Solvable Complemented Lie Algebras

In this paper a characterisation is given of solvable complemented Lie algebras. They decompose as a direct sum of abelian subalgebras and their ideals relate nicely to this decomposition. The class of such algebras is shown to be a formation whose residual is the ideal closure of the prefrattini subalgebras.


Prefrattini subalgebras
Throughout, L will denote a finite-dimensional solvable Lie algebra over a field F . We define the nilpotent residual, L ∞ , of L be the smallest ideal of L such that L/L ∞ is nilpotent. Clearly this is the intersection of the terms of the lower central series for L. The derived series for L is the sequence of ideals L (i) of L defined by L (0) = L, L (i+1) = [L (i) , L (i) ] for i ≥ 0; we will also write L 2 for L (1) . If L (n) = 0 but L (n−1) = 0 we say that that L has derived length n. We say that L is completely solvable if L 2 is nilpotent. Algebra direct sums will be denoted by ⊕, whereas vector space direct sums will be denoted by+.
The Frattini subalgebra of L, φ(L), is the intersection of the maximal subalgebras of L. When L is solvable this is always an ideal of L, by [ It was shown in [7] that the definition of prefrattini subalgebras does not depend on the choice of chief series. The Let Π(L) be the set of prefrattini subalgebras of L. Then the following results were established in [7].
in particular, all prefrattini subalgebras of L have the same dimension.
(iii) If A is an ideal of L and S ∈ Π(L) then (S + A)/A ∈ Π(L/A).
(vi) Suppose that F has characteristic p and that L ∞ has nilpotency class less than p. Then the elements of Π(L) are conjugate under inner automorphisms of the form exp(ad If L 2 is not nilpotent then Π(L) can contain more than one element (see [7]).

The Prefrattini Residual
Here we use the ideas of the previous section to re-examine complemented Lie algebras: that is, Lie algebras L for which [0 : L] is complemented, as studied in [5]. Results for groups similar to those in the next theorem were stated by Bechtell in [2].

an abelian Lie algebra and E is an elementary
Lie algebra of type I.
Proof. The equivalence of (i), (ii) and (iii) is [5, Theorem 1]. The equivalence of (iv) follows from [8, Theorem 2.5] (the requirement of a perfect field in that result is required only to establish that an elementary algebra is completely solvable, and that is not needed here).

Lemma 2.3
Let L be a solvable Lie algebra, let B, C be ideals of L with B ∩ C = 0, and suppose that L/B and L/C are complemented. Then L splits over B and over C.
Proof. We show that L splits over C.
A class H of finite-dimensional solvable Lie algebras is called a homomorph if it contains, along with an algebra L, all epimorphic images of L.
If H is a formation then, for every solvable Lie algebra L there is a smallest ideal R such that L/R ∈ H; this is called the H-residual of L. We denote the class of solvable complemented Lie algebras by C. Then we have the following result.

Theorem 2.4 C is a formation.
Proof. First note that C is a homomorph, by [5,Lemma 3]. Let B, C be distinct ideals of L with L/B, L/C ∈ C. We need to show that L/B ∩ C ∈ C.
Without loss of generality we may suppose that B ∩ C = 0. Let 0 < B k < . . . < B 1 = B be part of a chief series for L. We use induction on k. If k = 1 then B is minimal ideal of L and the result follows from Lemma 2.3 and [5, Lemma 4]. So suppose it holds whenever k < n and that we have k = n. Then B/B n , (C + B n )/B n are distinct ideals of L/B n and the corresponding factor algebras are isomorphic to L/B and (L/C)/((C +B n )/C) respectively. These are both complemented (by [5,Lemma 3] in the case of the second). It follows from the inductive hypothesis that L/B n is complemented. But now L is complemented by Lemma 2.3 and [5, Lemma 4], and the result follows.
We define the Prefrattini residual of a solvable Lie algebra L to be π(L) = {B : B is an ideal of L and L/B ∈ C}.
Clearly π(L) is the smallest ideal of L such that L/π(L) ∈ C. It is also the ideal closure of the prefrattini subalgebras of L, by Theorem 2.1.
We define the nilpotent series for L inductively by N 0 (L) = 0, N i+1 /N i = N (L/N i (L) for i > 0, where N (L) denotes the nilradical of L. Finally we have the following characterisation of solvable complemented Lie algebras which is analogous to a result of Zacher for groups (see [10]).
A consequence of the corresponding result for groups is that every normal subgroup of a complemented solvable group is itself complemented. The analogue of this holds for completely solvable Lie algebras, by Theorem 2.2. However, the analogue does not hold for all solvable Lie algebras as the following example shows.

Decomposition results for complemented algebras
A Lie algebra L is called an A-algebra if every nilpotent subalgebra of L is abelian. Here we have some basic structure theorems which mirror those obtained for solvable Lie A-algebras in [6]. Where proofs are very similar to the correponding one in [6] we will sketch the proof for the convenience of the reader and give a reference to [6] for more details. First we see that L splits over the terms in its derived series.
Proof. Suppose first that L is complemented. By Theorem 3.1 there is a subalgebra B n of L such that L = L (n)+ B n . Put A n = L (n) . Similarly Next we aim to show the relationship between ideals of L and the decomposition given in Corollary 3.2. First we need the following lemmas.    [6,Theorem 3.5]) Let L be a solvable complemented Lie algebra of derived length n + 1 with nilradical N , and let K be an ideal of L and A a minimal ideal of L. Then, with the same notation as Corollary 3.2, Proof. (i) We have that L = A n+ B n where A n = L (n) from the proof of Corollary 3.2. It follows from Lemma 3.4 that K = (K ∩ A n ) + (K ∩ B n ). But now K ∩ B n is an ideal of B n , which is complemented, so B n = A n−1+ B n−1 . Applying Lemma 3.4 again gives K ∩ B n = (K ∩ A n−1 )+(K ∩ B n−1 ). Continuing in this way gives the required result.
(ii) This is clear from It remains to show that N ∩ A i ⊆ Z(L (i) ); that is, [N ∩ A i , L (i) ] = 0. We use induction on the derived length of L. If L has derived length one the result is clear. So suppose it holds for Lie algebras of derived length ≤ k, and let L have derived length k + 1. Then B = A k−1 + · · · + A 0 ∼ = L/L (k) is a solvable complemented Lie algebra of derived length k, and, if N is the nilradical of L, then N ∩ A i is inside the nilradical of B for each 0 A Lie algebra L is called monolithic if it has a unique minimal ideal, called the monolith of L.  Given these shared properties between the classes of solvable Lie Aalgebras and solvable complemented Lie algebras it is natural to ask whether either class is contained in the other. This is not the case, as the following examples show. Examples of solvable complemented Lie algebras L that are not Aalgebras are a little harder to construct. In particular, if L is completely solvable and complemented then it is elementary, by Theorem 2.2, and so is an A-algebra. However, such algebras do exist in characteristic p as is shown below.
Example 3.2 Let F be an algebraically closed field of characteristic p, let L be the algebra described in Example 2.1 and let C be a faithful completely reducible L-module. Put X = C+L, where B 2 = 0 and L acts on B under the given L-module action. Then repeated use of [5,Lemma 4] shows that X is complemented. However, X is solvable of index four and so cannot be an A-algebra, by Drensky's Theorem (see [6,Theorem 6.2]).
Notice that an easy extension of the above construction shows that, over an algebraically closed field, there are solvable complemented Lie algebras of arbitrary solvable index, whereas solvable Lie A-algebras over such a field have solvable index at most three.