ON THE GROUP OF AUTOMORPHISMS OF THE BRANDT λ0 -EXTENSION OF
A MONOID WITH ZERO

arXiv:1609.06085v1 [] 20 Sep 2016

OLEG GUTIK

Abstract. The group of automorphisms of the Brandt λ0 -extension Bλ0 (S) of an arbitrary monoid S
with zero is described. In particular we show that the group of automorphisms Aut(Bλ0 (S)) of Bλ0 (S) is
isomorphic to a homomorphic image of the group defines on the Cartesian product Sλ × Aut(S) × H1λ
with the following binary operation:
[ϕ, h, u] · [ϕ′ , h′ , u′ ] = [ϕϕ′ , hh′ , ϕu′ · uh′ ],
where Sλ is the group of all bijections of the cardinal λ, Aut(S) is the group of all automorphisms of
the semigroup S and H1λ is the direct λ-power of the group of units H1 of the monoid S.

1. Introduction and preliminaries
Further we shall follow the terminology of [2, 21].
Given a semigroup S, we shall denote the set of idempotents of S by E(S). A semigroup S with
the adjoined unit (identity) [zero] will be denoted by S 1 [S 0 ] (cf. [2]). Next, we shall denote the unit
(identity) and the zero of a semigroup S by 1S and 0S , respectively. Given a subset A of a semigroup
S, we shall denote by A∗ = A \ {0S }.
If S is a semigroup, then we shall denote the subset of idempotents in S by E(S). If E(S) is closed
under multiplication in S and we shall refer to E(S) a band (or the band of S). If the band E(S) is a
non-empty subset of S, then the semigroup operation on S determines the following partial order 6 on
E(S): e 6 f if and only if ef = f e = e. This order is called the natural partial order on E(S).
If h : S → T is a homomorphism (or a map) from a semigroup S into a semigroup T and if s ∈ S,
then we denote the image of s under h by (s)h.
Let S be a semigroup with zero and λ a cardinal > 1. We define the semigroup operation on the set
Bλ (S) = (λ × S × λ) ∪ {0} as follows:
(
(α, ab, δ), if β = γ;
(α, a, β) · (γ, b, δ) =
0,
if β 6= γ,
and (α, a, β) · 0 = 0 · (α, a, β) = 0 · 0 = 0, for all α, β, γ, δ ∈ λ and a, b ∈ S. If S = S 1 then the
semigroup Bλ (S) is called the Brandt λ-extension of the semigroup S [4]. Obviously, if S has zero then
J = {0} ∪ {(α, 0S , β) : 0S is the zero of S} is an ideal of Bλ (S). We put Bλ0 (S) = Bλ (S)/J and the
semigroup Bλ0 (S) is called the Brandt λ0 -extension of the semigroup S with zero [8].
If I is a trivial semigroup (i.e. I contains only one element), then we denote the semigroup I with
the adjoined zero by I 0 . Obviously, for any λ > 2, the Brandt λ0 -extension of the semigroup I 0 is
isomorphic to the semigroup of λ×λ-matrix units and any Brandt λ0 -extension of a semigroup with
zero which also contains a non-zero idempotent contains the semigroup of λ×λ-matrix units. We shall
denote the semigroup of λ×λ-matrix units by Bλ . The 2 × 2-matrix semigroup with adjoined identity
B21 plays an impotent role in Graph Theory and its called the Perkins semigroup. In the paper [20]
Perkins showed that the semigroup B21 is not finitely based. More details on the word problem of the
Perkins semigroup via different graphs may be found in the works of Kitaev and his coauthors (see
[17, 18]).
Date: January 22, 2018.
2010 Mathematics Subject Classification. 20M15, 20F29 .
Key words and phrases. Semigroup, group of automorphisms, monoid, extension.
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OLEG GUTIK

We always consider the Brandt λ0 -extension only of a monoid with zero. Obviously, for any monoid
S with zero we have B10 (S) = S. Note that every Brandt λ-extension of a group G is isomorphic to the
Brandt λ0 -extension of the group G0 with adjoined zero. The Brandt λ0 -extension of the group with
adjoined zero is called a Brandt semigroup [2, 21]. A semigroup S is a Brandt semigroup if and only if
S is a completely 0-simple inverse semigroup [1, 19] (cf. also [21, Theorem II.3.5]). We shall say that
the Brandt λ0 -extension Bλ0 (S) of a semigroup S is finite if the cardinal λ is finite.
In the paper [14] Gutik and Repovš established homomorphisms of the Brandt λ0 -extensions of
monoids with zeros. They also described a category whose objects are ingredients in the constructions
of the Brandt λ0 -extensions of monoids with zeros. Here they introduced finite, compact topological
Brandt λ0 -extensions of topological semigroups and countably compact topological Brandt λ0 -extensions
of topological inverse semigroups in the class of topological inverse semigroups, and established the structure of such extensions and non-trivial continuous homomorphisms between such topological Brandt
λ0 -extensions of topological monoids with zero. There they also described a category whose objects
are ingredients in the constructions of finite (compact, countably compact) topological Brandt λ0 extensions of topological monoids with zeros. These investigations were continued in [10] and [9], where
established countably compact topological Brandt λ0 -extensions of topological monoids with zeros and
pseudocompact topological Brandt λ0 -extensions of semitopological monoids with zeros their corresponding categories. Some other topological aspects of topologizations, embeddings and completions
of the semigroup of λ×λ-matrix units and Brandt λ0 -extensions as semitopological and topological
semigroups were studied in [3, 5, 7, 11, 12, 13, 15, 16].
In this paper we describe the group of automorphisms of the Brandt λ0 -extension Bλ0 (S) of an arbitrary
monoid S with zero.
2. Automorphisms of the Brandt λ0 -extension of a monoid with zero
We observe that if f : S → S is an automorphism of the semigroup S without zero then it is obvious
that the map fb: S 0 → S 0 defined by the formula

(s)f, if s 6= 0S ;
(s)fb =
0S , if s = 0S ,

is an automorphism of the semigroup S 0 with adjoined zero 0S . Also the automorphism f : S → S of
the semigroup S can be extended to an automorphism fB : Bλ0 (S) → Bλ0 (S) of the Brandt λ0 -extension
Bλ0 (S) of the semigroup S by the formulae:
(α, s, β)fB = (α, (s)f, β),

for all α, β ∈ λ

and (0)fB = 0. We remark that so determined extended automorphism is not unique.
The following theorem describes all automorphisms of the Brandt λ0 -extension Bλ0 (S) of a monoid S.
Theorem 1. Let λ > 1 be cardinal and let Bλ0 (S) be the Brandt λ0 -extension of monoid S with zero.
Let h : S → S be an automorphism and suppose that ϕ : λ → λ is a bijective map. Let H1 be the group
of units of S and u : λ → H1 a map. Then the map σ : Bλ0 (S) → Bλ0 (S) defined by the formulae
(1)

((α, s, β))σ = ((α)ϕ, (α)u · (s)h · ((β)u)−1, (β)ϕ)

and

(0)σ = 0,

is an automorphism of the semigroup Bλ0 (S). Moreover, every automorphism of Bλ0 (S) can be constructed
in this manner.
Proof. A simple verification shows that σ is an automorphism of the semigroup Bλ0 (S).
Let σ : Bλ0 (S) → Bλ0 (S) be an isomorphism. We fix an arbitrary α ∈ λ.
Since σ : Bλ0 (S) → Bλ0 (S) is the automorphism and the idempotent (α, 1S , α) is maximal with the
respect to the natural partial order on E(Bλ0 (S)), Proposition 3.2 of [14] implies that ((α, 1S , α))σ =
(α ′ , 1S , α ′ ) for some α ′ ∈ λ.
Since (β, 1S , α)(α, 1S , α) = (β, 1S , α) for any β ∈ λ, we have that
((β, 1S , α))σ = ((β, 1S , α))σ · (α ′ , 1S , α ′ ),

ON THE GROUP OF AUTOMORPHISMS OF THE BRANDT λ0 -EXTENSION OF A MONOID WITH ZERO

3

and hence
((β, 1S , α))σ = ((β)ϕ, (β)u, α ′),
for some (β)ϕ ∈ λ and (β)u ∈ S. Similarly, we get that
((α, 1S , β))σ = (α ′ , (β)v, (β)ψ),
for some (β)ψ ∈ λ and (β)v ∈ S. Since (α, 1S , β)(β, 1S , α) = (α, 1S , α), we have that
(α ′ , 1S , α ′ ) = ((α, 1S , α))σ = (α ′ , (β)v, (β)ψ) · ((β)ϕ, (β)u, α ′) = (α ′ , (β)v · (β)u, α ′ ),
and hence (β)ϕ = (β)ψ = β ′ ∈ λ and (β)v · (β)u = 1S . Similarly, since (β, 1S , α) · (α, 1S , β) = (β, 1S , β),
we see that the element
((β, 1S , β))σ = ((β, 1S , α)(α, 1S , β))σ = (β ′ , (β)v · (β)u, β ′ )
is a maximal idempotent of the subsemigroup Sβ ′ ,β ′ of Bλ0 (S), and hence we have that (β)v · (β)u = 1S .
This implies that the elements (β)v and (β)u are mutually invertible in H1 , and hence (β)v = ((β)u)−1.
If (γ)ϕ = (δ)ϕ for γ, δ ∈ λ then
0 6= (α ′ , 1S , (γ)ϕ) · ((δ)ϕ, 1S , α ′ ) = ((α, 1S , γ))σ · ((δ, 1S , α))σ,
and since σ is an automorphism, we have that
(α, 1S , γ) · (δ, 1S , α) 6= 0
and hence γ = δ. Thus ϕ : λ → λ is a bijective map.
Therefore for s ∈ S \ {0S } we have
((γ, s, δ))σ = ((γ, 1S , α) · (α, s, α) · (α, 1S , δ))σ =
= ((γ, 1S , α))σ · ((α, s, α))σ · ((α, 1S , δ))σ =
= ((γ)ϕ, (γ)u, α ′) · (α ′ , (s)h, α ′ ) · (α ′ , ((δ)u)−1, (δ)ϕ)=
= ((γ)ϕ, (γ)u · (s)h · ((δ)u)−1 , (δ)ϕ).
Also, since 0 is zero of the semigroup Bλ0 (S) we conclude that (0)σ = 0.



Theorem 1 implies the following corollary:
Corollary 1. Let λ > 1 be cardinal and let Bλ (G) be the Brandt semigroup. Let h : G → G be an
automorphism and suppose that ϕ : λ → λ is a bijective map. Let u : λ → G be a map. Then the map
σ : Bλ (G) → Bλ (G) defined by the formulae
((α, s, β))σ = ((α)ϕ, (α)u · (s)h · ((β)u)−1, (β)ϕ)

and

(0)σ = 0,

is an automorphism of the Brandt semigroup Bλ (G). Moreover, every automorphism of Bλ (G) can be
constructed in this manner.
Also, we observe that Corollary 1 implies the following well known statement:
Corollary 2. Let λ > 1 be cardinal and ϕ : λ → λ a bijective map. Then the map σ : Bλ → Bλ defined
by the formulae
((α, β))σ = ((α)ϕ, (β)ϕ) and (0)σ = 0,
is an automorphism of the semigroup of λ×λ-matrix units Bλ . Moreover, every automorphism of Bλ
can be constructed in this manner.
The following example implies that the condition that semigroup S contains the identity is essential.
Example 1. Let λ be any cardinal > 2. Let S be the zero-semigroup of cardinality > 3 and 0S is
zero of S. It is easily to see that every bijective map σ : Bλ0 (S) → Bλ0 (S) such that (0)σ = 0 is an
automorphism of the Brandt λ0 -extension of S.

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OLEG GUTIK

Remark. By Theorem 1 we have that every automorphism σ : Bλ0 (S) → Bλ0 (S) of the Brandt λ0 extension of an arbitrary monoid S with zero identifies with the ordered triple [ϕ, h, u], where h : S → S
is an automorphism of S, ϕ : λ → λ is a bijective map and u : λ → H1 is a map, where H1 is the group
of units of S.
Lemma 1. Let λ > 1 be cardinal, S be a monoid with zero and let Bλ0 (S) be the Brandt λ0 -extension
of S. Then the composition of arbitrary automorphisms σ = [ϕ, h, u] and σ ′ = [ϕ′ , h′ , u′ ] of the Brandt
λ0 -extension of S defines in the following way:
[ϕ, h, u] · [ϕ′ , h′ , u′] = [ϕϕ′ , hh′ , ϕu′ · uh′ ].
Proof. By Theorem 1 for every (α, s, β) ∈ Bλ0 (S) we have that

(α, s, β)(σσ ′) = (α)ϕ, (α)u·(s)h·((β)u)−1, (β)ϕ σ ′ =



−1
= ((α)ϕ)ϕ′, ((α)ϕ)u′ · (α)u · (s)h · ((β)u)−1 h′ · (((β)ϕ)u′ ) , ((β)ϕ)ϕ′ =

and since h′ is an automorphism of the monoid S we get that this is equal to
= ((α)ϕ)ϕ′ , ((α)ϕ)u′ · ((α)u) h′ · ((s)h) h′ · (((β)u)h′ )

−1

· (((β)ϕ)u′)

−1
= (α)(ϕϕ′ ), (α)(ϕu′ · uh′ ) · ((s)h) h′ · (β) (ϕu′ · uh′ ) , (β)(ϕϕ′) .

−1


, ((β)ϕ)ϕ′ =

This completes the proof of the requested equality.



Theorem 2. Let λ > 1 be cardinal, S be a monoid with zero and let Bλ0 (S) be the Brandt λ0 -extension
of S. Then the group of automorphisms Aut(Bλ0 (S)) of Bλ0 (S) is isomorphic to a homomorphic image
of the group defines on the Cartesian product Sλ × Aut(S) × H1λ with the following binary operation:
(2)

[ϕ, h, u] · [ϕ′ , h′ , u′] = [ϕϕ′ , hh′ , ϕu′ · uh′ ],

where Sλ is the group of all bijections of the cardinal λ, Aut(S) is the group of all automorphisms of
the semigroup S and H1λ is the direct λ-power of the group of units H1 of the monoid S. Moreover, the
inverse element of [ϕ, h, u] in the group Aut(Bλ0 (S)) is defined by the formula:


[ϕ, h, u]−1 = ϕ−1 , h−1 , ϕ−1 u−1 h−1 .

Proof. First, we show that the binary operation defined by formula (2) is associative. Let [ϕ, h, u],
[ϕ′ , h′ , u′] and [ϕ′′ , h′′ , u′′ ] be arbitrary elements of the Cartesian product Sλ × Aut(S) × H1λ . Then we
have that

[ϕ, h, u] · [ϕ′ , h′ , u′ ] · [ϕ′′ , h′′ , u′′ ] = [ϕϕ′ , hh′ , ϕu′ · uh′ ] · [ϕ′′ , h′′ , u′′] =
= [ϕϕ′ ϕ′′ , hh′ h′′ , ϕϕ′ u′′ · (ϕu′ · uh′ )h′′ ] =
= [ϕϕ′ ϕ′′ , hh′ h′′ , ϕϕ′ u′′ · ϕu′ h′′ · uh′ h′′ ]
and
[ϕ, h, u] · ([ϕ′ , h′ , u′] · [ϕ′′ , h′′ , u′′ ]) = [ϕ, h, u] · [ϕ′ ϕ′′ , h′ h′′ , ϕ′ u′′ · u′ h′′ ] =
= [ϕϕ′ ϕ′′ , hh′ h′′ , ϕ(ϕ′ u′′ · u′ h′′ ) · uh′ h′′ ] =
= [ϕϕ′ ϕ′′ , hh′ h′′ , ϕϕ′ u′′ · ϕu′ h′′ · uh′ h′′ ],
and hence so defined operation is associative.
Theorem 1 implies that formula (1) determines a map F from the Cartesian product Sλ ×Aut(S)×H1λ
onto the group of automorphisms Aut(Bλ0 (S)) of the Brandt λ0 -extension Bλ0 (S) of the monoid S,
and hence the associativity of binary operation (2) implies that the map F is a homomorphism from
Sλ × Aut(S) × H1λ onto the group Aut(Bλ0 (S)).
Next we show that [1Sλ , 1Aut(S) , 1H1λ ] is a unit element with the respect to the binary operation (2),
where 1Sλ , 1Aut(S) and 1H1λ are units of the groups Sλ , Aut(S) and H1λ , respectively. Then we have

ON THE GROUP OF AUTOMORPHISMS OF THE BRANDT λ0 -EXTENSION OF A MONOID WITH ZERO

that

5


 

[ϕ, h, u] · 1Sλ , 1Aut(S) , 1H1λ = ϕ1Sλ , h1Aut(S) , ϕ1H1λ · u1Aut(S) =


= ϕ, h, ϕ1H1λ · u1Aut(S) =


= ϕ, h, 1H1λ · u =
= [ϕ, h, u]

and





1Sλ , 1Aut(S) , 1H1λ · [ϕ, h, u] = 1Sλ ϕ, 1Aut(S) h, 1Sλ u · 1H1λ h = [ϕ, h, u],

because every automorphism h ∈ Aut(S) acts on the group H1λ by the natural way as a restriction of
global automorphism of the semigroup S on every factor, and hence we get that 1H1λ h = 1H1λ .
Also, similar arguments imply that


[ϕ, h, u] · [ϕ, h, u]−1 = [ϕ, h, u] · ϕ−1 , h−1 , ϕ−1 u−1 h−1 =


= ϕϕ−1 , hh−1 , (ϕϕ−1 )u−1h−1 · uh−1 =


= ϕϕ−1 , hh−1 , (1Sλ )u−1 h−1 · uh−1 =


= ϕϕ−1 , hh−1 , u−1h−1 · uh−1 =


= 1Sλ , 1Aut(S) , 1H1λ
and



[ϕ, h, u]−1 · [ϕ, h, u] = ϕ−1 , h−1 , ϕ−1 u−1 h−1 · [ϕ, h, u]=


= ϕ−1 ϕ, h−1 h, ϕ−1 u · ϕ−1 u−1 h−1 h =


= ϕ−1 ϕ, h−1 h, ϕ−1 u · ϕ−1 u−1 =


= 1Sλ , 1Aut(S) , 1H1λ .

This implies that the elements [ϕ−1 , h−1 , ϕ−1 u−1 h−1 ] and [ϕ, h, u] are invertible in Sλ × Aut(S) × H1λ,
and hence the set Sλ × Aut(S) × H1λ with the binary operation (2) is a group.
Let Id : Bλ0 (S) → Bλ0 (S) be the identity automorphism of the semigroup Bλ0 (S). Then by Theorem 1
there exist some automorphism h : S → S, a bijective map ϕ : λ → λ and a map u : λ → H1 into the
group H1 of units of S such that
(α, s, β) = (α, s, β)Id = ((α)ϕ, (α)u · (s)h · ((β)u)−1, (β)ϕ),
for all α, β ∈ λ and s ∈ S ∗ . Since Id : Bλ0 (S) → Bλ0 (S) is the identity automorphism we conclude that
(α)ϕ = α for every α ∈ λ. Also, for every s ∈ S ∗ we get that s = (α)u · (s)h · ((β)u)−1 for all α, β ∈ λ,
and hence we obtain that
1S = (α)u · (1S )h · ((β)u)−1 = (α)u · ((β)u)−1
for all α, β ∈ λ. This implies that (α)u = (β)u = u
e is a fixed element of the group H1 for all α, β ∈ λ.
We define
n
o
λ
−1
ker N = [ϕ, h, u
e] ∈ Sλ × Aut(S) × H1 : ϕ : λ → λ is an idemtity map, u
e(s)he
u
= s for any s ∈ S .

It is obvious that the equality u
e(s)he
u−1 = s implies that (s)h = u
e−1 se
u for all s ∈ S. The previous
arguments implies that [ϕ, h, u
e] ∈ ker N if and only if [ϕ, h, u
e]F is the unit of the group Aut(Bλ0 (S)),
and hence ker N is a normal subgroup of Sλ × Aut(S) × H1λ . This implies that the quotient group
(Sλ × Aut(S) × H1λ )/ ker N is isomorphic to the group Aut(Bλ0 (S)).


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OLEG GUTIK

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Faculty of Mathematics, National University of Lviv, Universytetska 1, Lviv, 79000, Ukraine
E-mail address: o gutik@franko.lviv.ua, ovgutik@yahoo.com

