On some non-nilpotent Leibniz algebras of dimension 3 and their automorphism groups
Let \(L\) be an algebra over a field \(F\) with the binary operations \(+\) and \([,]\). Then \(L\) is called a (left) Leibniz algebra if it satisfies the (left) Leibniz identity: \([a,[b,c]]=[[a,b],c]+[b,[a,c]]\) for all \(a,b,c\in L\). A linear transformation \(f\) of \(L\) is called an endomorphi...
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| author | Kurdachenko, Leonid A. Pypka, Oleksandr O. Subbotin, Igor Ya. |
| author_facet | Kurdachenko, Leonid A. Pypka, Oleksandr O. Subbotin, Igor Ya. |
| author_institution_txt_mv | [
{
"author": "Leonid A. Kurdachenko",
"institution": null
},
{
"author": "Oleksandr O. Pypka",
"institution": "Oles Honchar Dnipro National University"
},
{
"author": "Igor Ya. Subbotin",
"institution": null
}
] |
| author_sort | Kurdachenko, Leonid A. |
| baseUrl_str | https://admjournal.luguniv.edu.ua/index.php/adm/oai |
| collection | OJS |
| datestamp_date | 2026-07-08T07:55:33Z |
| description | Let \(L\) be an algebra over a field \(F\) with the binary operations \(+\) and \([,]\). Then \(L\) is called a (left) Leibniz algebra if it satisfies the (left) Leibniz identity: \([a,[b,c]]=[[a,b],c]+[b,[a,c]]\) for all \(a,b,c\in L\). A linear transformation \(f\) of \(L\) is called an endomorphism of \(L\) if \(f([a,b])=[f(a),f(b)]\) for all \(a,b\in L\). A bijective endomorphism of \(L\) is called an automorphism of \(L\). The main goal of this article is to describe the structure of automorphism groups of certain types of non-nilpotent three-dimensional Leibniz algebras over an arbitrary field \(F\). |
| doi_str_mv | 10.12958/adm2499 |
| first_indexed | 2026-07-09T01:00:10Z |
| format | Article |
| fulltext |
© Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 41 (2026). Number 2, pp. 219–234
DOI:10.12958/adm2499
On some non-nilpotent Leibniz algebras of
dimension 3 and their automorphism groups
Leonid A. Kurdachenko, Oleksandr O. Pypka,
and Igor Ya. Subbotin
Abstract. Let L be an algebra over a field F with the binary
operations + and [, ]. Then L is called a (left) Leibniz algebra if it
satisfies the (left) Leibniz identity: [a, [b, c]] = [[a, b], c] + [b, [a, c]]
for all a, b, c ∈ L. A linear transformation f of L is called an endo-
morphism of L if f([a, b]) = [f(a), f(b)] for all a, b ∈ L. A bijective
endomorphism of L is called an automorphism of L. The main
goal of this article is to describe the structure of automorphism
groups of certain types of non-nilpotent three-dimensional Leibniz
algebras over an arbitrary field F .
Introduction
Let L be an algebra over a field F with binary operations + and [, ].
We say that L is a (left) Leibniz algebra if it satisfies the (left) Leibniz
identity
[a, [b, c]] = [[a, b], c] + [b, [a, c]]
for all elements a, b, c ∈ L [2]. The term “Leibniz algebra” first appeared
in the book [14] and in the article [15]. The modern theory of Leibniz
algebras includes numerous results concerning their structure, classifi-
cation, and properties; these results can be found, for instance, in the
books [1] and [10].
We recall that Leibniz algebras form a broad generalization of Lie al-
gebras. On the other hand, if L is a Leibniz algebra such that [a, a] = 0
2020 Mathematics Subject Classification: 17A32, 17A36.
Key words and phrases: Leibniz algebra, automorphism group, Leibniz kernel,
center.
https://doi.org/10.12958/adm2499
220 On Leibniz algebras and their automorphism groups
for every element a ∈ L, then L is a Lie algebra. Equivalently, Lie
algebras can be characterized as anticommutative Leibniz algebras. Al-
though many results for Lie algebras have analogues in the theory of
Leibniz algebras, there are also substantial differences between these two
classes of non-associative algebras; see, for example, the survey artic-
les [3, 4, 12, 17].
Let f be a linear transformation of a Leibniz algebra L. The map f
is called an endomorphism of L if f([a, b]) = [f(a), f(b)] for all elements
a, b ∈ L. A bijective endomorphism of L is called an automorphism of L.
It is well known that the set Aut[,](L) of all automorphisms of L forms
a group under composition; see, for example, [9].
Describing the automorphism groups of algebraic structures is a fun-
damental and important problem in algebra, and Leibniz algebras are no
exception. It is natural to begin with Leibniz algebras whose structure
is already well understood. The first such class consists of one-generated
Leibniz algebras. The automorphism groups of infinite-dimensional one-
generated Leibniz algebras were described in [13], while the finite-dimen-
sional case was considered in [9].
The next natural step is to consider low-dimensional Leibniz algebras.
The case of dimension 2 is rather simple. Up to isomorphism, there are
only two non-Lie Leibniz algebras of dimension 2 (see [4]); only non-zero
products are displayed:
• Lei1(2, F ) = Fa1 ⊕ Fa2, [a1, a1] = a2;
• Lei2(2, F ) = Fa1 ⊕ Fa2, [a1, a1] = [a1, a2] = a2.
The automorphism groups of these Leibniz algebras were described in
the paper [11].
The situation for three-dimensional Leibniz algebras is considerably
more complicated. The most comprehensive description of such algeb-
ras over arbitrary fields can be found in [8]. Below we list the three-
dimensional Leibniz algebras of the form Fa1 ⊕ Fa2 ⊕ Fa3 whose auto-
morphism groups have already been determined. Only non-zero products
are displayed:
• Lei1(3, F ) : [a1, a1] = a2, [a1, a2] = a3 [11];
• Lei2(3, F ) : [a1, a1] = a3 [11];
• Lei3(3, F ) : [a1, a1] = [a1, a2] = a3 [6];
• Lei4(3, F ) : [a1, a1] = a3, [a2, a2] = λa3, λ ̸= 0 [7];
• Lei5(3, F ) : [a1, a1] = a3, [a1, a2] = a2 + λa3 [5];
L. A. Kurdachenko, O. O. Pypka, I. Ya. Subbotin 221
• Lei6(3, F ) : [a1, a1] = [a2, a1] = a3 [16].
In this paper, we extend our investigation to several further types of
three-dimensional Leibniz algebras.
1. Preliminaries and remarks
We recall some necessary definitions.
Let L be a Leibniz algebra over a field F . The algebra L is called
abelian if [a, b] = 0 for all a, b ∈ L. If A and B are subspaces of L, then
[A,B] denotes the subspace generated by all products [a, b], where a ∈ A
and b ∈ B.
A subspace S of L is called a subalgebra of L if [a, b] ∈ S for all
a, b ∈ S.
A subalgebra I of L is called a left ideal of L if [b, a] ∈ I for all a ∈ I
and b ∈ L. Similarly, I is called a right ideal of L if [a, b] ∈ I for all a ∈ I
and b ∈ L. A subalgebra I of L is called an ideal of L if it is both a left
and a right ideal.
Denote by Leib(L) the subspace generated by all elements [a, a],
where a ∈ L. It is easy to verify that Leib(L) is an ideal of L. This
ideal is called the Leibniz kernel of L.
The left center ζ left(L) and the right center ζright(L) of L are defined
as follows:
ζ left(L) = {a ∈ L | [a, b] = 0 for each b ∈ L},
ζright(L) = {a ∈ L | [b, a] = 0 for each b ∈ L}.
We observe that the left center of L is an ideal, whereas this is not
necessarily true for the right center. The right center is a subalgebra of
L; in general, the left and right centers are distinct.
The center ζ(L) of L is defined as follows:
ζ(L) = {a ∈ L | [a, b] = 0 = [b, a] for each b ∈ L}.
The center is an ideal of L.
The lower central series of L is defined by
L = γ1(L) ≥ γ2(L) ≥ . . . γα(L) ≥ γα+1(L) ≥ . . . γτ (L),
where γ1(L) = L, γ2(L) = [L,L], γα+1(L) = [L, γα(L)] for all ordinals α,
and
γλ(L) =
⋂
µ<λ
γµ(L)
222 On Leibniz algebras and their automorphism groups
for every limit ordinal λ. We say that L is nilpotent if there exists a
positive integer k such that γk(L) = ⟨0⟩.
Let L be a Leibniz algebra over a field F , let A be a subalgebra of L,
and let G = Aut[,](L). We put
CG(A) = {α ∈ G | α(x) = x for every x ∈ A}.
Let L be a non-nilpotent three-dimensional Leibniz algebra. As usual,
we assume that L is not a Lie algebra, so that Leib(L) is nonzero. Thus,
there are two possible cases: dimF (Leib(L)) = 2 or dimF (Leib(L)) = 1.
Consider the first case. LetK = Leib(L). Since L is not a Lie algebra,
there exists an element a1 such that [a1, a1] = a3 ̸= 0. By the structure
of two-dimensional Leibniz algebras, either [a1, a3] = [a3, a1] = 0, or one
can choose an element a1 such that [a3, a1] = 0 and [a1, a3] = a3. Put
A1 = Fa1 ⊕ Fa3. The case in which the subalgebra A1 is nilpotent was
studied in [5]. We now suppose that the subalgebra A1 is non-nilpotent,
that is, [a3, a1] = 0 and [a1, a3] = a3.
Since the subalgebra K is abelian of dimension 2, we have K =
Fa3 ⊕ Fb for some element b. It is not hard to see that the subalgebra
⟨a3⟩ is an ideal of K. Since an abelian Leibniz algebra is a Lie algeb-
ra, the assumption that the factor algebra L/Fa3 is abelian leads to a
contradiction with the equality Leib(L) = Fa3 ⊕ Fb. Hence, the factor
algebra L/Fa3 is non-abelian. Therefore, L/Fa3 has a coset c + Fa3
such that ⟨c, Fa3⟩ = Leib(L) and
[a1 + Fa3, c+ Fa3] = c+ Fa3.
Put c = a2. Then [a1, a2] = a2 + λa3 for some scalar λ ∈ F . Since
a2 ∈ Leib(L), we have [a2, a1] = 0.
If λ = 0, then the subalgebra Fa2 is an ideal of L, and we obtain the
following type of Leibniz algebras:
Lei7(3, F ) = Fa1 ⊕ Fa2 ⊕ Fa3, where
[a1, a1] = [a1, a3] = a3, [a1, a2] = a2,
[a2, a1] = [a2, a2] = [a2, a3] = [a3, a1] = [a3, a2] = [a3, a3] = 0.
In other words, Lei7(3, F ) = L is the sum of the ideal A2 = Fa2 and
the non-nilpotent one-generated Leibniz algebra A1 = Fa1 ⊕ Fa3 of
dimension 2. Moreover,
[A1, A2] = A2, Leib(L) = Fa2 ⊕ Fa3 = [L,L] = ζ left(L),
ζright(L) = ζ(L) = ⟨0⟩.
L. A. Kurdachenko, O. O. Pypka, I. Ya. Subbotin 223
If λ ̸= 0, then we obtain the following type of Leibniz algebras:
Lei8(3, F ) = Fa1 ⊕ Fa2 ⊕ Fa3, where
[a1, a1] = [a1, a3] = a3, [a1, a2] = a2 + λa3, 0 ̸= λ ∈ F,
[a2, a1] = [a2, a2] = [a2, a3] = [a3, a1] = [a3, a2] = [a3, a3] = 0.
In other words, Lei8(3, F ) = L is the sum of the ideal Leib(L) = Fa2 ⊕
Fa3 and the non-nilpotent one-generated Leibniz algebra A1 = Fa1⊕Fa3
of dimension 2. Moreover,
[A1, Leib(L)] = Leib(L), Leib(L) = [L,L] = ζ left(L),
ζright(L) = ζ(L) = ⟨0⟩.
In the next section, we discuss the structure of the automorphism
groups of these Leibniz algebras.
2. The automorphism groups of Lei7(3, F ) and Lei8(3, F )
Denote by Ξ the canonical monomorphism of Aut[,](L) in M3(F ).
Theorem 1. Let G be an automorphism group of the Leibniz algebra
Lei7(3, F ). Then G is isomorphic to a subgroup of GL3(F ) that consists
of matrices of the form: 1 0 0
α2 β2 α2
α3 β3 1 + α3
,
α2, α3, β2, β3 ∈ F . Furthermore, G is isomorphic to the group GL2(F ).
Moreover, it is a product of two subgroups C1 = CG(a1), C2 = CG(a2),
and Ξ(C1) is the set of matrices of the form 1 0 0
0 β2 0
0 β3 1
,
β2, β3 ∈ F , and Ξ(C2) is the set of matrices of the form 1 0 0
α2 1 α2
α3 0 1 + α3
,
224 On Leibniz algebras and their automorphism groups
α2, α3 ∈ F . A subgroup C1 is a semidirect product of a normal sub-
group C3, which is isomorphic to the additive group of a field F , and a
subgroup C4, which is isomorphic to the multiplicative group of a field
F ; a subgroup C2 is a semidirect product of a normal subgroup, which
is isomorphic to the additive group of a field F , and a subgroup that is
isomorphic to the multiplicative group of a field F .
Proof. Let L = Lei7(3, F ) and let f ∈ Aut[,](L). By [9, Lemma 2.1],
f([L,L]) = [L,L], so that
f(a1) = α1a1 + α2a2 + α3a3,
f(a2) = β2a2 + β3a3,
f(a3) = γ2a2 + γ3a3.
Then
f(a3) = f([a1, a1]) = [f(a1), f(a1)]
= [α1a1 + α2a2 + α3a3, α1a1 + α2a2 + α3a3]
= α2
1[a1, a1] + α1α2[a1, a2] + α1α3[a1, a3]
= α2
1a3 + α1α2a2 + α1α3a3
= α1α2a2 + (α2
1 + α1α3)a3,
f(a3) = f([a1, a3]) = [f(a1), f(a3)]
= [α1a1 + α2a2 + α3a3, γ2a2 + γ3a3]
= α1γ2[a1, a2] + α1γ3[a1, a3]
= α1γ2a2 + α1γ3a3,
f(a2) = f([a1, a2]) = [f(a1), f(a2)]
= [α1a1 + α2a2 + α3a3, β2a2 + β3a3]
= α1β2[a1, a2] + α1β3[a1, a3]
= α1β2a2 + α1β3a3.
Thus, we can see that
α1α2a2 + (α2
1 + α1α3)a3 = α1γ2a2 + α1γ3a3 = γ2a2 + γ3a3,
α1β2a2 + α1β3a3 = β2a2 + β3a3.
It follows that
α1α2 = α1γ2 = γ2, α2
1 + α1α3 = α1γ3 = γ3, α1β2 = β2, α1β3 = β3.
L. A. Kurdachenko, O. O. Pypka, I. Ya. Subbotin 225
If we suppose that α1 = 0, then f(a1) ∈ Fa2 ⊕ Fa3. Since we
have f(a2), f(a3) ∈ Fa2 ⊕ Fa3, then f(L) ≤ Fa2 ⊕ Fa3 – in particular,
f(L) ̸= L – and we obtain a contradiction with the fact that f is an
automorphism of L. Hence, α1 ̸= 0. It follows that α2 = γ2.
If we suppose that both β2 = γ2 = 0, then f(a2) ∈ Fa3 and f(a3) ∈
Fa3, so that f([L,L]) ≤ Fa3. On the other hand, dimF ([L,L]) = 2, and
we, again, obtain a contradiction with the fact that f is an automorphism
of L. Hence, (β2, γ2) ̸= (0, 0). It follows that α1 = 1 and then 1+α3 = γ3.
Denote by Ξ the canonical monomorphism of Aut[,](L) in M3(F ).
Then Ξ(f) is the following matrix: 1 0 0
α2 β2 α2
α3 β3 1 + α3
,
α2, α3, β2, β3 ∈ F .
Conversely, let f be a linear transformation of L, having in a basis
{a1, a2, a3} the matrix above. Let x, y be the arbitrary elements of L,
x = ξ1a1+ξ2a2+ξ3a3, y = η1a1+η2a2+η3a3, where ξ1, ξ2, ξ3, η1, η2, η3 ∈
F . Then
[x, y] = [ξ1a1 + ξ2a2 + ξ3a3, η1a1 + η2a2 + η3a3]
= ξ1η1[a1, a1] + ξ1η2[a1, a2] + ξ1η3[a1, a3]
= ξ1η1a3 + ξ1η2a2 + ξ1η3a3
= ξ1η2a2 + (ξ1η1 + ξ1η3)a3 = ξ1η2a2 + ξ1(η1 + η3)a3,
f(x) = f(ξ1a1 + ξ2a2 + ξ3a3) = ξ1f(a1) + ξ2f(a2) + ξ3f(a3)
= ξ1(a1 + α2a2 + α3a3) + ξ2(β2a2 + β3a3) + ξ3(α2a2 + (1 + α3)a3)
= ξ1a1 + (ξ1α2 + ξ2β2 + ξ3α2)a2 + (ξ1α3 + ξ2β3 + ξ3 + ξ3α3)a3,
f(y) = η1a1 + (η1α2 + η2β2 + η3α2)a2 + (η1α3 + η2β3 + η3 + η3α3)a3.
Therefore,
f([x, y]) = f(ξ1η2a2 + ξ1(η1 + η3)a3) = ξ1η2f(a2) + ξ1(η1 + η3)f(a3)
= ξ1η2(β2a2 + β3a3) + ξ1(η1 + η3)(α2a2 + (1 + α3)a3)
= (ξ1η2β2 + ξ1(η1 + η3)α2)a2 + (ξ1η2β3 + ξ1(η1 + η3)(1 + α3))a3,
[f(x), f(y)]
= [ξ1a1 + (ξ1α2 + ξ2β2 + ξ3α2)a2 + (ξ1α3 + ξ2β3 + ξ3 + ξ3α3)a3,
η1a1 + (η1α2 + η2β2 + η3α2)a2 + (η1α3 + η2β3 + η3 + η3α3)a3]
226 On Leibniz algebras and their automorphism groups
= ξ1η1[a1, a1] + ξ1(η1α2 + η2β2 + η3α2)[a1, a2]
+ξ1(η1α3 + η2β3 + η3 + η3α3)[a1, a3]
= ξ1η1a3 + ξ1(η1α2 + η2β2 + η3α2)a2 + ξ1(η1α3 + η2β3 + η3 + η3α3)a3
= ξ1(η1α2 + η2β2 + η3α2)a2 + ξ1(η1 + η1α3 + η2β3 + η3 + η3α3)a3.
We can see that f([x, y]) = [f(x), f(y)]. It follows that a linear transfor-
mation f having in a basis {a1, a2, a3} the above matrix is an automor-
phism of a Leibniz algebra.
Denote by Φ the mapping of Ξ(G) in GL2(F ) defined by the rule: 1 0 0
α2 β2 α2
α3 β3 1 + α3
→
(
β2 α2
β3 1 + α3
)
.
We have: 1 0 0
α2 β2 α2
α3 β3 1 + α3
1 0 0
γ2 κ2 γ2
γ3 κ3 1 + γ3
=
1 0 0
α2 + β2γ2 + α2γ3 β2κ2 + α2κ3 β2γ2 + α2(1 + γ3)
α3 + β3γ2 + (1 + α3)γ3 β3κ2 + (1 + α3)κ3 β3γ2 + (1 + α3)(1 + γ3)
and (
β2 α2
β3 1 + α3
)(
κ2 γ2
κ3 1 + γ3
)
=
(
β2κ2 + α2κ3 β2γ2 + α2(1 + γ3)
β3κ2 + (1 + α3)κ3 β3γ2 + (1 + α3)(1 + γ3)
)
.
These equalities show that a mapping Φ is a homomorphism. Clearly,
Ker(Φ) consists only of the identity matrix. Finally, let(
σ11 σ12
σ21 σ22
)
be the arbitrary element of GL2(F ). Consider the matrix 1 0 0
σ12 σ11 σ12
σ22 − 1 σ21 σ22
.
L. A. Kurdachenko, O. O. Pypka, I. Ya. Subbotin 227
This matrix is a preimage of the matrix above by the mapping Φ. More-
over, both these matrices have the same determinants. It follows that
the last matrix belongs to Ξ(G), so that a mapping Φ is an isomorphism.
Let C1 = CG(a1), C2 = CG(a2). Then Ξ(C1) is the set of matrices of
the form 1 0 0
0 β2 0
0 β3 1
,
β2, β3 ∈ F , and Ξ(C2) is the set of matrices of the form 1 0 0
α2 1 α2
α3 0 1 + α3
,
α2, α3 ∈ F . We have: 1 0 0
σ2 1 σ2
σ3 0 1 + σ3
1 0 0
0 κ2 0
0 κ3 1
=
1 0 0
σ2 κ2 + σ2κ3 σ2
σ3 (1 + σ3)κ3 1 + σ3
.
If 1 0 0
α2 β2 α2
α3 β3 1 + α3
is an arbitrary matrix of Ξ(G), then we obtain 1 0 0
α2 β2 α2
α3 β3 1 + α3
=
1 0 0
σ2 1 σ2
σ3 0 1 + σ3
1 0 0
0 κ2 0
0 κ3 1
,
where σ2 = α2, σ3 = α3, κ3 = β3(1 + α3)
−1, κ2 = β2 − α2β3(1 + α3)
−1.
Thus, we can see that a group G is a product of the subgroups C1 and C2.
Furthermore, a subgroup C1 has a normal subgroup C3 such that
Ξ(C3) is the subset of matrices of the form 1 0 0
0 1 0
0 β3 1
,
which is isomorphic to the additive group of a field F , and a subgroup
C4 such that Ξ(C4) is the subset of matrices of the form 1 0 0
0 β2 0
0 0 1
,
228 On Leibniz algebras and their automorphism groups
which is isomorphic to the multiplicative group of a field F . Moreover,
C1 is clearly a semidirect product of C3 and C4.
Now, consider the group Ξ(C2). It is the set of matrices of the form 1 0 0
α2 1 α2
α3 0 1 + α3
,
α2, α3 ∈ F . Denote by Φ the mapping of Ξ(C2) in GL2(F ) defined by
the rule: 1 0 0
α2 1 α2
α3 0 1 + α3
→
(
1 α2
0 1 + α3
)
.
We have: 1 0 0
α2 1 α2
α3 0 1 + α3
1 0 0
γ2 1 γ2
γ3 0 1 + γ3
=
1 0 0
α2 + γ2 + α2γ3 1 γ2 + α2(1 + γ3)
α3 + (1 + α3)γ3 0 (1 + α3)(1 + γ3)
and (
1 α2
0 1 + α3
)(
1 γ2
0 1 + γ3
)
=
(
1 γ2 + α2(1 + γ3)
0 (1 + α3)(1 + γ3)
)
.
These equalities show that a mapping Φ is a homomorphism. Clearly,
Ker(Φ) consists only of identity matrix. Finally, let(
1 σ12
0 σ22
)
be the arbitrary element of GL2(F ). Consider the matrix 1 0 0
σ12 1 σ12
σ22 − 1 0 σ22
.
This matrix is a preimage of the matrix above by Φ. Moreover, both
these matrices have the same determinants. The last matrix belongs to
Ξ(G), so a mapping Φ is an isomorphism.
L. A. Kurdachenko, O. O. Pypka, I. Ya. Subbotin 229
In turn, clearly Ξ(C2) is a semidirect product of a normal subgroup
that consists of matrices of the form(
1 σ12
0 1
)
,
(i.e., a subgroup UT2(F )) that is isomorphic to the additive group of a
field F and a subgroup that consists of matrices of the form(
1 0
0 σ22
)
,
which is isomorphic to the multiplicative group of a field F .
Theorem 2. Let G be an automorphism group of the Leibniz algebra
Lei8(3, F ). Then G is isomorphic to a subgroup of GL3(F ) that consists
of matrices of the form 1 0 0
0 β1 0
β1 − 1 β2 β1
,
β1, β2 ∈ F . Furthermore, G is abelian, G = C ×A such that Ξ(C) is the
set of matrices of the form 1 0 0
0 1 0
0 σ 1
,
σ ∈ F , and Ξ(A) is the set of matrices of the form 1 0 0
0 β 0
β − 1 0 β
,
β ∈ F . A subgroup C is isomorphic to the additive group of a field F ,
and a subgroup A is isomorphic to the multiplicative group of a field F .
Proof. Let L = Lei8(3, F ) and let f ∈ Aut[,](L). By [9, Lemma 2.1],
f([L,L]) = [L,L], so that
f(a1) = α1a1 + α2a2 + α3a3,
f(a2) = β2a2 + β3a3,
f(a3) = γ2a2 + γ3a3.
230 On Leibniz algebras and their automorphism groups
Then
f(a3) = f([a1, a1]) = [f(a1), f(a1)]
= [α1a1 + α2a2 + α3a3, α1a1 + α2a2 + α3a3]
= α2
1[a1, a1] + α1α2[a1, a2] + α1α3[a1, a3]
= α2
1a3 + α1α2(a2 + λa3) + α1α3a3
= α1α2a2 + (α2
1 + λα1α2 + α1α3)a3,
f(a3) = f([a1, a3]) = [f(a1), f(a3)]
= [α1a1 + α2a2 + α3a3, γ2a2 + γ3a3]
= α1γ2[a1, a2] + α1γ3[a1, a3]
= α1γ2(a2 + λa3) + α1γ3a3
= α1γ2a2 + (λα1γ2 + α1γ3)a3,
f([a1, a2]) = [f(a1), f(a2)] = [α1a1 + α2a2 + α3a3, β2a2 + β3a3]
= α1β2[a1, a2] + α1β3[a1, a3]
= α1β2(a2 + λa3) + α1β3a3
= α1β2a2 + (λα1β2 + α1β3)a3,
f([a1, a2]) = f(a2 + λa3) = f(a2) + λf(a3)
= β2a2 + β3a3 + λ(γ2a2 + γ3a3)
= (β2 + λγ2)a2 + (β3 + λγ3)a3.
Thus, we can see that
α1α2a2 + (α2
1 + λα1α2 + α1α3)a3 =
α1γ2a2 + (λα1γ2 + α1γ3)a3 = γ2a2 + γ3a3,
α1β2a2 + (λα1β2 + α1β3)a3 = (β2 + λγ2)a2 + (β3 + λγ3)a3.
It follows that
α1α2 = α1γ2 = γ2, α2
1 + λα1α2 + α1α3 = λα1γ2 + α1γ3 = γ3,
α1β2 = β2 + λγ2, λα1β2 + α1β3 = β3 + λγ3.
If we suppose that α1 = 0, then f(a1) ∈ Fa2 ⊕ Fa3. Since we
have f(a2), f(a3) ∈ Fa2 ⊕ Fa3, then f(L) ≤ Fa2 ⊕ Fa3 – in particular,
f(L) ̸= L – and we obtain a contradiction with the fact that f is an
automorphism of L. Hence, α1 ̸= 0. It follows that α2 = γ2.
If we suppose that both β2 = γ2 = 0, then f(a2) ∈ Fa3, and
f(a3) ∈ Fa3. We obtain that f([L,L]) ≤ Fa3. On the other hand,
L. A. Kurdachenko, O. O. Pypka, I. Ya. Subbotin 231
dimF ([L,L]) = 2, and we, again, obtain a contradiction with the fact
that f is an automorphism of L. Hence, (β2, γ2) ̸= (0, 0). Then, if we
suppose that β2 = 0, we will see that λγ2 = 0. Since λ ̸= 0, we obtain
that γ2 = 0, which is impossible. It follows that β2 ̸= 0. Suppose that
γ2 = 0. Since β2 ̸= 0, we obtain that α1 = 1. If γ2 ̸= 0, then from
α1γ2 = γ2, we obtain that α1 = 1. Thus, in every case, α1 = 1. Then we
obtain an equation β2 = β2+λγ2 or λγ2 = 0. Since λ ̸= 0, we obtain that
γ2 = 0. Then α2 = 0, and we obtain 1 + α3 = γ3, λβ2 + β3 = β3 + λγ3.
The last equation implies that λβ2 = λγ3. Since λ ̸= 0, we obtain that
β2 = γ3.
Denote by Ξ the canonical monomorphism of Aut[,](L) in M3(F ).
Then Ξ(f) is the following matrix 1 0 0
0 β2 0
β2 − 1 β3 β2
,
β2, β3 ∈ F . Since this matrix must be non-degenerate, β2 ̸= 0.
Conversely, let f be a linear transformation of L, having in a basis
{a1, a2, a3} the matrix above. Let x, y be the arbitrary elements of L,
x = ξ1a1+ξ2a2+ξ3a3, y = η1a1+η2a2+η3a3, where ξ1, ξ2, ξ3, η1, η2, η3 ∈
F . Then
[x, y] = [ξ1a1 + ξ2a2 + ξ3a3, η1a1 + η2a2 + η3a3]
= ξ1η1[a1, a1] + ξ1η2[a1, a2] + ξ1η3[a1, a3]
= ξ1η1a3 + ξ1η2(a2 + λa3) + ξ1η3a3
= ξ1η2a2 + (ξ1η1 + λξ1η2 + ξ1η3)a3
and
f(x) = f(ξ1a1 + ξ2a2 + ξ3a3) = ξ1f(a1) + ξ2f(a2) + ξ3f(a3)
= ξ1(a1 + (β2 − 1)a3) + ξ2(β2a2 + β3a3) + ξ3(β2a3)
= ξ1a1 + ξ2β2a2 + (ξ1(β2 − 1) + ξ2β3 + ξ3β2)a3,
f(y) = η1a1 + η2β2a2 + (η1(β2 − 1) + η2β3 + η3β2)a3.
Therefore,
f([x, y]) = f(ξ1η2a2 + (ξ1η1 + λξ1η2 + ξ1η3)a3)
= ξ1η2f(a2) + (ξ1η1 + λξ1η2 + ξ1η3)f(a3)
= ξ1η2(β2a2 + β3a3) + (ξ1η1 + λξ1η2 + ξ1η3)β2a3
232 On Leibniz algebras and their automorphism groups
= ξ1η2β2a2 + (ξ1η2β3 + β2ξ1η1 + λξ1η2β2 + ξ1η3β2)a3,
[f(x), f(y)] = [ξ1a1 + ξ2β2a2 + (ξ1(β2 − 1) + ξ2β3 + ξ3β2)a3,
η1a1 + η2β2a2 + (η1(β2 − 1) + η2β3 + η3β2)a3]
= ξ1η1[a1, a1] + ξ1η2β2[a1, a2] + ξ1(η1(β2 − 1) + η2β3 + η3β2)[a1, a3]
= ξ1η1a3 + ξ1η2β2(a2 + λa3) + ξ1(η1(β2 − 1) + η2β3 + η3β2)a3
= ξ1η2β2a2 + (ξ1η1 + ξ1η2β2λ+ ξ1η1β2 − ξ1η1 + ξ1η2β3 + ξ1η3β2)a3
= ξ1η2β2a2 + (ξ1η2β2λ+ ξ1η1β2 + ξ1η2β3 + ξ1η3β2)a3.
We can see that f([x, y]) = [f(x), f(y)]. It follows that a linear transfor-
mation f having in a basis {a1, a2, a3} the matrix above, is an automor-
phism of a Leibniz algebra L.
The equality 1 0 0
0 β 0
β − 1 γ β
1 0 0
0 σ 0
σ − 1 κ σ
=
1 0 0
0 βσ 0
β − 1 + β(σ − 1) γσ + βκ βσ
=
1 0 0
0 βσ 0
βσ − 1 γσ + βκ βσ
shows that a group G is abelian.
According to the proof above, a subspace Fa3 is G-invariant. Let
C = CG(Fa3). Then we can see that Ξ(C) is the set of matrices of the
form 1 0 0
0 1 0
0 β 1
,
β ∈ F . It is not hard to see that C is isomorphic to the additive group
of a field F .
Denote by A the preimage by Ξ of the set of matrices of the form 1 0 0
0 β 0
β − 1 0 β
,
β ∈ F . We have: 1 0 0
0 β 0
β − 1 0 β
1 0 0
0 σ 0
σ − 1 0 σ
L. A. Kurdachenko, O. O. Pypka, I. Ya. Subbotin 233
=
1 0 0
0 βσ 0
β − 1 + β(σ − 1) 0 βσ
=
1 0 0
0 βσ 0
βσ − 1 0 βσ
.
We can now see that Ξ(A) is a subgroup of GL3(F ), so that A is a
subgroup of G.
Moreover, the mapping, defined by the rule
β →
1 0 0
0 β 0
β − 1 0 β
,
β ∈ F , is an isomorphism of the multiplicative group of F on Ξ(A).
Thus, we obtain that A is isomorphic to the multiplicative group of a
field F .
Furthermore, the equality 1 0 0
0 β 0
β − 1 σ β
=
1 0 0
0 β 0
β − 1 0 β
1 0 0
0 1 0
0 β−1σ 1
shows that G is a product of the subgroups C and A. Moreover, this
product is direct.
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Contact information
L. A. Kurdachenko,
O. O. Pypka
Oles Honchar Dnipro National University,
Nauky Av. 72, Dnipro, 49045, Ukraine
E-Mail: lkurdachenko@gmail.com,
sasha.pypka@gmail.com
I. Ya. Subbotin National University, 5245 Pacific Concourse
Drive, Los Angeles, CA 90045-6904, USA
E-Mail: isubboti@nu.edu
Received by the editors: 09.06.2026.
https://doi.org/10.1007/s11253-024-02367-y
https://doi.org/10.12958/adm2036
https://doi.org/10.22108/IJGT.2021.130057.1735
https://doi.org/10.22108/IJGT.2021.130057.1735
https://doi.org/10.1007/978-3-031-58148-9
https://doi.org/10.1007/s11253-023-02153-2
https://doi.org/10.32037/agta-2020-004
https://doi.org/10.1142/S0219498824500026
https://doi.org/10.1007/978-3-662-11389-9
https://doi.org/10.1007/978-3-662-11389-9
https://doi.org/10.12958/adm2264
https://doi.org/10.15421/242108
Leonid A. Kurdachenko, Oleksandr O. Pypka,and Igor Ya. Subbotin
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| id | admjournalluguniveduua-article-2499 |
| institution | Algebra and Discrete Mathematics |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-07-09T01:00:10Z |
| publishDate | 2026 |
| publisher | Lugansk National Taras Shevchenko University |
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| resource_txt_mv | admjournalluguniveduua/61/1af9db4204d16e624960e25fe604f661.pdf |
| spelling | admjournalluguniveduua-article-24992026-07-08T07:55:33Z On some non-nilpotent Leibniz algebras of dimension 3 and their automorphism groups Kurdachenko, Leonid A. Pypka, Oleksandr O. Subbotin, Igor Ya. Leibniz algebra, automorphism group, Leibniz kernel, center 17A32, 17A36 Let \(L\) be an algebra over a field \(F\) with the binary operations \(+\) and \([,]\). Then \(L\) is called a (left) Leibniz algebra if it satisfies the (left) Leibniz identity: \([a,[b,c]]=[[a,b],c]+[b,[a,c]]\) for all \(a,b,c\in L\). A linear transformation \(f\) of \(L\) is called an endomorphism of \(L\) if \(f([a,b])=[f(a),f(b)]\) for all \(a,b\in L\). A bijective endomorphism of \(L\) is called an automorphism of \(L\). The main goal of this article is to describe the structure of automorphism groups of certain types of non-nilpotent three-dimensional Leibniz algebras over an arbitrary field \(F\). Lugansk National Taras Shevchenko University 2026-07-08 Article Article Peer-reviewed Article application/pdf https://admjournal.luguniv.edu.ua/index.php/adm/article/view/2499 10.12958/adm2499 Algebra and Discrete Mathematics; Vol 41, No 2 (2026) 2415-721X 1726-3255 en https://admjournal.luguniv.edu.ua/index.php/adm/article/view/2499/pdf https://admjournal.luguniv.edu.ua/index.php/adm/article/downloadSuppFile/2499/1392 Copyright (c) 2026 Algebra and Discrete Mathematics |
| spellingShingle | Leibniz algebra automorphism group Leibniz kernel center 17A32 17A36 Kurdachenko, Leonid A. Pypka, Oleksandr O. Subbotin, Igor Ya. On some non-nilpotent Leibniz algebras of dimension 3 and their automorphism groups |
| title | On some non-nilpotent Leibniz algebras of dimension 3 and their automorphism groups |
| title_full | On some non-nilpotent Leibniz algebras of dimension 3 and their automorphism groups |
| title_fullStr | On some non-nilpotent Leibniz algebras of dimension 3 and their automorphism groups |
| title_full_unstemmed | On some non-nilpotent Leibniz algebras of dimension 3 and their automorphism groups |
| title_short | On some non-nilpotent Leibniz algebras of dimension 3 and their automorphism groups |
| title_sort | on some non-nilpotent leibniz algebras of dimension 3 and their automorphism groups |
| topic | Leibniz algebra automorphism group Leibniz kernel center 17A32 17A36 |
| topic_facet | Leibniz algebra automorphism group Leibniz kernel center 17A32 17A36 |
| url | https://admjournal.luguniv.edu.ua/index.php/adm/article/view/2499 |
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