arXiv:1712.01542v1 [math.RA] 5 Dec 2017

CAPABLE LIE ALGEBRAS WITH THE DERIVED
SUBALGEBRA OF DIMENSION TWO OVER AN ARBITRARY
FIELD
PEYMAN NIROOMAND, FARANGIS JOHARI, AND MOHSEN PARVIZI
Abstract. In this paper, we classify all capable nilpotent Lie algebras with
the derived subalgebra of dimension 2 over an arbitrary field. Moreover, the
explicit structure of such Lie algebras of class 3 is given.

1. Introduction and Motivation
The concept of capability was introduced by P. Hall in [9]. Recall that a group G
is called capable if there exists some group E such that G ∼
= E/Z(E), where Z(E)
denotes the center of E. There are some fundamental known results concerning
capability of p-groups. For instance, in [3, Corollary 4.16], it is shown the only capable extra-special p-groups (the p-groups with Z(G) = G′ and |G′ | = p) are those
of order p3 and exponent p. In the case that G′ = Z(G) and Z(G) is elementary
abelian p-group of rank 2, Heineken in [11] proved that the capable ones has order
at most p7 .
By some results due to Lazard, we may associate a p-group to a Lie algebra. Therefore some results of Lie algebras and p-groups have similarities in the structure. But
in this way not every thing are the same and there are differences between groups
and Lie algebras, so that most of time the proofs are different. Similar to the
concept of the capability for groups, a Lie algebra is called capable provided that
L∼
= H/Z(H) for a Lie algebra H. Beyl et al. in [3] introduced the epicenter Z ∗ (G)
of a group G that plays an important role in the capability of G. The analogous
notion of the epicenter, Z ∗ (L) for a Lie algebra L was defined in [19]. It is shown
that L is capable if and only if Z ∗ (L) = 0.
Another notion having relation to the capability is the concept of exterior square of
Lie algebras, L ∧ L, which was introduced in [6]. Our approach is on the concept of
the exterior center Z ∧ (L), the set of all elements l of L for which l ∧l′ = 0L∧L for all
l′ ∈ L. Niroomand et al. in [17] showed Z ∧ (L) = Z ∗ (L) for any finite dimensional
Lie algebra L. In [17], the last two authors obtained the structure of a capable
nilpotent Lie algebra L when dim L2 ≤ 1. It developed the result of [3, Corollary
4.16] for groups to the class of Lie algebras.
Recall that from [18], a Lie algebra H is called generalized Heisenberg of rank n if
H 2 = Z(H) and dim H 2 = n. If n = 1, then H is a Heisenberg Lie algebra that
is more well-known. Such algebras are odd dimension and have the presentation
H(m) ∼
= ha1 , b1 , . . . , am , bm , z [al , bl ] = z, 1 ≤ l ≤ mi.
Date: December 6, 2017.
Key words and phrases. Capability, Schur multiplier, generalized Heisenberg Lie algebras, stem
Lie algebras.
Mathematics Subject Classification 2010. Primary 17B30; Secondary 17B05, 17B99.
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P. NIROOMAND, F. JOHARI, AND M. PARVIZI

Recently, Niroomand et al. in [18] proved the capable generalized Heisenberg Lie
algebras of rank 2 have dimension at most 7 over a filed of characteristic not equal
to 2. They developed the result of Heineken [11] for groups to the area of Lie algebras. They also characterized the structure of all capable nilpotent Lie algebras
of class two when dim L2 = 2. In virtue of the recent results in [18], in this paper
we intend to classify the structure of all capable nilpotent Lie algebras of class two
with the derived subalgebra of dimension 2 over an arbitrary filed. Furthermore,
we determine the structure of all nilpotent Lie algebras of class 3 with the derived
subalgebra of dimension 2 and then we specify which one of them are capable.
2. Preliminaries
All Lie algebras in this paper are finite dimensional. The Schur multiplier of a
Lie algebra L, M(L), is defined as M(L) ∼
= F/R and F
= R ∩ F 2 /[R, F ] where L ∼
is a free Lie algebra. It can be shown that the Lie algebra M(L) is abelian and
independent of the choice of the free Lie algebra F (see [1, 2, 4, 10, 13, 14, 15, 16, 17]
for more information on this topics).
Throughout the paper ⊗ used to denote the operator of usual tensor product of
algebras. For a Lie algebra L, we denote the factor Lie algebra L/L2 by L(ab) . Also
we denote an abelian Lie algebra of dimension n by A(n).
The following proposition plays a key role in detecting the capability of Lie algebras.
Proposition 2.1. Let L be a finite dimensional Lie algebra with a central ideal I.
Then
(i) dim(M(L)) ≥ dim(M(L/I)) − dim(L2 ∩ I),
(ii) dim(M(L)) = dim(M(L/I)) − dim(L2 ∩ I) if and only if I ⊆ Z ∗ (L).
Proof. The result follows from [19, Proposition 4.1(iii) and Theorem 4.4 ].



The following lemma from [19] is a useful instrument in the next investigations.
Lemma 2.2. Let I be an ideal of a Lie algebras L such that L/I is capable. Then
Z ∗ (L) ⊆ I.
The next corollary shows that the epicenter of a Lie algebra is contained in its
derived subalgebra.
Corollary 2.3. Let L be a finite dimensional non-abelian nilpotent Lie algebra.
Then Z ∗ (L) ⊆ L2 .
Proof. Since dim L/L2 ≥ 2, L/L2 is capable, using [17, Theorem 3.3]. Therefore
the result follows by Lemma 2.2.

We need the notion of a central product of Lie algebras, it is defined as follows.
Definition 2.4. The Lie algebra L is a central product of A and B, if L = A + B,
where A and B are ideals of L such that [A, B] = 0 and A ∩ B ⊆ Z(L). We denote
the central product of two Lie algebras A and B by A ∔ B.
The Heisenberg Lie algebra can be presented in terms of central products.
Lemma 2.5. [12, Lemma 3.3] Let L be a Heisenberg Lie algebra of dimension
2m + 1. Then L is central products of its ideals Bj for all 1 ≤ j ≤ m such that Bj
is the Heisenberg Lie algebra of dimension 3.

LIE ALGEBRAS WITH THE DERIVED SUBALGEBRA OF DIMENSION TWO

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It is not an easy matter to determine the capability of a central product in
general. The next result gives the answer in a particular case.
Proposition 2.6. [18, Proposition 2.2] Let L be a Lie algebra such that L = A ∔ B
with A2 ∩ B 2 6= 0. Then A2 ∩ B 2 ⊆ Z ∧ (L) and so L is non-capable.
Following [18, Propositions 2.4 and 2.6], for determining the capable nilpotent
Lie algebras of class 2 with the derived subalgebra of dimension 2, it is enough to
consider a generalized Heisenberg Lie algebra H when
5 ≤ dim H ≤ 7.
Throughout the paper multiplication tables with respect to a fixed basis with trivial
products of the form [x, y] = 0 omitted, when x, y belongs to the Lie algebra.
Nilpotent Lie algebras of dimension at most 5 can be described uniformly over all
fields in [5, 7, 8]. For the dimension 6, 7 over an algebraic closed filed the structure
are known in [5, 7, 8]. Using notation and terminology of [5, 7, 8], we list the
generalized Heisenberg Lie algebras of rank two of dimension at most 6 and 7 over
field of characteristic 2 and characteristic different 2, respectively. Recall that in a
field F of characteristic 2, ω denotes a fixed element from F \ {x2 + x|x ∈ F}.
Theorem 2.7. Let H be a generalized Heisenberg Lie algebra of rank 2. Then
(i) over a field F of characteristic 2, the list of the isomorphism types of
generalized Heisenberg Lie algebras of dimension at most 6 are the fol(2)
lowing: L5,8 = hx1 , . . . , x5 [x1 , x2 ] = x4 , [x1 , x3 ] = x5 i and L6,7 (η) =
hx1 , . . . , x6 |[x1 , x2 ] = x5 , [x1 , x3 ] = x6 , [x2 , x4 ] = ηx6 , [x3 , x4 ] = x5 + x6 i
where η ∈ {0, ω},
(ii) over a field F of characteristic different from 2, the list of the isomorphism
types of generalized Heisenberg Lie algebras of dimension at most 7 are
L5,8 = hx1 , . . . , x5 [x1 , x2 ] = x4 , [x1 , x3 ] = x5 i,
L6,22 (ǫ) = hx1 , . . . , x6 [x1 , x2 ] = x5 = [x3 , x4 ], [x1 , x3 ] = x6 , [x2 , x4 ] = ǫx6 i,
∗
where ǫ ∈ F/(∼) and char F 6= 2,
L1 = 27A = hx1 , . . . , x7 [x1 , x2 ] = x6 = [x3 , x4 ], [x1 , x5 ] = x7 = [x2 , x3 ]i,
L2 = 27B = hx1 , . . . , x7 [x1 , x2 ] = x6 , [x1 , x4 ] = x7 = [x3 , x5 ]i.
3. capable generalized Heisenberg Lie algebras of rank two
Here, we are going to determine the structures of capable generalized Heisenberg
Lie algebras of rank two. By [18, Proposition 2.6], generalized Heisenberg Lie
algebras with the derived subalgebra of dimension 2 are capable if their dimension
lies between 5 and 7. According to Theorem 2.7, we have the presentation of all
capable generalized Heisenberg Lie algebras of rank two over a filed F with char F 6=
2. But when char F = 2 the structure of them is unknown. Therefore, at first we
intend to find the structure of them on an arbitrary filed and then we determine
which ones are capable.
Theorem 3.1. Let L be a 7-dimensional generalized Heisenberg Lie algebra over
a filed F with char F = 2 of rank two. Then L ∼
= hx1 , . . . , x7 [x1 , x2 ] = x6 =
[x3 , x4 ], [x1 , x5 ] = x7 = [x2 , x3 ]i or L ∼
= hx1 , . . . , x7 [x1 , x2 ] = x6 , [x1 , x4 ] = x7 =
[x3 , x5 ]i.

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P. NIROOMAND, F. JOHARI, AND M. PARVIZI

Proof. Let L2 = hz1 i ⊕ hz2 i. By [17, Theorem 3.6], we have L/hz2 i ∼
= H(2) ⊕ A(1)
or L/hz2 i ∼
= H(1) ⊕ A(3). First suppose that L/hz2 i ∼
= H(2) ⊕ A(1). There exist
two ideals I1 /hz2 i and I2 /hz2 i of L/hz2 i such that
I1 /hz2 i ∼
= H(2) and I2 /hz2 i ∼
= A(1).
Clearly, L = I1 + I2 , I1 ∩ I2 = hz2 i, [I1 , I2 ] ⊆ hz2 i and [L, I2 ] ⊆ hz2 i. Thus
I1 /hz2 i = hx1 + hz2 i, y1 + hz2 i, x2 + hz2 i, y2 + hz2 i, z1 + hz2 i|[x1 , y1 ] + hz2 i = [x2 , y2 ] +
hz2 i = z1 + hz2 ii and I2 = hqi ⊕ hz2 i ∼
= A(2) for some q ∈ L. Hence the set
{x1 , y1 , x2 , y2 , z1 , z2 , q} is a basis of L and
[x1 , y1 ] = z1 + α1 z2 , [x2 , y2 ] = z1 + α2 z2 ,
[x1 , y2 ] = α3 z2 , [x1 , x2 ] = α4 z2 ,
[x2 , y1 ] = α5 z2 , [y1 , y2 ] = α6 z2 ,
[x1 , q] = α7 z2 , [y1 , q] = α8 z2 ,
[x2 , q] = α9 z2 , [y2 , q] = α10 z2 .
By changing variable, we assume that α1 = α2 = 0. Since q ∈
/ L2 = Z(L) =
hz1 i⊕hz2 i, q is not central. Thus [L, I2 ] = [hqi, I2 ] = hz2 i. Without loss of generality,
assume that α7 6= 0. By [17, Theorem 3.6], we have L/hz1 i ∼
= H(2) ⊕ A(1) or
L/hz1 i ∼
= H(2) ⊕ A(1). There exist two
= H(1) ⊕ A(3). First suppose that L/hz1 i ∼
ideals I3 /hz1 i and I4 /hz1 i of L/hz1 i such that
∼ H(2) and I4 /hz1 i =
∼ A(1).
I3 /hz1 i =
Clearly, L = I3 + I4 and [I3 , I4 ] ⊆ hz1 i.
We claim that q, x1 ∈ I3 and [x2 , q] = [y1 , q] = [y2 , q] = [x1 , x2 ] = [x1 , y2 ] = 0. Let
a+hz1 i ∈ I3 /hz1 i and I4 /hz1 i = hb+hz1 ii such that q+hz1 i = (a+hz1 i)+(αb+hz1 i).
If a + hz1 i = 0, then since [L, I2 ] = [hqi, I2 ] = hz2 i, we have [hqi, L] ∈ hz1 i ∩ hz2 i = 0.
So q ∈ Z(L) = L2 = hz1 i ⊕ hz2 i. It is a contradiction. Thus q − a − αb ∈ hz1 i and
a + hz1 i 6= 0. We have
a = η1 x1 + η2 x2 + η3 y1 + η4 y2 + η5 z1 + η6 z2 + η7 q,
αb = η1′ x1 + η2′ x2 + η3′ y1 + η4′ y2 + η5′ z1 + η6′ z2 + η7′ q,
and so
q = a + αb + γz1 = η1 x1 + η2 x2 + η3 y1 + η4 y2 + η5 z1 + η6 z2 + η7 q+
η1′ x1 + η2′ x2 + η3′ y1 + η4′ y2 + η5′ z1 + η6′ z2 + η7′ q + γz1 .
Since the set {x1 , y1 , x2 , y2 , z1 , z2 , q} is linearly independent and L is a Lie algebra
on filed over characteristic two, η1 = η1′ , η2 = η2′ , η3 = η3′ , η4 = η4′ , η5 = −η5′ −γ, η6 =
η6′ , η7 = 1 − η7′ . Thus
q + hz1 i = a + hz1 i + αb + hz1 i = (η1 + η1′ )x1 + (η2 + η2′ )x2 + (η3 + η3′ )y1 +
(η4 + η4′ )y2 + (η6 + η6′ )z2 + (η7 + η7′ )q + hz1 i.
We conclude that q + hz1 i = (η7 + η7′ )q + hz1 i so η7 6= 0 or η7′ 6= 0. Since q + hz1 i ∈
/
I4 /hz1 i, we have η7 6= 0 and η7′ = 0. Thus q + hz1 i ∈ I3 /hz1 i. Hence q ∈ I3 . Now,
we prove that x1 ∈ I3 . By contrary, assume that x1 ∈
/ I3 . We may assume that
x1 + hz1 i = a1 + hz1 i + α′ b + hz1 i, where a1 + hz1 i ∈ I3 /hz1 i and I4 /hz1 i = hb + hz1 ii.
Now, if a1 + hz1 i = 0, then x1 + hz1 i = α′ b + hz1i ∈ I3 /hz1 i and so [q, x1 ] + hz1 i = 0.
Since [q, x1 ] = z2 , we have [q, x1 ] ∈ hz1 i ∩ hz2 i = 0 and hence [q, x1 ] = 0. It is a
contradiction. Thus a1 + hz1 i =
6 0 and x1 − a1 − α′ b ∈ hz1 i. Taking into account the

LIE ALGEBRAS WITH THE DERIVED SUBALGEBRA OF DIMENSION TWO

5

basis of L, we can write a1 = β1 x1 + β2 x2 + β3 y1 + β4 y2 + β5 z1 + β6 z2 + β7 q and
α′ b = β1′ x1 + β2′ x2 + β3′ y1 + β4′ y2 + β5′ z1 + β6′ z2 + β7′ q. Therefore x1 = a1 + α′ b + γz1 =
(β1 + β1′ )x1 + (β2 + β2′ )x2 + (β3 + β3′ )y1 + (β4 + β4′ )y2 + (β5 + β5′ )z1 + (β6 + β6′ )z2 +
(β7 + β7′ )q. Now, since the set {x1 , y1 , x2 , y2 , z1 , z2 , q} is linearly independent and
the characteristic of the field is 2, we have β1 = 1 − β1′ , β2 = β2′ , β3 = β3′ , β4 = β4′ ,
β5 = β5′ + γ, β6 = β6′ and β7 = β7′ . Now we have x1 + hz1 i = a1 + hz1 i + α′ b + hz1 i =
(β1 +β1′ )x1 +(β2 +β2′ )x2 +(β3 +β3′ )y1 +(β4 +β4′ )y2 +(β6 +β6′ )z2 +(β7 +β7′ )q +hz1 i =
(β1 + β1′ )x1 + hz1 i. Therefore β1 6= 0 or β1′ 6= 0. But x1 + hz1 i ∈
/ I4 /hz1 i and hence
β1 6= 0 and β1′ = 0 which implies x1 + hz1 i ∈ I3 /hz1 i. Thus x1 ∈ I3 .
Now, we want to show [x1 , x2 ] = [x2 , q] = [y2 , q] = [y1 , q] = [x1 , y2 ] = 0. By
contrary, let α9 6= 0. Using the same procedure as used in the previous step, we
can show x2 ∈ I3 . Because I3 /hz1 i ∼
= H(2), we have [x2 , q] + hz1 i = 0 and so
[x2 , q] ∈ hz1 i ∩ hz2 i = 0 which is a contradiction. Similarly, one can show that each
of the mentioned brackets vanishes.
Here, we want to prove that exactly one of the brackets [x2 , y1 ] or [y1 , y2 ] vanish.
Note that the case [x2 , y1 ] = [y1 , y2 ] = 0 leads to L/hz1 i ∼
= H(1) ⊕ A(3) which is a
contradiction since L/hz1 i ∼
= H(2) ⊕ A(1). Now, without loss of generality, assume
that [y1 , y2 ] 6= 0. Using the same process, we may prove [x2 , y1 ] = 0 and y1 , y2 ∈ I3
and x2 ∈ I4 . Thus L = hx1 , y1 , x2 , y2 , z1 , z2 , q|[y1 , y2 ] = z2 = [x1 , q], [x1 , y1 ] = z1 =
[x2 , y2 ]i. Similarly if either L/hz2 i ∼
= H(2) ⊕ A(1) and L/hz1 i ∼
= H(1) ⊕ A(3) or
∼
∼
L/hz2 i = H(1) ⊕ A(3) and L/hz1 i = H(1) ⊕ A(3), then L ∼
= hx1 , . . . , x7 [x1 , x2 ] =
x6 , [x1 , x4 ] = x7 = [x3 , x5 ]i. The result follows.

Corollary 3.2. Let L be a 7-dimensional generalized Heisenberg Lie algebra over
any filed F. Then L ∼
= hx1 , . . . , x7 [x1 , x2 ] = x6 = [x3 , x4 ], [x1 , x5 ] = x7 = [x2 , x3 ]i ∼
=
L1 or L ∼
= L2 .
= hx1 , . . . , x7 [x1 , x2 ] = x6 , [x1 , x4 ] = x7 = [x3 , x5 ]i ∼


Proof. The result follows from Theorems 2.7 (ii) and 3.1.

The following result gives the Schur multipliers of Lie algebras L6,22 (ǫ) and
It helps to determine the capability of these Lie algebras in the next proposition.
(2)
L6,7 (η).

(2)

Proposition 3.3. The Schur multiplier of Lie algebras L6,22 (ǫ) and L6,7 (η) are
abelian Lie algebras of dimension 8.
Proof. Using the method of Hardy and Stitzinger in [10], we can obtain that in
both cases, the dimension of the Schur multipliers are 8.

In the characterizing of capable generalized Heisenberg Lie algebra of rank two
of dimension 6 in [18, Theorem 2.12], the Lie algebra L6,22 (ǫ) is missing. Here, we
improve this result as bellow.
(2)

Proposition 3.4. L6,22 (ǫ) and L6,7 (η) are capable.
Proof. Let L ∼
= L6,22 (ǫ) is capable. Then by using Theorem 2.7, we have L6,22 (ǫ)2 =
Z(L6,22 (ǫ)) = hx5 i ⊕ hx6 i. By Proposition 2.1 (ii) and [19, Corollary 4.6], it is
enough to show that dim M(L6,22 (ǫ)/hxi i) − 1 < dim M(L6,22 (ǫ)) for i = 5, 6.
Clearly, L6,22 (ǫ)/hxi i ∼
= H(2) or L6,22 (ǫ)/hxi i ∼
= H(1) ⊕ A(2) for i = 5, 6, by using
[17, Theorem 3.6]. Thus for i = 5, 6, we have dim M(L6,22 (ǫ)/hxi i) = 5 or 8, by
[17, Lemma 2.6 and Theorem 2.7]. Since dim M(L6,22 (ǫ)) = 8, by Proposition 3.3,

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P. NIROOMAND, F. JOHARI, AND M. PARVIZI

we conclude that
dim M(L6,22 (ǫ)/hxi i) − 1 < dim M(L6,22 (ǫ)) for i = 5, 6.
(2)

Therefore L6,22 (ǫ) is capable. By a similar way, we can see that L6,7 (η) is also
capable. The proof is completed.

The next result is useful.
Lemma 3.5. [18, Lemma 2.11] L5,8 and L1 are capable while L2 is not.
We are ready to summarize our results to show that which ones of generalized
Heisenberg Lie algebras of rank 2 is capable.
Theorem 3.6. Let H be an n-dimensional generalized Heisenberg Lie algebra of
rank 2. Then H is capable if and only if H is isomorphic to one of Lie algebras
(2)
L5,8 , L6,22 (ǫ), L6,7 (η) or L1 .
Proof. Let H be capable. Then [18, Proposition 2.6] implies 5 ≤ dim H ≤ 7.
Using Corollary 3.2 and Theorem 2.7, H is isomorphic to one of Lie algebras
(2)
L5,8 , L6,22 (ǫ), L6,7 (η), L1 or L2 . By Proposition 3.4 and Lemma 3.5, all of them
are capable while L2 is non-capable. The converse is held by Proposition 3.4 and
Lemma 3.5.

4. Stem nilpotent Lie algebras of class 3 with the derived
subalgebra of dimension 2
We know that every nilpotent Lie algebra with the derived subalgebra of dimension 2 is of class two and three. In this section, we are going to obtain the structure
of stem Lie algebras of class 3 with the derived subalgebra of dimension 2. Then we
determine which of them is capable. Moreover, we show that all such Lie algebras
of dimension greater than 6 are unicentral.
Recall that an n-dimensional nilpotent Lie algebra L is said to be nilpotent of maximal class if the class of L is n − 1. If L is of maximal class, then dim(L/L2 ) = 2,
Zi (L) = Ln−i and dim(Lj /Lj+1 ) = 1 for all 0 ≤ i ≤ n − 1 and 2 ≤ j ≤ n − 1 (see
[4] for more information).
From [8], the only Lie algebra of maximal class of dimension 4 is isomorphic to
L4,3 = hx1 , . . . , x4 |[x1 , x2 ] = x3 , [x1 , x3 ] = x4 i.
We say a Lie algebra L is a semidirect sum of an ideal I and a subalgebra K if
L = I + K, I ∩ K = 0. The semidirect sum of an ideal I and a subalgebra K is
denoted by K ⋉ I.
Let cl(L) denotes the nilpotency class of a Lie algebra L. The following two
lemmas characterize the structure of all stem Lie algebras L of dimensions 5 and
6, when cl(L) = 3 and dim L2 = 2.
Lemma 4.1. Let L be a nilpotent stem Lie algebra of dimension 5 such that
dim L2 = 2 and cl(L) = 3. Then
∼ L5,5 = hx1 , . . . , x5 |[x1 , x2 ] = x3 , [x1 , x3 ] = x5 , [x2 , x4 ] = x5 i.
L=
Moreover, L5,5 = I ⋊ hx4 i, where
I = hx1 , x2 , x3 , x5 |[x1 , x2 ] = x3 , [x1 , x3 ] = x5 i ∼
= L4,3 , and [I, hx4 i] = hx5 i.

LIE ALGEBRAS WITH THE DERIVED SUBALGEBRA OF DIMENSION TWO

7

Proof. By the classification of 5-dimensional nilpotent Lie algebras in [8], we get
L∼
= L5,5 . It is easy to check that L5,5 = I⋊hx4 i such that I = hx1 , x2 , x3 , x5 |[x1 , x2 ] =
x3 , [x1 , x3 ] = x5 i ∼

= L4,3 and [I, hx4 i] = hx5 i.
Lemma 4.2. Let L be a nilpotent stem Lie algebra of dimension 6 such that
dim L2 = 2 and cl(L) = 3. Then L ∼
= L6,10 = hx1 , . . . , x6 |[x1 , x2 ] = x3 , [x1 , x3 ] =
x6 , [x4 , x5 ] = x6 i. Moreover, L6,10 = I ∔ hx4 , x5 , x6 |[x4 , x5 ] = x6 i = I ∔ K such that
I = hx1 , x2 , x3 , x6 |[x1 , x2 ] = x3 , [x1 , x3 ] = x6 i ∼
= L4,3 and K = hx4 , x5 , x6 |[x4 , x5 ] =
x6 i ∼
= H(1).
Proof. By the classification of 6-dimensional nilpotent Lie algebras in [8], we get
L∼
= L6,10 . Clearly Z(L) = hx6 i and L6,10 = I+K, where I = hx1 , x2 , x3 , x6 |[x1 , x2 ] =
x3 , [x1 , x3 ] = x6 i ∼
= L4,3 and K = hx4 , x5 , x6 |[x4 , x5 ] = x6 i ∼
= H(1). Since I ∩ K =
hx6 i = Z(I) = Z(L) and [I, K] = 0, we can see L6,10 = I ∔ K.

The following proposition is an useful instrument in the next.
Proposition 4.3. Let L be an n-dimensional nilpotent stem Lie algebra of class 3
(n ≥ 5) and dim L2 = 2 such that L = I + K, where I and K are two subalgebras
of L, I ∼
= L4,3 is the maximal class Lie algebra of dimension 4 and [I, K] ⊆ Z(I) =
Z(L). Then
(i) If K is a non-trivial abelian Lie algebra such that K ∩ I = 0, then [I, K] =
Z(L) and K ∼
= A(1). Moreover, L = I ⋊ K ∼
= L5,5 .
(ii) Assume dim K 2 = 1 and I ∩ K = K 2 = Z(L).
(a) If K 2 = Z(K), then L = I ∔ K, where n = 2m + 4. Moreover, for
m = 1, we have L = I ∔ K ∼
= L6,10 , where K ∼
= H(1). For m ≥ 2, we
∼
have L = I ∔ K = I1 ∔ I2 , where I1 ∼
= H(m − 1).
= L6,10 and I2 ∼
(b) If K 2 6= Z(K), then L = (I ⋊A)∔K, where K ∼
= H(m) and A ∼
= A(1),
[I, A] = Z(L) = Z(I) = K 2 and n = 2m + 5. Moreover, I ⋊ A ∼
= L5,5 .
Proof.

(i) We have [I, K] ⊆ Z(I) = Z(L), so I is an ideal of L. Since K ∩I = 0
and I ∼
= L4,3 , dim K = dim L − dim I = n − 4 and so L = I ⋊ K, where
K ∼
= A(n − 4). We claim that [I, K] = Z(I). By contrary, assume that
[I, K] = 0. Since K is abelian, we have K ⊆ Z(I) = Z(L) ⊆ I. Now
I∩K 6= 0, so we have a contradiction. Thus [I, K] = Z(I) = Z(L). We know
that I is a Lie algebra of maximal class of dimension 4 so Z(I) = Z(L) = I 3 .
We also have dim L2 = dim I 2 = 2. Therefore L2 = I 2 . We claim that
K∼
= A(1).
First assume that n = 5. Lemma 4.1 implies L ∼
= I ⋊K ∼
= L5,5 and
since I ∼
= L4,3 , we have I = hx1 , . . . , x4 |[x1 , x2 ] = x3 , [x1 , x3 ] = x4 i and
Ln−4
Z(I) = hx4 i = I 3 . Now let n ≥ 6 and K = i=1 hzi i. In this case we show
that K ∩ I 6= 0, which is a contradiction. So this case does not occur. By
using Jacobian identity, for all 1 ≤ i ≤ n − 4, we have
[zi , x3 ] = [zi , [x1 , x2 ]] = [zi , x1 , x2 ] + [x2 , zi , x1 ] = 0
since [zi , x1 ] and [x2 , zi ] are central. Thus [zi , x3 ] = 0 for all 1 ≤ i ≤ n − 4.
Now, let [zi , x1 ] = αi x4 with αi 6= 0. Putting zi′ = zi + αi x3 , we have
′
[zi , x1 ] = [zi + αi x3 , x1 ] = [zi , x1 ] + αi [x3 , x1 ] = αi x4 − αi x4 = 0. So we
also obtain [zi′ , x3 ] = 0. Thus [zi′ , x3 ] = [zi′ , x1 ] = 0 for all i, 1 ≤ i ≤ n − 4.
Now let [zi′ , x2 ] = αx4 and [zj′ , x2 ] = βx4 , in which α 6= 0 and β 6= 0 for
i 6= j and fixed i, j. Put di = βzi′ − αzj′ . We have [di , x2 ] = [βzi′ − αzj′ , x2 ] =

8

P. NIROOMAND, F. JOHARI, AND M. PARVIZI

βαx4 −βαx4 = 0 and so [di , x2 ] = 0. On the other hand, [di , x1 ] = [di , x2 ] =
[di , x3 ] = 0. Therefore [di , I] = 0 and hence di ∈ Z(L) = Z(I) = hx4 i. Since
di = βzi′ − αzj′ = β(zi − αi x3 ) − α(zj − αj x3 )
= βzi − αzj + (ααj − βαi )x3 ∈ Z(I)
so 0 6= βzi − αzj ∈ K ∩ I = 0, which is a contradiction. Thus n = 5,
K ∼
= A(1) and L = I ⋊ hz1 i and [x2 , z1 ] = x4 , as required. Considering
the classification of nilpotent Lie algebras of dimension 5 with dim L2 = 2
given in [8] and Lemma 4.1, we should have L ∼
= L5,5 .
(ii) Since I ∩ K = K 2 = Z(L) = Z(I) ∼
= A(1), dim(K) = dim(L) − dim(I) +
dim(I ∩ K) = n − 4 + 1 = n − 3. We know that dim K 2 = 1, so [17, Theorem
3.6] implies K ∼
= H(m) and A ∼
= A(n − 2m − 4). If
= K1 ⊕ A, in which K1 ∼
∼
A = 0, then K = H(m). Since I ∩K = K 2 = Z(L) = Z(I) = Z(K) ∼
= A(1),
dim(K) = dim(L) − dim(I) + dim(I ∩ K) = n − 4 + 1 = n − 3. Now
since 2m + 1 = dim(K) = n − 3, we have n = 2m + 4. We are going to
show that [I, K] = 0. In fact, we show that there exists I1 ∼
= L3,4 and
K2 ∼
= H(m) with [I1 , K2 ] = 0 and L = I1 ∔ K2 . First let m = 1. We have
dim L = 6 and K = hx, y, x4 |[x, y] = x4 i, since K ∼
= H(1). By looking
the classification of nilpotent Lie algebras of dimension 6 with dim L2 = 2
given in [5] and Lemma 4.2, we should have L ∼
= L6,10 . Now, let m ≥ 2 and
H(m) = ha1 , b1 , . . . , am , bm , z [al , bl ] = z, 1 ≤ l ≤ mi. Lemma 2.5 implies
that H(m) = T1 ∔ . . . ∔ Tm , in which Ti ∼
= H(1) for all 1 ≤ i ≤ m. With the
same procedure as case in (i) and changing the variables we can see that
[Ti , I] = 0 for all i, 1 ≤ i ≤ m. So [I, K] = 0 and hence L = I ∔ K. Since
m ≥ 2, we have L = (I ∔ T1 ) ∔ (T2 ∔ . . . ∔ Tm ) such that I ∔ T1 ∼
= L6,10
and T2 ∔ . . . ∔ Tm ∼
= H(m − 1), as required. The case (a) is completed.
Now, let A 6= 0 and so n 6= 2m − 4. Thus L = I + (K1 ⊕ A) such that
[I, K] ⊆ Z(L) = Z(I). We are going to show that A ∼
= A(1), [I, K1 ] = 0 and
[I, A] = Z(I) = Z(L). Similar to the part (ii), we can see that [I, K1 ] = 0.
We claim that [I, A] 6= 0. By contrary, let [A, K1 ] = [I, A] = 0 and so A ⊆
Z(L) = Z(I). Since A ∩ I = 0, we have A = 0, which is a contradiction. So
we have [I, A] = Z(L). We claim that dim A = 1. Let dim A ≥ 2. Similar to
the proof of the part (i), we have [a1 , x1 ] = [a2 , x1 ] = [a1 , x3 ] = [a2 , x3 ] = 0
where a1 , a2 ∈ A and a1 6= a2 . Now let [a1 , x2 ] = αx4 and [a2 , x2 ] = βx4
such that α 6= 0 and β 6= 0. Putting a′1 = βa1 − αa2 . We have
[a′1 , x2 ] = [βa1 − αa2 , x2 ] = βαx4 − βαx4 = 0
and so [a′1 , x2 ] = 0. Hence [a′1 , x1 ] = [a′1 , x2 ] = [a′1 , x3 ] = 0. Therefore
[a′1 , I] = 0 and hence a′1 ∈ Z(L) = Z(I) = hx4 i = K12 . So a′1 ∈ K1 and since
K1 ∩ A = 0, we have a contradiction. Hence A ∼
= A(1) and so n = 2m + 5.
Thus L = (I ⋊ A) ∔ K1 such that [I, A] = Z(L) = Z(I). By part (i), we
have I ⋊ A ∼
= L5,5 . The case (b) is completed. The result follows.

We need the following lemma for the next investigation.
Lemma 4.4. [20, Lemma 1] Let L be a nilpotent Lie algebra and H be a subalgebra
of L such that L2 = H 2 + L3 . Then Li = H i for all i ≥ 2. Moreover, H is an ideal
of L.

LIE ALGEBRAS WITH THE DERIVED SUBALGEBRA OF DIMENSION TWO

9

In the following, we determine the central factor of all stem Lie algebras T such
that cl(T ) = 3 and dim T 2 = 2.
Lemma 4.5. Let T be an n-dimensional stem Lie algebra such that cl(T ) = 3 and
dim T 2 = 2. Then Z(T ) = T 3 ∼
= A(1) and T /Z(T ) ∼
= H(1) ⊕ A(n − 4).
Proof. Since T is stem, we have A(1) ∼
= A(1).
= T 3 ⊆ Z(T ) $ T 2 . Thus Z(T ) = T 3 ∼
2
∼
This follows T /Z(T ) = A(1). Since T /Z(T ) is capable, [17, Theorem 3.6] implies
that T /Z(T ) ∼

= H(1) ⊕ A(n − 4). It completes the proof.
In the following theorem, we determine the structure of all stem Lie algebras of
class 3 with the derived subalgebra of dimension 2.
Theorem 4.6. Let T be an n-dimensional stem Lie algebra such that cl(T ) = 3
and dim T 2 = 2. Then one of the following holds
(a) T ∼
= L4,3 .
(b) T ∼
=I ⋊K ∼
= L5,5 where K ∼
= A(1), I ∼
= L4,3 and Z2 (T ) = Z2 (I) ⋊ K.
∼
∼
(c) T = I ∔I1 , where I1 = H(m), Z2 (T ) = Z2 (I)+I1 , I ∼
= L4,3 and n = 2m+4.
Moreover if m ≥ 2, then L ∼
= H(m − 1), if m = 1,
= L6,10 ∔ I2 , where I2 ∼
then L ∼
= L6,10 .
(d) T ∼
= H(m), K ∼
= A(1), I ∼
= L4,3 ,
= L5,5 ∔ I1 , where I1 ∼
= (I ⋊ K) ∔ I1 ∼
Z2 (T ) = (Z2 (I) ⋊ K) ∔ I1 and n = 2m + 5.
Moreover, in the cases (b), (c) and (d), Z(T ) = Z(I) = I12 = [I, K].
Proof. Since cl(T ) = 3, we have dim T ≥ 4. If dim T = 4, then T must be a Lie
algebra of maximal class and hence T ∼
= L4,3 . Assume that dim T ≥ 5. We have
T /Z(T ) ∼
= A(1), by Lemma 4.5. There exist
= H(1) ⊕ A(n − 4) and Z(T ) = T 3 ∼
ideals I1 /Z(T ) and I2 /Z(T ) of T /Z(T ) such that
I1 /Z(T ) ∼
= H(1) and I2 /Z(T ) ∼
= A(n − 4).

2
2
2
Since T /Z(T ) = I1 +Z(T ) /Z(T ), we have T = I12 +Z(T ) and Z(T ) = T 3 . Using
Lemma 4.4, we have T 2 = I12 and so cl(T ) = cl(I1 ) = 3. Hence I1 is a Lie algebra
of maximal class and since dim I1 = 4, we have I1 ∼
= L4,3 . Now, Z(T ) = Z(I1 )
because Z(T ) ∩ I1 ⊆ Z(I1 ) and dim Z(T ) = 1. Since Z(T ) ⊆ I1 ∩ I2 ⊆ Z(T ),
we have I1 ∩ I2 = Z(T ) = Z(I1 ). Now we are going to determine the structure
of I2 . We have I2 /Z(T ) ∼
= A(n − 4) so I22 ⊆ Z(T ) ∼
= A(1), and hence cl(I2 ) ≤ 2
and [I1 , I2 ] ⊆ I1 ∩ I2 = Z(T ) = Z(I1 ) ∼
= A(1). We have dim T /Z(T ) ≥ 4 and
so dim I2 ≥ 2. Let cl(I2 ) = 1. Therefore [I1 , I2 ] = I1 ∩ I2 = Z(T ), otherwise
[I1 , I2 ] = 0 and since I2 is abelian, I2 ⊆ Z(T ) ∼
= A(1). It is a contradiction, since
dim I2 ≥ 2. Hence I2 = Z(T ) ⊕ A, where A ∼
= A(n − 4) and [I1 , I2 ] = Z(T ). Now
Z(T ) ⊆ I1 , A ∩ I1 = 0 and I1 ∩ I2 = Z(T ) so T = I1 + I2 = I1 + Z(T ) + A = I1 ⋊ A.
Using the proof of Proposition 4.3 (i), we have T ∼
= I1 ⋊K ∼
= L5,5 in which K ∼
= A(1)
and [K, I1 ] = Z(T ). This is the case (b).
Now, let cl(I2 ) = 2. Since I22 = I1 ∩ I2 = Z(T ) = Z(I1 ) ∼
= A(1), by [17, Theorem
3.6], we have I2 ∼
= H(m) ⊕ A(n − 2m − 4). First assume that A(n − 2m − 4) = 0.
Then n = 2m − 4 and I2 ∼
= H(m). Using Proposition 4.3 (ii)(a), we can similarly
prove [I1 , I2 ] = 0 and T = I1 ∔ I2 where I2 ∼
= H(m). This is the case (c). Now, let
A(n − 2m − 4) 6= 0. Then n 6= 2m − 4 and hence T = I1 + (K ⊕ A) where K ∼
= H(m)
and A ∼
= A(n − 2m − 4) and [I1 , K ⊕ A] ⊆ Z(T ) = Z(I1 ). Similar to the case (c), we
have [I1 , K] = 0. Now we claim that [I1 , A] = Z(T ) ∼
= A(1). Let [I1 , A] = 0. Since
[K, A] = 0, we have A ⊆ Z(T ) = Z(I1 ) = Z(K) = K 2 ∼
= A(1). It is a contradiction,

10

P. NIROOMAND, F. JOHARI, AND M. PARVIZI

since A ∩ K = 0. Therefore [I1 , A] = Z(T ) and hence T ∼
= (I1 ⋊ A) ∔ K where
A∼
= A(n − 2m − 4) and [I1 , A] = Z(T ) = Z(I1 ). Similar to the case (b), one can
obtain that A ∼
= A(1), so n − 2m − 4 = 1, n = 2m + 5 and [I1 , A] = Z(T ). So
T = (I1 ⋊ A) ∔ K in which A ∼
= A(1) and [I1 , A] = Z(T ). This is the case (d).
Now, we have
Z2 (T )/Z(T ) = Z(T /Z(T )) = Z(I1 /Z(T )) ⊕ I2 /Z(T ) and Z(T ) = Z(I1 )
also Z(I1 /Z(T )) = I12 /Z(T ), so Z2 (T )/Z(T ) = I12 /Z(T ) ⊕ I2 /Z(T ). Since I1 is
maximal class of dimension 4, we have Z2 (T ) = Z2 (I1 ) + I2 = I12 + I2 . The result
follows.

In the following theorem, we classify all non-capable stem Lie algebras of class
3 with the derived subalgebra of dimension 2.
Theorem 4.7. Let T be an n-dimensional stem Lie algebra such that cl(T ) =
dim T 2 = 2 and n ≥ 6. Then T ∼
= (I ⋊ K) ∔ H or T ∼
= I ∔ H such that H
∼
∼
H(m), K = A(1), I = L4,3 and [K, I] = Z(T ) = Z(I) = H 2 . Moreover, T
non-capable.

3,
∼
=
is

Proof. By Theorem 4.6 (c) and (d), we obtain (I ⋊ A) ∔ H or I ∔ H such that
H ∼
= H(m), A ∼
= A(1), I ∼
= L4,3 and [A, I] = Z(T ) = Z(I) = H 2 . By using
Proposition 2.6, T is non-capable. The result follows.

The capable stem Lie algebras of class 3 with the derived subalgebra of dimension
2 are characterized as following.
Lemma 4.8. L4,3 and L5,5 are capable.
Proof. From [8], let L5,7 = hx1 , . . . , x5 |[x1 , x2 ] = x3 , [x1 , x3 ] = x4 , [x1 , x4 ] = x5 i and
L6,13 = hx1 , . . . , x6 |[x1 , x2 ] = x3 , [x1 , x3 ] = x5 , [x2 , x4 ] = x5 , [x1 , x5 ] = x6 , [x3 , x4 ] =
x6 i. We have Z(L5,7 ) = hx5 i and Z(L6,13 ) = hx6 i, so L5,7 /hx5 i ∼
= L4,3 and
L6,13 /hx6 i ∼

= L5,5 . Thus L4,3 and L5,5 are capable.
We are in a position to characterize the capability of an n-dimensional stem Lie
algebra T such that cl(T ) = 3 and dim T 2 = 2.
Theorem 4.9. Let T be an n-dimensional stem Lie algebra such that cl(T ) = 3
and dim T 2 = 2. Then T is capable if and only if T ∼
= L5,5 .
= L4,3 or T ∼
Proof. Let T be capable. By Theorems 4.6, 4.7 and Lemma 4.8, T is isomorphic to
L4,3 or L5,5 . The converse holds by Lemma 4.8.

The next theorem gives a necessary and sufficient condition for detecting the
capability of stem Lie algebras of class 3 with the derived subalgebra of dimension
2.
Theorem 4.10. Let T be an n-dimensional stem Lie algebra such that cl(T ) = 3
and dim T 2 = 2. Then T is capable if and only if 3 ≤ dim(T /Z(T )) ≤ 4.
Proof. The result follows from Lemma 4.5 and Theorem 4.9.



Recall that a Lie algebra L is called unicentral if Z ∗ (L) = Z(L).
Corollary 4.11. Let T be an n-dimensional stem Lie algebra such that cl(T ) = 3
and dim T 2 = 2. Then T is non-capable if and only if n ≥ 6. Moreover, T is
unicentral.
Proof. The result follows from Theorems 4.7 and 4.9.



LIE ALGEBRAS WITH THE DERIVED SUBALGEBRA OF DIMENSION TWO

11

5. Nilpotent Lie Algebras with the derived subalgebra of dimension
two
In this section, we are going to determine all capable nilpotent Lie algebras
with the derived subalgebra of dimension 2. At first we show that every finite
dimensional nilpotent Lie algebra of class 3 with derived subalgebra of dimension 2
can be considered as a direct sum of a non-abelian stem Lie algebra of class 3 and
an abelian Lie algebra.
The following result shows that the capability of the direct product of a non-abelian
Lie algebra and an abelian Lie algebra depends only on the capability of its nonabelian factor.
Theorem 5.1. Let L be a finite dimensional nilpotent Lie algebra of class 3 and
dim L2 = 2. Then L = T ⊕ A such that Z(T ) = L2 ∩ Z(L) = L3 = T 3 and
Z ∗ (L) = Z ∗ (T ), where A is an abelian Lie algebra.
Proof. By using [12, Proposition 3.1], L = T ⊕ A such that Z(T ) = L2 ∩ Z(L) and
Z ∗ (L) = Z ∗ (T ), where A is an abelian Lie algebra. Since T is stem, Lemma 4.5
implies Z(T ) = T 3 , as required.

In the following theorem, all capable nilpotent Lie algebras of class 2 with the
derived subalgebra of dimension 2 are classified.
Theorem 5.2. Let L be an n-dimensional nilpotent Lie algebra of nilpotency class
2 and dim L2 = 2. Then L is capable if and only if L ∼
=
= L5,8 ⊕ A(n − 5), L ∼
(2)
∼
∼
L6,22 (ǫ) ⊕ A(n − 6), L = L6,7 (η) ⊕ A(n − 6) or L = L1 ⊕ A(n − 7).
Proof. This is immediately obtained from [18, Propositions 2.4 and 2.6] and Theorem 3.6.

We are ready to determine all capable Lie algebras of class 3 when its derived
subalgebra is of dimension 2.
Theorem 5.3. Let L be an n-dimensional Lie algebra such that cl(L) = 3 and
dim L2 = 2. Then L is capable if and only if L ∼
= L4,3 ⊕ A(n − 4) or L ∼
= L5,5 ⊕
A(n − 5).
∼ T ⊕ A, where A is an abelian Lie algebra and
Proof. Theorem 5.1 implies L =
∼ A(1) and Z ∗ (L) = Z ∗ (T ). Now the result follows
Z(T ) = T 2 ∩ Z(L) = L3 = T 3 =
from Theorem 4.9.

The following result is obtained from Theorems 4.10 and 5.3.
Corollary 5.4. Let L be a finite dimensional Lie algebra of class 3 and dim L2 = 2.
Then L is capable if and only if 3 ≤ dim(L/Z(L)) ≤ 4.
We summarize all results to classify all capable nilpotent Lie algebras with the
derived subalgebra of dimension at most two.
Theorem 5.5. Let L be an n-dimensional nilpotent Lie algebra with dim L2 ≤ 2.
Then L is capable is if and only if L is isomorphic to one the following Lie algebras.
(i) If dim L2 = 0, then L ∼
= A(n) and n > 1.
(ii) If dim L2 = 1, then L ∼
= H(1) ⊕ A(n − 3).
(2)
2
(iii) If dim L = 2 and cl(L) = 2, then L ∼
= L5,8 ⊕ A(n − 5), L = L6,7 (η) ⊕ A(n −
∼
∼
6), L = L6,22 (ǫ) ⊕ A(n − 6), or L = L1 ⊕ A(n − 7).

12

P. NIROOMAND, F. JOHARI, AND M. PARVIZI

(iv) If dim L2 = 2 and cl(L) = 3, then L ∼
= L4,3 ⊕A(n−4) or L ∼
= L5,5 ⊕A(n−5).
Proof. The result follows from [17, Theorems 3.3 and 3.6], Theorems 5.2 and 5.3.

References
[1] P. Batten, K. Moneyhun, E. Stitzinger, On characterizing nilpotent Lie algebras by their
multipliers. Comm. Algebra 24 (1996) 4319-4330.
[2] P. Batten, E. Stitzinger, On covers of Lie algebras, Comm. Algebra 24 (1996) 4301-4317.
[3] F. R. Beyl, U. Felgner, and P. Schmid, On groups occurring as center factor groups, J. Algebra
61 (1970) 161-177.
[4] L. Bosko, On Schur multiplier of Lie algebras and groups of maximal class, Internat. J.
Algebra Comput. 20 (2010) 807-821.
[5] S. Cicalò, W. A. de Graaf, C. Schneider, Six-dimensional nilpotent Lie algebras, Linear
Algebra Appl. 436 (2012), no. 1, 163-189.
[6] G. Ellis, A non-abelian tensor product of Lie algebras, Glasg. Math. J. 39 (1991) 101-120.
[7] M. P. Gong, Classification of nilpotent Lie Algebras of dimension 7 (over Algebraically closed
fields and R), A thesis in Waterloo, Ontario, Canada, 1998.
[8] W. A. de Graaf, Classification of 6-dimensional nilpotent Lie algebras over fields of characteristic not 2, Algebra 309 (2007) 640-653.
[9] P. Hall, The classification of prime power groups, J. Reine Angew. Math. 182, (1940) 130-141.
[10] P. Hardy, E. Stitzinger, On characterizing nilpotent Lie algebras by their multipliers t(L) =
3; 4; 5; 6; Comm. Algebra, 1998, 26(11), 3527-3539.
[11] H. Heineken, Nilpotent groups of class 2 that can appear as central quotient groups, Rend.
Sem. Mat. Univ. Padova 84 (1990), 241-248.
[12] F. Johari, M. Parvizi, P. Niroomand, Capability and Schur multiplier of a pair of Lie algebras,
J. Geometry Phys 114 (2017), 184-196.
[13] K. Moneyhun, Isoclinisms in Lie algebras. Algebras Groups Geom. 11 (1994) 9-22.
[14] P. Niroomand, F. G. Russo, A note on the Schur multiplier of a nilpotent Lie algebra, Comm.
Algebra 39 (2011) 1293-1297.
[15] P. Niroomand, F. G. Russo, A restriction on the Schur multiplier of nilpotent Lie algebras,
Electron. J. Linear Algebra 22 (2011) 1-9.
[16] P. Niroomand, On the dimension of the Schur multiplier of nilpotent Lie algebras, Cent. Eur.
J. Math. 9 (2011) 57-64.
[17] P. Niroomand, M. Parvizi, F. G. Russo, Some criteria for detecting capable Lie algebras, J.
Algebra 384 (2013) 36-44.
[18] P. Niroomand, F. Johari, M. Parvizi, On the capability and Schur multiplier of nilpotent Lie
algebra of class two, Proc. Amer. Math. Soc. 144 (2016), 4157-4168.
[19] A. R. Salemkar, V. Alamian and H. Mohammadzadeh, Some properties of the Schur multiplier
and covers of Lie Algebras, Comm. Algebra 36 (2008) 697-707.
[20] L. M. Zack, Nilpotent Lie algebras with a small second derived quotient, Comm. Algebra, 36
(2008) 460-4619.
School of Mathematics and Computer Science, Damghan University, Damghan, Iran
E-mail address: p niroomand@yahoo.com, niroomand@du.ac.ir
Department of Pure Mathematics, Ferdowsi University of Mashhad, Mashhad, Iran
E-mail address: farangis.johari@mail.um.ac.ir, farangisjohary@yahoo.com
Department of Pure Mathematics, Ferdowsi University of Mashhad, Mashhad, Iran
E-mail address: parvizi@math.um.ac.ir

