Nilpotent Groups and Identities of Automorphisms

Home , Nilpotent group

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4 Examples 13 4.1 r(t)= t3 − 2t − 1...... 13 4.2 Φn(t) and Ψn(t)...... 15

5 Structure theorems for Lie rings 18 5.1 Constructing an embedding of Lie rings ...... 18 5.2 Bounded nilpotency of graded Lie rings ...... 21 5.3 Proof of theorem 1.2.6 ...... 23

6 Structure theorems for groups 23 6.1 Proof of theorem 1.2.4 ...... 23 6.1.1 Solvable case ...... 23 6.1.2 Generalcase ...... 26 6.2 Proof of theorem 1.2.5 ...... 26 6.3 Proof of corollary 1.2.8 ...... 27 6.4 Proof of corollary 1.1.6 ...... 27

7 Applications 27 7.1 Linear identities ...... 27 7.2 Cyclotomic identities ...... 28 7.2.1 Proof of theorems 7.2.5 and 7.2.6 ...... 29 7.2.2 Proof of corollaries 7.2.7 and 7.2.8 ...... 30 7.3 Anosovidentities ...... 31

8 Closing remarks 31

1 Introduction 1.1 Motivation In this text, we study the structure of finite groups with a fix-point-free automorphism satisfying an “identity.” Before clarifying what we mean by “identity,” we recall three classic results on finite groups with a fix-point-free automorphism. Theorem 1.1.1 (Rowley [36]). If a finite group G admits a fix-point-free automorphism, then G is solvable. We recall that an automorphism is called fix-point-free (or regular) if it fixes only the trivial element of the group. Theorem 1.1.1 has a long history, going back to at least Gorenstein—Herstein [15], and it was finally confirmed by means of the classification of the finite simple groups. We refer to Rowley’s paper for a particularly short proof. By considering a special case, we can hope to obtain a stronger conclusion. Theorem 1.1.2 (J. Thompson [41]). If a finite group G admits a fix-point-free automorphism of prime order, then G is nilpotent. Such automorphisms naturally appear in the study of groups acting simply-transitively on finite sets, and theorem 1.1.2 gives a positive answer to (what is generally known as) the Frobenius conjecture. Thompson’s proof used the celebrated p-complement theorem [42]

2 and an earlier result of Witt and Higman, but it did not require the classification of the finite simple groups. A follow-up result is: Theorem 1.1.3 (Higman [18]; Kreknin—Kostrikin [28,29]). If a nilpotent group G admits (p−1) a fix-point-free automorphism of prime order p, then its class satisfies c(G) ≤ (p−1)2 . In [18], Higman proved that there exists some upper bound for c(G) that depends only (p−1) on p. Kreknin and Kostrikin later proved in [28,29] an upper bound of (p − 1)2 . But it is conjectured that the minimal upper bound h(p) on c(G) satisfies h(p)= ⌈(p2 − 1)/4⌉. This problem has been referred to as the Higman conjecture, and it still open for primes p ≥ 11. We next make some elementary observations. Remark 1.1.4. We suppose, as in the above theorems, that the group G is finite or nilpotent, and that α : G −→ G is a fix-point-free automorphism of finite order n. Then 2 n 1 2 n 1 the transformation 1+α+α +···+α − of G, defined by x 7→ x·α(x)·α (x) ··· α − (x), vanishes identically. If, moreover, n is a prime, then the group G will have no n-torsion. In view of these impressive results, we propose the following family of problems. d Meta-Problem 1.1.5. We are given a polynomial r(t) := a0 + a1 · t + ··· + ad · t with integer coefficients, and we are told that some finite group G admits a fix-point-free automorphism α : G −→ G such that the map G −→ G, defined by

x 7→ xa0 · α(xa1 ) · α2(xa2 ) ··· αd(xad ), vanishes identically. Prove that the group G has a “good structure” modulo “bad torsion.” The main objective of this text is to reﬁne theorem 1.1.1 and to generalize theorems 1.1.2 and 1.1.3 in the spirit of Meta-Problem 1.1.5. We will do this in theorem 1.2.8, theorem 1.2.4, and theorem 1.2.5 respectively. This will then allow us to generalize other classic results of Hughes—Thompson, Kegel and E. Khukhro on the structure of ﬁnite groups with a p-splitting automorphism. We will do this in corollary 7.2.5 and corollary 7.2.6.

In order to state our results in full generality, we will have to introduce some notation and terminology. But we can already formulate a corollary that captures the spirit of our main results. d Corollary 1.1.6 (Irreducible case). Let r(t) = a0 + a1 · t + ··· + ad · t ∈ Z[t] be a polynomial that is irreducible over Q. Then there exist constants a,b,c ∈ N with the following property. Consider a ﬁnite group G with a ﬁx-point-free automorphism α : G −→ G and suppose that for all x ∈ G we have the equality

a0 a1 2 a2 d ad x · α(x ) · α (x ) ··· α (x )=1G.

Then G is solvable and of the form A · (B ⋊ (C × D)), where A is an a-group, B is a d b-group, C is a nilpotent c-group, and D is a nilpotent group of class at most d2 . In particular: • If G has no (a · b)-torsion, then G is nilpotent. d • If G has no (a · b · c)-torsion, then G is nilpotent of class at most d2 . Here, we recall terminology from [4]. Let n ∈ Z. A group H is said to be an n-group if the order of every element of H divides some natural power of n. We say that a group H n has n-torsion if some h ∈ H \{1H } satisﬁes h =1H . Otherwise, we say that H has no n-torsion. This naturally generalizes the notion of p-groups and p-torsion (for p a prime).

3 1.2 Main results 1.2.1 Invariants of good identities Definition 1.2.1 (Identities). We consider a group (G, ·), together with an endomorphism d γ : G −→ G, and a polynomial r(t) := a0 + a1 · t + ··· + ad · t ∈ Z[t]. We say that r(t) is a monotone identity of γ if and only if the map r(γ): G −→ G : x 7−→ xr(γ) := xa0 · a1 d ad γ(x ) ··· γ (x ) vanishes identically. In this case, we will simply write r(γ)=1G. More generally, we say that r(t) is an identity of γ if and only if there exists a decomposition r(t)= r1(t)+ ··· + rk(t) of r(t) into polynomials r1(t),...,rk(t) ∈ Z[t], such that the map r1(γ) rk (γ) r1(γ) ··· rk(γ): G −→ G : x 7−→ x ··· x vanishes identically. We will abbreviate this to r1(γ) ··· rk(γ)=1G. Definition 1.2.2 (Good polynomials). A polynomial r(t) ∈ Z[t] is said to be bad if and only if there exists some u ∈ N and some s(t) ∈ Z[t] \ Z such that s(tu+1) | r(0) · r(1) · r(t) in the ring Z[t]. Otherwise, we say that r(t) is good. For each polynomial r(t) ∈ Z[t], we will eventually introduce integer invariants Cong(r(t)), Discr (r(t)), and Prod(r(t)). We postpone their precise definitions to definition 3.2.1 and 3.3.1.∗ For now, it suffices to know that good polynomials have non-zero invariants: Proposition 1.2.3. Consider a good polynomial r(t) ∈ Z[t]. Then r(1) · Cong(r(t)) · Discr (r(t)) · Prod(r(t)) 6=0. ∗ 1.2.2 Structure Theorem 1.2.4. Let G be a finite group admitting a fix-point-free automorphism with a good identity r(t). Then either G has Cong(r(t))-torsion or G is nilpotent. We will prove the theorem by extending classic techniques of Higman [18, 19] and J. Thompson [41] in the context of the Frobenius conjecture. Theorem 1.2.5. Let G be a nilpotent group admitting an endomorphism with a good identity r(t), say of degree d. Then either G has (Discr (r(t)) · Prod(r(t)))-torsion or G d ∗ is nilpotent of class c(G) ≤ d2 . We will prove the theorem with various techniques from the theory of Lie rings. We will prove, in particular, the following auxiliary result. Theorem 1.2.6. Consider a Lie ring L, together with an endomorphism γ : L −→ L of the Lie ring, and a good polynomial r(t) ∈ Z[t] satisfying r(γ)=0L. Suppose (L, +) d has no (Discr (r(t)) · Prod(r(t)))-torsion. Then G is nilpotent of class c(G) ≤ d2 , where d = deg(r(t)).∗ By combining theorems 1.2.4 and 1.2.5, we immediately obtain: Corollary 1.2.7 (Bad torsion or bounded nilpotency). Let G be a finite group admitting a fix-point-free autorphism with a good identity r(t), say of degree d. Then either G has Cong(r(t))-torsion or G is of the form G =∼ C × D, where C is a nilpotent (Discr (r(t)) · d ∗ Prod(r(t)))-group and D is a nilpotent group of class at most d2 . None of the above results depends on the classification of the finite simple groups. But, by using the classification, we can obtain a stronger result. Theorem 1.2.8 (Bounded nilpotency modulo bad torsion). Let G be a finite group admitting a fix-point-free automorphism with a good identity r(t). Then G is solvable and of the form G =∼ B ⋊ (C × D), where B is a solvable Cong(r(t))-group, C is a nilpotent d (Discr (r(t)) · Prod(r(t)))-group, and D is a nilpotent group of class at most d2 . ∗

4 With a little more work, we can extend the theorem to identities that are irreducible over Q: see corollary 1.1.6. Since not all irreducible polynomials are good, one could hope to further extend theorem 1.2.8 all polynomials. But this is not possible. We will comment on this in the closing remarks.

1.2.3 Existence Meta-Problem 1.1.5 raises two obvious questions: which fix-point-free automorphisms of a group with “good structure” have a (monic) identity, and conversely, which (monic) polynomials are an identity of a fix-point-free automorphism of a group with “good structure”? It is easy to answer these questions if we interpret “good structure” to mean “finitely-generated and nilpotent.” In fact, a straightforward generalization of the theorem of Cayley—Hamilton shows: Proposition 1.2.9 (Existence of r(t)). Consider a finitely-generated, nilpotent group G. Then every endomorphism γ : G −→ G has a monic identity r(t) of degree deg(r(t)) ≤ Hirsch(G)+ | Tors(G)|. Here, Hirsch(G) is the Hirsch-length of G and Tors(G) is the torsion-subgroup of G. Lie-theoretic techniques of Higman and the classic Mal′cev-correspondence give a quantitative answer to the second question: Proposition 1.2.10 (Existence of G and γ). Consider a monic polynomial r(t) ∈ Z[t] \ {1}. Let λ and µ be roots of r(t) in Q, and let k be any natural number satisfying 2 k 1 r(λ)= r(λ · µ)= r(λ · µ )= ··· = r(λ · µ − )=0. (i.) Then there is a finitely-generated, torsion-free, k-step nilpotent group G admitting an endomorphism γ : G −→ G that has r(t)k as an identity. (ii.) For every prime p, there is a finite, k-step nilpotent p-group G admitting an endomorphism γ : G −→ G that has r(t)k as an identity. If r(0) · r(1) 6≡ 0 mod p, then such a γ must be a fix-point-free automorphism. Note that we can always make the minimal choice k := 1. And, since natural powers r(t)k of good polynomials r(t) are again good, we may conclude that every good polynomial of positive degree is an identity of some fix-point-free automorphism of some finite, nilpotent group of non-trivial class and of almost-arbitrary torsion.

1.3 Outline In section 2, we prove the existence of identities and endomorphisms in the context of nilpotent groups. In section 3, we deﬁne the invariants Cong(r(t)), Discr (r(t)), Prod(r(t)) and we study their basic properties. We then illustrate these results in section∗ 4 by means of examples. Next, we prove our auxiliary results about Lie rings in section 5. In section 6, we prove our main results and corollaries. We then specialize our results to linear, cyclotomic, and Anosov polynomials in section 7. We conclude with some remarks.

2 Existence 2.1 Preliminaries We begin with a simple observation.

Lemma 2.1.1 (Composition of polynomial maps). Let m, k1,...,km ∈ N and let

u(1,1)(t),...,u(1,k1)(t),...,u(m,1)(t),...,u(m,km)(t)

5 be polynomials with integer coeﬃcients. Then there exists an n ∈ N and polynomials s1(t),...,sn(t) ∈ Z[t] such that

u(j,i)(t)= s1(t)+ ··· + sn(t)

1 j m 1 i kj ≤Y≤ ≤X≤ and such that for all group endomorphisms γ : G −→ G, we have the equality of maps

(u(m,1)(γ) ··· u(m,km)(γ)) ◦···◦ (u(1,1)(γ) ··· u(1,k1)(γ)) = s1(γ) ··· sn(γ). (1)

In particular: if (1) vanishes identically, then j i u(j,i)(t) is an identity of γ. The proof is a straight-forward induction onQmP∈ N, and we leave it to the reader. As an immediate consequence, we obtain: Proposition 2.1.2 (Ideal of identities). Let G be a group and let γ : G −→ G be an endomorphism of G. Then the identities of γ form an ideal of Z[t].

2.2 Proof of proposition 1.2.9 Lemma 2.2.1. Consider a group G with an endomorphism γ : G −→ G. Suppose that G admits a subnormal series G = G1 D G2 D ··· D Gl D Gl+1 = {1G} of γ-invariant subgroups and identities r1(t),...,rl(t) ∈ Z[t] of the induced endomorphisms γGi/Gi+1 : Gi/Gi+1 −→ Gi/Gi+1 on the factors Gi/Gi+1. Then r1(t) ··· rl(t) is an identity of γ.

Proof. For each rj (t), there exists a kj ∈ N and polynomials u(j,1)(t),...,u(j,kj ) ∈ Z[t] such that i u(j,i) = rj (t) and such that uj,1(γGj /Gj+1 ) ··· uj,kj (γGj /Gj+1 )=1Gj /Gj+1 . So the map u(j,1)(γ) ··· u(j,k )(γ) sends Gj into Gj+1. The composition of these maps there- P j fore vanishes on all of G. Lemma 2.1.1 now implies that j i u(j,i)(t)= r1(t) ··· rl(t) is an identity of γ. Q P Proposition 2.2.2 (Cayley—Hamilton). Consider a group G with an endomorphism γ : G −→ G. If G admits a subnormal series G = G1 D G2 D ··· D Gl D Gl+1 = {1G} of γ-invariant subgroups such that every factor Gi/Gi+1 is free-abelian of ﬁnite rank or elementary-abelian of ﬁnite rank, then the endomorphism γ has a monic identity χ(t) of degree d(G1/G2)+ ··· + d(Gl/Gl+1).

Proof. If Gi/Gi+1 is free-abelian, then we may compute the characteristic polynomial

χi(t) = det(t · ½Gi/Gi+1 − γGi/Gi+1 ) ∈ Z[t] of the induced endomorphism γGi/Gi+1 . Else, k the factor Gi/Gi+1 is elementary-abelian, say isomorphic to (Zp, +), so that we may compute the characteristic polynomial χi(t) ∈ Fp[t] of γGi/Gi+1 over the ﬁeld Fp. There then exists a monic polynomial χi(t) ∈ Z[t] of the same degree k as χi(t) such that χi(t) mod p = χi(t). According to the (classic) theorem of Cayley—Hamilton and lemma 2.2.1, the product χ(t) := χ1(t) · χ2(t) ··· χl(t) ∈ Z[t], is an identity of γ, and χ(t) clearly has degree d(G1/G2)+ ··· + d(Gl/Gl+1). Since each χi(t) is monic, so is χ(t).

If all the factors Gi/Gi+1 in proposition 2.2.2 are free-abelian of ﬁnite rank, then the polynomial χ(t) is uniquely determined by this construction, so that we may refer to χ(t) as the characteristic polynomial of the endomorphism with respect to the series (Gi)i. Corollary 2.2.3. Consider a ﬁnite, solvable group G. Then every endomorphism γ : G −→ G of S admits a monic identity r(t) of degree deg(r(t)) ≤ |G|.

Proof. Since G is ﬁnite and solvable, it admits a fully-invariant series (Gi)i with elementary- abelian factors, so that we may apply proposition 2.2.2. It then suﬃces to observe that

i d(Gi/Gi+1) ≤ |G|. P 6 Proof. (Proposition 1.2.9) Let us consider the (well-defined) torsion subgroup T := Tors(G) of G, the fully-invariant series G ≥ T ≥{1G} of G, and the induced endomorphisms on the factors. The subgroup T is finite and nilpotent so that, it has a fully-invariant series with elementary-abelian factors. So proposition 2.2.2 provides a monic identity χT (t) of degree at most |T | for the naturally induced automorphism γT : T −→ T . Let us now consider the induced endomorphism γG/T : Q −→ Q on the factor Q := G/T . This Q is finitely- generated, torsion-free, and nilpotent. Moreover, its Hirsch-length coincides with that of G. For each term Γi(Q) of the lower central series of Q, we consider the (fully-invariant) isolator subgroup Qi := Γi∗(Q) in Q. Then (Qi)i satisfies the conditions of proposition 2.2.2, so that we obtain a monic identity χG/T (t) of γG/T of degree equal to the Hirsch- length of Q. Lemma 2.2.1 now shows that the product r(t) := χG/T (t) · χT (t) is a monic identity of γ with the correct degree. Similarly, lemma 2.2.1 shows that γ admits the (not necessarily monic) identity s(t) := |T | · χG/T (t) satisfying deg(s(t)) ≤ Hirsch(G). We conclude with a remark that will be useful later on. Remark 2.2.4. Assume that the γ of theorem 2.2.2 is an automorphism. If χ(0) = ±1, then every γ-invariant subgroup M of G is also hγi-invariant, so that the induced map γM : M −→ M is an automorphism of G.

1 Proof. It suﬃces to show that γ− (M) ⊆ M. One can use induction on l ∈ N to show χ(0) that for every x ∈ M, we have (χl(γ) ◦···◦ χ1(γ))(x) ∈ x · γ(M). Then theorem 2.2.2 χ(0) 1 gives 1G ∈ x · γ(M). So, if χ(0) = ±1, then γ− (x) ∈ M.

2.3 Proof of proposition 1.2.10 We begin by proving the corresponding statement for Lie algebras. Proposition 2.3.1. Consider a monic polynomial r(t) ∈ Z[t] \{1} with r(0) 6=0. Let λ and µ be roots of r(t) in Q, and let k be any natural number satisfying r(λ · µ) = ··· = k 1 r(λ · µ − )=0. Then there is a ﬁnitely-generated, k-step nilpotent Lie algebra L over the rational numbers with an automorphism α : L −→ L such that r(α)=0L.

Proof. According to proposition 2.1.2, the identities of an endomorphism form an ideal of Z[t]. So we may assume that r(t) is square-free and that Discr r(t) 6=0. If k = 1, then we simply consider the companion operator γ : Qdeg(r(t)) −→ Qdeg(r(t)) of r(t) on the abelian deg(r(t)) group (Q , +). It is well-known that this endomorphism γ satisﬁes r(γ)=0Qdeg(r(t)) .

So we may further assume that k ≥ 2. We let F be the free k-step nilpotent Lie algebra (over the rational numbers) on the generators x1,1,...,x1,d, x2,1,...,x2,d. Let C ∈ GLd(Q) be the companion operator of r(t) and let A be the direct sum C ⊕ C ∈ GL2d(Q) ∩ Mat2d,2d(Z). This A deﬁnes a linear transformation of the Q-span of the generators of F (in the obvious way) and A extends (in a unique way) to an automorphism α : F −→ F of the Lie algebra F . We now consider the ideal I of F that is generated by the subset (r(α))(F ) of F . This ideal is hαi-invariant, so that we may consider the quotient Lie algebra L := F/I with the induced automorphism α : L −→ L. By construction, we have r(α)=0L.

In order to prove that c(L) ≥ k, we may assume that L has coeﬃcients in the com- C plex numbers. Indeed, the larger Lie algebra L := L ⊗Q C over the complex numbers C C satisﬁes c(L) = c(L ) and it naturally admits the automorphism α := α ⊗ ½ satisfying C r(α )=0LC .

Let V be the (complex) span of the generators x1,1,...,x2,d. Since r(t) has no repeated roots, the operator C ∈ GLd(Q) can be diagonalised over C. So we may choose an ordered

7 eigenbasis (y1,1,...,y1,d,...,y2,d) of V and scalars λ1,...,λd,µ1,...,µd ∈ C such that, for each i ∈{1, 2} and j ∈{1,...,d}, we have α(y1,j )= λj · y1,j and α(y2,j )= µj · y2,j . It is clear that {λ1,...,λd} = {µ1,...,µd} is the set of roots of r(t). After permuting these basis vectors, may further assume that k 1 r(µ2)= r(λ1)= r(λ1 · µ2)= ··· = r(λ1 · µ2− )=0. (2)

Let us define a partial order on the elements of Mat2,d(Z). For a = (ai,j )i,j ,b = (bi,j )i,j ∈ Mat2,d(Z) we write a ≤ b if and only if a1,1 ≤ b1,1,...,a2,d ≤ b2,d. For each element a ∈ Mat2,d(Z) satisfying 0 ≤ a, we define the family B(a) of all left-normed Lie monomials in the eigenvectors y1,1,...,y2,d such that each yi,j appears with multiplicity exactly ai,j . For the remaining a, we define B(a) := ∅. If F (a)= hB(a)i denotes the (complex) splan of B(a), then we naturally obtain the grading F = F (a) (3) a Mat (Z) ∈ M2,d of the Lie algebra F by the grading group (Mat2,d(Z), +).

In order to understand the structure of the ideal I, we introduce some notation. For left-normed monomials [v1,...,vi] and [w1,...,wj ], we deﬁne the expression

[[v1,...,vi]; [w1,...,wj ]] := [v1,...,vi, w1,...,wj ].

For each a ∈ Mat2,d(Z), we deﬁne the C-span

I(a)= r(Λb) · h[v; w] | v ∈B(b), w ∈B(c)i, (4)

0 b,c Mat2,d(Z) ≤ c

b1,j b2,j where Λb := 1 j d λj · 1 j d µj ∈ C. By construction, we have I = a Mat (Z) I(a). ≤ ≤ ≤ ≤ ∈ 2,d Since we alsoQ have the inclusions Q I(a) ⊆ F(a), we derive from (3) the direct sumP decomposition I = I(a). Since I(a) ⊆ F (a), we conclude that the Lie algebra L is a Mat2,d(Z) also graded: L =∈ L(a), with homogeneous components L(a) := F (a)/I(a). L a Mat2,d(Z) Since ∈ L Γk(L)= L(a), 0 a Mat (Z) ≤ ∈M2,d Pi,j ai,j =k we need only show that there exists an a ∈ Mat2,d(Z) with i,j ai,j = k and I(a) ( F (a). We claim that 1 0 0 ··· 0 P a := ∈ Mat (Z) 0 k − 1 0 ··· 0 2,d is such an element. If we deﬁne the monomials

v1 := [y1,1,y2,2,...,y2,2]

v2 := [y2,2,y1,1,y2,2,...,y2,2] . .

vk := [y2,2,...,y2,2,y1,1], of length k, then B(a)= {v1, v2,...,vk}. Now (4) implies that k 1 I(a) = hr(λ1) · v1, r(λ1 · µ2) · v1,...,r(λ1 · µ2− ) · v1, k 1 r(µ2) · v2, r(λ1 · µ2) · v2,...,r(λ1 · λ2− ) · v2, 2 k 1 r(µ2) · v3, r(µ2) · v3,...,r(λ1 · µ2− ) · v3, ..., 2 k 1 r(µ2) · vk, r(µ2) · vk,...,r(λ1 · µ2− ) · vki.

8 Since the anti-symmetry of the Lie bracket implies v3 = v4 = ··· = vk = 0, we may use (2) in order to conclude that I(a) = {0} ( hv1, v2,...,vki = F (a). This ﬁnishes the proof.

Remark 2.3.2. If λ 6= µ, then the above proof can be simpliﬁed by replacing the free k-step nilpotent Lie algebra on 2d generators with the free k-step nilpotent Lie algebra on d generators, and by replacing the operator C := A⊕A with the operator C := A. In this case, the resulting Lie algebra will have all the correct properties, but it will have a strictly smaller dimension. The details are straightforward and we omit them. Cf. example 4.1.2. Remark 2.3.3. This proof was inspired, in part, by Higman’s construction in [18] of ﬁx-point-free automorphisms of prime order on groups of prescribed class. But it is also closely related to the so-called Auslander—Scheuneman relations for the construction of semi-simple Anosov automorphisms (cf. Payne’s construction in [35]).

We now consider the Mal′cev-correspondence: Proposition 2.3.4. Consider a ﬁnite-dimensional, nilpotent Lie algebra L over the rational numbers, together with an automorphism γ : L −→ L. Suppose that for every lower central factor Γi(L)/Γi+1(L), we are given a monic polynomial ri(t) ∈ Z[t] such that the induced automorphism γi : Γi(L)/Γi+1(L) −→ Γi(L)/Γi+1(L) satisﬁes ri(γi) =

0Γi(L)/Γi+1(L). (i.) Then s(t) := r1(t) ··· rc(L)(t) is a monic identity of the automorphism exp(γ) : exp(L) −→ exp(L). (ii.) Then the characteristic polynomial χ(t) of γ divides a natural power of s(t) and χ(t) has integer coeﬃcients.

Proof. (i.): Let us abbreviate G := exp(L) and β := exp(γ). We recall that the Baker— Campbell—Hausdorﬀ formula deﬁnes the group operation on G. This formula implies, in particular, that the induced automorphisms γi : Γi(L)/Γi+1(L) −→ Γi(L)/Γi+1(L) and βi : Γi(G)/Γi+1(G) −→ Γi(G)/Γi+1(G) on the lower central factors coincide. So ri(t) is an identity of βΓi(G)/Γi+1(G). We may now apply lemma 2.2.1. (ii.): Since ri(γi) =

0Γi(L)/Γi+1(L), the characteristic polynomial χi(t) = det(t · ½Γi(G)/Γi+1(G) − γi) ∈ Q[t] of γi divides a natural power of ri(t). Since ri(t) is monic with integer coefficients, Gauss’ lemma tells us that χi(t) has integer coefficients as well. Since each term Γi(L) of the lower central series of L is invariant under γ, the characteristic polynomial χ(t) of γ is just the product χ1(t) ··· χc(L)(t). This suffices to prove the second claim. Proposition 2.3.5. Consider a monic polynomial r(t) ∈ Z[t]\{1}. Let λ and µ be roots of k 1 r(t) in Q, and let k be any natural number satisfying r(λ·µ)= ··· = r(λ·µ − )=0. Then r(t)k is an identity of an endomorphism β : N −→ N of a finitely-generated, torsion-free, k-step nilpotent group N.

Proof. We may assume that r(0) 6= 0, since otherwise we may simply consider a finitely- generated, free nilpotent group F of class k and the endomorphism γ : F −→ F : x 7−→ 1F . So we may apply proposition 2.3.1 in order to find a finitely-generated, k-step nilpotent Lie algebra L over the rationals and automorphism α : L −→ L satisfying r(α)=0L. Let us consider the torsion-free, k-step nilpotent, divisible group G := exp(L) corresponding with L, together with the automorphism β := exp(α) of G corresponding with α. According to proposition 2.3.4, the polynomial r(t)k is a monic identity of β. We see, in particular, that if N is any β-invariant, full subgroup of G, then r(t)k is an identity of the restriction βN : N −→ N. Now, since the characteristic polynomial χ(t) of α has integer coefficients (cf. proposition 2.3.4), we may apply theorem 6.1 of [9] in order to obtain the desired subgroup N of G. This finishes the proof.

Proof. (Proposition 1.2.10) (i.) We simply apply proposition 2.3.5. (ii.) As before, we may assume that r(0) 6= 0. We ﬁrst construct the ﬁnitely-generated, torsion-free, k-step nilpotent group N and automorphism β : N −→ N of proposition 2.3.5. A well-known

9 result of Gruenberg then tells us that the group N is residually-(a finite p-group). Since N is finitely-generated, there exists a characteristic subgroup S of p-power index in N such that P := N/S is a finite p-group of class k. Let γ : P −→ P be the induced automorphism. Since r(t)k is an identity of β, it is clear that r(t)k is also an identity of γ. Finally, suppose that p does not divide r(0) · r(1). If γ(x)=1P resp. γ(x) = x, r(0)k r(1)k then x = 1P resp. x = 1P , and therefore x = 1P . We conclude that γ is a fix-point-free automorphism of P .

3 The invariants 3.1 Preliminaries Lemma 3.1.1. Let a(t),b(t) ∈ Q[t] and u ∈ N. Deﬁne A(t) := a(tu) and B(t) := b(tu). u Deﬁne h(t) := gcdQ[t](a(t),b(t)) and H(t) := gcdQ[t](A(t),B(t)). Then h(t )= H(t). Proof. There exists some f(t),g(t) ∈ Q[t] such that a(t)= h(t)·f(t) and b(t)= h(t)·g(t). By substituting t 7→ tu, we obtain A(t) = h(tu) · f(tu) and B(t) = h(tu) · g(tu). So h(tu)|H(t) in Q[t]. According to Bezout’s theorem, there exist v(t), w(t) ∈ Q[t] such that h(t)= v(t)·a(t)+w(t)·b(t). By substitution, we obtain h(tu)= v(tu)·A(t)+w(tu )·B(t). So H(t)|h(tu) in Q[t]. Since h(tu) and H(t) are monic, we conclude h(tu)= H(t).

Lemma 3.1.2 (Gauss). Let a(t) ∈ Z[t] be primitive, b(t) ∈ Z[t]. If a(t)|b(t) in Q[t], then also a(t)|b(t) in Z[t].

3.2 The invariant Cong(r(t)) d Definition 3.2.1 (Cong(r(t))). Consider a polynomial r(t) := a0+a1·t+···+ad·t ∈ Z[t]. i For all integers u > j ≥ 0, we define the partial sum ru,j (t) := i j mod u ai · t , so that ≡ we obtain the periodic decomposition r(t)= ru,0(t)+···+ru,u 1(t) of r(t). We define the periodic congruence number Cong(r(t)) of r(t) to be the (unique)− P non-negative generator of the (principal) Z-ideal

Z ∩ (ru,0(t) · Z[t]+ ··· + ru,u 1(t) · Z[t]). − 1

Proof. (ii.) =⇒ (i.) Suppose ﬁrst that s(tu+1)|r(t). Then there exists a v(t) ∈ Z[t] such u+1 that r(t)= s(t ) · v(t). Let v(t)= j vu+1,j (t) be the corresponding decomposition of u+1 u+1 u+1 v(t). Then r(t)= vu+1,0(t)·s(t )+vu+1,1(t)·s(t )+···+vu+1,u(t)·s(t ) is the partial u+1 P decomposition of r(t). So s(t ) divides each ru+1,j (t) and therefore their polynomial combination Cong(r(t)). Since deg(s(tu+1)) > 0 and deg(Cong(r(t))) = 0, we conclude that Cong(r(t)) = 0. (i.) =⇒ (ii) Suppose next that Cong(r(t)) = 0 and r(0) 6= 0. Then there is a u ∈ N such that h(t) := gcdQ[t](ru+1,0(t),...,ru+1,u(t)) has degree > 0. Since 0 r(0) 6= 0, we see that t does not divide h(t). So h(t) = gcdQ[t](ru+1,0(t) · t− , ru+1,1(t) · 1 u i u+1 t− ...,ru+1,u(t)·t− ). But each of the ru+1,i(t)·t− is in Z[t ]. Lemma 3.1.1 now implies that h(t) ∈ Q[tu+1] \ Q, say h(t) = S(tu+1) for some S(t) ∈ Q[t] \ Q. We can therefore write S(t) = s(t)/m, where m ∈ N and s(t) ∈ Z[t] is primitive. Then s(tu+1) = m · h(t) divides each ru+1,j (t), and therefore their sum ru+1,0(t)+ ··· + ru+1,u(t) = r(t) in Q[t]. Gauss’ lemma now implies that also s(tu+1)|r(t) in Z[t].

10 Corollary 3.2.3. Let r(t) be a good polynomial. Then Cong(r(t)) 6=0.

3.3 The invariants Discr∗(r(t)) and Prod(r(t)) Definition 3.3.1 (Discr (r(t)) and Prod(r(t))). Consider a polynomial r(t) ∈ Z[t]. If r(t) is constant, we define∗ Discr (r(t)) := r(t) and Prod(r(t)) := 1. Else, we let a be the ∗ leading coefficient of r(t), we let λ1,...,λl be the distinct roots of r(t) with corresponding multiplicities m1,...,ml, and we set m := max{m1,...,ml}. We then define

1+2d2 m Discr (r(t)) := a · (m − 1)! · (λi − λj ) ∗ 1 i,j l ≤Yi=j≤ 6 and 3 3 2d 2d mk Prod(r(t)) := a · r(λi · λj )= a · a · (λi · λj − λk) . 1 i,j l 1 i,j,k l ≤ ≤ ≤ ≤ r(λiYλj )=0 r(λiYλj )=0 · 6 · 6 Let us now verify that these invariants are non-zero integers (for r(t) ∈ Z[t] \{0}) and let us show how they can be computed algorithmically.

Discr∗(r(t)). When considering Discr (r(t)), we may assume that r(t) has degree at least 2, since otherwise the computation is quite∗ straight-forward. We may then use the stan- 1 2 n dard algorithms to compute the square-free factorisation r(t) = u1(t) · u2(t) ··· un(t) of r(t) in Z[t], with the convention that n be minimal. Then u(t) := u1(t) ··· un(t) is a greatest square-free factor u(t) := u1(t) ··· un(t) of r(t) in Z[t], and it is unique up to its sign. Then m = n, l = deg(u(t)), and the leading coeﬃcienta ¯ of u(t) divides a in Z. So the polynomial v(t) := (a/a¯) · u(t) = a · (t − λ1) ··· (t − λl) has integer coeﬃcients. Let Syl(v(t), v′(t)) be the Sylvester matrix of v(t) and its formal derivative v′(t). Lemma 3.3.2. We have 1+2d2 2m(l 1) m m Discr (r(t)) = a − − − · (m − 1)! · (det (Syl(v(t), v′(t)))) (5) ∗ and Discr (r(t)) ∈ Z \{0}. ∗ l(l 1) 2l 2 Proof. By using the formula Discr (v(t)) = (−1) − ·a − · 1 i=j l(λi −λj ), we obtain ∗ ≤ 6 ≤ m ml(l 1)/2 1+d2 2m(l 1) Discr (r(t))/((m − 1)! · (Discr(v(t))) ) = (−1) − · a Q− − . Since m,l ≤ d, we ∗ 2 1+2d2 2m(l 1) have 0 ≤ 1+2d − 2m(l − 1) and therefore a − − ∈ Z \{0}. Since v(t) ∈ Z[t], we also have Discr(v(t)) ∈ Z \{0}. So we may indeed conclude that Discr (r(t)) ∈ Z \{0}. l(l 1)/2 1 ∗ Finally, by using the formula Discr(v(t)) = (−1) − · a− · det (Syl(v(t), v′(t))) we obtain (5).

Remark 3.3.3. Formula (5) allows us to compute Discr (r(t)) without having to extract roots. Indeed: the invariants m, v(t), and l are given by the∗ algorithm for the square-free factorisation of r(t).

Prod(r(t)). Let us now ﬁnd a similar formula for Prod(r(t)). Let C be the companion matrix of (1/a) · v(t) and let us consider its Kronecker square C ⊗ C. The corresponding characteristic polynomial is given by χC C (t) = 1 i,j l(t − λi · λj ). We may next use ⊗ ≤ ≤ the Euclidean algorithm in order to compute a greatest factor w(t) of χC C(t) (in the ring Q[t]) that is co-prime to r(t). Since such aQ factor w(t) is determined⊗ only up to a (non-zero) rational number, we may choose the unique w(t) with leading coeﬃcient a2 deg(w(t)). We then have the factorisation 2 w(t)= a · (t − λi · λj ). 1 i,j l ≤ ≤ r(λiYλj )=0 · 6

11 Lemma 3.3.4. We have

2(d2 deg(w(t)))d Prod(r(t)) = a − · det (Syl(r(t), w(t))) (6) and Prod(r(t)) ∈ Z \{0}.

Proof. We deﬁne the auxiliary, monic polynomials

2 2 r¯(t) := (t − a · λi · λj ) and r¯(t) := (t − a · λi · λj ). 1 i,j l 1 i,j l ≤ ≤ ≤ ≤ r(λiYλj )=0 r(λiYλj )=0 · 6 · These polynomials have rational coefficients since their roots are permuted by every automorphism σ ∈ Gal(Q(λ1,...,λl): Q). In order to prove thatr ¯(t) ∈ Z[t], we next consider the auxiliary polynomial s(t; t1,...,tl) := 1 i,j l(t − ti · tj ) in the variable t ≤ ≤ with coefficients in the domain Z[t1,...,tl]. According to Vietà’s formula, we have Q 2 k [l ] l2 k s(t; t1,...,tl)= (−1) · ek (t1 · t1,t1 · t2, ··· ,tl · tl) · t − , 0 k l2 ≤X≤ where each e[i](t ,...,t ) := t t ··· t is the elementary symmetric j 1 i 1 n1<

[l] [l] l2 k s(t; t1,...,tl)= Pk(e1 (t1,...,tl),...,el (t1,...,tl)) · t − . 0 k l2 ≤X≤

By evaluating ti 7→ a · λi, we obtain

[l] l [l] l2 k s(t; a · λ1,...,a · λl) = Pk(a · e1 (λ1,...,λl),...,a · el (λ1,...,λl)) · t − . 0 k l2 ≤X≤

k [l] l k Since u(t)= (−1) · a · ek (λ1,...,λl) · t − and P1(t1,...,tl),...,Pl2 (t1,...,tl) all have integer coefficients, we see that the monic polynomial s(t; a · λ1,...,a · λl) also has integer P coefficients. Since s(t; a · λ1,...,a · λl) is a product s(t; a · λ1,...,a · λl)=¯r(t) · r¯(t) of two monic polynomials with rational coefficients, we may use Gauss’ lemma to conclude that alsor ¯(t), r¯(t) ∈ Z[t]. We see, in particular, that w(t)=¯r(a2 · t) ∈ Z[t], so that Res(r(t), w(t)) ∈ Z. We now observe that

2d3 Prod(r(t)) := a · r(λi · λj ) 1 i,j l ≤ ≤ r(λiYλj )=0 · 6 2(d2 deg(w(t)))d = a − · Res(r(t), w(t)) is a non-zero integer. By using the formula Res(r(t), w(t)) = det (Syl(r(t), w(t))), we ﬁnally obtain formula (7).

Remark 3.3.5. Formula (7) allows us to compute Prod(r(t)) without having to extract roots. Indeed: we only need the algorithm for the square-free factorisation of r(t) in Z[t] and the Euclidean algorithm in the ring Q[t].

12 3.4 Arithmetically-free sets of roots Deﬁnition 3.4.1 (Arithmetically-free subsets [34]). Let (A, ·) be an abelian group and let X be a ﬁnite subset of A. We say that X is arithmetically-free subset of A if and 2 X only if X contains no arithmetic progressions of the form λ, λ · µ, λ · µ ,...,λ · µ| | with λ, µ ∈ X. Proposition 3.4.2 (Arithmetically-free sets of roots). Consider a polynomial r(t) ∈ Z[t] satisfying r(0) 6=0. Let X be the set of roots in Q×. Then the following are equivalent.

(i.) X is not an arithmetically-free subset of (Q×, ·). u (ii.) There is some u ∈ N and some s(t) ∈ Z[t] \ Z such that Φu(t) and s(t ) divide r(t) in Z[t].

Proof. (ii.) =⇒ (ii.) Suppose that the second claim holds. Let λ be any root of s(tu) and let µ be any primitive u’th root of unity. Then r(µ) = 0 and r(λ)= r(λ·µ)= r(λ·µ2)= ··· is an arithmetic progression of roots in X with µ ∈ X. (i.) =⇒ (ii.) Now suppose that X is not an arithmetically-free subset of Q×. Then there exist λ, µ ∈ X such that X λ,...,λ · µ| | ∈ X. So µ has ﬁnite order, say u ≥ 1, and Φu(t)|r(t) in Z[t]. We now consider the partial decomposition r(t)= ru,0(t)+ ··· + ru,u 1(t) of r(t) and evaluate in the terms of the progression. We obtain the linear Vandermonde− system

1 1 ··· 1 ru,0(λ) 0 u 1 1 µ ··· µ − ru,1(λ) 0  . . . .  ·  .  =  .  ......  u 1 u 1 u 1       1 µ − ··· (µ − ) −   ru,u 1(λ)   0     −          Since the order of µ is u, the determinant of this system is non-zero. So the partial sums 0 ru,j (λ) all vanish. Since λ 6= 0, we see that λ is a common root of ru,0(t) · t− , ru,1(t) · 1 (u 1) t− ,...,ru,u 1(t) · t− − and their greatest common divisor, say h(t). According to lemma 3.1.1−, there is an S(t) ∈ Q[t] \ Q such that h(t) = S(tu). Write S(t) as s(t)/m, for some m ∈ N and some primitive s(t) ∈ Z[t]. Gauss’ lemma then implies that s(tu) divides r(t) in Z[t].

Corollary 3.4.3. Let r(t) ∈ Z[t] be a good polynomial. Then its roots form an arithmetically- free subset X of (Q×, ·).

Proof. Suppose that X is not arithmetically-free. Since r(t) is good, we have r(0) 6= 0. u So proposition 3.4.2 gives a u ∈ N and s(t) ∈ Z[t] \ Z such that Φu(t),s(t )|r(t). Since r(t) is good we know that u < 2. Since r(t) is good, we have r(1) 6= 0 and therefore Φu(1) 6=0. So u ≥ 2. This contradiction ﬁnishes the proof.

4 Examples 4.1 r(t)= t3 − 2t − 1 We now illustrate the methods of the previous sections by means of single, concrete example. We begin by illustrating the construction of 2.2. Example 4.1.1. We consider the discrete Heisenberg group H, deﬁned as the subgroup:

1 x z H := 0 1 y |x,y,z ∈ Z ⊆ GL (Z).    3  0 0 1      13 Let γ : H −→ H be the automorphism that is given by

y (y 1) 1 x z 1 y x · y + · 2− − z 0 1 y 7→ 0 1 x + y ,     0 0 1 00 1    

∼ 2 ∼ ½ and let us use the series H ≥ [H,H] ≥{½3} with factors H/[H,H] = Z and [H,H]/{ 3} = Z. A straight-forward computation gives us the characteristic polynomials χ1(t)= −1 − 2 t+t and χ2(t)=1+t, so that the characteristic polynomial of γ with respect to the series is given by χ(t) := (−1−t+t2)·(1+t). By substitution, as in the proof of Proposition 2.2.2, 3 2 1 1 2 1 1 we obtain ∀v ∈ H : γ (v)·γ (v− )·γ(v− )·γ (v)·γ(v− )·v− = ½3, so that the inverse auto- 1 1 2 1 1 1 1 morphism γ− : H −→ H is given by the formula γ− (v)= γ (v)·γ (v− )·(v− )·γ(v)·v− , and therefore by

1 x z x (1+x) 1 −x + y x · y − · 2 − z 0 1 y 7→ 0 1 x .     0 0 1 00 1     3 1 1 3 We emphasize that the map −1 − 2γ + γ : H −→ H : v 7−→ v− · γ(v− ) · γ (v) does not vanish. So r(t) is an identity of γ, but it is not a monotone identity of γ. We now use (a minor variation on) the construction of 2.3 to go in the other direction: Example 4.1.2. Let r(t) := (t2 −t−1)·(t+1) be the polynomial of example 4.1.1. Let us construct a ﬁx-point-free automorphism β : N −→ N on a ﬁnitely-generated, torsion-free, two-step nilpotent group N such that r(t) is the characteristic polynomial of β.

1 √5 1+√5 The roots of r(t) are −1, −2 , and 2 , and the product of the latter two roots is the ﬁrst root (cf. remark 2.3.2). So we consider the free two-step nilpotent Lie algebra F = Q · x1 + Q · x2 + Q · [x1, x2] on the generators x1 and x2. Then the companion matrix

0 1 A := ∈ GL (Z) 1 1 4 2 of (t − t − 1) deﬁnes a linear transformation of the Q-span of the generators: α(x1) := x2 and α(x2) := x1 +x2. This map extends (in a unique way) to an automorphism α : F −→ F of the Lie algebra F : α([x1, x2]) = [α(x1), α(x2)] = −[x1, x2]. We let I be the ideal of F

2 ½ that is generated by the subset (α −α− ½F )(Q·x1+Q·x2)+(α+ F )(Q·[x1, x2]) of F . Then the induced automorphism α : L −→ L on the quotient Lie algebra L := F/I satisﬁes

[x1,x2] r(α)=0L. In fact: I = {0F } and, with respect to the ordered basis (x1, x2, 2 ), the automorphism of L is given by the matrix

0 1 0 1 1 0 .   0 0 −1   We may now use the Baker—Campbell—Hausdorff formula to define the group operation ∗ on L. For rational numbers c1,c2,c12 and C1, C2, C12, we define (c1 · x1 + c2 · x2 + c12 · [x1,x2] [x1,x2] 2 ) ∗ (C1 · x1 + C2 · x2 + C12 · 2 ) to be (c1 + C1) · x1 + (c2 + C2) · x2 + (c12 + C12 + [x1,x2] [x1,x2] (c1 · C2 − c2 · C1)) · 2 . One can then verify that N := Z · x1 + Z · x2 + Z · 2 is an α- invariant subgroup of (L, ∗) of class two and Hirsch-length 3. The restriction β : N −→ N of α to N is an endomorphism of N and r(t) is the characteristic polynomial of β (with respect to the series of the isolators of the lower central series of N). Since also r(0) = −1, we may use remark 2.2.4 in order to conclude that β is, in fact, an automorphism of N.

14 Since r(1) = −2 and since N is torsion-free, we know that all ﬁx-points of β are trivial, so that β is ﬁx-point-free. In fact, the group N is a twisted Heisenberg group:

z 1 x 2 N =∼ 0 1 y |x,y,z ∈ Z ⊆ GL (Q).    3  0 0 1    And, under this identiﬁcation, the automorphism β : N−→ N is given by:

2 1 x z 2 x y+y z 2 1 y · · 2 − 0 1 y 7→ 0 1 x + y .     0 0 1 00 1     This group N is not the discrete Heisenberg group H of example 4.1.1. But H is a normal subgroup of index 2 in N. Example 4.1.3. The polynomial r(t) := t3 − 2t − 1 is good and its roots form an arithmetically-free subset of (Q×, ·).

Proof. Clearly, r(0) · r(1) = 2. We next observe that r(t) factorizes as (t + 1) · (t2 − t − 1). So (t) is good and we may apply proposition 3.4.2.

We conclude, in particular, that the invariants r(1), Cong(r(t)), Discr (r(t)), and Prod(r(t)) are non-zero. Let us now compute these invariants. ∗ Example 4.1.4. For r(t) := t3−2t−1, we have r(1)·Cong(r(t))·Discr (r(t))·Prod(r(t)) = (−2) · (2) · (−5) · (−27 · 5). ∗

Proof. (i.) It is clear that r(1) = −2. (ii.) In order to compute Cong(r(t)), we use the 2 Euclidean algorithm. We see that 1 = −r2,0(t)+0 · ru,1(t)and2=(−t ) · r3,0(t) + (−2) · r3,1(t)+0 · r3,2(t). So Cong(r(t))|2. Suppose that Cong(r(t)) = 1. Then there exist a(t),b(t),c(t) ∈ Z[t] with a(t) · r3,0(t)+ b(t) · r3,1(t)+ c(t) · r3,2(t) = 1. By evaluating in 1 and reducing modulo 2, we obtain the contradiction 0 = 1. (iii.) In order to compute the invariants Discr (r(t)) and Prod(r(t)), we assume the notation of 3.3. The roots of this ∗ 1 √5 1+√5 polynomial are simple and given by λ1 := −1, λ2 := −2 and λ3 := 2 . By deﬁnition, 2 2 2 we therefore have Discr (r(t)) := −(λ1 − λ2) · (λ2 − λ3) · (λ3 − λ1) = −5. We next note ∗ that r(λi · λj ) = 0 if and only if {i, j} = {2, 3}. So deﬁnition 3.3.1 gives us

7 Prod(r(t)) := r(λi · λj )= −2 · 5. 1 i,j 3 i,j≤Y= ≤2,3 { }6 { } Let us now come to the same conclusions without using the polynomial’s roots. Since r(t) is monic, we do not have to keep track of leading coeﬃcients. The square-free factorisation 1 of the monic polynomial r(t) is given by r(t) = u1(t) , so that r(t) = u(t) = v(t) and m = 1. By using formula (5), we obtain Discr (r(t)) = −5. Now let C be the companion ∗ 2 2 2 2 matrix of r(t). We then have χC C(t)= −(−1+ t)(1 + t) (1 − 3t + t )(−1+ t + t ) and w(t) = (t2 + t − 1)2(t2 − 3t + 1)(t⊗− 1). Formula (6) allows us to conclude once more that Prod(r(t)) = −27 · 5.

4.2 Φn(t) and Ψn(t)

We now compute the invariants of the cyclotomic polynomials Φn(t) and the polynomials n Ψn(t) = (t − 1)/(t − 1).

15 Deﬁnition 4.2.1. For r(t) ∈ Z[t] and u ∈ N, we deﬁne RResu(r(t)) to be the (unique) non-negative generator of the (principal) Z-ideal Z ∩ (ru,0(t) · Z[t]+ ··· + ru,u 1(t) · Z[t]). − Then clearly Cong(r(t)) is the least common multiple of RRes2(r(t)),..., RResu+1(r(t)). We will need two elementary properties of reduced resultants. Lemma 4.2.2 (Division). Consider integer polynomials a(t),b(t),c(t) ∈ Z[t] with a(t) · b(t)= c(t). For every integer u ≥ 2, we have RResu(a(t))| RResu(c(t)).

Proof. We ﬁrst note that for every natural i ≥ 0, we have

cu,i(t)= au,j(t) · bu,k(t). (7) j+k i mod u ≡X

By deﬁnition, there exist polynomials C0(t),...,Cu 1(t) ∈ Z[t] that give us the equality − RResu(c(t)) = 0 i

j m Proof. Let the polynomial be given by r(t) = 0 j d aj · t and deﬁne s(t) := r(t ). Then for all 0 ≤ i

By deﬁnition, there exist cu,0(t),...,cu,u 1(t) ∈ Z[t] such that RResu(r(t)) = j cu,j (t) · m − m m ru,j (t). By substituting t 7→ t , we obtain RResu(r(t)) = cu,j (t ) · ru,j (t ) ∈ su,0(t) · j P Z[t]+ ··· + su,u 1(t) · Z[t]. We conclude that RResu(r(t)) ∈ (su,0(t) · Z[t]+ ··· + su,u 1(t) · Z[t]) ∩ Z. − P −

Before computing Cong(Φn(t)), we record some elementary properties of the cyclotomic polynomials.

Lemma 4.2.4. Let n> 1 be a natural number and let m be its radical. Then deg(Φn(t)) = φ(n), where φ is the Euler totient-function. Then

n/m Φn(t)=Φm(t ). (8)

Let p be an odd prime that does not divide m. Then

p Φp m(t) · Φm(t)=Φm(t ). (9) · If m is odd, then Φ2 m(t)=Φm(−t). (10) · If n = m, then Φn(1) = m. If n 6= m, then Φn(1) = 1. Lemma 4.2.5. For every square-free natural number n and natural number u ≥ 2, we have RResu(Φn(t))=1.

Proof. Deﬁne r(t) := Φn(t). Since RRes2(−1+ t) = 1, we may assume that n> 1.

Case: n is a prime. Suppose ﬁrst that u ≥ φ(n)+1=(n − 1)+1 = n. Then ru,0(t) = 1, so that 1 = 1 · ru,0(t)+0 · ru,1(t) + ··· + 0 · ru,u 1(t), and therefore −

16 RResu(r(t)) = 1. Next, we suppose that 2 ≤ u

Case: n is square-free and odd. Let us proceed by induction on the number l of distinct prime factors of n. The base of the induction, l = 1, is given by the previous paragraph. So we suppose that l > 1. Let n = p1 ··· pl be the decomposition of n into (distinct, odd) primes. We may, as before, suppose that u is an integer satisfying 2 ≤ u ≤ φ(n)+1 = (p1 − 1) ··· (pl − 1) + 1. Then we note that there is at least one i ∈ {1,...,l} such that pi does not divide u. Formula (9) tells us that pi Φn(t)|Φn/pi (t ). Lemma 4.2.2, lemma 4.2.3, and the induction hypothesis then imply pi that RResu(Φn(t))| RResu(Φn/pi (t ))| RResu(Φn/pi (t))=1.

Case: n is square-free and even. Formula (10) gives us the equality Φn(t)=Φn/2(−t). The odd case then tells us that RResu(Φn(t)) = RResu(Φn/2(−t)) = RResu(Φn/2(t))=1. This ﬁnishes the proof.

Proposition 4.2.6 (Cong(Φn(t)) and Cong(Ψn(t))). Let n ∈ N. Then we have:

1 if n is square-free, Cong(Φn(t)) = (0 if n is not square-free.

Moreover: Cong(Ψn(t))=0 if and only if n is composite.

n/m Proof. (i) Let m be the radical of n. If n is not square-free, then Φn(t) ∈ Z[t ], so that RResn/m(Φn(t)) = 0, and therefore Cong(Φn(t)) = lcmu>1 RResu(Φn(t)) = 0. Else, n is square-free, and we may apply proposition 4.2.5 to conclude that Cong(Φn(t)) = lcmu>1 RResu(Φn(t)) = 1. (ii.) If n has a proper, non-trivial divisor u, then we have n/u 1 Ψn/u(t)+Ψn/u(t) · t + ··· +Ψn/u(t) · t − = Ψn(t), so that RResn/u(Ψn(t))=0. If n has no proper, non-trivial divisor, then n = 1 or a prime, in which case Ψn(t)=Φn(t). So we are in the previous case.

Proposition 4.2.7 (Discr (Φn(t)) · Prod(Φn(t))). Consider a natural number n> 1 and ∗ the corresponding cyclotomic polynomial Φn(t). Then Discr (Φn(t)) · Prod(Φn(t)) divides a natural power of n. ∗

Proof. If n = 2, then Φ2(t)=1+ t, so that we are in the linear case again. So we assume that n> 2. Since the cyclotomic field Kn corresponding with Φn(t) is monogenic, ϕ(n)/2 we know that Discr(Φn(t)) coincides with the field discriminant ∆Kn = (−1) · ϕ(n) ϕ(n)/ϕ(p) n n / p Pp of Kn. We see, in particular, that Discr (Φn(t)) divides n . Now p∈n ∗ | set r(tQ) := Φn(t). We use the notation of 3.3. By construction, for each root λ of the monic polynomial r(t), there exists a natural m (properly) dividing n, such that λ is an m’th root of unity. Since also r(t) ∈ Z[t], there exist non-negative integers am such that am am r(t) = m n Φm(t) . So Prod(r(t)) = Res(r(t), r(t)) = m n Res(Φn(t), Φm(t)) . m=| n m=| n The factorsQ 6 in this expression were computed explicitly by E.Q Lehmer6 [30], Apostol [3], Dresden [12], and several others [6]. We see, in particular, that also Prod(Φn(t)) divides a natural power of n. This finishes the proof.

2 We have, for example: Discr (Φ6(t)) · Prod(Φ6(t)) = (3) · (2 ) and Discr (Φ15(t)) · 4 6 24 8∗ ∗ Prod(Φ15(t)) = (3 · 5 ) · (3 · 5 ).

Proposition 4.2.8 (Discr (Ψn(t)) · Prod(Ψn(t))). Consider a natural number n > 1 ∗ n 1 and the corresponding split polynomial Ψn(t)=1+ t + ··· + t − . Then Discr (Ψn(t)) · n 2 n 1 ∗ Prod(Ψn(t)) = n − · n − . 17 Proof. We may assume that n> 2. According to formula (5), we have:

Discr (Ψn(t)) = det (Syl(Ψn(t), Ψn′ (t))) ∗ = Res(Ψn(t), Ψn′ (t)) 1 n = Res(t − 1, Ψn′ (t))− · Res(t − 1, Ψn′ (t)) 1 n n − n = (−1) · · det (Syl(t − 1, Ψ′ (t))) . 2 n n By performing row and column operations on the matrix Syl(t − 1, Ψn′ (t)), we obtain the n n circulant determinant det (Syl(t − 1, Ψn′ (t))) = − det (Circ(0, 1, 2,...,n − 1)) = (−1) · n n 2 n 2 2 ·n − . So Discr ((t)) = n − . Now set r(t) := Φn(t). We use the notation of 3.3 again. ∗ i+j Let ωn be a primitive n’th root of unity. We then have r(t)= 0

5 Structure theorems for Lie rings 5.1 Constructing an embedding of Lie rings Lemma 5.1.1 (Binomial commutator formula). Consider a Lie ring L with coeﬃcients in a commutative ring R. Let γ : L −→ L be a Lie endomorphism of L, let λ, µ ∈ R be coeﬃcients, and let v, w ∈ L. For all m ∈ N, we then have

m m m i i i m i

½ ½ (γ − (λ · µ) · ½ ) ([v, w]) = · [λ − · (γ − λ · ) (v),γ ◦ (γ − µ · ) − (w)]. L i L L 0 i m ≤X≤

mλ mµ ½ If, for some mλ,mµ ∈ N, we have (γ − λ · ½L) (v)=0L = (γ − µ · L) (w), then we mλ+mµ also have (γ − λ · µ · ½L) ([v, w]) = 0L.

Proof. One can prove the first formula by a simple induction on m and Pascal’s binomial n n 1 n i identity i = i −1 + −i . By specializing to m = mλ + mµ, we also obtain the second formula. − For an integer h and a Lie ring M with coefficients in a ring R, we define the h- n torsion ideal Th(M) of M by Th(M) := {v ∈ M|∃n ∈ N : h · v = 0M }. It is clear that such a set Th(M) is a Lie ideal of M that is invariant under R-multiplications and M-endomorphisms. Theorem 5.1.2. Consider a Lie ring L, together with an endomorophism γ : L −→ L, and a polynomial r(t) ∈ Z[t] such that r(γ)=0L. Suppose that L has no (Discr (r(t)) · Prod(r(t)))-torsion. Then the Lie ring Discr (r(t)) · L can be embedded into a Lie∗ ring K that is graded by the semi-group (Q, ·) and supported∗ by the root set X of r(t) in Q. Let us use the abbreviations δ := Discr (r(t)) and π := Prod(r(t)). We may assume ∗ that r(t) is not a constant polynomial, since otherwise we trivially have δ · L = 0L, and there is nothing to prove. Let λ1,...,λl be the distinct roots with respective multiplicities m1,...,ml and let a be the leading coefficient, so that

mi r(t)= a · (t − λi) . 1 i l ≤Y≤

Let R := Z[λ1,...,λl] be the ring generated by the roots and let F be the field of fractions of R. We next introduce a new Lie ring L := R ⊗Z L with coefficients in R. This new Lie ring L naturally admits the Lie ring endomorphism γ : L −→ L : j aj ⊗ vj 7−→ e P e e e e 18 j aj ⊗ (γ(vj )) and this map inherits the property r(γ)=0 . We further define the ideal Le PT := Ta δ π(L)= Tδ π(L) of L. And for each λ ∈ R, we define the R-submodule of L: · · · e n eEλ := e{v ∈ Le|∃n ∈ N : (γ − λ · ½ ) v ∈ T } e Le n1 n2 = {v ∈ L|∃n1,n2 ∈ N : (δ · π) · (γ − λ · ½L) v =0L}. e e e e Claim 1: For all λ, µ ∈ R, we have e e a. T ⊆ ν R Eν . ∈ b. [Eλ, Eµ] ⊆ Eλ µ. T · c. If r(λ)=0= r(µ) and r(λ · µ) 6=0, then [Eλ, Eµ] ⊆ T .

Proof. (a.) This property holds trivially.

Proof. (b.) Select arbitrary v ∈ Eλ and w ∈ Eµ. By deﬁnition, there exist n1,n2,n3,n4 ∈

n1 n2 n3 n4 ½ N such that (δ · π) · (γ − λ · ½L) v = 0L = (δ · π) (γ − µ · L) w. By using the e e n1+n3 e n2+n4 commutator formula of lemma 5.1.1, we get: (δ · π) · (γ − λ · µ · ½ ) [v, w]=0 . Le Le e b Proof. (c.) In view of the above, it suﬃces to show that Eλ µ ⊆ T . Let us abbreviate i e · ν := λ·µ. Let r(t) be given by r(t) := 0 i d ai ·t ∈ Z[t]. Then r(t)−r(ν)= s(t)·(t−ν), j k ≤ ≤ where s(t)= 0 i d ai ·( j+k=i t ·ν ) ∈ R[t]. Select an arbitrary v ∈ Eν . By deﬁnition, ≤ ≤ P n1 n2 there exist n1,n2 ∈ N such that (a · δ · π) · (γ − ν · ½ ) v =0 . Then also P P Le Le

n1 n2 n2 n1 n2 ½ (δ · π) · (r(γ) − r(ν) · ½ ) v = s(γ) (δ · π) · (γ − ν · ) v =0 . (11) Le e Le Le But, since r(γ) vanishese on L, we also have:e e

n1 n1 j n2 n1 (r(γ) − r(eν) · ½ ) v = e · r(γ) ((−r(ν)) · v) = (−r(ν)) · v. (12) Le j, n j+n =n 2 X2 1

e ne2 n1 By combining (11) and (12), we obtain (δ · π) · r(ν) · v =0L. Since r(λ)=0= r(µ) n2+n3 e and r(λ · µ) 6= 0, there exists a n3 ∈ N such that (δ · π) · v =0 . So v ∈ T . Le

Claim 2: The R-submodule K := λ X Eλ is a Lie subring of L. ∈ P Proof. According to the Jacobi-identity and the bi-linearity of thee Lie bracket, the Lie R-subalgebra of L generated by the R-submodule K of L is the R-span of left-normed words w of the form w := [v1,...,vn], where each vi is contained in some Kµi . So it suﬃces to show thate for such a word w, we have w ∈ K. Lete us do this. We may suppose that r(µ1)= ··· = r(µn) = 0, since otherwise w is contained in the ideal T and therefore in K. If µ1,µ1 · µ2,...,µ1 ··· µn := λ are all roots of r(t), then we need only apply claim 1.b (n−1)-times in order to conclude that w ∈ Kλ and therefore w ∈ K. Else, there exists an index n0 ∈{1,...,n−1}, such that µ1 ··· µn0 := µ is a root, but µ·µn0+1 is not. Deﬁne u := [v1,...,vn0 ]. By applying claim 1.b (n0 − 1)-times, we see that u ∈ Kµ. By applying claim 1.c, we see that [u, vn0+1] ∈ T . So also w = [[u, vn0+1], vn0+2,...,vn] ∈ T ⊆ K.

n1 We now note that T = {v ∈ K|∃n ∈ N : (δ ·π) ·v =0K }. So we consider the quotient

K := K/T = ( Eλ)/T = (Eλ/(Eλ ∩ T )) = (Eλ/T ). λ X λ X λ X X∈ X∈ X∈

For each λ ∈ Q, we deﬁne the R-submodule Kλ of K by:

Eλ/T if λ ∈ X, Kλ := (T/T if λ 6∈ X.

19 × Claim 3: K = λ Q Kλ is a grading of the Lie ring K by (Q , ·) and its support is contained in X. ∈ L

Proof. In view of claims 1 and 2, we need only show that the decomposition K = λ Q Kλ ∈ of K into R-submodules is direct. We select arbitrary v ∈ K ,...,v ∈ K such that λ1 λ1 λl λl P vλ1 + ··· + vλl ∈ T and we then need to show that vλ1 ,...,vλl ∈ T . So we select an arbitrary i ∈{1,...,l} and will show that vλi ∈ T . By deﬁnition, there exist n1,n2 ∈ N such that for all j ∈{1,...,l}, we have:

n2 n1 (δ · π) · (γ − λj · ½K ) vλj =0K (13)

n1 Deﬁne the auxiliary polynomial s(t) = j=i(t − λj ) ∈ R[t]. Then the theory of resultants tells us that there exist polynomialse g6 (t),h(t) ∈ R[t] such that Q n1 n1 n1 g(t) · s(t)+ h(t) · (t − λi) = Res(s(t), (t − λi) )= s(λi) . (14)

Since T is invariant under γ and multiplication by elements of R, we see that (δ · π)n2 · n2 s(γ)(vλi ) = (δ · π) · s(γ)(vλ1 + ··· + vλl ) ∈ T. So, by deﬁnition, there exists a n3 ∈ N such that e n2+n3 e e (δ · π) · s(γ)(vλi )=0K. (15)

By ﬁrst evaluating (14) in γ and then in vλi , and by substituting (13) and (15), we obtain: e n2+n3 n1 n2+n3 (δ · π) · s(λi) vλ = (δ · π) · g(γ) (s(γ)vλ ) e i i n2+n3 n1 + (δ · π) · h(γ) ((γ − λi · ½K ) vλi ) e e = 0K. e e Since the factor s(λi) divides δ in the ring R, there exists some n4 ∈ N such that also n2+n3+n2 n4 (δ · π) · · vλi =0K . We may therefore conclude that vλi ∈ T . Claim 4: We have the inclusion 1 ⊗ (δ · L) ⊆ K.

Proof. For each i ∈ {1,...,l} and each j ∈ {0,...,mi − 1}, we define the polynomial mi j Pi,j (t) := r(t)/(t − λi) − with coefficients in the ring R := Z[λ1,...,λl]. Let us first show that, for each polynomial Pi,j (t), there exists a coefficient θi,j ∈ R such that

Discr (r(t)) = θi,j · Pi,j (t). (16) ∗ 1 i l 0 j mi 1 X≤ ≤ ≤ X≤ − The partial fraction decomposition of a/r(t) is given by

(j) a 1 a j mi = · (λi) · (t − λi) − . r(t) j! Pi,0(t) 1 i l 0 j mi 1 ! X≤ ≤ ≤ X≤ − For each i ∈{1,...,l}, we deﬁne the auxiliary polynomial si(t) := 1 j l(t − λj ) ∈ R[t]. ≤j=≤i 6 2m Then the j’th derivative of a/(Pi,0(t)) is clearly of the form bi,j (Qt)/(si(t)) , for some explicitly computable bi,j (t) ∈ R[t]. We see, in particular, that

bi,j(λi) a = 2m · Pi,j (t). j! · (si(λi)) 1 i l 0 j mi 1 X≤ ≤ ≤ X≤ − After multiplying both sides of this equality by Discr (r(t))/a, we see that ∗

(m − 1)! m(l 1) 2d2 m θi,j := · (−1) − · a · (λk − λn)  · bi,j (λi) j! 1 k,n l  i=≤kY=n≤=i   6 6 6    20 is a solution to (16) in the ring of coeﬃcients R, where m := max{m1,...,ml}.

We now select an arbitrary v ∈ L. Corresponding with the (i, j)’th term of 16, we mi j deﬁne vi,j := Pi,j (γ)(1⊗v) ∈ L. Then we observe that (γ−λi · ½K ) − vi,j = r(γ)(1⊗v)=

0K , so that vi,j ∈ Kλi . By evaluating the expression (16) in γ and then in 1 ⊗ v, we see that 1 ⊗ (δ · v)= θ · v ∈ R · K ⊆ K. So we e 1 i l 0ej mi 1 i,j i,j 1 i el 0 j mi 1 λi e may indeed conclude≤ that≤ 1 ⊗≤(≤δ · L−) ⊆ K. ≤ ≤ ≤ ≤ − P P P P e Claim 5: The Lie ring δ · L embeds into the Lie ring K = K/T .

Proof. We consider the embedding E : δ · L −→ K : δ · v 7−→ 1 ⊗ (δ · v) and the projection P : K −→ K/T : w 7−→ w mod T . Since L has no (δ · π)-torsion, neither has δ · L. So the composition P ◦ E : δ · L −→ K/T is the required embedding of Lie rings.

5.2 Bounded nilpotency of graded Lie rings

Theorem 5.2.1. Let F be a ﬁeld. If a Lie ring K is graded by (F×, ·) with ﬁnite, |X| arithmetically-free support X, then K is nilpotent of class at most |X|2 .

We now fix a finite, arithmetically-free subset X of the multiplicative group (F×, ·) of a field F. We want to show that every Lie ring L that is graded by F× and supported |X| by X is nilpotent of class at most |X|2 . If |X| = 0, then L = 0, and we indeed have 20 21 c(L)=0 ≤ 0 . If |X| = 1, then [L,L] ⊆ LG X = {0L}, so that we have c(L)=1 ≤ 1 . We may therefore assume from now on that \x := |X|≥ 2. We begin by proving a useful property of this arithmetically-free set X. Lemma 5.2.2 (Growth or progression). Consider an abelian group (F, ·). For every k ∈ N, and elements b1,...,bk of F \{1F }, we can define the set S := S(b1,...,bk) := 0 i k{bσ(1) ··· bσ(i)|σ ∈ Sym(k)} of partial products. Then: |S|≥ k +1, or S contains ≤ ≤ a progression g,g · c,g · c2,...,g · ck with c ∈{b ,...,b }. S 1 k Proof. Let us prove the lemma by induction on k ∈ N. If k = 1, then the statement is clearly true. So we may suppose that k > 1. We suppose that the second conclusion fails for S and we will prove that |S| ≥ k + 1. Define S′ := S(b1,...,bk 1) and S := − S(b1,...,bk). Then S = S′ ∪ (S′ · bk) and the second conclusion also fails for S′. The 2 induction hypothesis now tells us that |S′|≥ k. If S = S′ · bk, then S = S′ · bk = S′ · bk = k ··· = S′ · bk. But then S′ would satisfy condition 2 (contradicting our assumption). So S′ · bk 6= S, and therefore |S| = |S′ ∪ (S′ · bk)|≥ 1+ |S′|≥ 1+ k.

Proposition 5.2.3 (Escape). For all elements a,b1,...,b X of X, there exists a permu- | | tation σ ∈ Sym(|X|) and a cut-oﬀ k ∈{1,..., |X|} such that a · bσ(1) ··· bσ(k) ∈ G \ X. Proof. We suppose that the conclusion is false, and we will derive a contradiction. Con- sider S := S(b1,...,b X ). Then a · S ⊆ X and |S| = |a · S| ≤ |X|. Lemma 5.2.2 now | | 2 X implies that S contains a progression b,b · c,b · c ,...,b · c| | with c ∈ X. Then a · S, and 2 X therefore X, contains the progression a · b, (a · b) · c, (a · b) · c ,..., (a · b) · c| | with c ∈ X. We conclude that X is not arithmetically-free. This is a contradiction.

We will break down our proof of theorem 5.2.1 into several steps (that are directly inspired by the work of Kreknin and Kostrikin (cf. E. Khukhro’s book [26])). Lemma 5.2.4. Consider a Lie ring L. For every permutation σ ∈ Sym(X) and for all elements v, w1,...,wx ∈ L, we have [v, w1,...,wx] ∈ [v, wσ(1),...,wσ(x)]+idL([v, [L,L]]).

21 Proof. A repeated application of the Jacobi-identity shows that the diﬀerence vector [v, w1,...,wx] − [v, wσ(1),...,wσ(x)] is in the Z-linear span of [v, [L,L]], [v, [L,L],L], ..., [v, [L,L],L,...,L], and therefore in the ideal of L that is generated by [v, [L,L]].

x times

Proposition| 5.2.5.{z }Consider a Lie ring L that is graded by (F×, ·) and supported by X. For every homogeneous ideal I of L, we have [I,L,...,L] ⊆ [I, [L,L]].

x times

Proof. Let the gradings be given by L = λ |F×{zLλ }and I = λ F× Iλ. Since the Lie-bracket is bi-linear, it suﬃces to show that∈ every left-normed bracket∈ of the form L L u := [va, wb1 ,...,wbx ], with va ∈ Ia and wb1 ∈ Lb1 ,...,vbx ∈ Lbx , is contained in the ideal [I, [L,L]]. If any of the a,b1,...,bx are in F× \ X, then we indeed have u =0 ∈ [I, [L,L]]. So we may assume that a,b1,...,bx ∈ X. According to proposition 5.2.3, there exists a permutation σ ∈ Sym(X) and a cut-oﬀ k ∈ {1,...,x}, such that c := a · bσ(1) ··· bσ(k) ∈ F× \ X, and therefore Lc = {0L}. We then apply lemma 5.2.4 to the bracket u and σ in order to obtain u ∈ [[va, wσ(1),...,wσ(k)], wσ(k+1),...,wσ(x)] + [I, [L,L]]. The grading property implies that the sub-bracket [va, wσ(1),...,wσ(k)] is contained in the homogeneous component Lc = {0L} of L. So u ∈ [0, wσ(k+1),...,wσ(x) ] + [I, [L,L]] = [I, [L,L]].

Proposition 5.2.6. Consider a Lie ring L that is graded by (F×, ·) and supported by X. For every homogeneous ideal I of L and l ∈ N, we have [I,L,...,L] ⊆ [I, [L,L],..., [L,L]].

lx times l times Proof. We proceed by induction on l ∈ N. The base case| {zl =} 1 corresponds| {z with the} previous proposition. So we may assume that l> 1. Then J := [I,L,...,L ] is a homo-

(l 1)x times − geneous ideal of L and proposition 5.2.5 gives us the inclusion [J,L,...,L] ⊆ [J, [L,L]]. | {z } x times The induction hypothesis for l − 1 also gives the inclusion J ⊆ [I, [L,L],..., [L,L]]. These | {z } (l 1) times − two combine to give [I,L,...,L] ⊆ [[I, [L,L],..., [L,L]], [L,L]] = [I, [L,L],..., [L,L]]. | {z } lx times (l 1) times l times − | {z } | {z } | {z } Proposition 5.2.7. Consider a Lie ring L that is graded by (F×, ·) and supported by X. For each l ∈ N, we have the inclusion of ideals: Γ2+lx(L) ⊆ Γl+1(∆1(L)).

Proof. We need only specialize the previous proposition to the homogeneous ideal I := [L,L] of L.

Proposition 5.2.8. Let s ∈ N. For every F×-graded Lie ring L that is supported by X, we have the inclusion of ideals Γ1+(xs 1)/(x 1)(L) ⊆ ∆s(L). − − Proof. We proceed by induction on s. If s = 1, then the statement is true by deﬁnition: s 1 Γ2(L) = [L,L] = ∆1(L). Now suppose that s> 1. We then deﬁne l := (x − − 1)/(x − 1) and we observe that (xs −1)/(x−1)= 1+xl. According to proposition 5.2.7, we then have Γ2+xl(L) ⊆ Γ1+l(∆1(L)). We note that the Lie ring ∆1(L) is naturally graded by (F×, ·) and supported by X. So the induction hypothesis for s − 1 tells us that Γl+1(∆1(L)) ⊆ ∆s 1(∆1(L)) = ∆s(L). We therefore obtain Γ2+xl(L) ⊆ Γ1+l(∆1(L)) ⊆ ∆s(L). −

Corollary 5.2.9. If a Lie ring L is graded by (F×, ·) and supported by X, then it is |X| nilpotent of class c(L) ≤ |X|2 .

22 Proof. We may suppose that x ≥ 2. Since X is an arithmetically-free subset of F×, it does not contain 1. So, according to Shalev’s generalization of Kreknin’s theorem in [38], the Lie ring L is solvable of derived length dl(L) ≤ 2x =: s. Proposition 5.2.7 now gives Γx2x +1(L) ⊆ Γ1+(xs 1)/(x 1)(L) ⊆ ∆s(L) ⊆ ∆dl(L)(L)= {0L}. − − This ﬁnishes the proof of theorem 5.2.1.

5.3 Proof of theorem 1.2.6 Proof. We ﬁrst use theorem 5.1.2 to embed Discr (r(t))·L into a Lie ring K that is graded ∗ by (Q×, ·) and supported by the root set X of r(t). According to corollary 3.4.3, X is an arithmetically-free subset of (Q×, ·). So theorem 5.2.1 implies that Discr (r(t)) · L is d ∗ nilpotent of class c(Discr (r(t)) · L) ≤ d2 . Since (L, +) has no Discr (r(t))-torsion, also ∗ d ∗ L is nilpotent of class at most d2 .

6 Structure theorems for groups 6.1 Proof of theorem 1.2.4 Preliminaries. We recall some basic terminology. Let G be a group and let α be one of its automorphisms. A subgroup H of G is α-invariant if α(H) ⊆ H. An α- invariant section of G is a quotient A/B of a α-invariant subgroup A of G by a α- invariant, normal subgroup B of A. A characteristic section of G is a quotient A/B of a G-characteristic group A by a G-characteristic subgroup B of A. A section A/B of G is proper if |A/B| < |G|. The following result is well-known. Lemma 6.1.1. Let G be a finite group and let α be a fix-point-free automorphism. (i.) 1 The map τ : G −→ G : x 7−→ x− · α(x) is a bijection. (ii.) If A/B is an α-invariant section then the corresponding automorphism α : A/B −→ A/B : a · B 7−→ α(a) · B is also fix-point-free. (iii.) If p is a prime, then G has a p-Sylow subgroup that is α- invariant. (iv.) If G is solvable and H is a Hall-subgroup of G, then some conjugate of H is α-invariant.

Let m ∈ Z. A group H is said to be an m′-group if H has no m-torsion (i.e.: only the m trivial element h ∈ H satisﬁes h =1H ).

6.1.1 Solvable case Theorem 6.1.2. Consider a ﬁnite, solvable group G, together with a ﬁx-point-free automorphism α : G −→ G, and an identity r(t) of the automorphism. Then G has a characteristic Cong(r(t))-subgroup S with nilpotent Cong(r(t))′-quotient G/S. We suppose that the statement is false and we will then deduce a contradiction. We may well suppose that (G, α, r(t)) is a counter-example of minimal order |G|. We then see that G is not nilpotent and not a Cong(r(t))-group. We also see that all proper, α-invariant sections of G satisfy the conclusion of the theorem. We will use these observations repeatedly but not always explicitly.

Claim 1: If a characteristic subgroup K of G is a Cong(r(t))-group, then K = {1G}.

Proof. Suppose that the characteristic subgroup K is non-trivial. Then the proper, α- invariant section G/K of G has a characteristic, Cong(r(t))-subgroup S/K with nilpotent Cong(r(t))′-quotient (G/K)/(S/K) =∼ G/S. But then S is a characteristic Cong(r(t))- subgroup of G with nilpotent Cong(r(t))′-quotient, contradicting our initial assumption.

23 Claim 2: If a characteristic subgroup K of G is proper, then K is a nilpotent Cong(r(t))′-group.

Proof. The proper section K of G is an extension of a characteristic Cong(r(t))-group L by a nilpotent Cong(r(t))′-group K/L. This L is also characteristic in G. Claim 1 implies that L =1. So K =∼ K/L is a nilpotent Cong(r(t))′-group.

Claim 3: If a characteristic subgroup A of G is non-trivial, then the quotient G/A is a Cong(r(t))-group or a nilpotent Cong(r(t))′-group.

Proof. The proper section G/A of G has a characteristic Cong(r(t))-subgroup K/A with nilpotent Cong(r(t))′-quotient (G/A)/(K/A) =∼ G/K. By lifting, we obtain a characteristic subgroup K of G. If K = G, then G/A = K/A is a Cong(r(t))-group. So we may suppose that K is a proper subgroup. Claim 2 then implies that the group K, and therefore the section K/A, is a Cong(r(t))′-group. So K/A is both a Cong(r(t))-group and a Cong(r(t))′-group, and therefore the trivial group. We conclude that G/A = G/K is a nilpotent Cong(r(t))′-group.

Let F be the (ﬁrst upper) Fitting subgroup of G.

Claim 4: The quotient G/F is non-trivial and elementary-abelian. So there is a prime ∼ l p and a natural number l such that G/F = (Zp, +). Proof. As a counter-example to the theorem, G is not nilpotent. So F is a proper subgroup of G and G/F is non-trivial. Now every proper, characteristic subgroup S/F of G/F lifts to a proper, non-trivial, characteristic subgroup S of G. By claim 2, such an S is nilpotent, and therefore contained in F . So the solvable group G/F is characteristically simple, and therefore elementary-abelian.

Claim 5: The group F is a q-group, for some prime q not dividing Cong(r(t)).

Proof. Suppose that the nilpotent group F is not a q-group. Such a group F then de- composes as the direct product of proper, non-trivial, characteristic subgroups A and B: F = A × B. According to claim 2, the non-trivial section F/A =∼ B of G/A is a Cong(r(t))′-group. So G/A is not a Cong(r(t))-group. According to claim 3, the group G/A is then nilpotent. Similarly, we obtain that G/B is nilpotent. So G/(A ∩ B) is nilpotent (a contradiction).

Claim 6: F is non-trivial and elementary-abelian. So there is a natural number k ∼ k such that F = (Zq , +). Proof. Since G is solvable and non-trivial, so is F . Let Φ(F ) be the Frattini subgroup of F . We now consider the characteristic series 1 ≤ Φ(F ) ≤ F < G of G. We suppose that F is not elementary abelian and we will derive a contradiction. In this case, the inclusions of the series are all strict. Claim 2 implies that F is a Cong(r(t))′-group, so that G/Φ(F ) is not a Cong(r(t))-group. Claim 3 now implies that G/Φ(F ) is nilpotent, so that also G is nilpotent (a contradiction).

Since G is not nilpotent, p 6= q.

Claim 7: F has an α-invariant complement P in G.

Proof. Zassenhaus’ theorem implies that G =∼ F ⋊ (G/F ). By using lemma 6.1.1, we can ﬁnd a complement P of F in G that is α-invariant.

24 We now consider the k-dimensional vector space V := Fk over the algebraically-closed ﬁeld F := Fq.

Claim 8: The vector space V admits a non-trivial, elementary-abelian p-group of automorphisms R ≤ GL(V ) and an automorphism f ∈ NGL(V )(R) \ CGL(V )(R) such that r(f)= {0V }.

Proof. Since P is α-invariant, the group P ⋊ hαi is well-deﬁned and it acts naturally on F via conjugation within the larger group (F ⋊ P ) ⋊ hαi = F ⋊ (P ⋊ hαi). Let this action be given by the map θ : P ⋊ hαi −→ Aut(F ). Since the Fitting subgroup is self-centralizing, this action is faithful on P , so that θ(P ) is a non-trivial p-subgroup of Aut(F ). It is clear that θ(α) normalizes θ(P ). Since α is ﬁx-point-free on P , θ(α) does not centralize θ(P ). Trivially, r(t) is an identity of the automorphism θ(α) of F .

We now identify the Fitting subgroup F of G with the additive group of the vector k k space Fq . We then deﬁne R as the subgroup of GL(Fq ) corresponding with θ(P ) and f k as the automorphism of GL(Fq ) corresponding with θ(α). By extending the scalars from Fq to Fq, we naturally obtain the desired subgroup R of GL(V ) and the automorphism f ∈ GL(V ).

We had already observed that p 6= q. This implies that the operators of the abelian group R can be diagonalized simultaneously. So let χ1,...,χl be the distinct characters of R with coeﬃcients in F× and let Vχi := {v ∈ V |∀b ∈ R : b(v) = χi(b) · v} be the (by deﬁnition non-trivial) character-space corresponding with such a character χi : R −→ F×.

Then we have the decomposition V = Vχ1 ⊕···⊕ Vχl of V into its R-character-spaces. Since f normalises R, we can deﬁne, for each n ∈ Z and each character χi, a new character n n n (f ∗ χi) by the rule: ∀b ∈ R : (f ∗ χi)(b) := χi(f − ◦ b ◦ f ). So the group hfi acts as permutations on the set of character spaces according to the rule

n n f (Vχi )= V(f χi). ∗

Claim 9: The automorphism f maps some character space, say Vχm , onto a diﬀerent character space.

1 Proof. Since f does not centralise R, there exists an element b ∈ R such that f − ◦b◦f 6= b. 1 These operators f − ◦ b ◦ f and b must diﬀer in some element of some character-space, 1 say v ∈ Vχm . Now, if f(Vχm )= Vχm , then we obtain the contradiction (f − ◦ b ◦ f)(v)= 1 1 1 (f − ◦b)(f(v)) = f − (χm(b)·f(v)) = χm(b)·f − (f(v)) = b(v). So we may indeed conclude that f(Vχm ) 6= Vχm .

u Let u be the minimal natural number such that f (Vχm ) = Vχm . By the previous remark, we necessarily have u ≥ 2.

Claim 10: We have Vχm = {0V }.

Proof. Let v be any vector of Vχm and let us show that v = 0V . We ﬁrst consider the u-periodic decomposition r(t) = j ru,j (t) of r(t) into partial sums. By evaluating this decomposition in f, we obtain the equality of V -endomorphisms: r(f) = ru,0(f)+ P ru,1(f)+···+ru,u 1(f). By further evaluating this decomposition in the distinguished vec- − tor v ∈ Vχm , we obtain the decomposition: (r(f))(v) = (ru,0(f))(v)+ ···+ (ru,u 1(f))(v). By assumption, the map r(f) vanishes on all of V , so that the left-hand side of this− decomposition is the trivial vector, 0V . By construction, we also have the inclusions of vector u 1 spaces ru,0(f)(Vχm ) ⊆ Vχm ,...,ru,u 1(f)(Vχm ) ⊆ f − (Vχm ). Since the character-spaces −

25 u 1 Vχm , f(Vχm ), ...,f − (Vχm ) are are distinct, they are linearly independent. So we may conclude that all of the terms in this decomposition vanish:

(ru,0(f))(v)= ··· = (ru,u 1(f))(v)= {0V }. (17) − We had already observed that q does not divide Cong(r(t)). By Bezout’s theorem, there exist polynomials s0(t),...,su 1(t) ∈ Z[t] such that s0(t)·ru,0(t)+···+su 1(t)·ru,u 1(t) ≡ 1 mod q. By evaluating this− decomposition in f, we obtain the decomposition− − of V - endomorphisms

s0(f) ◦ ru,0(f)+ ··· + su 1(f) ◦ ru,u 1(f)= ½V . (18) − −

By further evaluating this expression in the distinguished vector v ∈ Vχm , we obtain

v = ½V (v) (18) = (s0(f) ◦ ru,0(f))(v)+ ··· + (su 1(f) ◦ ru,u 1(f))(v) − − = s0(f)(ru,0(f)(v)) + ··· + su 1(f)(ru,u 1(f)(v)) − − (17) = s0(f)(0) + ··· + su 1(f)(0) − = 0V .

But the character spaces are non-trivial by definition. This contradiction finishes the proof of theorem 6.1.2. Corollary 6.1.3. Consider a finite, solvable group G, together with a fix-point-free automorphism α : G −→ G, and an identity r(t) of the automorphism. If G has no Cong(r(t))- torsion, then G is nilpotent.

6.1.2 General case Proof. We suppose that the conclusion is false and we will deduce a contradiction (this is essentially Thompson’s proof of the Frobenius conjecture [41], and we include it here for completeness). Let (G, α, r(t)) be a counter-example of minimal order |G|. We note that every proper, α-invariant section of G is then nilpotent. If |G| is the power of a single prime, then G is nilpotent (a contradiction). So some odd prime p divides |G|. Lemma 6.1.1 gives us a p-Sylow subgroup satisfying α(P )= P . We now distinguish between two cases.

Suppose ﬁrst that P has a non-trivial, normal, α-invariant subgroup H with NG(H)= G. Then the section G/H is proper and α-invariant. So G is the extension of the nilpotent group H by the nilpotent group G/H. The group G is therefore solvable. Corollary 6.1.3 now implies that G is nilpotent (a contradiction). Suppose next that every non-trivial, normal, α-invariant subgroup H of P has a normalizer NG(P ) that is properly contained in G. As a proper, α-invariant section of G, this NG(H) is nilpotent. So theorem 1 of Thompson [42] gives us a normal complement K to P in G: G =∼ K ⋊ P . As a proper, α-invariant section of G, this K is nilpotent. As an extension of a nilpotent group K by a nilpotent group P , our group G is solvable. Corollary 6.1.3 now implies that G is nilpotent (a contradiction).

6.2 Proof of theorem 1.2.5 Proof. Let G be a nilpotent group having an endomorphism γ : G −→ G with a good identity r(t), say of degree d. Suppose that G has no (Discr (r(t)) · Prod(r(t)))-torsion. ∗ We then consider the Lie ring L(G) := n N Γn∗ (G)/Γn∗+1(G) that corresponds with the ∈ isolators of the lower central series (Γn(G))n N of G. This Lie ring has a number of good properties. It is also nilpotent andL its class∈ coincides with that of G. Moreover,

26 (L(G), +) has no (Discr (r(t)) · Prod(r(t)))-torsion. And the Lie ring naturally admits ∗ an automorphism α : L(G) −→ L(G) of Lie rings that satisﬁes r(α)=0L(G). So we need d only apply corollary 1.2.6 to conclude that c(G) ≤ d2 .

6.3 Proof of corollary 1.2.8 Proof. Let α : G −→ G be the ﬁx-point-free automorphism of the ﬁnite group G. Rowley’s theorem 1.1.1 implies that G is solvable. Corollary 6.1.3 implies that G is of the form B ⋊N, where B is a characteristic Cong(r(t))-subgroup and N is a nilpotent, Cong(r(t))′- subgroup. According to lemma 6.1.1, we may further assume that N is α-invariant. Corollary 1.2.7 implies that N is of the form C × D, where C is a nilpotent (Discr (r(t)) · d ∗ Prod(r(t)))-group and D is a nilpotent group of class at most d2 .

6.4 Proof of corollary 1.1.6 Definition 6.4.1 (a, b, c). Let r(t) ∈ Z[t] be irreducible over Q. Suppose first that r(0) = 0 or r(1) = 0. Then we define a := b := c := 1. Suppose next that r(0) · r(1) 6= 0. Then we define a := r(1). Let u be the largest natural number such that r(t) ∈ Z[tu]. Let s(t) be the unique s(t) ∈ Z[t] such that r(t)= s(tu). Then we define b := Cong(s(t)) and c := Discr (s(t)) · Prod(s(t)). ∗ Proof. (Corollary 1.1.6) We can now prove the theorem case-by-case. Suppose first that r(0) = 0. Then r(t) = 0 and we have, for all x ∈ G, the equality α(x)=1G. So G is trivial. Suppose next that r(1) = 0. Then r(t) = t − 1 and we have, for all x ∈ G, that α(x)= x. Since α is fix-point-free, we conclude again that G is the trivial group. Finally, suppose that r(0)·r(1) 6= 0. Then a is clearly non-zero. Since r(t) is irreducible, so is s(t). Since s(0) · s(1) · s(t) has no divisors of the form h(tv+1) for any v ∈ N, the polynomial is good. So proposition implies that b 6= 0. Finally, since s(t) 6= 0, we see that c 6= 0. Let A be some a-Hall subgroup A of G. Lemma 6.1.1 gives us some a′-Hall subgroup H of G that is α-invariant. Then G factors as A · H. Now we consider the restriction of α to H, and apply corollary 1.2.8 to obtain the desired subgroups B, C, and D of H.

7 Applications 7.1 Linear identities Definition 7.1.1 (n-abelian groups and power endomorphisms). Let n be a natural number. A group G is said to be n-abelian (or n-commutative) if the n’th power map n γn : G −→ G : x 7−→ x is an endomorphism. An endomorphism γ : G −→ G of the group G is said to be a (universal) power endomorphism if there is a natural number n such that, for all x ∈ G, we have γ(x)= xn. These n-abelian groups have been studied by various authors, including Levi [31], Baer [4], Schenkman—Wade [37], and J. Alperin [2]. They were ultimately classified by Alperin. Theorem 7.1.2 (Alperin [2]). Let n> 1 be a natural number. The finite n-abelian groups are the subdirect product of a finite abelian group, a finite group of exponent dividing n, and a finite group of exponent dividing (n − 1). 2 Now let G be an n-abelian group, for some n > 1. Then n − n 6= 0 and rn(t) := −n + t is clearly a monic and monotone identity of the n’th power endomorphism γn, i.e.: 2 rn(γn)=1G. Moreover, this γn is a fix-point-free automorphism if G has no (n − n)- torsion. More generally, we may consider endomorphisms with a linear identity that is not necessarily monic or monotone.

27 Corollary 7.1.3. Let G be a ﬁnite group with an endomorphism with a linear identity, say −n+m·t ∈ Z[t]. Suppose that (m·n·(m−n)) 6=0. Then either G has (m·n·(m−n))- torsion, or G is abelian.

Proof. Set r(t) := −n + m · t. It is not diﬃcult to verify that r(0) · r(1) · Cong(r(t)) · Discr (r(t)) · Prod(r(t)) divides a natural power of m · n · (m − n). So the endomorphism is a ﬁx-point-free∗ automorphism of the group and we may apply corollary 1.2.7.

7.2 Cyclotomic identities Deﬁnition 7.2.1 (Splitting and cyclotomic automorphisms). Let n > 1 be a natural number. An automorphism α : G −→ G is said to be n-splitting if the polynomial n 1 Ψn(t) := 1 + t + ··· + t − is a monotone identity of α, i.e.: Ψn(α)=1G. We say that the automorphism is n-cyclotomic if the n’th cyclotomic polynomial Φn(t) is a monotone identity of α, i.e.: Φp(α)=1G. More generally, we say that α is splitting (resp. cyclotomic) if it is m-splitting (resp. m-cyclotomic) for some natural m> 1.

It is well-known that Ψn(t)=Φn(t) if n is a prime. So the automorphism α in theorem 1.1.1 is an |α|-splitting automorphism, while the automorphism α in theorems 1.1.2 and 1.1.3 is an |α|-cyclotomic automorphism (cf. remark 1.1.4). We now recall three companions of theorems 1.1.1, 1.1.2, and 1.1.3. Theorem 7.2.2 (Ersoy [14]). Let n ∈ N be odd. If a finite group G has an n-splitting automorphism, then G is solvable. Theorem 7.2.2 is a (partial) generalization of theorem 1.1.1 and it based on the classification of the finite simple groups. Such automorphisms naturally appear in the study of automorphisms with finite order and with finite Reidemeister-number [5, 22]. Theorem 7.2.3 (Hughes—Thompson [20]; Kegel [24]). Let p be a prime number. If a finite group G has a p-cyclotomic automorphism, then G is nilpotent. Theorem 7.2.3 clearly generalizes theorem 1.1.2. The solvability of G was proven by Hughes—Thompson using the fundamental results of Hall and Higman about minimal polynomials of operators on finite-dimensional vector spaces [17]. The nilpotency of G is due to Kegel. These p-split automorphisms naturally appear in the study of almost-fix- point-free automorphisms of prime order by Bettio, Endimioni, Jabara, Wehrfritz, Zappa, and others [5, 13]. Theorem 7.2.4 (E. Khukhro [25]). For each natural number d and each prime p, there exists a natural number C(d, p) with the following property. If a finite p-group G on d generators has a p-cyclotomic automorphism, then c(G) ≤ C(d, p). In combination with theorem 1.1.3, this gives us a bound on the class of all finite, nilpotent groups with a p-split automorphism. One could hope to obtain an upper bound on c(G) that depends only on p (as in theorem 1.1.3), but examples show that this is not possible. The theorem applies, in particular, to finite p-groups with a partition. More- over, Kostrikin’s positive solution [27] of the restricted Burnside problem in exponent p corresponds with the automorphism x 7→ x.

Examples show that theorem 7.2.2 cannot be extended to all natural numbers n. But we can generalize theorem 7.2.3 of Hughes—Thompson and Kegel, theorem 1.1.3 of Higman and Kreknin—Kostrikin, as well as theorem 7.2.4 of Khukhro:. Theorem 7.2.5. Compact groups with a cyclotomic automorphism are locally-nilpotent. Theorem 7.2.6. Consider a locally-nilpotent group G with an n-cyclotomic automorphism. Let p be the largest prime divisor of n. Then every k-generated subgroup N of G (p−1) is nilpotent of class c(N) ≤ max{(p − 1)2 , C((2p − 3) · k,p)}.

28 Here, we always assume that compact groups are Hausdorff. These theorems allow us to to recover (and extend) two theorems of Jabara about automorphisms with a finite Reidemeister-number. Corollary 7.2.7 (Jabara [22]). Consider a residually-finite group G admitting an automorphism α : G −→ G of prime order p. If the Reidemeister-number of α is finite, then G has an α-invariant subgroup N of finite index that is nilpotent of class (p−1) c(N) ≤ (p − 1)2 . Corollary 7.2.8 (Jabara [22]). Consider a finitely-generated, solvable group G with an automorphism α : G −→ G of prime order p. If the Reidemeister-number n of α is finite, (p−1) then G has a finite-index subgroup N that is nilpotent of class c(N) ≤ (p − 1)2 and n (p−1) dl(G) ≤ 22 + A(p,n) + (p − 1)2 . Here, A : P × N −→ N is the map in J. Alperin’s theorem 1 of [1]. We note that Jabara’s proofs of these corollaries implicitly used the classification of the finite simple groups, as well as Zel′manov’s positive solution of the restricted Burnside problem in arbitrary exponent, and the theorems of Hartley, Hartley—Meixner, Fong, and Khukhro (cf. corollary 5.4.1 of [26]). Our proofs will be closely modeled on those of Jabara, but they will avoid all of these rather difficult results.

7.2.1 Proof of theorems 7.2.5 and 7.2.6 For a natural number n, we let n be the radical of n: the product of the distinct primes that divide n. Proposition 7.2.9. Let n > 1 be a natural number. Let p be the largest prime dividing n. Consider a ﬁnite group G on k generators with an n-cyclotomic automorphism α : G −→ G. (i.) Then G is nilpotent. (ii.) If n = p and G has no p-torsion, then c(G) ≤ (p−1) (p − 1)2 . (iii.) If n = p and if G is a p-group, then c(G) ≤ C(k,p). (iv.) If n 6= p, (p−1) (p−1) then c(G) ≤ (p − 1)2 . So c(G) ≤ max{(p − 1)2 , C(k,p)}.

Proof. The automorphism αn/n is n-cyclotomic. So, after replacing α with αn/n, we may assume that n = n. (i.) If n is a prime, then we apply theorem 7.2.3. If n is not a prime, then Φn(1) · Cong(Φn(t)) = 1 (cf. lemma 4.2.4 and proposition 4.2.6), so that we may apply corollary 1.2.7. In either case, G is nilpotent. (ii.) We may apply theorem 1.1.3 or corollary 1.2.8. (iii.) We may apply theorem 7.2.4. (iv.) We need only (p−1) show that an arbitrary q-Sylow subgroup Q of G satisfies c(Q) ≤ (p − 1)2 . We let L be the Lie ring of Q corresponding with the lower central series of Q. Then the natural Lie automorphism α : L −→ L satisfies Φn(α)=0L. Since n is composite, there exists a prime l, distinct from q, that divides n. We may then apply (8) of lemma 4.2.4 u in order to obtain a natural number u such that Φn(t) divides Φl(t ) in the ring Z[t]. u u Then the Lie automorphism α satisfies Φl(α )=0L. The second claim now gives us (l−1) (p−1) c(Q)=c(L) ≤ (l − 1)2 ≤ (p − 1)2 .

This already proves corollaries 7.2.5 and 7.2.6 in the finite case. In order to extend these results from finite groups to locally-(residually-finite) groups, we make some elementary observations. Lemma 7.2.10. Consider a group G with an automorphism α : G −→ G and a mono- n tone identity r(t) := a0 + a1 · t + ··· + an · t ∈ Z[t] of degree n ≥ 1. Suppose that a0,an ∈ {1, −1}. Let H be a finitely-generated subgroup of G and define the subgroup n+1 1 n 1 H := hα− (H),...,α− (H),H,α(H),...,α − (H)i of G. Then H is hαi-invariant and d(H) ≤ (2n − 1) · d(H). Suppose, moreover, that G is locally-(residually-finite). Then e e e 29 H admits a family (Hi)i of H-characteristic, finite-index subgroups with trivial intersection, so that r(t) is a monotone identity of all the induced automorphisms α : H/e Hei e e e H/Hi −→ H/Hi.

1 Proof.e e Let use showe that H is invariant under α and α− . Since an ∈{1, −1}, we observe n+2 n 1 n n 1 that α(H) ⊆ hα− (H),...,α − (H), α (H)i ⊆ hH, α − (H),...,α(H),Hi⊆ H. Since 1 a0 ∈{1, −1}, we similarlye observe that α− (H) ⊆ H. So H is hαi-invariant. By construction, wee also have d(H) ≤ (2n−1)·d(H). Finally, wee suppose that G is locally-(residually-e finite), so that the subgroup H is residually-finite.e e Let He= N1 D N2 D ··· be a normal series of finite-indexe subgroups with finite intersection. Since H is finitely-generated, every subgroup Ni contains theeH-characteristic subgroupeHi := β Aut(H) β(Ni) of finite e ∈ e index in H (the characteristic core). Since i Hi ⊆ i Ni = {1TH }, we see that (Hi)i is the desired series. e e e e T e T e Proposition 7.2.11. Consider a locally-(residually-finite) group G with an n-cyclotomic automorphism. Then every k-generated subgroup H of G is nilpotent and c(H) ≤ max{(p− (p−1) 1) , C((2p − 3) · k,p)}, where p is the largest prime dividing n.

Proof. Let α : G −→ G be the n-cyclotomic automorphism of the group. Let H be the hαi-invariant subgroup of G on k := (2ϕ(n) − 1) generators containing H from lemma 7.2.10. According to lemma 7.2.10, H is residually-(a ﬁnite group on k generatorse with an n-cyclotomic automorphism).e So proposition 7.2.9 implies that H is nilpotent of class (p−1) c(H) ≤ c(H) ≤ max{(p − 1)2 , C(ke(2ϕ(n) − 1),p)}. e

Proof. (Theoreme 7.2.5) The theoremse of Peter—Weyl and Mal′cev imply that G is locally- (residually-ﬁnite). So we need only apply proposition 7.2.11.

Proof. (Theorem 7.2.6) By Mal′cev’s theorems, the group is locally-linear and therefore locally-(residually-ﬁnite). So we need only apply proposition 7.2.11.

7.2.2 Proof of corollaries 7.2.7 and 7.2.8 Proof. (Corollary 7.2.7) It is easy to show that G has an hαi-invariant, finite-index subgroup M such that the induced automorphism αM : M −→ M is p-cyclotomic (cf. lemma 5 of [22]). It is also easy to see that αM fixes only finitely-many elements (cf. lemma 1 and 4 of [22]). As a subgroup of G, this M is residually-finite and therefore locally-nilpotent by proposition 7.2.11. So we may consider the p-Sylow subgroup P of the torsion subgroup T of G. This P is a characteristic subgroup of G such that the locally-nilpotent factor G/P has no p-torsion.

Let us first show that this P is finite. Since G is residually-finite, so is its subgroup P . Since α fixes only finitely-many elements of G, we may select a finite-index subgroup P0 of P such that P0 ∩ CG(α)= {1G}. After replacing P0 with the finite-index subgroup i 0 i p 1 α (P0) of P , we may further assume that P0 is α-invariant. So the restriction of ≤ ≤ − α to the locally-nilpotent p-group P0 is a fix-point-free, p-cyclotomic automorphism. But T it is well-known that such a group P0 is trivial (cf. [18] and [26]). We conclude that P is indeed a finite group. Since P is a finite p-group, P is nilpotent. Theorem 1.2.5 implies that also the quotient G/P is nilpotent. We see, in particular, that G is solvable, and therefore nilpotent (by Khukhro’s theorem of [21]).

Proof. (Corollary 7.2.8) We re-consider Jabara’s proof of theorem B in [22] and we replace theorem A of [22] with corollary 7.2.7. This way, we obtain the subgroup N without relying on the classiﬁcation. By assumption, G is ﬁnitely-generated. So, after replacing N with

30 β Aut(G) β(N), we may further assume that N is characteristic in G. Then the induced ∈ n automorphism α : G/N −→ G/N on the ﬁnite, solvable group G/N ﬁxes at most 22 T G/N elements (cf. lemma 1 and 4 of [22]) and its order divides p. If αG/N has order 1, then 2n |G/N|≤ 2 . Else, we may apply Alperin’s theorem 1 of [1] to G/N and αG/N in order to conclude that dl(G/N) ≤ A(p,m). Since dl(G) ≤ dl(G/N) + dl(N), we are done.

7.3 Anosov identities Definition 7.3.1 (Anosov polynomial). A monic polynomial r(t) ∈ Z[t] is said to be Anosov if and only if r(0) ∈ {−1, 1} and r(t) has no roots of modulus one. We recall that Anosov polynomials naturally appear in the study of Anosov-diffeomorphisms on compact manifolds (e.g. [7, 10, 35]) and we recall one such situation in particular: Theorem 7.3.2 (Manning – [33]). A nil-manifold M admits an Anosov diffeomorphism if and only if the fundamental group π1(M) of M admits an automorphism α : π1(M) −→ π1(M) with an Anosov identity. In fact: every known example of an Anosov-diffeomorphism on a compact manifold is topologically-conjugated to an infra-nil-manifold endomorphism (cf.[8]) and it is conjectured that there are no other examples (cf. Smale’s problem 3.5 in [40]). So it makes sense to ask which (fundamental) groups admit an automorphism with an Anosov identity. We have a partial rersult: Proposition 7.3.3. Consider a finitely-generated group G, together with an automorphism admitting an Anosov identity r(t), say of degree d. If G has a p-congruence system such that Discr (r(t)) · Prod(r(t)) 6≡ 0 mod p, then G has a nilpotent subgroup N of finite ∗ d index and of class c(N) ≤ d2 .

Proof. Here we do not assume that the p-congruence system comes with a bound: we are using the terminology of definition B.1 in [11], rather than the original terminology of Lubotzky in [32]. After replacing G with an appropriate characteristic subgroup of finite index, we may assume that G is finitely-generated and residually-(a finite p-group). Let us consider the Zassenhaus series (∆i(G, p))i N of G with respect to the prime p. Since ∈ i ∆i(G, p)= {1G}, we need only show that each factor G/∆i(G, p) is nilpotent of class d at most d2 . Let α : G −→ G be the automorphism. Since the series is characteristic, we T obtain the induced automorphism αG/∆i(G,p) : G/∆i(G, p) −→ G/∆i(G, p) with identity r(t). Since G/∆i(G, p) is a finite p-group and since the roots of an Anosov-polynomial form an arithmetically-free subset of (Q×, ·), we may apply theorem 1.2.5 in order to 2d conclude that c(G/∆i(G, p)) ≤ d . This finishes the proof.

8 Closing remarks

Our main theorem 1.2.8 roughly states that finite groups with a fix-point-free automorphism satisfying a good identity are nilpotent of bounded class modulo some bad torsion. But the theorem, as it is stated, cannot be extended to all polynomials. Example 8.0.1 (Gross [16]). Let n be a natural number and let k be the number of prime divisors of n, counted with multiplicity. Let p, q be distinct primes not dividing n. Then there are finite (p · q)-groups of Fitting-height k with a fix-point-free automorphism of order n. So we propose the following (alternative) interpretation of Meta-Problem 1.1.5.

31 d Problem. For every r(t) := a0 + a1 · t + ··· + ad · t ∈ Z[t] \{0}, there exists some h ∈ N with the following property. Consider a ﬁnite group G, together with a ﬁx-point- free automorphism α : G −→ G, and suppose that, for every x ∈ G, we have the equality a0 a1 2 a2 d ad x · α(x ) · α (x ) ··· α (x )=1G. Then G has Fitting-height h(G) ≤ h. Some evidence in favour of a positive solution exists. If r(t) is the constant polynomial ei n with prime factorization i pi , then h(G) ≤ i(2ei + 1) (cf. Shalev’s lemma [39]). If r(t) is the linear polynomial −n + t, then h(G) ≤ 1 (cf. Alperin’s theorem [2]). If Q n Q 2 r(t) is a splitting polynomial Ψn(t) = (t − 1)/(t − 1), then h(G) ≤ 7 · n . (cf. Jabara’s theorem [23]). If r(t) is a cyclotomic polynomial Φn(t), then h(G) ≤ 1 (cf. theorem 7.2.5).

Acknowledgements. The author would like to thank Efim Zel′manov and Lance Small for their hosting and their support during the first phase of the project at the Uni- versity of California, San Diego. The author also thanks Jonas Deré (for helpful discussions about the construction of automorphisms in subsection 2.3 and Anosov-diffeomorphisms in subsection 8), Evgeny Khukhro (for his very valuable feedback on an earlier draft of this paper), Gerry Myerson (for providing references to the relevant results about reduced resultants that informed subsection 4.2), John Thompson (for allowing the re-use of his proof in subsection 6.1), and Bertram Wehrfritz. The author also thanks Harald Schwab (for his generous help with the project) and the “Geometry and Analysis on Groups” seminar at the University of Vienna.

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