Frattini Properties and Nilpotency in Leibniz Algebras

Ideals that share properties with the Frattini ideal of a Leibniz algebra are studied. Similar investigations have been considered in group theory. However most of the results are new for Lie algebras. Many of the results involve nilpotency of these algebras.


I. INTRODUCTION
Frattini theory for algebras goes back at least 50 years. A general theory can be found in [14] and there are many works on special classes of algebras, especially Lie algebras. Leibniz algebras, as a generalization of Lie algebras, is a natural class to investigate and [2][3][4][5][6][7][8] contain results on Frattini subalgebras and ideals. Frattini theory for groups goes back to the nineteenth century, and there have been many results that are similar in groups and Lie algebras. Subgroups that have Frattini-like properties have been considered in [9] and related special types of subgroups have been studied in [11]. It is the purpose of this paper to view the Leibniz algebra analogues to some of these theories. Many of these results are new for Lie algebras as well. We consider only finite dimensional Leibniz algebras over a field F. The intersection of all maximal subalgebras of A is called the Frattini subalgebra of A and is denoted by F (A). Even in the solvable case, it need not be an ideal in A [5]. The maximal ideal of A contained in F (A) is called the Frattini ideal of A and is denoted by Φ (A). References for Leibniz algebras include [1], [2], [10] and [12].

II. GENERALIZED FRATTINI IDEALS
In [9], a proper subgroup, H, of a finite group G is called generalized Frattini if whenever G=HN G (P) for any Sylow subgroup P of any normal subgroup K of G, then G=N G (P). To consider such a property in Leibniz algebras, we replace Sylow subgroups with Cartan subalgebras. Unlike the group theory case, Frattini subalgebras do not have to be invariant, and we will use the Frattini ideal as our model to be formalized. To guarantee existence of Cartan subalgebras, we assume the algebras are over an infinite field, [2]. In this context, an ideal H of A is generalized Frattini in A if whenever A=H+N A (C), where C is a Cartan subalgebra of ideal K in A, it follows that A=N A (C). We will show H is a generalized Frattini ideal of A if and only if whenever D and B are ideals of A and D is contained in B∩H, then B/D nilpotent implies that B is nilpotent, a property possessed by Frattini ideals. We will find examples of this concept and conditions that guarantee that an ideal is generalized Frattini. Proposition 1. Let H be a generalized Frattini ideal in A. Then the following are true. 1. H is nilpotent. 2. Any ideal of A that is contained in H is also a generalized Frattini ideal in A 3. H+Φ(A) is a generalized Frattini ideal in A 4. H+Z(A) is a generalized Frattini ideal in A whenever H+Z(A) is a proper subalgebra of A Proof. 1. Let C be a Cartan subalgebra of H. Then A=H+N A (C) by Theorem 6.6 in [2]. H is generalized Frattini in A, hence A=N A (C). Therefore, H=N H (C). Since C is a Cartan subalgebra of H, N H (C)=C. Thus H=C and H is nilpotent.
2. Let N be an ideal of A such that N⊆ H. Let K be an ideal of A and let C be a Cartan subalgebra of K such that A=N+N A (C). Then A=H+N A (C) and hence, A=N A (C) since H is generalized Frattini in A. Thus by definition, N is also generalized Frattini in A.

Let K be an ideal in A with Cartan subalgebra C such that A=H+Φ(A)+N
4. Suppose that K is an ideal in A with Cartan subalgebra C such that A =H+Z(A)+N A (C). Then A=H+N A (C) and A=N A (C) since H is generalized Frattini in A. Therefore H+Z (A) is also generalized Frattini in A. Proof. Let K be an ideal in A with Cartan subalgebra C such that H+N A (C)=A. Then C=K and N A (C)=A.
The next result shows that an important property of the Frattini ideal is shared with any generalized Frattini ideal.
Theorem 4. Let H be generalized Frattini in A. If K is an ideal in A that contains H and K/H is nilpotent, then K is nilpotent.
Proof. Let K be as in the statement of the theorem and let C be a Cartan subalgebra of K. Then (C+H)/H is a Cartan subalgebra of K/H Since K/H is nilpotent, K/H=(C+H)/H and K=C+H. Furthermore A=K+N A (C) by Theorem 6.6 of [2]. Then Proof. If A 2 is generalized Frattini, then the result follows from Theorem 4. If A is nilpotent, then the result follows from Lemma 3.
Proof. Let σ be the natual mapping from K/K ω onto K+H/H. K+H/H is nilpotent and then K+H is nilpotent by Theorem 4.

Theorem 7 Let H be an ideal in a Leibniz algebra A. H is generalized Frattini in A if and only if for each ideal J of A that contains H, whenever J/H is nilpotent, then J is nilpotent.
Proof. If H is generalized Frattini, then the result is Theorem 4. Conversely, suppose that the condition on ideals J holds. Let K be an ideal of A, C a Cartan subalgebra of K with A=H+N A (C). Then (C+H)/H is an ideal in A/H, hence also in (K+H)/H. Since (C+H)/H is Cartan in (K+H)/H, (C+H)/H=(K+H)/H. Therefore (K+H)/H is nilpotent and K+H is nilpotent by hypothesis. Therefore, K=C. Hence N A (C)=N A (K)=A and H is generalized Frattini in A.
Example 8. Let A be a Leibniz algebra with basis x,y,z and multiplications xz=x=-zx, zy=y=-yz, and xy=yx= 0. Let H=(x) and K=(y). H and K are generalized Frattini in A but H+K is not. Thus the sum of two generalized Frattini ideals need not be generalized Frattini. Note that this example is Lie so the result stands in Lie algebras as well.  In [2] and [3] Barnes extends his theory of Engel subalgebras from Lie to Leibniz algebras.

For a∈ A, set E A (a) be the Fitting null component of left multiplication by a on A. This space is a subalgebra called the Engel subalgebra for a. He notes that although a may not be in E
. Hence when working with these subalgebras, we usually can assume that a is in E A (a). For a subalgebra U of A,if the Engel subalgebra for u in U both contains U and is minimal in the set of Engel subalgebras for all elements in U, then the Engel subalgebras for all elements in U contain U. He then shows C is a Cartan subalgebra of A if and only if C is minimal in the set of Engel subalgebras of A.

Theorem 16. Let H be an ideal in A. Then H is generalized Frattini in A if and only if for each ideal K of A and each Cartan subalgebra C of K, A=E
Proof. Let H be generalized Frattini in A. Let K, C be as in the theorem such that for each c∈C, A=H+E A (c). (C+H)/H is a Cartan subalgebra of (K+H)/H by Theorem 6.3 of [2]. Then, using Engel's theorem,

C acts nilpotently on A/H since A=H+E A (c) for all c, hence also on (K+H)/H. Then (C+H)/H also acts nilpotently on (K+H)/H. Then (C+H)/H =(K+H)/H. Therefore (K+H)/H is nilpotent, as is K+H since H is generalized Frattini in A. Hence K is nilpotent and K=C. Since c∈ C=K and K is an ideal, A=E A (c).
Conversely, suppose that H satisfies the conditions in the theorem. Let K be an ideal with H ⊆ K with K/H nilpotent. Let C be a Cartan subalgebra of K and let c ∈ C be an element such that E K (c) is minimal in the set of Engel subalgebras for c ∈C. Then A=K+E A (c)= H+C+E A (c) since K/H is nilpotent. Since C ⊆ E A (c), A=H+E A (c). Hence A=E A (c) by supposition and K= E K (c)= C. Therefore K is nilpotent and H is generalized Frattini by Theorem 7.

III. PRIMITIVE IDEALS
Example 17. Let A be the three dimensional cyclic Leibniz algebra generated by a with aa 3 =a 2 . Let K be the ideal with basis a 2 +a 3 . Then A/K has basis a,a 2 where we delete K from the notation. A/K is cyclic with generator a and aa 2 =-a 2 . The minimum polynomial for L a is x(x+1). Thus A/K has 2 maximal subalgeras and Φ(A/K)=0 using section 4 of [5] and has a unique minimal ideal. Hence K is a primitive ideal in A. Conversely, suppose that B=Nil (A). Since A ios solvable, Nil(A/K)=B/K=Nil(A)/K and K is a proper subalgebra of Nil (A). If N is any ideal of A with N/K nilpotent, then N is nilpotent. Hence K is generalized Frattini in A by Theorem 7.

Lemma 18. Let K be a primitive ideal in A. Then K contains Φ(A) and A/K is non nilpotent. Hence
Corollary 21. Let K be a primitive ideal of a solvable Leibniz algebra A. If K is generalized Frattini in A, then K is maximal with respect to the generalized Frattini property in A.
Proof. Suppose that H is generalized Frattini ideal in A such that K ⊆ H. Then H is nilpotent by Proposition 1 and hence, H is contained in Nil (A). Let B/K be the unique minimal ideal in A/K. By Theorem 20, B=Nil (A). Hence, either H=K or H=Nil(A) since K ⊆ H ⊆ Nil (A).Suppose that H=Nil (A). Then, by Proposition 10, every solvable ideal of A is nilpotent. In particular, A is nilpotent.This contradicts Lemma 18. Thus H=K and K is maximal with respect to the generalized Frattini property.

Let R(A) be the intersection of all maximal subalgebras that are ideals in A and T(A) be the intersection of all maximal subalgebras that are not ideals in A. As for F(A), T(A) may not be an ideal and we let τ (A) be the largest ideal of A contained in T(A). Of course Φ(A)=R(A)∩τ (A).
An algebra is power solvable if all subalgebras generated by a single element are solvable. For Leibniz algebras, these subalgebras are cyclic subalgebras in which A 2 =Leib ( Proof. If N is a maximal ideal of a solvable Leibniz algebra, then dim (A/N)=1. Thus every maximal ideal is a maximal subalgebra that is an ideal. The converse also holds. Hence the result holds. If A is not solvable, then Proposition 32 does not hold.
Example 33. Let A=sl(2,F). Since A is simple, nFrat(A)=0. Since A contains no maximal subalgebras that are ideals, A=R (A). A subset S of A that is closed under multiplications by elements of A is called a normal set in A. An element x ∈ A is a normal non-generator in A if whenever A= < x, T > it follows that A=< T >. Proof. Suppose that x in R (A) and A=< x, S > for a normal subset of S of A. Is < S > = A, then dim A= dim < S >+1, so < S > is a maximal subalgebra of A that is an