Fase IV: Comunicación de Resultado

H 1 + H3 ' SO that J Si H 1 + J. Since I L and H2 idealises «3, it follows that [J,H2] ç J. Therefore <J,H2> = J + h2 . But H2 si

H 2 + H3 and H 2 + J ç H 2 + H3 , hence: H, si h2 ♦ J. Also J <• H3 and

H 3 si H 2 + H 3 implies that J si H2 + H 3> Hence J si H2 + J. Therefore H 1# H2 , J are pairwise permutable cosubideals. But H 1 om-1 I = H 1 + J, hence by induction , H 2 and J are subideals of H, ♦ H2 ♦ J * Hj ♦ I» Since H 2 si H2 * H 3 and I <i L, it follows that Hj ♦ I si Hj ♦ Hj M > L. Hence H 2 si L which completes the proof. □

t H

Since H3 si H 2 + H3 and = H 1J « H 1 + H 3 , it follows that H^1 + H 3 si + H 2 + H 3 = L. But H^ and H 3 are subideals of H^1 + H 3 , hence H^ and H3 are subideals of L.

Suppose that H 3 <<m H2 + Hj. We prove that H 2 si L by induction on m. If m = 1, then H 3 <J H 2 + H 3 which implies that H 2 idealises H 3> But H 2 idealises by assumption, hence by Lemma 6.3.7, H_ si L. Now suppose that the result

h 2

is true for some m 2 1, and let I = H 3 o H 2 + H 3 , J = I n H 2 . Consider the subalgebras H 1, H 3 and J. We claim that H^, H 3 , J are pairwise permutable cosubideals and H 3 <im_1 j + h3 .

Since H2 idealises H 1, it follows that [JjH^ = [H2 n I,H^] c H 1

Proof of Theorem 6 . 3 .5 a n d < H ^ ,J > = H 1 + J. N o w J <J H 2 s i + H 2 i m p l i e s t h a t J s i H 1 + ; «2. B u t H 1 + J c H 1 ■f H 2 , h e n c e J s i H 1 + J. A l s o H 1 si H 1 + J . S i n c e I = i n (h2 + h3 ) = h3 + (h2 fl I) = + J a n d I 4 H 2 ♦ H 3 , i t f o l l o w s t h a t < H 3 , J > = H 3 ♦ J . A l s o H 3 si H 3 + J and J si H 3 + J. By assumption we have

■ H 1 ■*> H 3 and H 1 si H 1 + H3 , H 3 si H 1 + H.j. Therefore H ^ , J, H 3 are pairwise permutable cosubideals. But H 3 <am_1 I * H 3 + J, hence by induction it follows that J li H 1 ♦ J ♦ Hj > H 1 ♦ I, In particular , J, H 3 being subideals of ♦ I implies I * J + H 3 si ♦ I. Now consider the subalgebras H.^, H 2 and I. We know that H2

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H2 idealises I ]by the definition of I, so <H₂,I> = h2 + Also H2 si

H 2 + I# since H2 si H2 + Hg and Hg c H2 + I. Further, I 4 H, + Hg and I c H2 + I implies I 4 H2 + I. Thus H1 si

H 1 + I and I si H1 + I. Therefore by Lemma H2 si H1 +

H 2 + I = L which completes the proof. □ In the rest of this section we shall investigate ascendant subalgebras of the join of permutable subalgebras of ideally finite Lie algebras.

We start with the following which generalizes Stitzinger [28, Lemma 2] .

Theorem 6.3.6

Let L be an ideally finite Lie algebra over a field of characteristic zero and let A, H , K be subalgebras of L such that L = H + K, and A c H, A c K . If A ase H and A ase K, then A ase L.

Proof

By assumption we have L = UIa where 1^ is a finite-- dimensional ideal of L for each a. Let F^ = C^II^). Then since A 4 A + (L) , we may assume that C ^ L ) * 0 and

nF = 0 . Since L/F is finite-dimensional, it follows that

a a

(A + Fa >/F a si L^Fa by Theorem 6 *2 *1 * Hence A + Fa si L

for each a. Now given Ia we can find Fg such that Ia n F^ = 0 for if not then Ia n CL (Ig) + ₀, so that Ia n C, ( D >* ₀ which is a contradiction. Since A n Ia c A Fg and

[A n I , A + F„] c A n I , it follows that A n I <1 A + F

a 3 — a a 8

But A + F. si L, hence A n I si L and A n I si I for

0 a a a

each a. Let X be any finite-dimensional subalgebra of L. Then since L is ideally finite, it follows that X is contained in a finite-dimensional ideal of L. Therefore A n X si X. Hence by Lemma 5.3.11, A ser L. But A w <j L, so we may assume that A u = 0. Now A c p(L) by Lemma 5.3.8 By Lemma 5.3.9, p(L) is hypercentral. Therefore A ase L. Indeed A L. a

An immediate consequence of Theorem 6.3.6, we have the following:

Corollary 6.3.9

Let L be an ideally finite Lie algebra over a field of characteristic zero and let (i = 1,2,...,n) be subalgebras of L such that L = <H^ i ■ 1,...,n>. Suppose H ± ase <H± ,Hj> = + Hj for all i,j » 1,...,n. Then

ase L for all i. a

Next we prove the following: Theorem 6.3.10

Let L be an ideally finite Lie algebra over a field of characteristic zero and let be subalgebras of L such that L ■ and asc <H^,Hj> for all i,j « 1,2,3. If <H.,H2> is permutable with H3 , then

ase L for all i. Proof

That H., H2 are ascendant in L follows from Lemma 6.2.10 and Theorem 6.3.8. By assumption we have L = UI^, where I _a is a finite-dimensional ideal of L for each a.

Let F = C_ (I ). Then since H, < H, + c , (L) we may assume

oi L a 3 3 1

that t,(L) = 0 and nF = 0 . Since L/F is finite-dimensional,

* 1 a a

it follows that from Theorem 6.2.9 that (H^ + F^l/F^ si Hence + F^ si L. Now argue as in Theorem 6.3.8, we get

ase L. □

Finally we prove the following: Theorem b .3.11

Let L be an ideally finite Lie algebra over a field of characteristic zero and let < L such that

L = <H1 |i = 1,...,n> and H i asc <HifHj> for all i,j = 1,...,n. If H i+1 is permutable with H 1 + Hj + ... + f°r *■> then

ase L. Proof

By Lemma 6.2.10 and Theorem 6.3.8 we have , H2 ascendant in L. To show that ase L, i 2 3 argue as in Theorem 6.3.8 and apply Theorem 6.2.14. o