Remarks on Azarov’s work on soluble groups of finite rank

We present proofs of D. N. Azarov’s recent three theorems determining precisely when a soluble group of finite rank is residually a finite $$\pi $$π-group for a specified finite set $$\pi $$π of primes. Our proofs seem to be substantially shorter; they also apply to groups with a somewhat weaker notion of finite rank.


(d) G is a finite extension of a reduced soluble FATR group with no π-divisible elements of infinite order.
Corollary Suppose G is a finite extension of a soluble FAR group. The following are equivalent.
(a) There exists a finite set π of primes such that G is residually a finite π-group. (b) G is reduced and a finite extension of an FATR group.
Soluble FAR and FATR groups are defined in [3]. An equivalent definition, often more convenient, is the following. A soluble group G is FAR if it has finite Hirsch number and satisfies min-p for every prime p. (A group G has Hirsch number h if G has a series of finite length with exactly h of the factors infinite cyclic, the remaining factors of the series being locally finite; G satisfies min-p if it satisfies the minimal condition on p-subgroups.) It is elementary that to within some normal subgroup of finite index, locally finite factors can essentially be moved down a series past torsion-free abelian factors of finite rank and finite factors can be moved up past torsion-free abelian factors of finite rank and past periodic abelian factors satisfying min-p for all primes p. Further divisible abelian factors in a periodic FAR group sink to the bottom (e.g. [2] 3.18). Thus it is elementary to see that a group G is a finite extension of a soluble FAR group if and only if it has a characteristic series where G 1 is periodic, divisible, abelian and satisfies min-p for all primes p, G i+1 /G i for 1 ≤ i < r is infinite periodic abelian with all its primary components (i.e. its Sylow subgroups) finite, G i+1 /G i for r ≤ i < s is torsion-free abelian of finite rank and G/G s is finite. The soluble FATR groups are exactly those G above with G 1 involving only finitely many primes, with r = 1 and, of course, with G/G s soluble.
Suppose G is a finite extension of a soluble FAR group with its maximal periodic normal subgroup τ (G) finite. Then from the above, the following hold.
(a) G is (torsion-free)-by-finite and (b) G is a finite extension of a soluble FATR group. Further (c) FittG is nilpotent and G/FittG is abelian-by-finite (see [3] 5.2.2) and (d) G/FittG is a finite extension of a free abelian group of finite rank (see [3] Note that in general FittG s = G s ∩FittG, FittG/FittG s is finite and FittG is nilpotent if and only FittG s is nilpotent. Also the soluble groups G of finite rank discussed by Azarov in [1] are exactly FAR groups above with G/G s soluble and G r of finite rank.

The proofs
We use the following elementary results.

Lemma 1 Let A be a torsion-free abelian group.
(a) If π is any set of primes then A is residually a finite π-group if and only if A is π-reduced. (b) If A is reduced and of finite rank, then for some finite set π of primes, A is residually a finite π-group.

Lemma 2 Let G be a nilpotent FAR group and π any set of primes. Then G is residually a finite π-group if and only if ζ 1 (G) is π-reduced.
Thus in Lemma 2 if ζ 1 (G) is π-reduced, then G is residually a finite π-group, so each G/ζ i (G) is also residually a finite π-group by Learner's Lemma. Hence each ζ i+1 (G)/ζ i (G) is π-reduced, something that is easy to see directly.
The proof of the theorem (a) implies (b). Let H be a normal subgroup of G that is residually a finite π-group. Then ζ 1 (FittH) is π-reduced by Lemma 1. Also τ (H) is a π-group. If P is a p-subgroup of τ (H) then P is Chernikov, residually finite and hence finite, and π is finite. Consequently τ (H) is finite. Clearly τ (G)/τ (H) is finite, so τ (G) is finite.
(b) implies (c). (This is actually the core of the proof of the Theorem.) There exist normal subgroups N ≤ L of G with L ≤ H, G/L finite, L/N free abelian of finite rank and N = FittL torsion-free nilpotent of finite rank. Clearly N ≤ FittH. By (b) and Lemma 2 FittH is residually a finite π-group, so N is residually a finite π-group.
Set q = p∈π p and M = ∩ i C L (ζ i+1 (N ) q ζ i (N )/ζ i (N )). Clearly N ≤ M ≤ L and G/M is finite. We claim that M is residually a finite nilpotent π-group. If so then c) holds. Now M/N is free abelian, so M/N at least is residually a finite nilpotent π-group. Let x ∈ N\ 1 . Since N is residually a finite π-group, there exists a power m = q μ of q with x / ∈ N m . Now N/N m is finite and M/N is polycyclic, so there exists a torsion-free normal subgroup T/N m in M/N m with M/T finite. Also N/N m lies in the hypercentre of M/N m and M/N is abelian. Hence M/N m is nilpotent, as therefore is its finite image M/T. Let S/T denote the Hall π'-subgroup of M/T. Then M/S is a finite nilpotent π-group. Further x / ∈ S, since x is a non-trivial π-element modulo N m and S/N m is an extension of a torsion-free group by a π'-group. It follows that M is residually a finite nilpotent π-group.
(c) implies (a). This is trivial. Thus (a), (b) and (c) are equivalent. (a) and (b) imply (d). G is residually finite, so G is reduced. Also τ (G) is finite, so G is a finite extension of a residually finite-π, reduced FATR group H. Since H is residually finite-π, so H contains no π-divisible elements of infinite order. Consequently neither does G.
(d) implies (b). By (d) G has a reduced soluble normal FATR subgroup H of finite index. Then H is (torsion-free)-by-finite, τ (H) is finite and consequently τ (G) is finite. Further we may choose H torsion-free. Then H has no non-trivial π-divisible elements by (d) and hence ζ 1 (FittH) is π-reduced. Thus (b) holds.
The proof of the corollary If (a) holds, then so does (b) by the Theorem. Clearly (b) implies (c). Suppose (c) holds. By Lemma 1 there exists a finite set κ of primes such that ζ 1 (FittG) is κ-reduced. Hence by the Theorem, (b) implies (a), there exists a normal subgroup H of G of finite index that is residually a finite κ-group. But then G is residually a finite π-group for π = κ∪{all prime divisors of (G:H)}. Thus (a) holds.
Remark In a special but still quite general case there is a slicker but less elementary proof of (b) implies (c), the main implication of the theorem.
With N as in the original proof let N p denote the finite-p residual of N. Then ∩ p∈π N p = 1 . The upper central factors of N/N p are p-reduced (Lemma 2). If they are actually (torsionfree)-by-(a p'-group)-by-finite, then G/N p embeds into GL(n, J) for some integer n and J the integers localized at p. Thus G/N p is a finite extension of a residually finite p-group and consequently G is a finite extension of a residually finite nilpotent π-group.
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