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

Abstract

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

According to its English summary Azarov’s paper [1] is devoted to proving the following. If $\pi$ is a finite set of primes, then a soluble group of finite rank is a finite extension of a residually a finite $\pi$-group if and only if it is a finite extension of a residually finite nilpotent $\pi$-group, which happens if and only if it is a reduced FATR group with no $\pi$-divisible elements of infinite order (see below or see [3] for definitions). Further he proves that a soluble group of finite rank is residually a finite $\pi$-group for some finite set $\pi$ of primes if and only if it is a reduced FATR group. (This is effectively an old result of D. J. S. Robinson, see [3] 5.3.8 or [4] Page 138.) We give below what presumably are much shorter proofs of these interesting results. In fact we prove the following. (Below $\tau$(G) denotes the maximal periodic normal subgroup of a group G, FittG its Fitting subgroup and {$\zeta _{\alpha }$(G)} its upper central series.)

FormalPara Theorem

Let G be a finite extension of a soluble FAR group and let $\pi$ be some finite set of primes. The following are equivalent.

• (a) G is a finite extension of a residually finite $\pi$-group.

• (b) $\tau$(G) is finite and $\zeta _{1}$(FittH) is $\pi$-reduced for some normal subgroup H of G of finite index.

• (c) G is a finite extension of a residually finite nilpotent $\pi$-group.

• (d) G is a finite extension of a reduced soluble FATR group with no $\pi$-divisible elements of infinite order.

FormalPara Corollary

Suppose G is a finite extension of a soluble FAR group. The following are equivalent.

• (a) There exists a finite set $\pi$ of primes such that G is residually a finite $\pi$-group.

• (b) G is reduced and a finite extension of an FATR group.

• (c) $\tau$(G) is finite and $\zeta _{1}$(FittG) is reduced.

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

$$\begin{aligned} \langle 1\rangle =G_0 \le G_1 \le \cdots \le G_r \le \cdots \le G_s \le G, \end{aligned}$$

where G$_{1}$ is periodic, divisible, abelian and satisfies min-p for all primes p, G$_{i+1}$/G$_{i}$ for $1\le i<r$ is infinite periodic abelian with all its primary components (i.e. its Sylow subgroups) finite, G$_{i+1}$/G$_{i}$ for $r\le i< \mathrm{s}$ 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 $\tau$(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] 5.2.3).

Note that in general FittG$_{s}$ = G$_{s}\cap$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.

1 The proofs

We use the following elementary results.

Lemma 1

Let A be a torsion-free abelian group.

• (a) If $\pi$ is any set of primes then A is residually a finite $\pi$-group if and only if A is $\pi$-reduced.

• (b) If A is reduced and of finite rank, then for some finite set $\pi$ of primes, A is residually a finite $\pi$-group.

For example see [3] 5.3.4 and 5.3.5. We also use the following, see [4] 9.38.

Lemma 2

Let G be a nilpotent FAR group and $\pi$ any set of primes. Then G is residually a finite $\pi$-group if and only if $\zeta _{1}$(G) is $\pi$-reduced.

Thus in Lemma 2 if $\zeta _{1}$(G) is $\pi$-reduced, then G is residually a finite $\pi$-group, so each G/$\zeta _{i}$(G) is also residually a finite $\pi$-group by Learner’s Lemma. Hence each $\zeta _{i+1}$(G)/$\zeta _{i}($G) is $\pi$-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 $\pi$-group. Then $\zeta _{1}$(FittH) is $\pi$-reduced by Lemma 1. Also $\tau$(H) is a $\pi$-group. If P is a p-subgroup of $\tau$(H) then P is Chernikov, residually finite and hence finite, and $\pi$ is finite. Consequently $\tau$(H) is finite. Clearly $\tau$(G)/$\tau$(H) is finite, so $\tau$(G) is finite.

(b) implies (c). (This is actually the core of the proof of the Theorem.) There exist normal subgroups N $\le$ L of G with L $\le$ H, G/L finite, L/N free abelian of finite rank and N = FittL torsion-free nilpotent of finite rank. Clearly N $\le$ FittH. By (b) and Lemma 2 FittH is residually a finite $\pi$-group, so N is residually a finite $\pi$-group.

Set q = $\Pi _{p\in \pi }$ p and $M=\cap _i C_L (\zeta _{i+1} ( N)^q\zeta _i ( N)/\zeta _i ( N))$. Clearly N $\le$ M $\le$ L and G/M is finite. We claim that M is residually a finite nilpotent $\pi$-group. If so then c) holds. Now M/N is free abelian, so M/N at least is residually a finite nilpotent $\pi$-group.

Let x $\in$ N$\backslash \langle 1\rangle$. Since N is residually a finite $\pi$-group, there exists a power m = q$^{\mu }$ of q with x $\notin$ 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 $\pi$’-subgroup of M/T. Then M/S is a finite nilpotent $\pi$-group. Further x $\notin$ S, since x is a non-trivial $\pi$-element modulo N$^{m}$ and S/N$^{m}$ is an extension of a torsion-free group by a $\pi$’-group. It follows that M is residually a finite nilpotent $\pi$-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 $\tau$(G) is finite, so G is a finite extension of a residually finite-$\pi$, reduced FATR group H. Since H is residually finite-$\pi$, so H contains no $\pi$-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, $\tau$(H) is finite and consequently $\tau$(G) is finite. Further we may choose H torsion-free. Then H has no non-trivial $\pi$-divisible elements by (d) and hence $\zeta _{1}$(FittH) is $\pi$-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 $\kappa$ of primes such that $\zeta _{1}$(FittG) is $\kappa$-reduced. Hence by the Theorem, (b) implies (a), there exists a normal subgroup H of G of finite index that is residually a finite $\kappa$-group. But then G is residually a finite $\pi$-group for $\pi =\kappa \cup${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 $\cap _{p\in \pi } N_p =\langle 1\rangle$. The upper central factors of N/N$_{p}$ are p-reduced (Lemma 2). If they are actually (torsion-free)-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 $\pi$-group.