Computing the test rank of a free solvable Lie algebra

Abstract

We computed the test rank of a free solvable Lie algebra of finite rank.An element g of an algebra (group) G is a test element if each endomorphism ϕ of G keeping g invariant is an automorphism; i.e., ϕ(g) = g implies that ϕ is an automorphism.A natural generalization of the notion of test element is the notion of test set. Let G be an n-generated algebra (group). The set of elements {g 1 , . . . , g m }, m ≤ n, is called test if, for every endomorphism ϕ of G and i = 1, . . . , m, the conditions ϕ(g i ) = g i imply that ϕ is an automorphism.The test rank of an algebra (group) G is the minimal cardinality of a test set. The majority of articles on test elements and ranks considers groups. Also, the test ranks of Lie algebras are investigated.A question on existence of a group of rank r ≥ 3 and test rank 2 was posed by Fine in (Magnus program website; Open Problems, FP15, http://www.grouptheory.org). Answering this question, the first author computed in [1] the test rank of the free metabelian group of rank r. It happened that the test rank of this group is equal to r − 1 when r ≥ 2. In [2, 3] the test rank of a free metabelian Lie algebra was computed independently; it happened to be the same as in the group case.In [4], the test elements in color metabelian Lie superalgebras of rank 2 were studied. In [5], the admissible values for the test ranks of some free polynilpotent groups were founded. It was proved that the test rank of a free group of the products AN c 1 . . . N c l of varieties of abelian groups A and nilpotent groups N c i is equal either to the group rank r or r − 1. The analog of this result for Lie algebras was obtained in [6].Interest in the test ranks of free solvable groups is explained partially by Fine and Shpilrain's question posed in [7, Question 14.88] and at the Magnus program website. The authors of the question ask whether there exist test elements in the free solvable group F 2 (A l ) of rank 2 with l ≥ 3. The first author was successful in answering Question 14.88. Moreover, he proved in [8] that the test rank of the group F r (AN c 1 . . . N c l ) is equal to r − 1 for every r ≥ 2 and each set of the classes (c 1 , . . . , c l ).The aim of the present note is to compute the test rank of a free solvable Lie algebra. More exactly, the following theorem is proved which generalizes the results of [2, 3] on the test rank of a free metabelian Lie algebra.Theorem. The test rank of a free solvable Lie algebra G of rank r ≥ 2 is equal to r − 1. § 2. Preliminary ResultsDenote by a commutator the product in a Lie algebra G over a field k; by G i , the ith term of the lower central series; and by G (i) , the ith commutant. Abbreviate long commutators: [a, b; 0] = a, [a, b; m] = [[a, b; m − 1], b] when m > 0.

Algebraically and verbally closed subgroups and retracts of finitely generated nilpotent groups

“…The following more general result was proved in [18] Lemma 5.3. Let S 2d be a free solvable group of rank 2 and class d ≥ 2 with basis z 1 , z 2 , and let v ∈ S (d−1) 2d…”

“…It is likely that the answer to this question is negative. By Timoshenko's theorem (see [18] or [19]) the test rank of S rd , r, d ≥ 2, is r − 1. Recall, that test rank of a group G is the minimal number of elements of G such that every endomorphism fixing any of this elements is automorphism.…”

“…Is it true that for any two-generated subgroup H of any free polynilpotent group P of rank r ≥ 2 and class (1, c 1 , ..., c l ), l ≥ 1, the following conditions are equivalent: 1) H is a retract of P , 2) H is a algebraically closed subgroup of P , 3) H is an verbally closed subgroup of P ? By Timoshenko's theorem [18] (see also [19]) any free solvable group S 2d , d ≥ 2, contains test elements, all belong to S (d−1) rd . For the affimative solution to this problem sufficient to prove that any S 2d contains a test element with properties as in Lemma 5.2.…”

Test rank of an abelian product of a free Lie algebra and a free abelian Lie algebra

“…It turns out that the test rank of a free polynilpotent Lie algebra is either equal to the rank of the Lie algebra or is one less than the rank of it. Interest in the test ranks of free soluble Lie algebras is explained in [13] and [15]. Temizyurek and Ekici [13] showed that a free soluble Lie algebra of rank 2 with solvability class 3 has test rank 1 and pointed out the particular test element for such algebras.…”

“…Temizyurek and Ekici [13] showed that a free soluble Lie algebra of rank 2 with solvability class 3 has test rank 1 and pointed out the particular test element for such algebras. The test rank for free soluble Lie algebras of solvability class greater than 3 is calculated by Timoshenko and Shevelin in [15] and it is shown that the test rank of a free soluble Lie algebra of rank n ≥ 2 is equal to n − 1.…”