The structure of linear PF-Engel groups in characteristic zero

Abstract

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 ≤ ··· ≤ 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. 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.

If X is a class of groups, say that a group G is X-Engel if for each g ∈ G there is an X-subgroup E of G such that for all x in G there is a positive integer m such that for all n ≥ m we have [x, _ng] ∈ E. Let F denote the class of finite groups, Ch the class of Chernikov groups and P the class of polycyclic groups, so PF denotes the class of polycyclic-by-finite groups.

A special case of Shumyatsky’s main theorem in [4] says that a linear group G is F-Engel if and only if G is finite-by-hypercentral. By Theorem Ch of [7] a linear group G is Ch-Engel if and only if G is Chernikov-by-hypercentral and by Theorem PF of [7] a linear group G of positive characteristic is PF-Engel if and only if G is PF-by-hypercentral (Ch is exactly the class of linear groups satisfying the minimal condition on subgroups and PF is exactly the class of linear groups satisfying the maximal condition on subgroups, so for linear groups Ch and PF are fully duals of each other).

The situation for linear PF-Engel groups of characteristic zero is much more complex and in particular the obvious analogue of the positive characteristic case is false, see Example 2 below. In this present paper we study in detail the characteristic zero case.

Thus let G be a PF-Engel subgroup of GL(n, F) for n some positive integer and F a field of characteristic 0. Then G is soluble-by-finite by Theorem PF of [7]. Let T be a triangularizable normal subgroup of G of finite index (which always exists, e.g. use [6] 4.2) and denote the unipotent radical of T by U. Enlarging F if necessary and replacing G by a conjugate, we may assume that T ≤ Tr(n, F). Note that U is nilpotent, T/U is abelian and G/T is finite.

FormalPara Theorem

With the hypotheses and notation above G has a normal series

$$ < 1> = {\text{U}}_{0} \le {\text{V}}_{ 1} \le {\text{U}}_{ 1} \le {\text{V}}_{ 2} \le {\text{U}}_{ 2} \le \cdots \le {\text{V}}_{\text{t}} \le {\text{U}}_{\text{t}} = {\text{U}} \le {\text{N}} \le {\text{T}} \le {\text{G}}, $$

• where t = n(n–1)/2 and for each i ≥ 1,

• 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 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.

FormalPara Corollary

Suppose G is linear group of characteristic zero. Then G isPF-Engel if and only if G has a normal series < 1 > = T₀ ≤ T₁ ≤ ··· ≤ 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 > = U₀ ≤ V₁ ≤ ··· ≤ 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 $\zeta \left( {\text{G}} \right) = \bigcup\nolimits_{\text{i}} {\zeta_{{{\text{i}} \le {\text{s}}}} \left( {\text{G}} \right)}$ its hypercentre.

FormalPara Proposition 1

Let F be a field of characteristic 0, A a non-trivial subgroup of the multiplicative group F* of F and B a non-zero subgroup of the additive group F⁺of F such that BA = B. Suppose the split extension G of B by A isPF-Engel. Then G is (finitely generated abelian)-by-hypercentral. Further A is finitely generated and B is minimax. In fact B contains a finitely generated subgroup E normal in G such that rankE = rankB and B/E is Chernikov and G-hypercentral.

FormalPara 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 E₁ = < b(a – 1)ⁿ: n ≥ m(b), b ∈ B > ≤ E. Then E₁ is also finitely generated, so we may assume that E = E₁. 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

$$ \cdots \ge {\text{E}}\left( {{\text{a}} {-} 1} \right)^{{{-}{\text{i}}}} \ge \cdots \ge {\text{E}}\left( {{\text{a}} {-} 1} \right)^{ - 1} \ge {\text{E}} \ge {\text{E}}\left( {{\text{a}} {-} 1} \right) \ge \cdots \ge {\text{E}}\left( {{\text{a}} {-} 1} \right)^{\text{i}} \ge \cdots $$

where i runs over the positive integers. Set ${\text{E}}_{ 2} = \bigcup\nolimits_{{{\text{i}} \ge {\text{o}}}} {{\text{E}}\left( {{\text{a}} {-} 1} \right)^{{ - {\text{i}}}} } .$ If b ∈ B, then for some n ≥ 1 we have b(a – 1)ⁿ ∈ E, so b ∈ E(a – 1)⁻ⁿ. Therefore B ≤ E₂. Further E(a – 1) ≤ E yields that Ea ≤ E. Also BA = B, so ${\text{E}}_{ 3} = {\text{EA}} = \bigcup\nolimits_{{{\text{i}} \ge 0}} {{\text{Ea}}^{{ - {\text{i}}}} \le {\text{B}}.}$ Suppose b ∈ B with ba ∈ E. There exists n ≥ 1 with b(a – 1)ⁿ ∈ E. Then (–1)ⁿb ∈ E. If actually b ∈ E₃, 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 E₃ = E and hence E is a normal subgroup of G. Clearly B/E ≤ E₂/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 E₂ 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 = < a₁, a₂, …, a_t > , where we choose the generators a_i non-zero and such that for each i, a⁻¹_i 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_ia_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_* = E₁ + E₂ + ··· + 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

$$ {\text{E}}_{**} = < {\text{E}}_{*} {\text{a}}_{ 1}^{{{\text{e}}( 1)}} {\text{a}}_{ 2}^{{{\text{e}}( 2)}} \ldots {\text{a}}_{\text{t}}^{{{\text{e}}({\text{t}})}} :0 \le {\text{e}}\left( {\text{i}} \right) < {\text{m}}, 1\le {\text{i}} \le {\text{t}} > . $$

Then E_** is finitely generated. Also that E_*(a_i – 1)^m ≤ E_* implies E_*a^m_i ≤ < E_*a^e_i: 0 ≤ e < m > . Consequently E_**a_i ≤ E_** for each i. Further E_**a⁻¹_i ≤ E_** since a⁻¹_i = 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/E₁ is Chernikov, so B/E_** is too and rankE_** ≤ rankB = rankE₁ ≤ rankE_**, so rank E_** = rank B. This completes the proof of the A finitely-generated 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 B₀ of rank r = rankB with B/B₀ 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 Z[t^−1]B of Z[t⁻¹] by B over Z is equal to Z[t⁻¹]B₀ and hence is isomorphic to Z[t⁻¹]^(r), the free Z[t⁻¹]-module of rank r. Therefore J embeds into the endomorphism ring of this module and hence into the matrix ring Z[t⁻¹]^r×r.

There exists a finitely generated subring R ≥ Z[t⁻¹] of the complex numbers, an element x of GL(r, R) and a subgroup A₀ of A of finite index such that (A₀)^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 ∈ A₀ 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 (A₀)^x is trivial. It follows that A₀ 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 A₀ and A are also finitely generated. This completes the proof of Proposition 1.

FormalPara Lemma 1

Let A ≤ T be normal subgroups of thePF-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)).

FormalPara Proof

Let g₁, g₂, …, 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, _ng_i] ∈ A(r). Now A ≤ ζ(T) and ${\text{A}} = \bigcup\nolimits_{{{\text{j}} \ge 1}} {{\text{A}}\left( {\text{j}} \right),}$ so there is a series

$$ {\text{A}}\left( {\text{r}} \right) = {\text{A}}_{0} < {\text{A}}_{ 1} < \cdots < {\text{A}}_{\text{i}} < \cdots < {\text{A}}_{\omega } = {\text{A,}} $$

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.

FormalPara Lemma 2

Let A ≤ T be normal subgroups of thePF-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.

FormalPara Proof

Let g₁, g₂, …, g_t be a transversal of T to G. For each i there exists a finitely generated subgroup E_i of A such that for all a in A there exists m ≥ 1 with [a, _ng_i] ∈ E_i for all n ≥ m. Set

$$ {\text{B}} = < {\text{E}}_{\text{i}}^{\text{g}} :{\text{g}} \in {\text{G}}, 1\le {\text{i}} \le {\text{t}} > = < {\text{E}}_{\text{i}}^{\text{g}} :{\text{g}} = {\text{g}}_{ 1} , \ldots ,{\text{g}}_{\text{t}} , 1\le {\text{i}} \le {\text{t}} > . $$

Clearly B is finitely generated and normal in G. Now A/B consists of right Engel elements of the split extension of A/B by G/T. Consequently A/B ≤ ζ(G/B), e.g. [2] Theorems 1.2 and 1.4).

FormalPara Proof of the Theorem

There is a central series < 1 > = U₀ ≤ U₁ ≤ … ≤ 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> = U₀ ≤ V₁ ≤ U₁ ≤ V₂ ≤ U₂ ≤… ≤ 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 ${\text{N }} = {\text{I}}_{{\rm i}} {\text{C}}_{{\rm T}} \left( {{\text{U}}_{{\rm i}} /{\text{U}}_{{{{\rm i}}{-} 1}} } \right)).$ Then T ≥ N ≥ U, the group N is nilpotent (U₀ ≤ U₁ ≤ ··· ≤ 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.

Finally G/U is an abelian-by-finite PF-Engel group, so by Lemma 2 (with A = T) there is a normal subgroup M_o of G with U ≤ M_o ≤ T, M_o/U finitely generated abelian and T/M_o ≤ ζ(G/M_o). Then there exists M ≥ M_o normal in G with M/M_o finite and G/M hypercentral, see [1].

Suppose E ≤ H are subgroups of a group G and g ∈ G. If for each h in H there exists a positive integer m(h) such that [h, _ng] ∈ E for all n ≥ m(h), for brevity say E is a g-sink in H. (This is a slight modification of the terminology used in [4].) If this is so, note that E_o = < [h, _ng]: n ≥ m(h), h ∈ H > ≤ E is also a g-sink in H and it is easy to see that [E_o, g] ≤ E_o ≤ [H, g]. (If a, b ∈ E_o with [a, g] and [b, g] in E_o, then [ab, g] = [a, g]^b[b, g] ∈ E_o and [a⁻¹, g] = a[a, g]⁻¹a⁻¹ ∈ E_o.)

FormalPara 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.

FormalPara Proof

We induct on s; if s = 0 the claim is trivial. Suppose s ≥ 1, let g ∈ G and assume there exists T₁ ≤ X ≤ G with X/T₁ in PF and a g-sink in G/T₁. We choose X, as we may, with [X, g] ≤ X, so also X^g ≤ X. If T₁ is PF, then X is PF and also a g-sink in G. From now on assume that T₁ is abelian.

There is a finitely generated subgroup K of X with T₁K = X. If K = < k₁, k₂, …, k_r > , then each [k_i, g] ∈ u_iK for some u_i in T₁. Set R = < u₁, u₂, …, 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]⁻¹x⁻¹ ∈ RK. Therefore [K, g] ≤ RK and hence [RK, g] ≤ RK.

Now K is finitely generated, K/(K ∩ T₁) ∈ PF and K ∩ T₁ is abelian. By a theorem of P. Hall (e.g. 3.10 of [6]) K ∩ T₁ is finitely K-generated. If [T₁, T] = < 1 > , since (G: T) is finite, so K ∩ T₁ and R are finitely generated (as groups note, not just as normal subgroups). Therefore RK ∈ PF. If T₁ is Chernikov (and abelian), then T₁ 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 ∩ T₁ and R are finite. Again we obtain that RK ∈ PF.

Now assume T₁ is G-hypercentral. Thus let < 1 > = Z₀/R ≤ Z₁/R ≤ ··· ≤ Z_τ/R = T₁/R be an ascending central series of G in T₁/R, so all the Z_σ are normal in G. If k ∈ K, then [k, g] ∈ RK, see above. Let u ∈ T₁\R with σ the least ordinal with u ∈ Z_σ. Then σ–1 exists and [uk, g] = [u, g]^k[k, g] ∈ Z_σ–1RK = Z_σ–1K. Consequently there exists a positive integer n with [uk, _ng] ∈ RK. We have now shown that if T₁ ∈ PF, then X is PF and a g-sink in G and if T₁ ∉ PF, the RK is PF and a g-sink in X and hence in G. Therefore G is PF-Engel.

FormalPara Example 1

Let a be the companion matrix of the polynomial X² – 5X + 1 in GL(2, Z). The split extension G₀ of Q⁽²⁾ by < a > embeds into GL(3, Q) and also into Tr(2, C). Set E = Z⁽²⁾ ≤ Q⁽²⁾, ${\text{B}} = \varSigma_{{{\text{i}} \in {\mathbf{Z}}}} {\text{E}}\left( {{\text{a}} - 1} \right)^{{ - {\text{i}}}}$ and G = < a, B > ≤ G₀. (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 ${\text{B}} = {\bigcup } _{{{\text{i}} \ge 0}} {\text{E}}\left( {{\text{a}} - 1} \right)^{{ - {\text{i}}}} .$. 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 ζ₁(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 α² = 5α – 1. Set E = Z ⊕ Zα ≤ C*. Then Eα⁻ⁱ⁺¹ ≤ Eα⁻ⁱ for all i and if ${\text{B}} = {\bigcup } _{{{\text{i}} \ge 0}} {\text{E}}\left( {\alpha - 1} \right)^{{ - {\text{i}}}} ,$ 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 G₀ also embeds into Tr(2,C). The construction of Example 1 is complete.

FormalPara Example 2

Let a = diag(1, –1, 1), T = Tr₁(3, Z[1/2]) and G = < a > T ≤ Tr(3, Z[1/2]) ≤ GL(3, Q). Then G is PF-Engel, linear of characteristic zero, but not PF-by-hypercentral.

For brevity represent (x_ij) ∈ T by the triple (x₂₁, x₃₂, x₃₁). Let x = (ξ, η, ζ) and y = (λ, μ, ν) be elements of T. Then an elementary calculation in G yields that

$$ \left[ {{\text{y}}, {\text{ax}}} \right] \, = \, \left( {{-} 2\lambda , {-} 2\mu ,{ 2}\lambda \mu {-} \mu \xi + \eta \lambda } \right), $$

and then a simple induction yields for all d ≥ 0 that

$$ \left[ {{\text{y}}, _{{{\text{d}} + 1}} {\text{ax}}} \right] \, = \, \left( { \pm 2^{{{\text{d}} + 1}} \lambda , \pm 2^{{{\text{d}} + 1}} \mu , \pm 2^{{ 2 {\text{d}} + 1}} \lambda \mu \pm 2^{\text{d}} \mu \xi \pm 2^{\text{d}} \eta \lambda } \right). $$

Thus for all large enough d we have [y, _d+1ax] ∈ E = Tr₁(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, ₃x] = 1. Consequently G is PF-Engel.

Let N be a normal, finitely generated subgroup of T. We claim that N ≤ Z = ζ₁(T). Hence suppose x = (ξ, η, ν) ∈ N\Z, so at least one of ξ and η is non-zero. Since T is nilpotent N ∩ Z ≠ < 1 > , so N ∩ Z is infinite cyclic and NZ/N is a Prüfer 2-group. If ξ ≠ 0, then

$$ \left[ {\left( {0, 2^{{{-}{\text{h}}}} , 0} \right),{\text{ x}}} \right] \, = \, \left( {0, 0, 2^{{{-}{\text{h}}}} \xi } \right), $$

which does not lie in N ∩ Z for large enough h. Also if η ≠ 0, then

$$ \left[ {{\text{x}}, \left( { 2^{{{-}{\text{h}}}} , 0, 0} \right)} \right] \, = \, \left( {0, 0, 2^{{{-}{\text{h}}}} \eta } \right), $$

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.

FormalPara 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.