The structure of linear PF-Engel groups in characteristic zero

Suppose G is linear group. If G has characteristic zero, we prove that G is (polycyclic-by-finite)-Engel if and only if G has a normal series < 1 > = T0 ≤ T1 ≤ ··· ≤ Ts = T ≤ G with s and the index (G:T) finite and each Ti/Ti–1 either polycyclic-by-finite, or G-hypercentral with [Ti, T] ≤ Ti–1, or G-hypercentral, abelian and Chernikov. This is much more complex than the positive characteristic case where G is (polycyclic-by-finite)-Engel if and only if G is (polycyclic-by-finite)-by-hypercentral.


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conjugate, we may assume that T ≤ Tr(n, F). Note that U is nilpotent, T/U is abelian and G/T is finite.
Theorem With the hypotheses and notation above G has a normal series where t = n(n-1)/2 and for each i ≥ 1,

is finitely generated abelian, U i /V i is G-hypercentral and either U i /U i-1 is T-central or U i /V i is divisible, abelian and Chernikov.
Further N is nilpotent of class max{1, n-1} ≤ n, T/N is finitely generated abelian, G/N is polycyclic-by-finite and there exists M a normal subgroup of G such that U ≤ M, M/U is (finitely generated abelian)-by-finite and G/M is hypercentral.

Corollary Suppose G is linear group of characteristic zero. Then G is PF-Engel if and
only if G has a normal series < 1 > = T 0 ≤ T 1

≤ ··· ≤ T s = T ≤ G with s and the index (G:T) finite and each T i /T i-1 either polycyclic-by-finite, or G-hypercentral with [T i , T] ≤ T i-1 , or G-hypercentral, abelian and Chernikov.
For if G is PF-Engel take the series < 1 > = U 0 ≤ V 1 ≤ ··· ≤ U t ≤ M ∩ T ≤ T ≤ G of the Theorem with s = 2t + 2. For the converse see Proposition 2 below.
We give examples showing in particular that the group G in the theorem above need not be PF-by-hypercentral and, unlike the positive characteristic case (see [7]), need not be nilpotent-by-finite.
Below if G is any group {ζ i (G): 0 ≤ i ≤ s}, s some ordinal, denotes the upper central series of G with (G) = ⋃ i i≤s (G) its hypercentre. Proof Suppose first that A = <a> is cyclic. By hypothesis there is a finitely generated (and abelian note) subgroup E of B such that for all b in B there exists an integer m(b) ≥ 1 such that for all n ≥ m(b) we have b(a -1) n ∈ E. Subject to this we choose our E of least possible rank. Set E 1 = < b(a -1) n : n ≥ m(b), b ∈ B > ≤ E. Then E 1 is also finitely generated, so we may assume that E = E 1 . Clearly now E(a -1) ≤ E. Since a ≠ 1, so a -1 ∈ F* is an (additive) automorphism of F + . Thus E(a -1) ≅ E and where i runs over the positive integers. Set E 2 = ⋃ i≥o E(a−1) −i . If b ∈ B, then for some n ≥ 1 we have b(a -1) n ∈ E, so b ∈ E(a -1) −n . Therefore B ≤ E 2 . Further E(a -1) ≤ E yields that Ea ≤ E. Also BA = B, so E 3 = EA = ⋃ i≥0 Ea −i ≤ B. Suppose b ∈ B with ba ∈ E. There exists n ≥ 1 with b(a -1) n ∈ E. Then (-1) n b ∈ E. If actually b ∈ E 3 , then ba m ∈ E for some m ≥ 1, so (ba m−1 )a ∈ E and thus ba m−1 ∈ E. A trivial induction yields that b ∈ E. This shows that E 3 = E and hence E is a normal subgroup of G. Clearly B/E ≤ E 2 /E is A-hypercentral (recall here A = < a >) and therefore G/E is a hypercentral group.
Since we chose E of least rank, so E/E(a -1) finite. But E(a -1) −i /E(a -1) −i+1 is isomorphic to E/E(a -1) for each i. Therefore B/E is a periodic π-group for π the finite set of prime divisors of the index (E: E(a -1)). Also E 2 and B have finite rank equal to the rank of E and in particular B/E has finite rank. Therefore B/E is Chernikov. We have now completed the proof of the case where A is cyclic. Now assume A is finitely generated, say A = < a 1 , a 2 , …, a t > , where we choose the generators a i non-zero and such that for each i, a i −1 equals some a j . By the cyclic case above B is minimax and for each i there is a finitely generated subgroup E i of B such that E i a i = E i , E i (a i -1) ≤ E i B/E i is Chernikov and < a i > -hypercentral and rankE i is minimal with these properties and in fact equals rankB.
Set E * = E 1 + E 2 + ··· + E t . Since t is finite, E * is finitely generated and there exists an integer m ≥ 1 such that for all i and all n ≥ m we have E * (a i -1) n ≤ E i ≤ E * . Set Then E ** is finitely generated. Also that E * (a i -1) m ≤ E * implies E * a i m ≤ < E * a i e : 0 ≤ e < m > . Consequently E ** a i ≤ E ** for each i. Further E ** a i −1 ≤ E ** since a i −1 = a j for some j. Therefore E ** is an A submodule of B. Now X = G/E ** is the split extension of B/E ** by A and A and B are abelian. If b ∈ B, then b(a i -1) s ∈ E i < E * ≤ E ** for some s ≥ 1. Thus the a i are left Engel elements of X, as clearly so are the elements of B/E ** . Trivially X is abelian-by-polycyclic. Therefore X is hypercentral by Gruenberg's Theorems 1.2 and 1.4 in [2]. That is, G is (finitely generated abelian)-by-hypercentral. Finally B/E 1 is Chernikov, so B/E ** is too and rankE ** ≤ rankB = rankE 1 ≤ rankE ** , so rank E ** = rank B. This completes the proof of the A finitelygenerated case.
Finally it remains to prove that A is always finitely generated. Let J be the subring of F generated by A. Then J is a domain and B is a faithful J-module. Also B is torsion-free and, by the A cyclic case, minimax. Hence B has a free abelian subgroup B 0 of rank r = rankB with B/B 0 a periodic π-group for some finite set π of primes. Let t denote the product of all the elements of π. Then the tensor product Therefore J embeds into the endomorphism ring of this module and hence into the matrix ring There exists a finitely generated subring R ≥ Z[t −1 ] of the complex numbers, an element x of GL(r, R) and a subgroup A 0 of A of finite index such that (A 0 ) x ≤ Tr(r, R). (We now regard J ⊇ A as a subring of R r×r via the obvious embeddings J ≤ Z[t −1 ] r×r ≤ R r×r .) If a ∈ A 0 with a x unipotent, then (a x -1) r = 0 in R r×r and hence (a -1) r = 0 in the domain J. Consequently a = 1 and the unipotent radical of (A 0 ) x is trivial. It follows that A 0 is isomorphic to a subgroup of the full diagonal group D(r, R) and the latter is isomorphic to the direct product of r copies of the group U of units of R. But R is a finitely generated domain so U is finitely generated (e.g. [3]). Therefore A 0 and A are also finitely generated. This completes the proof of Proposition 1.

Lemma 1 Let A ≤ T be normal subgroups of the PF-Engel group G such that A is abelian and Chernikov, A ≤ ζ(T) and G/T is finite. Then there exists an integer r ≥ 1 such that A(r) = {a ∈ A: a r = 1} is finite with A/A(r) ≤ ζ ω (G/A(r)).
Proof Let g 1 , g 2 , …, g t be a transversal of T to G. Clearly PF subgroups of A are finite.
Hence there exists an integer r ≥ 1 such that for each a in A and each i there exists n ≥ 0 such that [a, n g i ] ∈ A(r). Now A ≤ ζ(T) and A = ⋃ j≥1 A(j), so there is a series of length the first infinite ordinal ω of normal subgroups of G such that each A i /A i-1 is finite and T-central. Now A i /A i-1 consists of right Engel elements of the (finite) split extension of A i /A i-1 by G/T. Therefore A i /A i-1 ≤ ζ(G/A i-1 ) and the claim of the lemma follows.

Lemma 2 Let A ≤ T be normal subgroups of the PF-Engel group G such that [A, T] = < 1 > and G/T is finite. Then there is a finitely generated subgroup B of A such that B is normal in G and A/B ≤ ζ(G/B).
Note that necessarily A is abelian so B is necessarily polycyclic.

Proof of the Theorem
There is a central series < 1 > = U 0 ≤ U 1 ≤ … ≤ U t = U of U and normal in T, where t = n(n-1)/2 and such that each U i /U i-1 embeds into F + so that the action of T on U i /U i-1 is given by a subgroup of F* and a homomorphism of it onto T/ C T (U i /U i-1 ). Thus for each i Proposition 1 is applicable. Therefore we have a normal series <1> = U 0 ≤ V 1 ≤ U 1 ≤ V 2 ≤ U 2 ≤… ≤ V t ≤ U t = U of T such that for each i U i /U i-1 is central in U, T/C T (U i /U i-1 ) is finitely generated, V i /U i-1 is finitely generated abelian, U i /V i is T-hypercentral and either U i /U i-1 is T-central or U i /V i is divisible, abelian and Chernikov. Now G/T is finite. Replacing each subgroup in the above series by its normal closure in G produces a new series that preserves the conditions above. Let us keep the same notation for this new series. Thus we now have the series exactly as above but with the added bonus that the terms are all normal in G.
Suppose U i /V i is divisible, abelian, Chernikov and T-hypercentral. Then by Lemma 1 there exists V i ≤ W i ≤ U i with W i normal in G, W i /V i finite and U i /W i G-hypercentral. Then W i /U i-1 is finitely generated abelian. Replace V i by W i ; that is, assume V i = W i . If U i /U i-1 is T-central then by Lemma 2 there exists U i-1 ≤ W i ≤ U i with W i normal in G, W i /U i-1 finitely generated and U i /W i G-hypercentral. Again set V i = W i . Now T/C T (U i /U i-1 ) is finitely generated for each i. Set N = I i C T U i ∕U i−1 ). Then T ≥ N ≥ U, the group N is nilpotent (U 0 ≤ U 1 ≤ ··· ≤ U t ≤ N is a central series of N) and normal in G, T/N is finitely generated abelian and G/N ∈ PF. Also U is the unipotent radical of N. Hence N embeds into U* × D, where U* is a unipotent linear group of degree n and characteristic zero and D is abelian (diagonalizable in fact). Therefore U* × D is nilpotent of class max{1, n-1} and consequently so is N.

Proposition 2 Let < 1 > = T 0 ≤ T 1 ≤ ··· ≤ T s = T ≤ G be a normal series of the group G with s and (G: T) finite and each T i /T i-1 either polycyclic-by-finite, or G-hypercentral with [T i , T] ≤ T i-1 , or G-hypercentral, abelian and Chernikov.
Then G is PF-Engel.
Proof We induct on s; if s = 0 the claim is trivial. Suppose s ≥ 1, let g ∈ G and assume there exists T 1 ≤ X ≤ G with X/T 1 in PF and a g-sink in G/T 1 . We choose X, as we may, with [X, g] ≤ X, so also X g ≤ X. If T 1 is PF, then X is PF and also a g-sink in G. From now on assume that T 1 is abelian. There is a finitely generated subgroup K of X with T 1 K = X. If K = < k 1 , k 2 , …, k r > , then each [k i , g] ∈ u i K for some u i in T 1 . Set R = < u 1 , u 2 , …, u r > G , a normal subgroup of G. If x, y ∈ K with [x, g] and [y, g] both in RK, then [xy g] = [x, g] y [y, g] ∈ (RK) y (RK) = RK and [x −1 , g] = x[x, g] −1 x −1 ∈ RK. Therefore [K, g] ≤ RK and hence [RK, g] ≤ RK. Now K is finitely generated, K/(K ∩ T 1 ) ∈ PF and K ∩ T 1 is abelian. By a theorem of P. Hall (e.g. 3.10 of [6]) K ∩ T 1 is finitely K-generated. If [T 1 , T] = < 1 > , since (G: T) is finite, so K ∩ T 1 and R are finitely generated (as groups note, not just as normal subgroups). Therefore RK ∈ PF. If T 1 is Chernikov (and abelian), then T 1 has an ascending series of normal subgroups of G of length at most the first infinite ordinal ω and with all its factors finite. Hence here K ∩ T 1 and R are finite. Again we obtain that RK ∈ PF. Now assume T 1 is G-hypercentral. Thus let < 1 > = Z 0 /R ≤ Z 1 /R ≤ ··· ≤ Z τ /R = T 1 /R be an ascending central series of G in T 1 /R, so all the Z σ are normal in G. If k ∈ K, then [k, g] ∈ RK, see above. Let u ∈ T 1 \R with σ the least ordinal with u ∈ Z σ . Then σ-1 exists and [uk, g] = [u, g] k [k, g] ∈ Z σ-1 RK = Z σ-1 K. Consequently there exists a positive integer n with [uk, n g] ∈ RK. We have now shown that if T 1 ∈ PF, then X is PF and a g-sink in G and if T 1 ∉ PF, the RK is PF and a g-sink in X and hence in G. Therefore G is PF-Engel.
Example 1 Let a be the companion matrix of the polynomial X 2 -5X + 1 in GL (2, Z). The split extension G 0 of Q (2) by < a > embeds into GL(3, Q) and also into Tr(2, C). Set E = Z (2) ≤ Q (2) , B = i∈ E(a − 1) −i and G = < a, B > ≤ G 0 . (Note that det(a-1) = -3, so a-1 is invertible.) Then G is linear of characteristic zero and polycyclic-by-hypercentral and hence PF-Engel, but is not PF, is not hypercentral and is not (locally nilpotent)-by-finite.
For E ≥ E(a-1); thus E(a-1) −i−1 ≥ E(a-1) −i for all i and hence B = ⋃ i≥0 E(a − 1) −i . . Also Ea = E = Ea −1 , so Ba = B = Ba −1 ; hence B is an < a > -submodule of Q (2) and G = < a > B. Now E is normal in G and, being finitely generated abelian, polycyclic. Clearly G/E is hypercentral. Thus in particular G is PF-Engel.
The eigenvalues of a have infinite order (as does a of course). Thus a, and also a r for each r ≠ 0, have no non-zero fixed points in Q (2) and hence none in B. It follows that ζ 1 (G) = < 1 > = ζ(G). In particular G is not hypercentral. If G ∈ PF, then B is free abelian, necessarily of rank 2, and then (B:E) is finite. Hence there exists i with E(a-1) −i = E(a-1) −i−1 , But then E = E(a-1) −1 and consequently det(a-1) = ± 1. This contradiction shows that G is not polycyclic-by-finite.
If G is (locally nilpotent)-by-finite, there exists r > 0 with < a r > B locally nilpotent. Locally nilpotent linear groups are hypercentral ( [5] 8.2). Hence a r has a non-trivial fixed point in B. This contradiction shows that G is not (locally nilpotent)-by-finite. Finally if α = (5 + √21)/2 ∈ C, then α 2 = 5α -1. Set E = Z ⊕ Zα ≤ C*. Then Eα −i+1 ≤ Eα −i for all i and if B = ⋃ i≥0 E( − 1) −i , then Bα = B and the split extension of B by < α > embeds into Tr(2, C). It is also isomorphic to the group G above. Similarly the group G 0 also embeds into Tr(2,C). The construction of Example 1 is complete. For brevity represent (x ij ) ∈ T by the triple (x 21 , x 32 , x 31 ). Let x = (ξ, η, ζ) and y = (λ, μ, ν) be elements of T. Then an elementary calculation in G yields that and then a simple induction yields for all d ≥ 0 that Thus for all large enough d we have [y, d+1 ax] ∈ E = Tr 1 (3, Z). Note that E is polycyclic, being finitely generated nilpotent. If g ∈ G, then y = [g, ax] lies in T, as does [g, x] and then [g, 3 x] = 1. Consequently G is PF-Engel.
0, 2 −h , 0 , x = 0, 0, 2 −h , x, 2 −h , 0, 0 = 0, 0, 2 −h , which again does not lie in N ∩ Z for large enough h. Thus xZ and hence no non-trivial element of NZ/Z lies in the centre of the nilpotent group T/Z. Therefore no such x exists and N ≤ Z as claimed. Finally, suppose M is a normal PF subgroup of G. If ax ∈ M, where x ∈ T, then [G, ax] ≤ M ∩ T and clearly [G, ax]Z = [G, a]Z = T. But M ∈ PF, so M ∩ T is finitely generated. Hence M ∩ T ≤ Z by the previous paragraph and so Z = T, which is false. Therefore no such ax exists, M ≤ T and M ≤ Z, again by the above. But then G/M contains sections that are infinite dihedral and consequently G/M cannot be hypercentral. This completes the proof that G is not PF-by-hypercentral.

Remark
The fact that the degree is 3 in Example 2 is significant. If G is a PF-Engel subgroup of GL(2, F), then G is always PF-by-hypercentral. For if G is irreducible, then G is abelian-by-finite and Lemma 2 and the theorem of [1] apply and if G is reducible, then G is triangularizable and the Theorem applies with t = 1.