Automorphisms of the endomorphism semigroup of a free abelian diband
We determine all isomorphisms between the endomorphism semigroups of free abelian dibands and prove that all automorphisms of the endomorphism semigroup of a free abelian diband are inner.
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irk-123456789-1883672023-02-26T01:26:55Z Automorphisms of the endomorphism semigroup of a free abelian diband Zhuchok, Y.V. We determine all isomorphisms between the endomorphism semigroups of free abelian dibands and prove that all automorphisms of the endomorphism semigroup of a free abelian diband are inner. 2018 Article Automorphisms of the endomorphism semigroup of a free abelian diband / Y.V. Zhuchok // Algebra and Discrete Mathematics. — 2018. — Vol. 25, № 2. — С. 322–332. — Бібліогр.: 19 назв. — англ. 1726-3255 2010 MSC: 08B20, 17A30, 08A30, 08A35. http://dspace.nbuv.gov.ua/handle/123456789/188367 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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We determine all isomorphisms between the endomorphism semigroups of free abelian dibands and prove that all automorphisms of the endomorphism semigroup of a free abelian diband are inner. |
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Zhuchok, Y.V. Automorphisms of the endomorphism semigroup of a free abelian diband Algebra and Discrete Mathematics |
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Zhuchok, Y.V. |
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Zhuchok, Y.V. |
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Automorphisms of the endomorphism semigroup of a free abelian diband |
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Automorphisms of the endomorphism semigroup of a free abelian diband |
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Automorphisms of the endomorphism semigroup of a free abelian diband |
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Automorphisms of the endomorphism semigroup of a free abelian diband |
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Automorphisms of the endomorphism semigroup of a free abelian diband |
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automorphisms of the endomorphism semigroup of a free abelian diband |
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Інститут прикладної математики і механіки НАН України |
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Automorphisms of the endomorphism semigroup of a free abelian diband / Y.V. Zhuchok // Algebra and Discrete Mathematics. — 2018. — Vol. 25, № 2. — С. 322–332. — Бібліогр.: 19 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT zhuchokyv automorphismsoftheendomorphismsemigroupofafreeabeliandiband |
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2025-07-16T10:23:43Z |
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“adm-n2” — 2018/7/24 — 22:32 — page 322 — #160
Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 25 (2018). Number 2, pp. 322–332
c© Journal “Algebra and Discrete Mathematics”
Automorphisms of the endomorphism semigroup
of a free abelian diband∗
Yurii V. Zhuchok
Communicated by V. A. Artamonov
Abstract. We determine all isomorphisms between the
endomorphism semigroups of free abelian dibands and prove that
all automorphisms of the endomorphism semigroup of a free abelian
diband are inner.
1. Introduction
The problem of the description of automorphisms of the endomorphism
semigroup for free algebras in a certain variety was raised by B. I. Plotkin
in his papers on universal algebraic geometry (see, e.g., [1], [2]). At the
present time there are many papers devoted to describing automorphisms
of endomorphism semigroups of free finitely generated universal algebras
of different varieties: groups, (inverse) semigroups and monoids [3–5],
modules and semimodules [6], Lie algebras and associative algebras [7],
[8], dimonoids and g-dimonoids [9], [10] and some others [11], [12].
As it is well-known, the notion of a dimonoid, and constructions of a
free associative dialgebra and a free dimonoid were defined by J.-L. Loday
[13]. Later on, free dimonoids and free commutative dimonoids were
investigated in detail in [14] and [15], respectively. Free abelian dimonoids
(this class does not coincide with the class of commutative dimonoids) were
∗The publication is based on the research provided by the grant support of the
State Fund For Fundamental Research (project F83/43909).
2010 MSC: 08B20, 17A30, 08A30, 08A35.
Key words and phrases: dimonoid, free abelian diband, endomorphism semi-
group, automorphism.
“adm-n2” — 2018/7/24 — 22:32 — page 323 — #161
Yu. V. Zhuchok 323
described in [16]. The structure of free normal dibands, free (lr, rr)-dibands
and free (ln, rn)-dibands was considered in [17], [18]. For the variety of
abelian idempotent dimonoids it was shown [19] that it coincides with
the variety of (ln, rn)-dibands. In this paper, we consider the mentioned
above problem of the description of automorphisms of endomorphism
semigroups for free algebras in the variety of abelian dibands.
The paper is organized as follows. In Sections 2 and 3, we give necessary
definitions and prove auxiliary statements, respectively. In Section 4, we
describe stable automorphisms of the endomorphism semigroup of a free
abelian diband. In Section 5, we establish that all automorphisms of the
endomorphism semigroup of a free abelian diband are inner and show that
the automorphism group of such endomorphism semigroup is isomorphic
to the symmetric group.
2. Preliminaries
Recall that a nonempty set D with two binary associative operations
⊣ and ⊢ is called a dimonoid if for all x, y, z ∈ D,
(D1) (x ⊣ y) ⊣ z = x ⊣ (y ⊢ z),
(D2) (x ⊢ y) ⊣ z = x ⊢ (y ⊣ z),
(D3) (x ⊣ y) ⊢ z = x ⊢ (y ⊢ z).
A dimonoid (D,⊣,⊢) is called abelian [16] if for all x, y ∈ D,
x ⊣ y = y ⊢ x.
A band is a semigroup whose elements are idempotents. If for a
dimonoid (D,⊣,⊢) the semigroups (D,⊣) and (D,⊢) are bands, then this
dimonoid is called idempotent (or simply a diband).
For example, any non-singleton left zero and right zero dimonoid
(D,⊣,⊢), that is, (D,⊣) is a left zero semigroup and (D,⊢) is a right zero
semigroup, is an abelian diband. More examples of abelian dibands can
be found, e.g., in [18], [19].
An idempotent semigroup S is called a left regular band if aba = ab
for all a, b ∈ S. If instead of the last identity, aba = ba holds, then S is a
right regular band.
A dimonoid (D,⊣,⊢) is called a (lr, rr)-diband [18] if (D,⊣) is a left
regular band and (D,⊢) is a right regular band.
Let X be a nonempty set and FS(X) the free semilattice of all
nonempty finite subsets of X with respect to the operation of the set
“adm-n2” — 2018/7/24 — 22:32 — page 324 — #162
324 Automorphisms of the endomorphism semigroup
theoretical union. Define two binary operations ⊣ and ⊢ on the set
Blz,rz(X) = {(a,A) ∈ X × FS(X)|a ∈ A} as follows:
(x,A) ⊣ (y,B) = (x,A ∪B)
and
(x,A) ⊢ (y,B) = (y,A ∪B).
Proposition 1. [18] The algebraic system (Blz,rz(X),⊣,⊢) is the free
(lr, rr)-diband.
A semigroup S is called left commutative (respectively, right commu-
tative) if it satisfies the identity xya = yxa (respectively, axy = ayx).
An idempotent semigroup S is called a left (respectively, right) normal
band if it is right (respectively, left) commutative.
A dimonoid (D,⊣,⊢) is called a (ln, rn)-diband [18] if (D,⊣) is a left
normal band and (D,⊢) is a right normal band.
Theorem 1. [19] A dimonoid (D,⊣,⊢) is abelian idempotent if and only
if (D,⊣,⊢) is a (ln, rn)-diband.
From Theorem 1 we immediately obtain that the variety of abelian
dibands coincides with the variety of (ln, rn)-dibands. It is well-known also
[18] that the variety of (ln, rn)-dibands and the variety of (lr, rr)-dibands
coincide. So, (Blz,rz(X),⊣,⊢) is the free abelian diband.
Further, we will denote (Blz,rz(X),⊣,⊢) simply by Blz,rz(X). We note
that the cardinality of X is the rank of the diband Blz,rz(X) and this
diband is uniquely determined up to an isomorphism by |X|.
Obviously, operations of the free abelian diband Blz,rz(X) coincide if
and only if |X| = 1. In this case Blz,rz(X) is singleton.
For every (u1, U) ∈ Blz,rz(X), where U = {u1, u2, . . . , un}, we have
(u1, U) = (u1, {u1}) ⊣ (u2, {u2}) ⊣ · · · ⊣ (un, {un}).
This representation we call a canonical form of elements of the diband
Blz,rz(X). It is clear that such representation is unique for n ∈ {1, 2} and
it is unique up to an order of (ui, {ui}), 2 6 i 6 n, for n > 3. Moreover,
〈{(x, {x})|x ∈ X}〉 = Blz,rz(X),
that is, X ′ = {(x, {x})|x ∈ X} is a generating set of Blz,rz(X).
“adm-n2” — 2018/7/24 — 22:32 — page 325 — #163
Yu. V. Zhuchok 325
3. Auxiliary statements
We begin this section with the following lemma.
Lemma 1. Let Blz,rz(X) and Blz,rz(Y ) be free abelian dibands defined on
X and Y , respectively. Every bijection ϕ : X → Y induces an isomorphism
εϕ : Blz,rz(X) → Blz,rz(Y ) such that
(u, U)εϕ = (uϕ,Uϕ)
for all (u, U) ∈ Blz,rz(X).
Proof. The proof is obvious.
Let Blz,rz(X) be the free abelian diband. Every endomorphism ξ of
Blz,rz(X) is uniquely determined by a mapping ξ′ : X ′ → Blz,rz(X).
Indeed, to determine ξ, it suffices to set
(u1, U)ξ = (u1, {u1})ξ
′ ⊣ (u2, {u2})ξ
′ ⊣ · · · ⊣ (un, {un})ξ
′
for all (u1, U) ∈ Blz,rz(X), U = {u1, u2, . . . , un}.
In particular, the endomorphism ξ of Blz,rz(X) is an automorphism if
and only if a restriction ξ on X ′ belongs to the symmetric group S(X ′).
Therefore, the group Aut(Blz,rz(X)) is isomorphic to S(X).
Let F (X) be a free algebra in a variety V with a generating set X
and u ∈ F (X). An endomorphism θu ∈ End(F (X)) is called constant if
xθu = u for all x ∈ X.
We denote by N the set of all positive integers. For every nonempty
subset G of Blz,rz(X) and i ∈ N we put
ΘG = {θ(a,A)|(a,A) ∈ G}
and
UX
i = {(a,A) ∈ Blz,rz(X) : |A| = i}.
Lemma 2. Let Ψ : End(Blz,rz(X)) → End(Blz,rz(Y )) be an isomorphism.
For every i ∈ N there exists j ∈ N such that ΘUX
i
Ψ = ΘUY
j
.
Proof. For |X| = 1 the statement is trivial. Let |X| > 2 and θ(a,A) ∈
End(Blz,rz(X)), where (a,A) ∈ UX
i . It is not hard to see that all elements
θ(u,U), (u, U) ∈ Blz,rz(X), and only they have the property that ξθ(u,U) =
θ(u,U) for all ξ ∈ End(Blz,rz(X)). Therefore, θBlz,rz(X)Ψ = θBlz,rz(Y ), so
that θ(a,A)Ψ = θ(b,B) for some (b, B) ∈ Blz,rz(Y ).
“adm-n2” — 2018/7/24 — 22:32 — page 326 — #164
326 Automorphisms of the endomorphism semigroup
Take arbitrary (a′, A′) ∈ UX
i , (a
′, A′) 6= (a,A), and an automorphism
ϕ : Blz,rz(X) → Blz,rz(X) such that (a,A)ϕ = (a′, A′). It is clear that
there exists a such automorphism, since |A| = |A′|. Then
θ(a′,A′)Ψ = θ(a,A)ϕΨ = (θ(a,A)ϕ)Ψ = (θ(a,A)Ψ)(ϕΨ)
= θ(b,B)(ϕΨ) = θ(b,B)(ϕΨ).
Since ϕΨ is an automorphism of Blz,rz(Y ), then putting (b, B)(ϕΨ) =
(b′, B′) we obtain |B| = |B′|. Hence (b′, B′) ∈ UY
j , where j = |B|. It means
that ΘUX
i
Ψ ⊆ ΘUY
j
. Since Ψ−1 is an isomorphism and obviously |Y | > 2,
we can analogously obtain the inclusion ΘUY
j
Ψ−1 ⊆ ΘUX
i
, consequently
ΘUX
i
Ψ = ΘUY
j
.
The following statement shows that any isomorphism of the endomor-
phism monoids of free abelian dibands induces a uniquely determined
bijection between the generating sets of these dibands.
Lemma 3. Every isomorphism Ψ : End(Blz,rz(X)) → End(Blz,rz(Y ))
implies that ΘUX
1
Ψ = ΘUY
1
.
Proof. Let Ψ be an arbitrary isomorphism of monoids End(Blz,rz(X)) and
End(Blz,rz(Y )), and |X| > 2 (the case |X| = 1 is trivial). By Lemma 2,
ΘUX
1
Ψ = ΘUY
α
for some α ∈ N.
Assume that α 6= 1, then there exist some (a′, C) and (b′, C) of UY
α
with distinct a′ and b′. Using Lemma 2, there exists β ∈ N such that
ΘUX
β
Ψ = ΘUY
1
. It is easy to see that β 6= 1. Let (s, S) and (t, T ) be two
distinct pairs from UX
β . Then
θ(a,{a})Ψ = θ(a′,C), θ(b,{b})Ψ = θ(b′,C),
θ(s,S)Ψ = θ(x,{x}) and θ(t,T )Ψ = θ(y,{y})
for some a, b ∈ X and x, y ∈ Y . Evidently, a 6= b and x 6= y.
Now we take an arbitrary mapping η′ : UX
1 → UX
β such that
(a, {a})η′ = (s, S) and (b, {b})η′ = (t, T ).
This mapping is uniquely extended (see reasoning after Lemma 1) to an
endomorphism of Blz,rz(X) that we will denote by η. Thus,
θ(x,{x}) = θ(s,S)Ψ = (θ(a,{a})η)Ψ = (θ(a,{a})Ψ)(ηΨ)
= θ(a′,C)(ηΨ) = θ(a′,C)(ηΨ).
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Yu. V. Zhuchok 327
Analogously, we can obtain θ(y,{y}) = θ(b′,C)(ηΨ). From here it follows
that (a′, C)(ηΨ) = (x, {x}) and (b′, C)(ηΨ) = (y, {y}).
Further, we have
((a′, C) ⊣ (b′, C))(ηΨ) = (a′, C)(ηΨ) = (x, {x})
and
(a′, C)(ηΨ) ⊣ (b′, C)(ηΨ) = (x, {x}) ⊣ (y, {y}) = (x, {x, y}),
that contradicts the fact that ηΨ is an endomorphism of Blz,rz(Y ). Hence
α = 1, that is, ΘUX
1
Ψ = ΘUY
1
.
Let Ψ : End(Blz,rz(X)) → End(Blz,rz(Y )) be an arbitrary isomor-
phism,X ′ and Y ′ generating sets ofBlz,rz(X) and ofBlz,rz(Y ), respectively.
By Lemma 3, for every x = (x, {x}) ∈ X ′ there exists y = (y, {y}) ∈ Y ′
such that θxΨ = θy. Define a bijection ψ : X → Y putting xψ = y if
θxΨ = θy. In this case we say that ψ is induced by the isomorphism Ψ.
4. Stable automorphisms of End(Blz,rz(X))
Let F (X) be a free algebra in a variety V over a set X. An automor-
phism Ψ of the endomorphism monoid End(F (X)) is called stable if Ψ
induces the identity permutation of X, that is, θxΨ = θx for all x ∈ X.
An endomorphism θ of the free algebra F (X) is called linear if xθ ∈ X
for all x ∈ X.
Throughout this section, X denotes an arbitrary set with |X| > 2.
Lemma 4. Let Ψ be a stable automorphism of End(Blz,rz(X)), (a,A),
(b, B) ∈ Blz,rz(X) and ξ ∈ End(Blz,rz(X)). Then
(i) ξΨ = ξ, if ξ is linear;
(ii) A = B, if θ(a,A)Ψ = θ(b,B).
Proof. (i) Using the linearity of ξ and the stability of Ψ, we obtain
θ(x,{x})(ξΨ) = θ(x,{x})(ξΨ) = (θ(x,{x})Ψ)(ξΨ)
= (θ(x,{x})ξ)Ψ = θ(x,{x})ξΨ = θ(x,{x})ξ, x ∈ X.
From here (x, {x})(ξΨ) = (x, {x})ξ for all x ∈ X.
(ii) Suppose that A \ B 6= ∅ and take z ∈ A \ B, x ∈ X,x 6= z and
ξ ∈ End(Blz,rz(X)) such that (z, {z})ξ = (x, {x}) and (y, {y})ξ = (y, {y})
“adm-n2” — 2018/7/24 — 22:32 — page 328 — #166
328 Automorphisms of the endomorphism semigroup
for all y ∈ X, y 6= z. So ξ is linear, in addition, (b, B)ξ = (b, B). By (i) of
this lemma, ξΨ = ξ and
θ(a,A)Ψ = θ(b,B) = θ(b,B)ξ = θ(b,B)ξ = (θ(a,A)Ψ)(ξΨ)
= (θ(a,A)ξ)Ψ = θ(a,A)ξΨ.
Using the injectivity of Ψ, we have θ(a,A) = θ(a,A)ξ. From here (a,A) =
(a,A)ξ which contradicts to the definition of ξ, so A \ B = ∅. In the
similar way we can prove that B \A = ∅. Therefore, A = B.
Corollary 1. Let Ψ be a stable automorphism of End(Blz,rz(X)) and
a, b ∈ X, a 6= b. Then
θ(a,{a,b})Ψ = θ(a,{a,b}) or θ(a,{a,b})Ψ = θ(b,{a,b}).
Proof. From Lemma 2 it follows that θ(a,{a,b})Ψ = θ(u,U) for some (u, U)
of Blz,rz(X). By (ii) of Lemma 4, U = {a, b}. Thus, (u, U) = (a, {a, b})
or (u, U) = (b, {a, b}).
Corollary 2. Let Ψ be a stable automorphism of End(Blz,rz(X)). Then
for all x ∈ X, θ(x,{x})ξΨ = θ(x,{x})(ξΨ).
Proof. It follows from the proof of the condition (i) of Lemma 4.
We denote by Φ0 the identity automorphism of End(Blz,rz(X)).
Lemma 5. Let Ψ be a stable automorphism of End(Blz,rz(X)) and
a, b ∈ X, a 6= b. If θ(a,{a,b})Ψ = θ(a,{a,b}), then Ψ = Φ0.
Proof. Show firstly that θ(u,U)Ψ = θ(u,U) for all (u, U) ∈ Blz,rz(X) by
induction on |U |. Since Ψ is stable, θ(u,{u})Ψ = θ(u,{u}) for all u ∈ X.
Assume that θ(v,V )Ψ = θ(v,V ) for all (v, V ) ∈ Blz,rz(X) with |V | < n, and
let (u, U) ∈ Blz,rz(X), where U = {u1, u2, . . . , un}, u = u1 and n > 2. We
put v1 = (u, U \ {un}), v2 = (un, {un}) and take the endomorphism f of
Blz,rz(X) such that (a, {a})f = v1, (b, {b})f = v2 and (y, {y})f = (y, {y})
for all y ∈ X \ {a, b}. Then for all x ∈ X,
(x, {x})(θ(a,{a,b})f) = ((a, {a}) ⊣ (b, {b}))f = (a, {a})f ⊣ (b, {b})f
= (u, U) = (x, {x})θ(u,U).
Therefore, θ(a,{a,b})f = θ(u,U) for all (u, U) ∈ Blz,rz(X) with |U | > 2.
Using the induction hypothesis and Corollary 2, we have
θ(a,{a})(fΨ) = θ(a,{a})fΨ = θv1Ψ = θv1 = θ(a,{a})f ,
θ(b,{b})(fΨ) = θ(b,{b})fΨ = θv2Ψ = θv2 = θ(b,{b})f
“adm-n2” — 2018/7/24 — 22:32 — page 329 — #167
Yu. V. Zhuchok 329
and
θ(y,{y})(fΨ) = θ(y,{y})fΨ = θ(y,{y})Ψ
= θ(y,{y}) = θ(y,{y})f , y ∈ X \ {a, b}.
Thus, fΨ = f and for all (u, U) ∈ Blz,rz(X) with |U | > 2,
θ(u,U)Ψ = (θ(a,{a,b})f)Ψ = (θ(a,{a,b})Ψ)(fΨ) = θ(a,{a,b})f = θ(u,U).
So, θ(u,U)Ψ = θ(u,U) for all (u, U) ∈ Blz,rz(X). Moreover, for all x ∈ X
and ϕ ∈ End(Blz,rz(X)) we obtain
θ(x,{x})(ϕΨ) = θ(x,{x})ϕΨ = θ(x,{x})ϕ.
This means that ϕΨ = ϕ for all ϕ ∈ End(Blz,rz(X)) and so Ψ = Φ0.
Lemma 6. Let a, b ∈ X be distinct. There is no a stable automorphism
Ψ of End(Blz,rz(X)) such that θ(a,{a,b})Ψ = θ(b,{a,b}).
Proof. Assume that there exists a stable automorphism Ψ of the monoid
End(Blz,rz(X)) such that θ(a,{a,b})Ψ = θ(b,{a,b}). Using the condition (ii)
of Lemma 4, θ(b,{a,b})Ψ = θ(a,{a,b}).
Let g ∈ End(Blz,rz(X)) such that (a, {a})g = (a, {a}), (b, {b})g =
(a, {a, b}) and (x, {x})g = (x, {x}) for all x ∈ X \ {a, b}. It is easy to see
that θ(a,{a})g = θ(a,{a}) and θ(b,{b})g = θ(a,{a,b}). Then
θ(a,{a}) = θ(a,{a})Ψ = (θ(a,{a})g)Ψ = θ(a,{a})(gΨ) = θ(a,{a})(gΨ)
and
θ(b,{a,b}) = θ(a,{a,b})Ψ = (θ(b,{b})g)Ψ = θ(b,{b})(gΨ) = θ(b,{b})(gΨ).
From here (a, {a})(gΨ) = (a, {a}) and (b, {b})(gΨ) = (b, {a, b}).
Using the equality θ(b,{a,b})g = θ(a,{a,b}), on the one hand we obtain
θ(b,{a,b}) = θ(a,{a,b})Ψ = (θ(b,{a,b})g)Ψ = θ(a,{a,b})(gΨ) = θ(a,{a,b})(gΨ),
and so (a, {a, b})(gΨ) = (b, {a, b}). On the other hand,
(a, {a, b})(gΨ) = ((a, {a}) ⊣ (b, {b}))(gΨ)
= (a, {a}) ⊣ (b, {a, b}) = (a, {a, b})
that contradicts the previous expression for (a, {a, b})(gΨ).
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330 Automorphisms of the endomorphism semigroup
5. The automorphism group of End(Blz,rz(X))
Firstly, we describe all isomorphisms between the endomorphism semi-
groups of free abelian dibands.
Recall that εϕ denotes the isomorphism Blz,rz(X) → Blz,rz(Y ) which
is induced by the bijection ϕ : X → Y (see Lemma 1).
Theorem 2. Every isomorphism Φ : End(Blz,rz(X)) → End(Blz,rz(Y ))
is induced by the isomorphism εf of Blz,rz(X) to Blz,rz(Y ) for a uniquely
determined bijection f : X → Y .
Proof. The case |X| = 1 is trivial, so that we suppose further |X| > 1.
Let Φ : End(Blz,rz(X)) → End(Blz,rz(Y )) be an arbitrary isomorphism.
By Lemma 3, Φ induces a uniquely determined bijection f : X → Y such
that θ(x,{x})Φ = θ(xf,{xf}) for every x ∈ X. By Lemma 1, f induces the
isomorphism εf : Blz,rz(X) → Blz,rz(Y ). It is not hard to check that the
mapping
Ef : End(Blz,rz(X)) → End(Blz,rz(Y )) : η 7→ ε−1
f ηεf
is an isomorphism. From here it follows that Ω = ΦE−1
f is an automor-
phism of End(Blz,rz(X)). Since for all x ∈ X we have
θ(x,{x})Ω = (θ(x,{x})Φ)E
−1
f = θ(xf,{xf})E
−1
f = θ(xff−1,{xff−1}) = θ(x,{x}),
then Ω is stable.
By Corollary 1, Lemma 5 and Lemma 6, Ω is an identity automor-
phism Φ0. From ΦE−1
f = Φ0 we obtain Φ = Ef , i.e., Φ is an isomorphism
induced by εf .
Let F (X) be a free algebra in a variety V over a set X. An automor-
phism Φ of End(F (X)) is called inner if there exists an automorphism α
of F (X) such that βΦ = α−1βα for all β ∈ End(F (X)).
Finally, we characterize the automorphism group of the endomorphism
monoid of a free abelian diband.
Theorem 3. All automorphisms of End(Blz,rz(X)) are inner. In addi-
tion, the automorphism group Aut(End(Blz,rz(X))) is isomorphic to the
symmetric group S(X).
Proof. Let X = Y in Theorem 2, then it will be the first part of the given
theorem. By Theorem 2, every automorphism Φ of End(Blz,rz(X)) has
a form Φ = Ef , where ηΦ = ε−1
f ηεf for all η ∈ End(Blz,rz(X)) and a
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Yu. V. Zhuchok 331
suitable bijection f : X → X. As follows from Lemma 1 (see Section 3),
εf ∈ Aut(Blz,rz(X)) for all f ∈ S(X), therefore all automorphisms of
End(Blz,rz(X)) are inner.
Define a mapping ζ : Aut(End(Blz,rz(X))) → S(X) as follows:
Efζ = f for all Ef ∈ Aut(End(Blz,rz(X))). An immediate check shows
that ζ is an isomorphism.
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Contact information
Yurii V. Zhuchok Luhansk Taras Shevchenko National University,
Gogol square 1, Starobilsk, Ukraine, 92703
E-Mail(s): zhuchok.yu@gmail.com
Received by the editors: 19.06.2018
and in final form 30.06.2018.
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