Complements of Intervals and Prefrattini Subalgebras of Solvable Lie Algebras

In this paper we study a Lie-theoretic analogue of a generalisation of the prefrattini subgroups introduced by W. Gasch\"utz. The approach follows that of P. Hauck and H. Kurtzweil for groups, by first considering complements in subalgebra intervals. Conjugacy of these subalgebras is established for a large class of solvable lie algebras.


Complements of subalgebra intervals
Throughout, L will denote a solvable Lie algebra over a field F . in particular, to show that, for a large class of solvable Lie algebras L, the minimal elements of this set, Ω(U, L) min , are conjugate in L. The development initially follows closely that of [3].
We denote by [U : L] max the set of maximal subalgebras in [U : L]; that is, the set of maximal subalgebras of L containing U . If L = A + B where A and B are subalgebras of L and A ∩ B = 0 we will write L = A ⊕ B. 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 [1,Lemma 3.4]. Extending this notion slightly we put φ(S, L) = {M : M ∈ [S : L] max }; clearly, φ(0, L) = φ(L). The above lemma shows that φ(U, L) ⊆ S for all S ∈ Ω(U, L). If Proof. This follows easily from Lemma 1.1.  Proof. It suffices to show that (S + A)/A ∈ Ω((U + A)/A, L/A) min and so we may suppose that A is a minimal ideal of L. The result is clear if A ⊆ S, since then U + A ⊆ S. So suppose that A ⊆ S.
Then there is a complement M of A in L containing S, by Lemma 1.5, and L = A⊕M . Moreover, S+A ∈ Ω(U +A, L). Choose C ∈ Ω(U +A, L) min At this point the theory starts to diverge from that for groups. We say that L is completely solvable if L 2 is nilpotent. For these algebras Ω(U, L) min takes on a particularly simple form.
Theorem 1.7 Let L be completely solvable and let U be a subalgebra of L.
Then So suppose now that no such minimal ideal exists. Then L is φ-free and so L is complemented, by [4,Theorem 1]. Thus there is a subalgebra V such that C, V = L and C ∩ V = 0. It follows that C, U + V = L and C ∩ (U + V ) = U + C ∩ V = U , whence C ∈ [U : L] and [C : L] is complemented. Thus C ∈ Ω(U, L) and the minimality of B yields that B = C.
If L is not completely solvable then Ω(U, L) min can contain more than one element as we shall see in the next section. However, we do have a conjugacy result in some cases. First we need to consider inner automorphisms of L. Let x ∈ L and let ad x be the corresponding inner derivation of L. If F has characteristic zero, suppose that (ad x) n = 0 for some n; if F has characteristic p, suppose that x ∈ I where I is a nilpotent ideal of L of class less than p. Put Then exp(ad x) is an automorphism of L.
where L i are the terms of the lower central series for L. Then we have conjugacy for the following metanilpotent Lie algebras. Theorem 1.8 Suppose that L is a solvable Lie algebra over a field F of characteristic p, and suppose further that L ∞ has nilpotency class less than p. Let U be a subalgebra of L. Then the elements of Ω(U, L) min are conjugate under I(L : L ∞ ).
Proof. We use induction on the dimension of L. It is clearly true if L has dimension one, so suppose it holds for such algebras with dimension smaller than that of L. We can assume that If C = 0, then (S 1 + C)/C and (S 2 + C)/C are conjugate under I(L/C : (L ∞ + C)/C), by the inductive hypothesis. It follows that there is an x ∈ L ∞ such that S 2 exp(ad x + C) ⊆ S 1 + C exp(ad a) ⊆ M 1 , which gives Then, for each s 2 ∈ S 2 , we have s 2 + s 2 ad x+ . . . + s 2 (ad x) n ∈ M 1 , which gives s 2 + s 2 ad a ∈ M 1 . Thus, again we have that S 2 exp(ad a) ⊆ M 1 for some a ∈ A.
So S 1 , S 2 exp(ad a) ⊆ M 1 for some a ∈ A. Now U ⊆ S 1 ⊆ M 1 and U exp(ad a) ⊆ S 2 exp(ad a) ⊆ M 1 , so, for each u ∈ U , u + [a, u] ∈ M 1 which gives [a, u] ∈ A ∩ M 1 = 0; that is, a ∈ C L (U ) and U exp(ad a) = U . Thus S 2 exp(ad a) ∈ Ω(U exp(ad a), L) min = Ω(U, L) min . But now Lemma 1.3 yields that S 1 , S 2 exp(ad a) ∈ Ω(U, M 1 ) min and the required conjugacy of S 1 and S 2 follows from the inductive hypothesis.

U-prefrattini subalgebras
be a fixed chief series for L. We say that When L is solvable it is easy to see that a chief factor is Frattini if and only if it is not complemented. This can be generalised as follows.
We have a sharpened form of the Jordan-Hölder Theorem in which U -Frattini chief factors correspond. First we need a lemma.

Lemma 2.1 Let A 1 , A 2 be distinct minimal ideals of the solvable Lie algebra L. Then there is a bijection
such that corresponding chief factors have the same dimension and U -Frattini chief factors correspond to one another.
Proof. Clearly we can assume that U = L. Put A = A 1 ⊕ A 2 . Suppose first that A 1 is a U -Frattini chief factor. Then A 1 ⊆ φ(U, L). Thus A ⊆ φ(U + A 2 , L) and A/A 2 is a U -Frattini chief factor. If A/A 1 is also a U -Frattini chief factor, then A ⊆ φ(U + A 1 , L), which yields that A ⊆ φ(U, L), and all four factors are U -Frattini. In this case we can choose θ so that θ(A 1 ) = A/A 2 and θ(A/A 1 ) = A 2 . If A/A 1 is not a U -Frattini chief factor, then nor is A 2 , by the same argument as above, and so the same choice of θ suffices; likewise if none of the factors are U -Frattini chief factors. The remaining case is where A 1 and A 2 are not U -Frattini chief factors but A/A 2 is. Then It follows that all of the chief factors have the same dimension.
If U + A 1 = L, then A/A 1 is a U -Frattini chief factor, so we can choose θ so that θ(A 1 ) = A 2 and θ(A/ L): a contradiction. We must, therefore, have A 2 ⊆ N and so A ⊆ N . Thus A ⊆ φ(U + A 1 , L); that is, A/A 1 is a U -Frattini chief factor. In this case we can again choose θ so that θ(A 1 ) = A 2 and θ(A/A 1 ) = A/A 2 .

Theorem 2.2 Let
be chief series for the solvable Lie algebra L. Then there is a bijection between the chief factors of these two series such that corresponding factors have the same dimension and such that the U -Frattini chief factors in the two series correspond.
Proof. These two series have the same length by a version of the Jordan Hölder Theorem. We induction on n. The result is clearly true if n = 1. So let n > 1 and suppose that the result holds for all solvable Lie algebras with chief series of length ≤ n − 1. If A 1 = B 1 , then applying the inductive hypothesis to L/A 1 gives a suitable bijection between the factors above A 1 , and then we can map A 1 to B 1 and we have the result. So suppose that A 1 and B 1 are distinct and put A = A 1 ⊕ B 1 . Then A/A 1 and A/B 1 are chief factors of L and there are chief series of the form Define an equivalence relation on the chief series of L by saying that two such series are equivalent if there is a bijection between their chief factors satisfying the requirements of the theorem. Since series (1) and (3) have a minimal ideal in common, they are equivalent. Similarly, sereis (2) and (4) are equivalent. Moreover, since series (3) and (4) coincide above A they are also equivalent, by Lemma 2.1. Hence the series (1) and (2) are equivalent, as required.
We define the set I by i ∈ I if and only if A i /A i−1 is not a U -Frattini chief factor of L. For each i ∈ I put If U = 0 we will refer to B simply as a prefrattini subalgebra of L. The Then we have the following important property of U -prefrattini subalgebras of L. (1) and avoids the rest.

Lemma 2.3 If B is a U -prefrattini subalgebra of L then it covers all U -Frattini chief factors of L in
Proof. Let B be a U -prefrattini subalgebra of L and let A i /A i−1 be a chief factor of L. If it is a U -Frattini chief factor then either A i ⊆ φ(U + A i−1 , L) or else U +A i−1 = L. In the former case, every maximal subalgebra of L that contains U + A i−1 also contains A i , and so A i ⊆ B. In either case, therefore, The next four results are dedicated to showing how the U -prefrattini subalgebras relate to the material in the previous section. The first lemma is helpful when trying to calculate U -prefrattini subalgebras.
in particular, all U -prefrattini subalgebras of L have the same dimension.
Proof. We use induction on dim L. The result is clear if L is abelian, so suppose it holds for Lie algebras of smaller dimension than L. It is easy to check that (B + A 1 )/A 1 is a ((U + A 1 )/A 1 )-prefrattini subalgebra of L/A 1 and so Let Π(U, L) be the set of U -prefrattini subalgebras of L.
Proof. (i) We use induction on dim L. The result is clear if L is abelian, so suppose it holds for Lie algebras of dimension less than that of L. Let B ∈ Π(U, L). Then Proof. Let B ∈ Ω(U, L) min and let A i /A i−1 be a chief factor of L. By Lemma 1.6, We now apply Lemma 1.5 to the minimal ideal Proof. This follows from Theorem 2.7 and Theorem 1.8.
If L 2 is not nilpotent then Π(U, L) can contain more than one element, as the following example shows.
is a chief series for L in which A 2 /A 1 is the only Frattini chief factor. It is, therefore, straightforward to see that the prefrattini subalgebras of L are the one-dimensional subalgebras F (αc + a) where a ∈ A 1 = L ∞ , α ∈ F . Note that these are all conjugate under inner automorphisms of the form 1+ ad a. This is not always the case, however. For, if B is a faithful completely reducible L-module and we form X = B⊕L, where B 2 = 0 and L acts on B under the given L-module action, then the prefrattini subalgebras are still of the form F (αc + a) where a ∈ A 1 . However, B is the unique minimal ideal of L and these subalgebras are not conjugate under inner automorphisms of the form 1+ ad b, b ∈ B. Since B is the nilradical of X, defining other inner automorphisms is problematic. Note that X ∞ = B +A 1 which is not nilpotent.