic to the algebraic group SO 2 (R) k .
If G is not definably connected, then by [8], G = F · G 0 , for some finite normal subgroup F . By the connected case, G 0 is linear algebraic. Since finite groups are linear algebraic, so is G. Therefore a definable real nilpotent G is Lie isomorphic to U × K, where U is a closed connected subgroup of some U T m (R), and K, the maximal compact subgroup of G, is isomorphic to F · SO 2 (R) k , for some finite nilpotent F . By Theorem 1.1 and results of Hrushovski, Peterzil and Pillay [9,10] on compact groups, the problem of determining whether a definable nilpotent group is elementarily equivalent to a linear algebraic group reduces to the torsion-free case.
By [12], every linearizable abelian torsion-free definable group can be decomposed into the product of definable 1-dimensional subgroups. This definable splitting has been proved also in [14] for groups definable in several o-minimal structures, and by an induction argument it reduces to the 2-dimensional case: Proof. We know by [15] that G has a 1-dimensional definable subgroup H. Suppose A is a different 1-dimensional definable subgroup of G. Then A is a definable complement of H, and we are done. This is because A ∩ H = {0}, as both A and H have no proper non-trivial definable subgroups, and H + A = G, because H + A is a definable subgroup of full dimension, and G is definably connected.
So assume for a contradition that H is the only non-trivial definable subgroup of G, and setḠ = G/H. Thus H is definably characteristic and for each x ∈ G, x / ∈ H, G = x and each definable homomorphism from G is determined by its value on x. Therefore no definable automorphism of G can send in H an element that is not in H already. Lemma 4.3. Let ϕ 1 , ϕ 2 ∈ Aut(G), and letφ 1 ,φ 2 ∈ Aut(Ḡ) be the induced maps on the quotientḠ. Thenφ Therefore Aut(G) ֒→ Aut(Ḡ).
Fix now x ∈ G, x / ∈ H, and consider the set Clearly X is a definable set, and dim X = dim Aut(G) = 1, because x is a generator of G. Moreover X ∩ H = ∅, because no element in H is a generator. We claim that K = X ∪ {0} is a subgroup: • a ∈ K ⇒ −a ∈ K, because if ϕ ∈ Aut(G), then −ϕ ∈ Aut(G).
• a, b ∈ K ⇒ a + b ∈ K: (i) If b = −a, then a + b = 0, and there is nothing to prove.
(ii) Let b = −a, with ϕ(x) = a and ψ(x) = b. We claim that ϕ + ψ ∈ Aut(G). Otherwise F = ker(ϕ + ψ) = {g ∈ G : ϕ(g) = −ψ(g)} would be a proper (because ϕ(x) = −ψ(x)) non-trivial definable subgroup of G, so H = F . Therefore f = (−ψ) −1 • ϕ would be a definable automorphism of G that is the identity on H, and is not the identity on G. So consider the set of all such automorphisms of G: Now Y would be an infinite (because it contains f and all its powers) definable subgroup of Aut(G). By dimension reasons Y 0 = Aut(G) 0 , which is impossible, because Aut(G) 0 contains all multiplications by positive rational numbers, none of which is the identity on H.
So we have proved that if dim Aut(G) > 0, then the definable 1-dimensional subroup H has a definable complement in G, as wanted.
We conclude with a remark and a question about the general case: Remark 4.6. Let G be a definably connected group in an o-minimal structure. Assume G is elementarily equivalent to a real algebraic group. Then N (G) is nilpotent and G has a definable Levi decomposition.
Proof. In real algebraic groups any normal closed connected simply-connected subgroup is nilpotent, so N (G) must be nilpotent. Moreover, in real algebraic groups the intersection between the solvable radical and any Levi subgroup is finite, therefore Levi subgroups of G from [6] must be definable.
Question 4.7. Let G be a definably connected group in an o-minimal structure such that N (G) is nilpotent and G has a definable Levi decomposition. Is G elementarily equivalent to a real algebraic group?