Variants of theorems of Schur, Baer and Hall

If a group G is ‘restricted’ modulo its hypercentre, then to what extent does G have an equally restricted normal subgroup L with G / L hypercentral? We consider these questions where restricted means ﬁnite- π , Chernikov, locally ﬁnite- π , polycyclic or polycyclic-by-ﬁnite.


Introduction
For any group G denote its centre by ζ 1 (G) and its hypercentre by ζ(G). If t is a positive integer, say t = Π primes p p e ( p) , let e(t) denote the maximum of the e( p) (so e(1) = 0) and h(t) the sum of all the e( p). Set a(t) = [e(t)/2] + 1, where [r ] denotes the integer part of a real number r , and set b(t) = t (e(t)+1)/2 . Obviously b(t) ≤ t a(t) . The following variant of theorems of Schur and Baer was essentially proved by de Falco et al. [2].
Theorem A (cf. [2,5]) Let G be a group with G/ζ (G) finite of order t. Then G has a normal subgroup L with G/L hypercentral and with L of finite order dividing t a(t) +1 and at most b(t)t. de Falco et al. [2] gives no specific bounds. The later paper [5] by Kurdachenko et al. contains two proofs of Theorem A, one shorter with no bounds and one with just a bound for |L| slightly larger than b(t)t.
There are variants of the classical Schur and Baer theorems where finite is replaced by notions like Chernikov, polycyclic or locally finite, see [7], especially Page 115.
Here we consider corresponding questions in the context of Theorem A. The following is the main result of this paper.
Theorem B Let G be a group with G/ζ (G) a Chernikov group. Then G has a normal Chernikov subgroup L with G/L hypercentral.
A minor variation to our proof of Theorem B gives yet another short proof of Theorem A. In fact we prove Theorem A with rather better bounds than those stated above, but with bounds less briefly explained. Let Z be a central subgroup of a group G of finite index dividing t. Then (Schur's theorem) the order |G | of the derived subgroup of G is finite and in fact boundedly so (e.g. the easy proof of [11] 1.18, yields that log t |G | ≤ (t − 1) 2 + 1. Given t define the integers c(t) and d(t) as follows: c(t) is the least integer such that for any G and Z as above (but with fixed t), . By Theorem 1 of [12] we have c(t) ≤ [e(t)/2]+1 and d(t) ≤ t (e(t)+1)/2 . Hence Theorem A follows from the following.

Theorem C Let G be a group with Z = ζ(G) of finite index in G dividing t. Then G has a normal subgroup L with G/L hypercentral and with L of finite order dividing t c(t)+1 and at most d(t)t.
Wiegold [13] has a different type of bound for d(t). Assume t > 1, let q be the least integer to divide t and set t = log q t and t = [t ]; clearly e(t) ≤ h(t) ≤ t ≤ t ≤ t. Then Wiegold proves that d(t) ≤ t (t −1)/2 . In fact one can do a little better than this (see [12] Theorems 2 and 3), namely that c(t) ≤ [t /2] and d(t) ≤ t (t −1)/2 unless t = p e q or pq e with p > q primes and e ≥ 1 when c(t) ≤ [t /2] + 1. Further if t = pq e with e ≥ 2 or if t = p e q with p e > q e+1 , then d(t) ≤ t (t −1)/2 . With the exceptional t = p e q (e.g. t = 6) we have of course Wiegold's bound d(t) ≤ t (t −1)/2 .
The obvious analogues of Theorem B, with Chernikov replaced by polycyclic or polycyclic-by-finite, are false, see Example 1 below. We do however have the following easier result.
Theorem D Let G be a group with G/ζ (G) a locally finite π -group for some set π of primes (e.g. π the set of all primes). Then G has a locally finite, normal π -subgroup L with G/L hypercentral.
Casolo, Dardano and Rinauro in their recent paper [1] prove the corresponding result to Theorem A in the context of Hall's theorem. Specifically they prove the following.
Theorem E (see [1] Theorem A) Let L be a finite normal subgroup of the group G such that G/L is hypercentral. Then the index (G:ζ(G)) is finite and divides |Aut L|.|ζ 1 Simple examples show that the corresponding statements are false with finite replaced by Chernikov, polycyclic, polycyclic-by-finite, or locally finite, see Examples 2, 3 and 4 below. Theorem E is a very easy consequence of our final theorem.

where for each i the Gcomposition factors of V i are all G-isomorphic. In particular if V as G-module has a non-trivial factor centralized by G, then V has a non-zero element fixed by G and a non-trivial image centralized by G.
To obtain such a decomposition of V , see [10] 7.15. Note that the hypothesis there that the field F is algebraically closed is only used to ensure that the Jordan decomposition of each g in G takes place in G L(n, F). If g ∈ G has finite order, then trivially g u and g d lie in g , so here, as H is finite, we can dispense with the algebraic closure hypothesis. (Actually it suffices just to have F perfect, e.g. see [9] 3.1.6, which of course automatically covers the F = G F( p) case.) Remark Suppose G is a nilpotent group and V a finite G-module such that V = [V, G]. If q is prime, Lemma 1 shows that V /qV has no trivial G-composition factors. Thus nor does any G-image of V /qV ; in particular nor does any q i V /q i+1 V . Applying this for every q dividing the order of V shows that V itself has no trivial G-composition factors.
As well as ζ(G) = ∪ w≥0 ζ w (G), the hypercentre of G we consider γ G = ∩ w≥0 γ w+1 G, the hypocentre G; here w runs over the ordinals, {ζ w (G)} is the upper central series of G and {γ w+1 (G)} is the lower central series of G. Let k ≥ 0 and t ≥ 1 be integers. If (G : ζ k (G)) = t, then clearly ζ(G) = ζ k+e(t) (G). Also by Baer's theorem |γ k+1 G| is finite (see [8] 14.5.1), so G/γ G is nilpotent. Then G/γ G.ζ k (G) is nilpotent of order dividing t, so G/γ G is nilpotent of class at most k + e(t) and γ G = γ k+e(t)+1 G. Suppose instead that |γ k+1 G| = d. Clearly then γ G = γ k+e(d)+1 (G). Also the upper central series of G intersected with γ k+1 G has length at most e(d) and ζ(G)/(ζ (G) ∩ γ k+1 G) embeds into G/γ k+1 G as a G-group and hence has G-central height at most k. Therefore ζ(G!) = ζ k+e(d) (G). The above might seem rather pedantic, but one needs to be slightly careful in dealing with γ G for infinite groups G. We use these remarks below.
We now start on the proofs of Theorem B and, indirectly, Theorem C. Thus below G denotes a group with G/ζ (G) a Chernikov group. Set Z = ζ(G) and Γ = γ G.
Suppose first that G is finite and that (G : Z ) divides t. Now G/C G (Z ) stabilizes the upper central series of G and hence is nilpotent. Therefore Γ ≤ C G (Z ), Γ ∩ Z ≤ ζ 1 (Γ ), (Γ : Γ ∩ Z ) divides t and |Γ | divides t c(t) and is at most d(t). Set V = Γ /Γ . Clearly Γ centralizes V , so G/C G (V ) is nilpotent. By the Remarks above V has no trivial G-composition factors, so Γ ∩ Z ≤ Γ . Thus (Γ : Γ ) divides t. Therefore |Γ | divides t c(t)+1 and is at most d(t)t. Now suppose that G is finitely generated. Again we have t = (G : Z ) finite. Also G is polycyclic-by-finite, so there exists an integer k with ζ k (G) = Z . By Baer's theorem γ k+1 G is finite, so Γ ≤ γ k+1 G is finite and G/Γ is nilpotent. Now G is residually finite. Hence there is a normal subgroup N of G of finite index with Γ ∩ N = 1 . Clearly Z N/N ≤ ζ(G/N ) and Γ ∼ = Γ N /N = γ (G/N ). By the finite case we have that |Γ | divides t c(t)+1 and is at most d(t)t. Also by the finite case we have The Proof of Theorem B. Suppose X ≤ Y are finitely generated subgroups of G.
Clearly γ X ≤ γ Y , so L = ∪ X γ X is a normal subgroup of G. By the finitely generated case above we have that [γ X, X ∩ Z ] = 1 and γ X ∩ Z ≤ (γ X ) ; further X/γ X is nilpotent. If x ∈ L and z ∈ Z , there exists an X with x ∈ γ X and z ∈ X ∩ Z . Then Now G/L is locally nilpotent since each X/γ X is nilpotent and locally nilpotent Chernikov groups are hypercentral. Hence G/L Z and G/L are hypercentral. Further L/(L ∩ Z ) is Chernikov and L ∩ Z ≤ ζ 1 (L). Therefore L is Chernikov by Polovickii's theorem (see [7] 4.23). Consequently L is Chernikov. The proof is complete.
The Proof of Theorem C. Here we have (G : Z ) dividing t. Let X be a finitely generated subgroup of G with X Z = G. By the finitely generated case we have that X/γ X is nilpotent and that |γ X | divides t c(t)+1 and is at most d(t)t. Choose X so that |γ X | is maximal. If Y is any finitely generated subgroup of G containing X , then γ X ≤ γ Y since [γ X, X ] = γ X . By the maximal choice of X we have γ X = γ Y . This is for all such Y and consequently L = γ X is normal in G. If ψ is the natural map of G onto G/L, then X ψ is nilpotent and Gψ = X ψ.Z ψ. Consequently Gψ is hypercentral. The proof is complete.
Comments on the above proofs. Notice that in general, unlike the finitely generated case, in Theorem C we cannot prove that γ G is finite; just consider the infinite locally dihedral 2-group. However, since L = γ X = [γ X, X ], so L = [L , G] ≤ γ G and G/γ G is hypercentral. Further L is actually the hypercentral residual of G and in particular L is fully invariant in G.
A similar remark applies to Theorem B. If A = ⊕ i≥1 a i is free abelian of infinite rank and x ∈ Aut A is given by a i x = a i−1 + a i for all i (with a 0 = 0), then the split extension G of A by x is hypercentral and yet γ G = A is not Chernikov. Suppose α = ch(G), the central height of G, and β = ch(G/L). Assuming (G : Z ) = t, if e = e(t), then β ≤ α + e. On the other hand if |L| = d and if f = e(d), then α ≤ f + β, so if either of α and β is infinite, they both are and α ≤ β ≤ α + e.

Lemma 2 Let G be a π -torsion-free group for π some set of primes. If G/ζ (G) is a locally finite π -group, then G = ζ(G).
Proof If X is a finitely generated subgroup of G, then X is nilpotent-by-finite, ζ(X ) = ζ k (X ) for some finite k and X/ζ k (X ) is a finite π -group. Hence γ k+1 (X ) is also a finite π -group (e.g. [7] Page 115 or use the above). But G is π -torsion-free; therefore γ k+1 (X ) = 1 and so G is locally nilpotent. But then ζ(G) is π -isolated in G (see [4] 4.8b). Therefore ζ(G) = G. (Alternatively, if T is the maximal periodic normal subgroup of G, then T is a π -group, so T ≤ ζ(G) and ζ(G/T ) is isolated in G/T by [6] 2.3.9i); thus again ζ(G) = G.) The Proof of Theorem D. Let X ≤ Y be finitely generated subgroups of G. Then X/ζ (X ) is a finite π -group, so by Theorem C there exists a finite normal π -subgroup L X of G with X/L X hypercentral and hence nilpotent. Clearly we may choose L X so that X/L X is π -torsion-free. Then L X = X ∩ L Y . Set L = ∪ X L X . Then L is a locally finite, normal π -subgroup of G with G/L π -torsion-free and locally nilpotent. By the lemma above G/L is hypercentral.
is polycyclic, there is no need for G to be (polycyclic-by-finite)by-hypercentral.
Let A be a divisible abelian 2-group of rank 2. Then Aut A ∼ = G L(2, Z 2 ). Let H = x, h ≤ G L(2, Z 2 ); here x = 1 permutes the standard basis of (Z 2 ) (2) and Let G = H A be the split extension of A by H . Then ζ(G) = A and G/ζ (G) ∼ = H is polycyclic. Suppose T is any polycyclic-by-finite normal subgroup of G. Then T ∩ A ≤ A i for some i. If m is a positive integer with h m ∈ T , then h m stabilizes the series 1 < A 1 < A 2 < · · · < A i < A and hence h mn centralizes A for some n ≥ 1 (e.g. [11] 1.21), contradicting h of infinite order. Consequently H ∩ T = 1 and so G/T cannot be hypercentral.
The Proof of Theorem F. Define K by K /A = (H/A) ∩ ζ(G/A). We induct on the exponent e of A. Suppose first that e = p, a prime and that ζ(G) = 1 . If K > A, then there exists k ∈ K \A with k A ∈ ζ 1 (G/A). Then V = k A is abelian and normal in G and clearly [v p , g] = [v, g] p = 1 for all v in V and g in G. Also V ≤ H , so G/C G (V ) is an image of G/C G (H ) and hence is nilpotent. It follows that V ∩ ζ 1 (G) = 1 , either because V p = 1 or by the Remark above, contradicting the assumption that ζ(G) = 1 . Thus in this case K = A. Applying this to G/ζ (G) yields that if A is elementary abelian, then Now suppose that p is just some prime dividing e and set B = A p . By the case above
Let L be a finite group of order d. Then any series of subgroups of L has length at most h(d), the minimal number of generators of L is at most h(d) and Aut L has order at most d h (d) . For example, appying this to Theorem E yields that (G : ζ(G)) is at most d h(d) +1 .
Assume k ≥ 0 and d > 1 are integers and suppose G is a group with L = γ k+1 G of order d. Then from Theorem 2 of [3] we have (G : ζ 2k (G)) ≤ d s , where s = r k + h(d) and r is the rank of Aut L. Note that r is bounded by a function of d only; for example r ≤ h(d) 2 . Also ζ(G) = ζ k+e(d) (G), see Remarks above; consequently We have proved the following (cf. [1] Corollary A ). If |γ k+1 G| = d, then (G : ζ 2k (G)) ≤ d u(d) for u as above, a function of d only.
For the analogues of Theorem E the results are negative.
Example 2 If G is (infinite cyclic)-by-hypercentral, then G/ζ (G) need not be polycyclic-by-finite. Let A = Z, B = Z[1/2], g the automorphism b → −b of B and G the split extension g B. Then A is infinite cyclic and normal in G and G/A is hypercentral, being an infinite locally dihedral 2-group. Finally if x ∈ G\B, then x acts fixed-point freely on B, so 1 = ζ 1 (G) = ζ(G). Clearly G is not polycyclic-by-finite.

Example 3
If G is (locally finite)-by-hypercentral, then G/ζ (G) need not be periodic.
Let G be the wreath product of a cyclic group of prime order p by an infinite cyclic group. Then G is an elementary abelian p-group and yet ζ(G) = ζ 1 (G) = 1 .

Example 4
If G is Chernikov-by-hypercentral, then G/ζ (G) need not be Chernikov or even periodic.
Let G be the split extension of the Prüfer p-group P for the odd prime p by the infinite cyclic group ab , where a is the inversion automorphism of P and b is an automorphism of P of infinite order that stabilizes the (only) composition series of P. Then G = P and so is Chernikov, but ζ(G) = 1 , so G/ζ (G) is not even periodic.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.