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Nilpotent Lie Algebras with a SmallSecond Derived QuotientLaurie M. Zack aa Department of Mathematics and Computer Science , High PointUniversity , High Point, North Carolina, USAPublished online: 12 Dec 2008.

To cite this article: Laurie M. Zack (2008) Nilpotent Lie Algebras with a Small Second DerivedQuotient , Communications in Algebra, 36:12, 4607-4619, DOI: 10.1080/00927870802186193

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Communications in Algebra®, 36: 4607–4619, 2008Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927870802186193

NILPOTENT LIE ALGEBRAS WITH A SMALL SECONDDERIVED QUOTIENT1

Laurie M. ZackDepartment of Mathematics and Computer Science,High Point University, High Point, North Carolina, USA

Here we find the structure of nilpotent Lie algebras L with dim�L′/L′′� = 3 and L′′ �= 0.Following the pattern of results of Csaba Schneider in p-groups, we show that L isthe central direct sum of ideals H and U , where U is the direct sum of a generalizedHeisenberg Lie algebra and an abelian Lie algebra. We then find over the complexnumbers that H falls into one of fourteen isomorphism classes.

Key Words: Nilpotent Lie algebra; Second derived quotient.

2000 Mathematics Subject Classification: 17B30.

1. INTRODUCTION

In a recent article, Schneider (2000) found the structure of finite p-groupswith properties �G′/G′′� = p3 and G′′ �= 1 for odd primes. He also found that�G2/G3� = p = �G3/G4� = �G4/G5�, and that G5 = G′′. Then he showed that G isa central product of two simpler groups, H and U , where H is generated by atmost five elements, and that U ′ ⊆ G5. Schneider remarked that to obtain the firstof these results, he solved the analogous Lie algebra problem and then followed asimilar pattern in p-groups. This article was inspired by Schneider’s remark that itwould be interesting to solve the Lie algebra analog to the central product result.After obtaining this Lie algebra decomposition, L = H + U , we have then classifiedthe possible H’s and U ’s over the complex numbers. H can be one of fourteenisomorphism classes and U is the direct sum of a generalized Heisenberg Lie algebraand an abelian Lie algebra. We list without proof the results which have proofscorresponding with Schneider’s work (Schneider, 2000, 2003). It will be clear that Uis the direct sum of a generalized Heisenberg Lie algebra and an abelian Lie algebra.Our emphasis is on the classification of H over the complex numbers.

We begin by introducing notation for the remainder of the article. Let L be afinite dimensional nilpotent Lie algebra of class t, Li the ith term of the lower centralseries, and L′ = �L� L�� L′′ = �L′� L′�, and L′′′ = �L′′� L′′�. Using notation from

Received August 17, 2007; Revised September 27, 2007. Communicated by K. C. Misra.1Dissertation results.Address correspondence to Laurie M. Zack, 1973 Sivclair Trace, Burlington, NC 27215, USA;

E-mail: lzack@highpoint.edu

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Blackburn (1958), we define L1 = �x ∈ L��x� L2� ⊆ L4�. Now we have a chain ofideals: L ⊃ L1 ⊃ L′ = L2 ⊃ L3 ⊃ L4 ⊃ L5 ⊃ · · · ⊃ Lt+1 = 0. To simplify the bracketstructure, we will use the notation ����x1� x2�� x3�� � � � xn�� = �x1� x2� x3� � � � � xn�.

The next results parallel the group theory results (Schneider, 2000, 2003), andwe omit any proofs which are similar to the group theory case.

Theorem 1. Let L be a nilpotent Lie algebra over a field of characteristic not equalto 2, dim�L′/L′′� = 3 and L′′ �= 0. Then dim�L2/L3� = 1, and L′′ = L5.

Lemma 1. Let L be a nilpotent Lie algebra and H be a subalgebra of L such thatL2 = H2 + L3. Then

Li = Hi + Li+1 for all i ≥ 2 and

Li = Hi for all i ≥ 2�

Moreover, H is an ideal of L.

Corollary 1. Let L be a nilpotent Lie algebra.

(i) If dim�L2/L3� = 1 then L has a 2 generator ideal H such that Li = Hi for all i ≥ 2�(ii) If dim�L2/L3� = 2 then L has a 3 generator ideal H such that Li = Hi for all i ≥ 2�

The next result due to Dixmier (1955) has a group theory analog due toHall and Blackburn (1987). We omit the proof but mention an interesting result,Lemma 9, that is used in the proof of the theorem.

Theorem 2. Suppose L is a nilpotent Lie algebra of characteristic not 2 such thatdim�L′/L′′� = 3 and dim�L′′� ≥ 1. Then dim�L′′� = 1.

The next result is the central decomposition theorem that L can be written asthe sum of two simpler Lie algebras, H and U , which has a proof similar to thegroup theory case (Schneider, 2003).

Theorem 3. Let L be a nilpotent Lie algebra such that dim�L′/L′′� = 3 and L′′ �= 0.Then L can be written as L = H + U , where:

(i) H is an ideal of L generated by at most five generators;(ii) Hi = Li for all i ≥ 2(iii) U is an ideal of L such that U ′ ⊆ L5;(iv) H and U centralize each other.

The remainder of the article, which has no analog in Schneider’s article,contains the classification of H over the complex numbers. We find that there arefourteen isomorphism classes for H . The classification of U is easily seen to be thedirect sum of a generalized Heisenberg Lie algebra and an abelian Lie algebra.

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2. DISCUSSION OF THEOREM 2

Dixmier (1955) proved Theorem 4. A group theory version was shown by Halland Blackburn (1987) and was used in Schneider (2003). For two generated groups,Schneider constructed a generating set with useful properties. In this section, weconstruct an analogous generating set for two generated Lie algebras which will beuseful throughout the article. This construction was used in Zack (2007) to give analternative proof of Theorem 2 using Lemma 4. These results are similar to those inBlackburn (1987) and therefore will be omitted.

Lemma 2. Let L be a nilpotent Lie algebra with dim�L′/L′′� = 3 over a field ofcharacteristic not equal to 2. Then L1 has codimension 1 in L.

Proof. Since every commutator is a linear combination of left justifiedcommutators, �L2� L2� ⊆ L4, and L2 ⊆ L1 by definition of L1. We may take L4 = 0and by Theorem 1, dim�L2/L3� = dim�L3/L4� = 1. Now let L2 = �y + L3 andL3 = �z + L4. Then f�x� = �x� y� = xz has a one dimensional image. Therefore thekernel of f has codimension 1 in L and clearly ker f = L1.

We immediately obtain the following corollary.

Corollary 2. Let L be a nilpotent Lie algebra with dim�L′/L′′� = 3 over a field ofcharacteristic not equal to 2. Suppose that dim�L/L′� = 2. Then L has generators a andb such that L = �a + L1 and L1 = �b + L2.

For the remainder of the article, when working with a two generator algebra,we will take generators a and b as stated in Corollary 2.

Lemma 3. Let L be a 2-generated finite dimensional nilpotent Lie algebra such thatdim�L′/L′′� = 3, dim�L′/L3� = 1, and L′′ �= 0. Then generators a and b of L can bechosen such that the following hold:

(i) L2 = ��a� b� + L3(ii) L3 = ��a� b� a� + L4 and �a� b� b� ∈ L4(iii) L4 = ��a� b� a� a� + L5 and �a� b� a� b� ∈ L5(iv) L5 = ��a� b� a� a� b� + L6 and �a� b� a� a� a� ∈ L6�

Proof. Let b ∈ L1 and a ∈ L be generators of L. In L ⊃ L1 ⊃ L2 ⊃ L3 ⊃ L4 ⊃ L5,where each Li+1 has codimension 1 in Li. Of course, L = �a + L1, L1 = �b + L2 andL2 = ��a� b� + L3. Since �a� b� b� ∈ �L2� L1� ⊆ L4, it follows thatL3 = ��a� b� a� + L4.Then �a� b� a� b� = �a� b� b� a�− ��a� b�� �b� a�� ∈ L5 and L4 = ��a� b� a� a� + L5.

We can assume L6 = 0 and since ��a� b� b�� �a� b�� ∈ �L4� L2� = 0, it followsthat ��a� b� a�� �a� b�� �= 0. Then �a� b� a� a� b� = �a� b� a� b� a�+ ��a� b� a�� �a� b�� =�a� b� a� �a� b��, and L5 = ��a� b� a� a� b�. Then �a� b� a� a� a� = �a� b� a� a� b�, andwe claim that a change in a allows to be 0. Let a′ = a− b. Then all of theabove equations hold by replacing a with a′. Also �a′� b� a′� a′� a′� = �a� b� a� a� a�−�a� b� a� a� b� = 0. Hence, the lemma is shown.

Central to the proof of Theorem 4 is the following analogue to a result ofBlackburn (1987).

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Lemma 4. If L is a nilpotent Lie algebra over a field of characteristic not equal to 2,with series L ⊃ L′ ⊃ L′′ ⊃ L′′′ = 0, where dim�L′/L′′� = 3 and dim�L′′� = 2. Then thefollowing hold:

(i) L′′ = Z�L′�(ii) L′ has a unique abelian subalgebra A of codimension 1 in L′(iii) If B is an ideal of L where dim�B� = 3 and L′ ⊃ B ⊃ L′′, then B ⊂ A.

The proof of Theorem 2 now follows as in Blackburn (1987). Since dim�L′�= 1and L′′ = L5, it follows that L6 = 0 along with the relations in Lemma 3, and wenow have the following properties.

Corollary 3. Assuming L is a nilpotent Lie algebra with the properties of Lemma 3,the following also hold:

(i) �a� b� b� a� = �a� b� a� b� = �a� b� a� a� a� = 0;(ii) �a� b� a� a� b� = �a� b� a� �a� b��

Proof. Since �a� b� b� = c�a� b� a� a�+ d�a� b� a� a� b�, it follows that �a� b� a� b� =�a� b� b� a� = c�a� b� a� a� a�+ d�a� b� a� a� b� a� = 0. Also, since �a� b� a� b� a� = 0, itfollows that �a� b� a� a� b� = �a� b� a� �a� b��.

3. DISCUSSION OF THEOREM 3

As we have already seen, L has a 2-generator ideal H , such that for all i≥ 2,Li = Hi. We have also found in the previous section, a particular generating setwhich satisfies some additional conditions. These results are used to obtain thedesired factorization. Our work parallels Schneider’s, so the proofs are omitted.

Lemma 5. Let L be nilpotent Lie algebra over a field of characteristic not equalto 2, such that dim�L′/L′′� = 3 and L′′ �= 0. Then L has a minimal generating set�a� b� u1� u2� � � � � ur� which is a basis for L/L′ such that for all ui and uj:

(i) H = �a� b is an ideal of L such that Hi = Li for all i ≥ 2. Furthermore, a and bare as in Lemma 2;

(ii) �a� ui� ∈ L5 for all ui;(iii) �b� ui� ∈ L4 for all ui;(iv) �ui� uj� ∈ L5 for all ui and uj .

The proof of Theorem 3 uses the above lemma and the results of Theorem 2.Theorem 3 is the Lie algebra analog to the Schneider’s central decomposition.

The nilpotent Lie algebra U in Theorem 3 has the multiplication of a skewform from U into �z, where z = �a� b� a� a� b�. U has a basis x1� y1� x2� y2� � � � �xr� yr� z� w1� � � � � ws where �xi� yj� = �ijz and Z�U� = �z� w1� � � � � ws by Theorem 6.3in Jacobson (1985). Hence U is a direct sum of a generalized Heisenberg made upby the x� y� and z′s and an abelian Lie algebra made up by the w′s.

For the remainder of this section, we will examine the possible structures ofthe Lie algebra H over the complex numbers. We consider the cases as the numberof generators for H increase. H is said to be degenerate if H = H1 + U1 where H1

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Table 1 Two generated Lie algebra: Case 1

a b �a� b� �a� b� a� �a� b� a� a� �a� b� a� a� b�

a 0 �a� b� −�a� b� a� −�a� b� a� a� 0 0b −�a� b� 0 0 0 −�a� b� a� a� b� 0

�a� b� �a� b� a� 0 0 −�a� b� a� a� b� 0 0�a� b� a� �a� b� a� a� 0 �a� b� a� a� b� 0 0 0

�a� b� a� a� 0 �a� b� a� a� b� 0 0 0 0�a� b� a� a� b� 0 0 0 0 0 0

and U1 are as in Theorem 3 and U1 �= 0, and in which case we do not consider themfurther.

4. H = �a�b�Let H be as in Lemma 8, and let b′ = b + �a� b�. Then �a� b′� =

�a� b�+ �a� �a� b�� and �a� b′� b′� = �a� b� b�+ �a� �a� b�� b�+ 2�a� �a� b�� �a� b�� =�a� b� b�− 2���a� b�� a�� �a� b��. Since �a� b� b� = c�a� b� a� a�+ d�a� b� a� a� b�, then�a� b′� b′� = c�a� b� a� a�+ d�a� b� a� a� b�− 2���a� b�� a�� �a� b��.

Now let = √d, then �a� b′� b′� = c�a� b� a� a� = c�a� b′� a� a�� If c = 0, then

�a� b′� b′� = 0. If c �= 0, replace b′ by b′′ = c−1b′ and then �a� b′′� b′′� = �a� b′′� a� a�.Thus H is isomorphic to one of the algebras whose multiplication is given in eitherTables 1 or 2.

Summarizing, if H is a two-generated Lie algebra with dim�H ′/H ′′� = 3,H ′′ �= 0 and with the same notation as before, the basis for H is given by�a� b� �a� b�� �a� b� a�, �a� b� a� a�� �a� b� a� a� b�� and dim�H� = 6. Also, H2 = H ′ =��a� b�, �a� b� a�� �a� b� a� a�� �a� b� a� a� b�, dim�H ′� = 4, H3 = ��a� b� a�� �a� b� a� a�,�a� b� a� a� b�, H4 = ��a� b� a� a�, �a� b� a� a� b�, H5 = H ′′ = ��a� b� a� a� b��,dim�H ′′� = 1, Z�H� = ��a� b� a� a� b��, and H6 = 0 so H is nilpotent of class 5.This forms the following theorem.

Theorem 4. Suppose H is two generated nilpotent Lie algebra with dim�H ′/H ′′�= 3and H ′′ �= 0� Then there exists generators a and b such that the multiplication tablebetween basis elements is given either in Tables 1 or 2. These algebras are notisomorphic since in one case H ′/H5 is abelian and in the other case it is not.

Table 2 Two generated Lie algebra: Case 2

a b �a� b� �a� b� a� �a� b� a� a� �a� b� a� a� b�

a 0 �a� b� −�a� b� a� −�a� b� a� a� 0 0b −�a� b� 0 −�a� b� a� a� 0 −�a� b� a� a� b� 0

�a� b� �a� b� a� �a� b� a� a� 0 −�a� b� a� a� b� 0 0�a� b� a� �a� b� a� a� 0 �a� b� a� a� b� 0 0 0

�a� b� a� a� 0 �a� b� a� a� b� 0 0 0 0�a� b� a� a� b� 0 0 0 0 0 0

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In the sections that follow, a change in generators is needed when discussingisomorphism classes. The following general remarks will help with these changes.Let H be the Lie algebra in Theorem 3 with subalgebra M = �a� b as inCorollary 1(i) and properties as in Lemma 3 and Corollary 3. Let �u1� u2� � � � � ur�be as in Lemma 5. Suppose that a� b� u are changed to a′� b′� u′ which are newgenerators with all the same conditions as the old generators. For �a′� b′� b′� ∈ M4 asin Lemma 3(ii), b′ ∈ �M2� b� u1� � � � � ur. For �a′� b′� a′� a′� a′� = 0 as in Lemma 3(iv),a′ ∈ �M2� a� u1� � � � � ur. Since �u′� H2� = 0, it follows that u′ ∈ �M4� u1� � � � � ur.

Summarizing, any such change of bases has

a′ = c0a+ c2�a� b�+ c3�a� b� a�+ c4y + c5z+r∑

i=1

ci+5ui

b′ = h1b + h2�a� b�+ h3�a� b� a�+ h4y + h5z+r∑

i=1

hi+5ui

ui = d4y + d5z+r∑

i=1

di+5ui

where y = �a� b� a� a� and z = �a� b� a� a� b�.

5. H = �a�b� u�Now consider H = �a� b� u where M = �a� b is the 2-generated algebra as in

the previous section and let y = �a� b� a� a� and z = �a� b� a� a� b�. From Lemma 5,�u� a� = z� �u� b� = �y + z and u � CH�M�,where � �� ∈ � and � is the complexnumbers. We will divide these algebras into isomorphism classes, each with asimple representation. Let u′ = �y + �u, with � �= 0, otherwise u′ ∈ M . Then �u′� a� =��u� a� = �z and �u′� b� = �z+ ���y + z�. Let � = −� . Then �u′� b� = ��y and�u′� a� = �z. We now alter � by looking at all the possible cases for and �,which are:

1. If = 0 and � = 0, then u′ ∈ CH�M� and H = M + U , therefore H is degenerate;2. If = 0 and � �= 0, then let � = 1

�. Then �u′� a� = 0 and �u′� b� = y;

3. If �= 0 and � = 0, then let � = 1. Then �u′� a� = z and �u′� b� = 0;

4. If �= 0 and � �= 0, then let � = 1. Then �u′� a� = z and �u′� b� = �y where � = �

.

In the last case, we want � = 1, so we rescale a, b, and u′ to produce afurther simplification. Letting a′ = da, b′ = eb, and u′′ = fu′ where d� e� and f arescalars in �, then �a′� b′� = de�a� b�� �a′� b′� a′� = d2e�a� b� a�� �a′� b′� a′� a′� = y′ =d3ey� �a′� b′� a′� a′� b′� = z′ = d3e2z, and �a′� b′� b′� = de2�a� b� b�. Also, �u′′� a′� =df�u′� a� = dfz = f

d2e2z′ and �u′′� b′� = f�

d3y′.

Either �a� b� b� = 0 or �a� b� b� = y from Section 4. In the first case, �a′� b′� b′� =de2�a� b� b� = 0. In the second case, �a′� b′� b′� = de2y = e

d2y′. Now we choose e� d,

and f such that f

d2e2= 1� f

d3� = 1, and e

d2= 1. We accomplish this by letting

d = 1�1/3

� f = 1�2

and e = 1�2/3

.Dropping all primes, we have six possible isomorphism classes (see Table 3)In this section, H is degenerate precisely when dim�Z�H�� > 1. Directly

checking the six cases, we find that they are all nondegenerate.

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Table 3 Possible isomorphism classes for threegenerated Lie algebra

1 �u� a� = z, �u� b� = 0, �a� b� b� = 02 �u� a� = 0, �u� b� = y, �a� b� b� = 03 �u� a� = z, �u� b� = y, �a� b� b� = 04 �u� a� = z, �u� b� = 0, �a� b� b� = y

5 �u� a� = 0, �u� b� = y, �a� b� b� = y

6 �u� a� = z, �u� b� = y, �a� b� b� = y

Next we examine which of these cases are isomorphic to each other. Noticethat H = �a� b� �a� b�, �a� b� a�� y� z� u�, H2 = ��a� b�� �a� b� a�� y� z�, H1 = ��a� b��

�a� b� a�� y� z� b� u�, and Z�H� = H5 = z. In cases 1, 2, and 3, CH�H2/H5� = �b� �a� b�,

�a� b� a�� y� z� u� and in cases 4, 5, and 6, CH�H2/H5� = ��a� b�� �a� b� a�� y� z� u�.

Hence, cases 1, 2, and 3 are not isomorphic to cases 4, 5, and 6. Now by examiningCH�H

1/H5� we get that in case 1, CH�H1/H5� = �b� �a� b�� �a� b� a�� y� z� u�, in

cases 2 and 3, CH�H1/H5� = ��a� b�� �a� b� a�� y� z�, in case 4, CH�H

1/H5� =��a� b� a�� y� z� u�, and in cases 5 and 6, CH�H

1/H5� = ��a� b� a�� y� z�. Again, bycounting the dimensions we see that case 1 is not isomorphic to case 2 or 3, andcase 4 is not isomorphic to case 5 or 6. It remains to show the relationship betweencases 2 and 3, and cases 5 and 6.

Suppose that case 2 is represented with primes, case 3 is not, andthat they are isomorphic. Let a′ = c0a+ c2�a� b�+ c3�a� b� a�+ c4y + c5z+ c6u andb′ = h1b + h2�a� b�+ h3�a� b� a�+ h4y + h5z+ h6u similar to equations (1) and (2)from the previous section. Let u′ = d4y + d5z+ d6u, and note that �u′� H2� = 0. Now0 = �u′� a′� = d6c0�u� a�.

Since c0 �= 0, then d6 = 0 and u′ ∈ �a� b which is a contradiction. Hence case2 is not isomorphic to case 3. By similar reasoning we can conclude cases 5 and 6 arealso not isomorphic. Therefore, we have six isomorphism classes when H = �a� b� u.This case is summarized in the following theorem.

Theorem 5. If H is a nilpotent Lie algebra such that dim�H ′/H ′′� = 3, H ′′ �= 0,H = �a� b� u with a and b as in the previous theorem, and u � CH��a� b�. Then thereare six isomorphism classes whose canonical algebras are given as in Table 3.

6. H = �a�b� u1� u2�Now consider the case H = �a� b� u1� u2, where H is as in Theorem 2 and

M = �a� b as in Tables 1 or 2. Again let y = �a� b� a� a� and z = �a� b� a� a� b�

for simplification. From �u1� a� = 1z, �u1� b� = �1y + 1z, �u2� a� = 2z, �u2� b� =�2y + 2z we will divide the algebras into isomorphism classes, each with a simplemultiplication between basis elements. Let

u′1 = c0y + c1u1 + c2u2�

u′2 = d0y + d1u1 + d2u2�

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4614 ZACK

where[ c1 c2d1 d2

]is nonsingular. Then

�u′1� a� = �c11 + c22�z = �1z

�u′2� a� = �d11 + d22�z = �2z

�u′1� b� = c0z+ c1��1y + 1z�+ c2��2y + 2z� = �1y + �1z

�u′2� b� = d0z+ d1��1y + 1z�+ d2��2y + 2z� = �2y + �2z�

Setting �1 = �2 = 0 and then letting c0 = −c1 1 − c2 2 and d0 = −d1 1 − d2 2, weobtain that �u′

1� b�� �u′2� b� ∈ �y. Also,

[ c1 c2d1 d2

] [1 �12 �2

]= � �1 �1

�2 �2 �, which we denote byAB = C.

If rank�B� = 2, then solve AB = I to obtain

�u′1� a� = z� �u′

1� b� = 0� �u′2� a� = 0� and �u′

2� b� = y�

If rank�B� = 0, then H is degenerate and we do not consider this case.If rank�B� = 1, we obtain the following three cases:

1. Solve AB = � 1 00 0 � to obtain

�u′1� a� = z� �u′

1� b� = 0� �u′2� a� = 0� and �u′

2� b� = 0

2. Solve AB = � 0 10 0 � to obtain

�u′1� a� = 0� �u′

1� b� = y� �u′2� a� = 0� and �u′

2� b� = 0

3. Solve AB = � 1 �0 0 � for some � �= 0 to obtain

�u′1� a� = z� �u′

1� b� = �y� �u′2� a� = 0� and �u′

2� b� = 0�

Another change will allow us to take � = 1. First, dropping all previousprimes to simplify the notation, and then let a′ = da, b′ = eb, and u′

1 = fu1 whered� e� f �= 0. Then

�a′� b′� = de�a� b�

�a′� b′� a′� = d2e�a� b� a�

y′ = �a′� b′� a′� a′� = d3ey

z′ = d3e2z

�a′� b′� b′� = de2�a� b� b�

�u′1� a

′� = df�u1� a� = dfz = df

d3e2z′ = f

d2e2z′

�u′1� b

′� = ef�u1� b� = ef�y = ef�

d3ey′ = f�

d3y′�

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NILPOTENT LIE ALGEBRAS 4615

Furthermore, �a� b� b� = 0 or y from Section 4. Hence, �a′� b′� b′� = 0 or �a′� b′� b′� =de2y = e

d2y′. Therefore, we seek a solution to

f

d2e2= 1�

f�

d3= 1� and

e

d2= 1�

Letting d = 1�1/3

, e = 1�2/3

, and f = 1�2

satisfies the equations. Hence, dropping allprimes again, the rank�B� = 1 third case becomes

�u1� a� = z� �u1� b� = y� �u2� a� = 0� �u2� b� = 0� and �a� b� b� = 0 or y�

Since u1 and u2 have been altered, now �u1� u2� = �z. A final change will allow� = 0 or � = 1 to be used. In all rank�B� = 1 cases � �= 0, otherwise these cases aredegenerate. A simple change of u′

2 = 1�u2 allows us to take �u1� u2� = z, and we have

finished the rank�B� = 1 situation.Now consider �u1� u2� = �z in the rank�B� = 2 case and let a′ = da� b′ = eb�

u′1 = f1u1� and u′

2 = f2u2. As before, y′ = d3ey and z′ = d3e2z, and we compute�u′

1� a′�, �u′

2� b′�, �u′

1� u′2�, �a

′� b′� b′� as done previously and find that we want

e

d2= 1�

f1f2d3e2

� = 1�f1d2e2

= 1 andf2d3

= 1�

Let d = 1�1/2

� e = 1�� f1 = 1

�3, and f2 = 1

�3/2to solve these equations. Then

�u′1� a

′� = z′� �u′1� b

′� = 0� �u′2� a

′� = 0� �u′2� b

′� = y′�

�u′1� u

′2� = z′ or 0� and �a′� b′� b′� = y′ or 0�

In summary, dropping all primes, we have the ten possible isomorphism classesin Table 4.

It remains to divide these ten cases into isomorphism classes. Similar to whenH = �a� b� u, we see cases 1 through 5 contain the multiplication �a� b� b� = 0, cases6 through 10 contain the multiplication �a� b� b� = y, and bases for H , H ′ = H2

and H5 are

H = �a� b� �a� b�� �a� b� a�� y� z� u1� u2�

H1 = �b� �a� b�� �a� b� a�� y� z� u1� u2�

Table 4 Possible isomorphism classes for four generated Lie algebra

1 �u1� a� = z, �u1� b� = 0, �u2� a� = 0, �u2� b� = y, �u1� u2� = z, �a� b� b� = 02 �u1� a� = z, �u1� b� = 0, �u2� a� = 0, �u2� b� = y, �u1� u2� = 0, �a� b� b� = 03 �u1� a� = z, �u1� b� = 0, �u2� a� = 0, �u2� b� = 0, �u1� u2� = z, �a� b� b� = 04 �u1� a� = 0, �u1� b� = y, �u2� a� = 0, �u2� b� = 0, �u1� u2� = z, �a� b� b� = 05 �u1� a� = z, �u1� b� = y, �u2� a� = 0, �u2� b� = 0, �u1� u2� = z, �a� b� b� = 06 �u1� a� = z, �u1� b� = 0, �u2� a� = 0, �u2� b� = y, �u1� u2� = z, �a� b� b� = y

7 �u1� a� = z, �u1� b� = 0, �u2� a� = 0, �u2� b� = y, �u1� u2� = 0, �a� b� b� = y

8 �u1� a� = z, �u1� b� = 0, �u2� a� = 0, �u2� b� = 0, �u1� u2� = z, �a� b� b� = y

9 �u1� a� = 0, �u1� b� = y, �u2� a� = 0, �u2� b� = 0, �u1� u2� = z, �a� b� b� = y

10 �u1� a� = z, �u1� b� = y, �u2� a� = 0, �u2� b� = 0, �u1� u2� = z, �a� b� b� = y

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4616 ZACK

H2 = ��a� b�� �a� b� a�� y� z�

H5 = �z��

Notice for cases 3 and 8, by letting a = a− u2, we obtain �u1� a� = �u1� a�−�u1� u2� = z− z = 0. Then u1� u2 ∈ U , and these two cases are degenerate. We nowcheck to see if any of the remaining eight cases are degenerate. If dim�Z�H�� = 2or 3, then clearly H is degenerate. Consider cases 1, 2, 6, and 7. Direct calculationshows that dim�Z�H�� = 1 and dim�Z�H/H5�� = 2.

Suppose that H = H1 + U1. If dim�U1� = 1, then dim�Z�H�� = 2, which is acontradiction. If dim�U1� = 2, then dimZ�H/H5� = 3, which is also a contradiction.Hence all four cases are nondegenerate. Cases 4, 5, 9, and 10 are handled in a similarmanner, and these four cases are also nondegenerate.

Now that we have determined which cases are degenerate, we now divide theeight remaining nondegenerate cases into isomorphism classes. In cases 1 through 5,dimCH�H

2/H5� = 7, in cases 6 through 10, dimCH�H2/H5� = 6. Therefore, cases 1

through 5 are not isomorphic to cases 6 through 10. Also, CH�H1� = �z� u1� in cases

2 and 7 and CH�H1� = �z� in cases 1, 4, 5, 6, 9, and 10. Thus the class defined by 2

and the one defined by 7 are not isomorphic to any of the other classes. In summary,cases 1, 2, 4, 5, are not isomorphic to cases 6, 7, 9, and 10, case 2 is not isomorphicto cases 1, 4, and 5, and case 7 is not isomorphic to cases 6, 9, and 10.

We claim that case 5 is isomorphic to case 4. First let a� b� u1� u2 give themultiplication in case 5. Letting a = a− u2, yields the multiplication in case 4.Similarly, using the same change, we find that case 10 is isomorphic to case 9.

We also claim that case 1 is isomorphic to case 4. Let a� b� u1� u2 give themultiplication in case 4. Letting a = a+ u1 − u2� b = b − �a� b�+ u2� u1 = −u2, andu2 = −y + u1 − u2. Then this change takes case 4 into an algebra with multiplication

�u1� a� = z� �u1� b� = 0� �u2� a� = 0� �u2� b� = y� �u1� u2� = z� �a� b� b� = 0�

Hence case 4 is isomorphic to case 1. The same transformation produces case 9 tobe isomorphic to case 6. Hence cases 1, 4, and 5 are isomorphic to each other, andcases 6, 9, and 10 are isomorphic to each other.

Therefore, there are four isomorphism classes when H = �a� b� u1� u2. Theyhave representations of type 1, 2, 6, and 7 shown in Table 4.

Theorem 6. Let H = �a� b� u1� u2 as in Theorem 2. Then there are four isomorphismclasses of types 1� 2� 6, and 7 in Table 4.

7. H = �a�b� u1� u2� u3�Finally, consider H = �a� b� u1� u2� u3, where H is as in Theorem 2 and

M = �a� b as in Tables 1 or 2. Again, let y = �a� b� a� a� and z = �a� b� a� a� b�.Similarly to Section 6, �u1� a� = 1z, �u1� b� = �1y + 1z, �u2� a� = 2z, �u2� b� = �2y + 2z, �u3� a� = 3z, �u3� b� = �3y + 3z where all of the scalars are in �. Let

u′1 = c0y + c1u1 + c2u2 ++c3u3

u′2 = d0y + d1u1 + d2u2 ++d3u3

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NILPOTENT LIE ALGEBRAS 4617

u′3 = e0y + e1u1 + e2u2 ++e3u3�

where[ c1 c2 c3d1 d2 d3e1 e2 e3

]is non-singular. Then

�u′1� a� = c11z+ c22z+ c33z = �1z

�u′2� a� = d11z+ d22z+ d33z = �2z

�u′3� a� = e11z+ e22z+ e33z = �3z

�u′1� b� = c0z+ c1��1y + 1z�+ c2��2y + 2z�+ c3��3y + 3z� = �1y + �1z

�u′2� b� = d0z+ d1��1y + 1z�+ d2��2y + 2z�+ d3��3y + 3z� = �2y + �2z

�u′3� b� = e0z+ e1��1y + 1z�+ e2��2y + 2z�+ e3��3y + 3z� = �3y + �3z�

Set �1 = �2 = �3 = 0. Letting c0 = −c1 1 − c2 2 − c3 3, d0 = −d1 1 − d2 2 − d3 3,and e0 = −e1 1 − e2 2 − e3 3 we obtain �u′

1� b�� �u′2� b�, and �u′

3� b� ∈ �y. Also

c1 c2 c3d1 d2 d3

e1 e2 e3

1 �1

2 �2

3 �3

=

�1 �1

�2 �2

�3 �3

which we denote by AB = C, where A is nonsingular. The analysis follows thesame path as in the last section. If rank�B� = 0, then u1� u2� u3 ∈ U and therefore isdegenerate.

If rank�B� = 2, then we can find A such that AB =[1 00 10 0

]to obtain

�u′1� a� = z� �u′

1� b� = 0� �u′2� a� = 0� �u′

2� b� = y� �u′3� a� = 0 and �u′

3� b� = 0�

If rank�B� = 1, as in the last section, we obtain 3 cases:

1. �u1� a� = z� �u1� b� = 0� �u2� a� = 0� �u2� b� = 0� �u3� a� = 0� and �u3� b� = 0;2. �u1� a� = 0� �u1� b� = y� �u2� a� = 0� �u2� b� = 0� �u3� a� = 0� and �u3� b� = 0;3. �u1� a� = z� �u1� b� = �y� �u2� a� = 0� �u2� b� = 0� �u3� a� = 0, and �u3� b� = 0 where

� �= 0�

In case 3, similar to Section 6, we change generators to make � = 1 byletting a′ = da, b′ = eb, and u′

1 = fu1. The same computations as in the last sectionshow that letting d = �1/3, e = 1

�2/3, and f = 1

�2changes case 3 to �u′

1� a′�= z and

�u′1� b

′�= y and the other products are 0. To proceed, dropping the primes, case 3becomes

�u1� a� = z� �u1� b� = y� �u2� a� = 0� �u2� b� = 0� �u3� a� = 0� and �u3� b� = 0�

Now for each of the above cases, �u1� u2� = �1z� �u1� u3� = �2z and �u2� u3� =�3z. For the rank 1 case, if �1 = 0, then interchanging u2 and u3 yields�1 �= 0. Now let u2 = 1

�1u2, then �u1� u2� = 1

�1�u1� u2� = z. Dropping the hat, we

have �u1� u2� = z� �u1� u3� = �2z� and �u2� u3� = �3z. Now let u3 = u3 − �2u2. Then�u1� u3� = �2z− �2z = 0 and �u2� u3� = �u2� u3�− �2�u2� u2� = �u2� u3�. Dropping the

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4618 ZACK

hat again, �u1� u2� = z� �u1� u3� = 0� �u2� u3� = �3z. Now �3 �= 0 otherwise this case isdegenerate, but letting u3 = 1

�3u3 and changing back to the u′s we have �u1� u2� =

z� �u1� u3� = 0� and �u2� u3� = z. Then letting u1 = u3 + u1 yields �u1� u3� = �u3� u3�+�u1� u3� = 0 and �u1� u2� = �u3� u2�+ �u1� u2� = −z+ z = 0. Hence, both u2 and u3

drop out and this case is degenerate. Therefore, all rank 1 cases are degenerate.For the rank�B� = 2 case, we have

�u1� a� = z� �u1� b� = 0� �u2� a� = 0� �u2� b� = y� �u3� a� = 0� �u3� b� = 0�

�u1� u2� = �1z� �u1� u3� = �2z� and �u2� u3� = �3z�

Note that �2 and �3 cannot both be 0, otherwise u3 ∈ U , and this case would bedegenerate.

Suppose that �3 = 0. Then �2 �= 0. Let a′ = da� b′ = eb� u′1 = f1u1� u

′2 = f2u2,

and u′3 = f3u3. Then �a′� b′� = de�a� b�� �a′� b′� a′� = d2e�a� b� a�� y′ = �a′� b′� a′� a′� =

d3ey� z′ = �a′� b′� a′� a′� b′� = d3e2z, and just as before, we compute �a′� b′� b′�,�u′

1� a′�, �u′

2� b′�, �u′

1� u′2�, and �u′

1� u′3�.

If �1 �= 0, we want ed2

= f1d2e2

= f2d3

= f1f2�1d3e2

= f1f3�2d3e2

= 1. Let d = 1√�1, e = 1

�1,

f1 = 1�31, f2 = 1

�3/21

, and f3 = 1��1�

1/2�2, and these equations are satisfied. Thus, dropping

the primes, �u1� u2� = z, �u1� u3� = z, and �u2� u3� = 0. Now let u′2 = u2 − u3. We get

that �u1� u′2� = 0� �u2� u

′2� = 0� �u3� u

′2� = 0, and �u2� b� = y. Then letting a′ = a− u3,

we also get �u1� a′� = 0� �u′

2� a′� = 0, and �u3� a

′� = 0. Hence, dropping the primes,u1� u3 ∈ U , and this case is degenerate.

If �1 = 0 (and �3 = 0), then we want ed2

= f1d2e2

= f2d3

= f1f3�2d3e2

= 1. Let f3 = 1�2,

f1 = f2 = d = e = 1, and by dropping the primes we obtain �u1� u2�= 0, �u1� u3�= z,and �u2� u3� = 0. Let a′ = a− u3, and we have �u1� a

′� = 0. Therefore u1� u3 ∈ U , andthis case is degenerate.

Thus far, all cases have been degenerate. There remains the rank�B� = 2 casewith �3 �= 0. So if �3 �= 0 and �1 �= 0, let u′

1 = �3u1 + �1u3. Then �u′1� u2� = ��3u1 +

�1u3� u2� = ��3�1 − �1�3�z = 0, so we can assume �1 = 0. Note that this changeforces �u′

1� a� = �3z, �3 �= 0. Therefore, dropping the prime, we have

�u1� a� = �3z� �u1� b� = 0� �u2� a� = 0� �u2� b� = y� �u3� a� = 0� �u3� b� = 0�

�u1� u2� = 0� �u1� u3� = �2z� and �u2� u3� = �3z�

Once again let a′ = da� b′ = eb� u′1 = f1u1� u

′2 = f2u2, and u′

3 = f3u3. As before, wehave y′ = d3ey, z′ = d3e2z, �a′� b′� b′� = e

d2y′ or = 0, �u′

1� a′� = �3f1

d2e2z′, �u′

2� b′� = f2

d3y′,

�u′1� u

′3� = f1f3

d3e2�2z

′ and �u′2� u

′3� = f2f3

d3e2�3z

′. Again we want to solve for the coefficients.First let �2 �= 0. Then we want

�3f1d2e2

= f2d3

= f1f3�2

d3e2= f2f3�3

d3e2= e

d2= 1�

Letting d = ( �23�2

)1/3, e = ( �23

�2

)2/3, f2 = �23

�2, f1 = �33

�22, and f3 = 1

�3

( �23�2

)4/3solves these

equations. Hence we get �u1� a� = z� �u1� b� = 0� �u2� a� = 0� �u2� b� = y� �u3� a� = 0��u3� b� = 0 �u1� u2� = 0� �u1� u3� = z� �u2� u3� = z, and �a� b� b� = 0 or y. Let a′ = a−u3 and u′

2 = u2 − u1. Then u1� u3 ∈ U , and this case is degenerate.

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NILPOTENT LIE ALGEBRAS 4619

Table 5

�u1� a� = z �u1� b� = 0�u2� a� = 0 �u2� b� = y

�u3� a� = 0 �u3� b� = 0�u1� u2� = 0 �u1� u3� = 0�u2� u3� = z �a� b� b� = 0 or y

Finally, let �2 = 0. Then we want

�3f1d2e2

= f2d3

= f2f3�3

d3e2= e

d2= 1

A solution is d = e = f2 = 1, f1 = 1�3, and f3 = 1

�3. Note that �3 is not 0. It

remains to show that this case is nondegenerate. Direct computation shows thatdimZ�H�= 1. Suppose that H is degenerate and H = H1 + U as in Theorem 3.Clearly, dimU = 2. Let v1� v2 ∈ U and M = �a� b in the usual 2 generatedsubalgebra. Since �U�M2� = 0, v1 = c5y + c6z+ c7u1 + c8u2 + c9u3 and v2 = d5y +d6z+ d7u1 + d8u2 + d9u3� Then 0 = �v1� a� = c7z and 0 = �v2� a� = d7z. Hencec7 = d7 = 0. Also, 0 = �v1� b� = c8y and 0 = �v2� b� = d8y, which implies c8 =d8 = 0.Then �v1� v2� = 0 and U ⊆ Z�H�. Hence dimZ�H� > 1 which is a contradiction,therefore this case is nondegenerate.

Hence we are left with only two isomorphism classes when dim�H� = 5.

Theorem 7. Let H = �a� b� u1� u2� u3 as in Theorem 2. Then there are twoisomorphism classes represented in Table 5.

In summary, there are fourteen isomorphism classes when L is a nilpotent Liealgebra such that dim�L′/L′′� = 3 and L′′ �= 0. There are two isomorphism classeswhen L is 6 dimensional, six isomorphism classes when L is 7 dimensional, fourisomorphism classes when L is 8 dimensional, and two isomorphism classes when Lis 9 dimensional.

REFERENCES

Blackburn, N. (1958). On a special class of p-groups. Acta Math. 100:45–92.Blackburn, N. (1987). The derived group of a 2-group. Math. Proc. Cambridge Phil. Soc.

101(2):193–196.Dixmier, J. (1955). Sur les algèbres dérivées d’une algèbre de lie. Proc. Cambridge Philos.

Soc. 51:541–544.Jacobson, N. (1985). Basic Algebra 1. 2nd ed. New York: W. H. Freeman and Company.Schneider, C. (2000). Some Results on the Derived Series of p-Groups. Ph.D. thesis,

Canberra, Austrailia: The Austrailian National University.Schneider, C. (2003). Groups of prime-power order with a small second derived quotient.

Journal of Algebra 266:539–551.Zack, L. (2007). Nilpotent Lie Algebras with a Small Second Derived Quotient. Ph.D

thesis, North Carolina State University. Available at http://www.lib.ncsu.edu/theses/available/etd-02212007-113353/

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