Groups whose lattices of normal subgroups are factorial
We prove that the groups G for which the lattice of normal subgroups N(G) is factorial are exactly the UND-groups, that is the groups for which every normal subgroup have a unique normal complement, with finite length.
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irk-123456789-1885672023-03-07T01:26:51Z Groups whose lattices of normal subgroups are factorial Rajhi, A. We prove that the groups G for which the lattice of normal subgroups N(G) is factorial are exactly the UND-groups, that is the groups for which every normal subgroup have a unique normal complement, with finite length. 2020 Article Groups whose lattices of normal subgroups are factorial / A. Rajhi // Algebra and Discrete Mathematics. — 2020. — Vol. 30, № 2. — С. 239–253. — Бібліогр.: 7 назв. — англ. 1726-3255 DOI:10.12958/adm1264 2010 MSC: 20E99,06B99 http://dspace.nbuv.gov.ua/handle/123456789/188567 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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We prove that the groups G for which the lattice of normal subgroups N(G) is factorial are exactly the UND-groups, that is the groups for which every normal subgroup have a unique normal complement, with finite length. |
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Rajhi, A. |
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Rajhi, A. Groups whose lattices of normal subgroups are factorial Algebra and Discrete Mathematics |
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Rajhi, A. |
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Rajhi, A. |
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Groups whose lattices of normal subgroups are factorial |
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Groups whose lattices of normal subgroups are factorial |
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Groups whose lattices of normal subgroups are factorial |
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Groups whose lattices of normal subgroups are factorial |
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Groups whose lattices of normal subgroups are factorial |
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groups whose lattices of normal subgroups are factorial |
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Інститут прикладної математики і механіки НАН України |
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2020 |
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Groups whose lattices of normal subgroups are factorial / A. Rajhi // Algebra and Discrete Mathematics. — 2020. — Vol. 30, № 2. — С. 239–253. — Бібліогр.: 7 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT rajhia groupswhoselatticesofnormalsubgroupsarefactorial |
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2025-07-16T10:40:30Z |
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© Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 30 (2020). Number 2, pp. 239–253
DOI:10.12958/adm1264
Groups whose lattices of normal subgroups
are factorial
A. Rajhi
Communicated by L. A. Kurdachenko
Abstract. We prove that the groups G for which the lattice
of normal subgroups N (G) is factorial are exactly the UND-groups,
that is the groups for which every normal subgroup have a unique
normal complement, with finite length.
Introduction
The link between the structure of a group and the structure of some
lattices of it’s subgroups constitutes an important domain of research in
group theory. The topic has enjoyed a rapid development starting with
the first half of the 20th century. Many classes of groups determined
by different properties of partially ordered subsets of their subgroups
(especially lattices of subgroups or more particulary lattices of normal
subgroups) have been identified. We refer to Schmidt’s book [7] for more
information about this theory. In this paper, we consider the lattice N (G)
of normal subgroups of an arbitrary group G as an arithmetic object. It
is not clear for the moment what we mean by this. In order to clarify our
idea, let’s take an example. Let n > 2 be an integer and D(n) the partially
ordered set (ordered by divisibility) of divisor’s of n. This partially ordered
set (for short poset) is actually a lattice in which the meet a ∧ b and the
join a ∨ b of two elements a, b ∈ D(n) are respectively the latest common
2010 MSC: 20E99,06B99.
Key words and phrases: lattice of normal subgroups, semilattices, idempotent
monoids, partial monoids.
https://doi.org/10.12958/adm1264
240 Factorial lattice of normal subgroups
multiple and the greatest common divisor of a and b. Note that D(n) is a
bounded lattice (in fact it is a finite lattice) in which the initial element is
1 and the final element is n. Moreover, the atoms of D(n) (ie. the minimal
elements of D(n)\{1}) are exactly the prime divisor’s of n and the coatoms
of D(n) (ie. the maximal elements of D(n)\{n}) are exactly the divisor’s
a of n such that the number n/a is a prime number. We will say that
the lattice D(n) is factorial (respectively cofactorial) if every element
a ∈ D(n)\{1} (respectively a ∈ D(n)\{n}) can be expressed uniquely (up
to permutation) as join (respectively meet) of a finite number of pairwise
distinct atoms (respectively coatoms). It is not hard to see that D(n) is
factorial (respectively cofactorial) if and only if n is a squarefree integer.
Now we can introduce this notion for different algebraic structures. For
example, let V be a finite dimensional vector space and L(V ) the lattice
of it’s subspaces Recall that the meet and the join of two subspaces F1
and F2 of V are given by
F1 ∧ F2 = F1 ∩ F2 and F1 ∨ F2 = F1 + F2.
Note that the atoms (respectively coatoms) of L(V ) are exactly the one
dimensional (respectively one codimension) subspaces of V . It is an easy
exercise in linear algebra to check that every subspace F ∈ L(V )\{0}
(respectively F ∈ L(V )\{V }) can be expressed as join (respectively meet)
of a finite number of pairwise distinct atoms (respectively coatoms) but
that such decomposition can not be unique. The lattice L(V ) of subspaces
of V is then atomic (respectively coatomic) but it is not factorial (re-
spectively cofactorial). Let G be an arbitrary group and N (G) it’s lattice
of normal subgroups. In this paper, we show that the lattice N (G) is
atomic (respectively coatomic) if and only if G is an ND-group, that is a
group in which every normal subgroup have a normal complement, with
finite length on normal subgroups and that the lattice N (G) is factorial
(respectively cofactorial) if and only if G is an UND-group, that is a group
in which every normal subgroup have a unique normal complement, with
finite length on normal subgroups.
1. Preliminaries
Recall that a join-semilattice is a partially ordered set (L,6) in which
the joint x∨y (the least upper bound) of any two elements exists. Similarly,
a meet-semilattice is a partially ordered set L in which the meet x∧y (the
greatest lower bound) of any two elements exists. If L is a join-semilattice
(respectively meet-semilattice), the zero element (respectively unit element)
A. Rajhi 241
of L, if it exists, is the least element (respectively the greatest element) of
L, we denote it by 0L (respectively 1L). If there is no risk of confusion, we
denote by 0 (respectively by 1) the zero element (respectively unit element)
of a join-semilattice (respectively meet-semilattice) L. For simplicity, a join-
semilattice (respectively meet-semilattice) with zero element (respectively
unit element) is called a (∨, 0)-semilattice (respectively (∧, 1)-semilattice).
Note that every commutative monoid (M,+, 0) in which every element is
idempotent, which we call a commutative idempotent monoid, endowed
with its algebraic partial ordering defined by x 6 y if and only if x+y = y,
is a (∨, 0)-semilattice and it is a (∧, 1)-semilattice if we endow M by
the reverse order. Conversely, every (∨, 0)-semilattice (respectively (∧, 1)-
semilattice) L can be viewed as a commutative idempotent monoid in
which the addition is given by x+ y = x ∨ y (respectively x+ y = x ∧ y),
for all x, y ∈ L. In this paper, by abuse of notation we don’t make
any difference between (∨, 0)-semilattices (respectively (∧, 1)-semilattices)
and commutative idempotent monoids. Let L be a (∨, 0)-semilattice. An
element p ∈ L\{0} is called an atom of L, if for all x, y ∈ L with x 6= y,
p = x ∨ y implies x = 0 or y = 0. Dually, if L is a (∧, 1)-semilattice, an
element q ∈ L\{1} is called an coatom of L, if for all x, y ∈ L with x 6= y,
q = x ∧ y implies x = 1 or y = 1.
Definition 1. Let L be a (∨, 0)-semilattice.
a) We say that L is atomic if every nonzero element of L is join of
a finite number of atoms, that is for every x ∈ L\{0} there exist
atoms p1, . . . , pn of L such that x = p1 ∨ · · · ∨ pn.
b) We say that L is factorial if it is atomic and if for every pairwise
distinct atoms p1, . . . , pn ∈ L and for every pairwise distinct atoms
q1, . . . , qm ∈ L, if p1 ∨ · · · ∨ pn = q1 ∨ · · · ∨ qm, then n = m and
there is a permutation σ ∈ Sn such that, for every i ∈ {1, . . . , n},
qi = pσ(i).
We define dually coatomic and cofactorial (∧, 1)-semilattices. Let
(L,∧,∨) be a bounded lattice in which the zero element is denoted by 0
and the unit element is denoted by 1. We write a ≺ b (a, b ∈ L) if a < b
and if a 6 c 6 b implies c = a or c = b, for all c ∈ L. An element p ∈ L is
called an atom (respectively coatom) of L if 0 ≺ p (respectively p ≺ 1). A
lattice L is called atomistic (respectively coatomistic) if every element of
L is a join of atoms (respectively meet of coatoms). For more details, see
[6] section (5.2). If L is a bounded lattice, we denote respectively by L∨
and L∧ the two corresponding (∨, 0)-semilattice and (∧, 1)-semilattice. It
is not hard to verify that an element of the (∨, 0)-semilattice (respectively
242 Factorial lattice of normal subgroups
(∧, 1)-semilattice) L∨ (respectively L∧) is irreducible if and only if it is an
atom (respectively coatom) of the (∨, 0)-semilattice (respectively (∧, 1)-
semilattice) L∨ (respectively L∧). This allows us to define the notion
of factorial and cofactorial lattices : A bounded lattice L will be called
factorial (respectively cofactorial) if the (∨, 0)-semilattice L∨ (respectively
L∧) is factorial (respectively cofactorial).
For an arbitrary group G, we denote by N (G) the bounded lattice of
normal subgroup of G. Recall that for every normal subgroups H and K
of G, the meet and the join of H and K in N (G) are defined by
H ∧K = H ∩K and H ∨K = HK.
Let denote respectively by N∧(G) and N∨(G) the two corresponding
(∧, 1)-semilattice and (∨, 0)-semilattice of the bounded lattice N (G). Let’s
recall firstly some notion of group theory which the reader can be found in
[4]. For a subgroups H and K of G, we write H 6 K if H is a subgroup
of K. A subgroup H of G is called a minimal normal subgroup if H is a
normal subgroup of G and if for every normal subgroup N of G, N 6 H
implies that N = H or N is trivial. If H is a subgroup of G, a complement
of H in G is any subgroup K of G such that G = HK and H ∩K = {1}.
We call normal complement of H in G any complement of H in G which
is normal in G. Note that if H is a normal subgroup of G, then H have
a normal complement in G if and only if H is direct factor in G, that is
G = H ×K for some subgroup K of G which is necessarily normal. The
group G is called a T -group if for every subgroups H and K of G, if H
is normal in K and K is normal in G, then H is normal in G. That is
every subnormal subgroup of G is normal. For every elements x, y ∈ G,
[x, y] = xyx−1y−1 denotes the commutator of x and y. If X and Y are
subsets of G, we denote by [X,Y ] the subgroup of G generated by the
commutators [x, y], with x ∈ X and y ∈ Y . The commutator subgroup of
G, denoted by D(G), is defined to be [G,G]. The group G is called perfect
if D(G) = G and it is called super-perfect if [G,H] = H for every normal
subgroup H of G.
Proposition 1. Let G be a group. Then the atoms of the idempotent
monoid N∨(G) are exactly the minimal normal subgroups of G.
Proof. Let H be a normal subgroup of G, with H 6= {1}. Suppose that H
is an atom of N∨(G) and let’s prove that H is a minimal normal subgroup
of G. Assume that H is not a minimal normal subgroup, then there exist
a non trivial normal subgroup K of G such that K1 < H. Denote by
A. Rajhi 243
K2 the subgroup of G generated by H\K1. We can check easily that
K2 is a normal subgroup of G contained in H, distinct to K1 and that
H = K1K2. Since H is an atom of N∨(G), then K1 = {1} or K2 = {1}.
But by hypothesis K1 6= {1}, then K2 = {1} which implies that K1 = H
which is impossible. Consquently, H is a minimal normal subgroup of G.
Conversely, assume that H is a minimal normal subgroup of G. If H1 and
H2 are two distinct normal subgroup of G such that H = H1H2, then
obviously H1 6 H and H2 6 H which implies by minimality of H that
H1 = {1} or H1 = H and H2 = {1} or H2 = H . The case where H1 = {1}
and H2 = {1} and likewise, the case where H1 = H and H2 = H are
impossible because H is not trivial and because H1 and H2 are distinct.
The two other cases implies that H1 = {1} or H2 = {1}. We deduce then
that H is an atom of N∨(G).
Remark 1.
1) In the same way, we can check easily that the atoms of the idempotent
monoid N∧(G) are exactly the maximal normal subgroups of G.
2) It is clear that any simple normal subgroup of a given group G is a
minimal normal subgroup, but the converse is not true in general.
Otherwise, for a T-group G, it is obviously true that any minimal
normal subgroup of G is simple.
Proposition 2. Let G be a group and let H1, H2, . . . , Hn are simple
normal subgroups of G. If G = H1H2 . . . Hn, then there exist some indices
i1, . . . , ik ∈ {1, . . . , n} such that
G = Hi1 × · · · ×Hik .
Proof. Let I = {1, . . . , n} and let E be the partially ordered set (ordered
by inclusion) of non empty subsets J of I, J = {j1, . . . , jr}, such that
G = Hj1 . . . Hjr . It is clear that E is non empty (it contain I itself).
Let J = {j1, . . . , jr} be an element of E of a minimal cardinal. Suppose
that the product G = Hj1 . . . Hjr is not direct. Then we have Hjℓ ∩
(Hj1 . . . Ĥjℓ . . . Hjr) 6= {1G}, for some index jℓ ∈ J , where the hat means
that the product is taken for all indices j1, . . . , jr except the index jℓ .
But since Hjℓ is simple, then we have Hjℓ ∩ (Hj1 . . . Ĥjℓ . . . Hjr) = Hjℓ
which implies that Hjℓ ⊂ (Hj1 . . . Ĥjℓ . . . Hjr). Therefore we have G =
Hj1 . . . Ĥjℓ . . . Hjr which is impossible.
Proposition 3 ([5], proposition (1.6.3)). Let G = G1 × · · · ×Gn and H
be a normal subgroup of G. If G1, . . . , Gn are non abelian simple groups,
244 Factorial lattice of normal subgroups
then there exists a subset J = {j1, . . . , jr} ⊂ {1, . . . , n}, such that
H = Gj1 × · · · ×Gjr .
Proposition 4. Let G be a group. If G = G1×· · ·×Gn, where G1, . . . , Gn
are non abelian simple subgroups, then G is super-perfect.
Proof. Let H be a normal subgroup of G. By the previous proposition,
we have H = Gi1 × . . . Gir for some indices i1, . . . , ir ∈ {1, . . . , n}. By
rearranging the indices, we can assume that H = G1 × · · · × Gr. The
subgroup [G,H] is a normal subgroup of G, thus it is a normal subgroup of
H . By a way of contraposition, assume that [G,H] < H which implies that
some factor Gi, for 1 6 i 6 r, is not appear in [G,H]. Say for example that
Gr is not appear in [G,H]. We have then [G,H] 6 G1 × · · · ×Gr−1. For
every x, y ∈ Gr, we have [x, y] = xyx−1y−1 ∈ Gr ∩ (G1 . . . Gr−1) = {1G}
which implies that Gr is abelain which is a contradiction. We deduce
then that [G,H] = H for every normal subgroup of G and then G is
super-perfect.
2. ND-groups and UND-groups
A group G is called an ND-group if every normal subgroup of G has a
normal complement. J. Weigold in [1], theorem (4.5), prove that a group
is an ND-group if and only if it is the restricted direct product of simple
groups. In order to characterize the groups G for which the idempotent
monoid N∨(G) is factorial we need to introduce a subclass of the class of
ND-groups. A group G is called an UND-group if every normal subgroup
of G has a unique normal complement. If H is a normal subgroup of an
UND-group G, we denote by H⊥ the unique normal complement of H in
G. Obviously, every UND-group is an ND-group but the converse is not
true in general. For example, we let the reader to check that the Klein
group Z/2Z× Z/2Z is an ND-group but it is not an UND-group.
The first thing what we can say about the ND-groups that they are
in fact T-groups. Indeed, let G be an ND-group and let N be a normal
subgroup of G. We will check that every normal subgroup of N is normal
in G. For that, take a normal subgroup M of N . Since G is an ND-group
and N is one of its normal subgroup, then G = R×N for some normal
subgroup R of G. For m ∈ M and for g = rn ∈ G, with n ∈ N and
r ∈ R, we have gmg−1 = r(nmn−1)r−1 = nmn−1 ∈ M, as M is normal
in N and [R,M ] = {1}. By the remark (2.3), we deduce then that for
an ND-group G, the atoms of the monoid N∨(G) are exactly the simple
A. Rajhi 245
normal subgroups of G. It has been proven, lemma (4.1) of [1], that every
normal subgroup of an ND-group is an ND-group. In the following result
we prove the same property for the UND-groups :
Lemma 1. Every normal subgroup of an UND-group is an UND-group.
Proof. Let G be an UND-group and let H be a normal subgroup of G.
We will show that every normal subgroup of H have a unique normal
complement in H. Let then K be a normal subgroup of H. As G is a
T-group, then K is a normal subgroup of G. Therefore since G is an UND-
group, there exist a unique normal subgroup L of G such that G = K ×L.
The subgroup L ∩H is normal in G and it is contained in H, hence it is
a normal subgroup of H . We will show that H = K × (L∩H). As K and
L∩H are normal subgroups of H , then K(L∩H) is a subgroup of H . Let
h ∈ H, then h = kl for some element (k, l) ∈ K × L as G = K × L. We
have l = hk−1 ∈ H , then l ∈ L∩H which implies that h = kl ∈ K(L∩H).
Hence H ⊂ K(L∩H) and then H = K(L∩H). Moreover, we have clearly
K ∩ (L∩H) = {1}, as K ∩L = {1}, which implies that H = K× (L∩H).
We deduce then that L ∩H is a normal complement of K in H. Suppose
that V is another normal complement of K in H, that is V is a normal
subgroup of H and H = K×V . We will prove that V = L∩H . As G is an
UND-group and since H is normal in G, then there exist a unique normal
subgroup H ′ of G such that G = H×H ′. But we have H = K×(L∩H) and
H = K×V , then we have K×((L∩H)×H ′) = G and K×(V ×H ′) = G.
This implies that (L ∩H)×H ′ and V ×H ′ are two normal complement
of K in G. Thus we have the equality (L ∩ H) × H ′ = V × H ′. Let’s
prove now that V = L ∩ H. Let v ∈ V , then v = xy for some element
(x, y) ∈ (L∩H)×H ′ because V is a subgroup of (L∩H)×H ′. But since
v ∈ V ⊂ H and since x ∈ L∩H ⊂ H , then x−1v ∈ H . Or x−1v = y ∈ H ′,
then x−1v ∈ H ∩H ′ = {1} which implies that v = x ∈ L∩H . We deduce
then that V ⊂ L∩H . Likewise, if ℓ ∈ L∩H , then ℓ = ab for some element
(a, b) ∈ V ×H ′. We have ℓ ∈ L∩H ⊂ H and a ∈ V ⊂ H , then a−1ℓ ∈ H .
But we have too a−1ℓ = b ∈ H ′, then a−1ℓ ∈ H ∩ H ′ = {1} and then
ℓ = a ∈ V . We deduce then that L∩H ⊂ V and consequently the equality
V = L ∩H is obtained which proves that the normal subgroup H of G is
an UND-group.
Lemma 2. Let G be an UND-group and let H be a simple normal subgroup
of G. Then for every normal subgroups K1 and K2 of G, with K1 ∩K2 =
{1}, if H 6 K1 ×K2 then H 6 K1 or H 6 K2.
246 Factorial lattice of normal subgroups
Proof. Let K1 and K2 two normal subgroups of G, with K1 ∩K2 = {1},
such that H is a subgroup of K1×K2. Denote by S the subgroup H ∩K1.
Since S is a normal subgroup of H , then S = {1} or S = H . If S = H , then
H 6 K1. Otherwise, H ∩K1 = {1} which implies that K1H = K1 ×H.
As H and K1 are two normal subgroups of K1 ×K2, then K1 ×H is a
normal subgroup of K1 ×K2. But since G is an UND-group, then by the
proposition (3.1) the normal subgroup K1×K2 of G is also an UND-group.
There exist then a normal subgroup M of K1 ×K2, which is normal in G
as G is a T -group, such that K1 × (H ×M) = (K1 ×H)×M = K1 ×K2.
The previous equality implies that the subgroups H × M and K2 are
two normal complement of K1 in the group K1 × K2. Then we have
H ×M = K2 which implies that H is a subgroup of K2. We deduce then
that H 6 K1 or H 6 K2.
Corollary 1. Let G be an UND-group and let H be a simple normal
subgroup of G. Then for every normal subgroups K1, . . . ,Kn of G, if
H 6 K1 ×K2 × · · · ×Kn,
then there exist a unique i ∈ {1, . . . , n} such that H 6 Ki.
Proof. Immediate by induction.
Corollary 2. Let G be an UND-group and let H be a simple normal
subgroup of G. Then for every simple normal subgroups K1, . . . ,Kn of G,
if
H 6 K1 ×K2 × · · · ×Kn,
then there exist a unique i ∈ {1, . . . , n} such that H = Ki.
Proof. Immediately follows from the previous corollary.
Lemma 3. Let G be a group. If G is an UND-group, then for every
pairwise distinct simple normal subgroups K1, . . . ,Kn of G, we have
K1K2 . . .Kn = K1 ×K2 × · · · ×Kn.
Proof. Assume to the contrary that the product K1K2 . . .Kn is not a
direct product, then there exist some index j ∈ {1, 2, . . . , n} such that Kj∩
(K1 . . . K̂j . . .Kn) 6= {1}. Since Kj is simple and as Kj∩(K1 . . . K̂j . . .Kn)
is a non trivial normal subgroup of Kj , then Kj ∩ (K1 . . . K̂j . . .Kn) = Kj
which implies that Kj ⊂ K1 . . . K̂j . . .Kn. By the proposition (2.4) we
A. Rajhi 247
can extract from the product K1 . . . K̂j . . .Kn a direct product, that is
for some indices i1, . . . , ik ∈ {1, 2, . . . , n}\{j} we have
K1 . . . K̂j . . .Kn = Ki1 × · · · ×Kik .
Therefore the simple normal subgroup Kj of G is in fact a simple normal
subgroup of the direct product Ki1 ×· · ·×Kik . Then by the previous corol-
lary Kj = Kiℓ for some index iℓ ∈ {1, 2, . . . , n}\{j} which is impossible
since the subgroups K1,K2, . . . ,Kn are pairwise distinct.
Recall that a group G is called of finite length if it satisfies both
ascending and descending chain conditions on normal subgroups. We
conclude this section by proving that an ND-group is of finite length if
and only if it is the direct product of a finite number of simple groups.
We will use this fact in the next section.
Lemma 4. Let G be an ND-group, H0 be a normal subgroup of G and
K0 be a normal complement of H0 in G. For every normal subgroup H1
of G, if H1 6 H0, then there exist a normal complement K1 of H1 in G
such that K0 6 K1.
Proof. Let H1 be a normal subgroup of G such that H1 6 H0. Since H0 is a
normal subgroup of the ND-group G, then by lemma (4.1) of [1], H0 is also
an ND-group. Therefore we have H0 = H1×L for some normal subgroup L
of H0 which is in fact a normal subgroup of G since G in an ND-group and
in particular a T -group. If we denote by K1 the normal subgroup LK0, then
it is clear that K0 6 K1. We will prove that K1 is a normal complement
of H1 in G. We have H1K1 = H1(LK0) = (H1L)K0 = H0K0 = G.
Furthermore, if x ∈ H1 ∩K1 then x ∈ H0 since x ∈ H1 6 H0 and x = uv
for some element (u, v) ∈ L × K0 as x ∈ K1 = LK0. As L 6 H0, then
u ∈ H0 and therefore u−1x = v ∈ H0∩K0 = {1}. Thus we have x = u ∈ L,
but x ∈ H1, then x ∈ H1 ∩ L = {1} which implies that x = 1. We have
then H1 ∩K1 = {1} and consequently K1 is a normal complement of H1
in G.
Corollary 3. Let G be an ND-group. Then for every decreasing sequence
(Hn)n∈N of normal subgroups of G, there exist a sequence (Kn)n∈N of
normal subgroups of G such that :
1) for every n ∈ N, Kn is a normal complement of Hn in G,
2) the sequence (Kn)n∈N is increasing.
Proof. Using the previous lemma, we can easily construct a such sequence
by induction.
248 Factorial lattice of normal subgroups
Corollary 4. Let G be an ND-group. Then G satisfy the descending chain
condition on normal subgroups if and only if it satisfy the ascending chain
condition on normal subgroups.
Proof. Suppose that G have the ascending chain condition on normal
subgroups. Let (Hn)n∈N be a decreasing sequence of normal subgroups
of G. By the previous corollary, there exist an increasing sequence (Kn)n∈N
of normal subgroups of G such that, for every n ∈ N, Kn is a normal
complement of Hn in G. Since G have the ascending chain condition, then
there exist an integer n0 ∈ N such that Kn = Kn0
, for every n > n0. We
will show that, for every n > n0, Hn = Hn0
. Since the sequence (Hn)n∈N is
decreasing, then Hn 6 Hn0
for every n > n0. Let n > n0 and let x ∈ Hn0
.
We have G = Hn × Kn, but Kn = Kn0
, then we have G = Hn × Kn0
.
Thus we have x = ab, for some element (a, b) ∈ Hn ×Kn0
which implies
that a−1x = b ∈ Kn0
. But since a ∈ Hn 6 Hn0
, then a ∈ Hn0
which
implies that a−1x ∈ Hn0
. Therefore a−1x ∈ Hn0
∩Kn0
= {1} and then
x = a ∈ Hn. We deduce then that Hn0
6 Hn. Consequently, for every
n > n0, Hn = Hn0
. Thus G have descending chain condition on normal
subgroups. By the same way, we can prove the other implication.
By the previous corollary, clearly an ND-group is of finite length if
it satisfies the ascending or the descending chain condition on normal
subgroups.
Proposition 5. Let G be an ND-group. Then G is of finite length if and
only if G is a direct product of a finite number of simple groups.
Proof. Suppose that G is of finite length. By the Krull-Schmidt theorem,
theorem (3.3) page 83 of [4], the group G is then the direct product of a
finite number of indecomposable subgroups, say G = G1 ×G2 × · · · ×Gr
where G1, . . . , Gr are indecomposable subgroups of G. Every factor Gi,
for 1 6 i 6 r, is a normal subgroup of G, thus Gi is an ND-group. But
obviously an indecomposable ND-group is simple. Therefore all the factor
G1, . . . , Gr are simple and hence G is a direct product of a finite number
of simple groups. The converse is immediate.
3. Main results
Theorem 1. Let G be group. Then the commutative idempotent monoid
N∨(G) is atomic if and only if the group G is an ND-group with finite
length.
A. Rajhi 249
Proof. Suppose that N∨(G) is atomic. Then every normal subgroup of G
can be expressed as product of pairwise distinct simple normal subgroups.
In particular, there exist some pairwise distinct simple normal subgroups
G1, G2, . . . , Gn such that G = G1 . . . Gn. By the proposition (2.4), we
can assume that G = G1 × · · · ×Gn. Let H be a normal subgroup of G.
We will show that H is a direct factor in G. There are three cases : For
the first case, assume that all the simple normal subgroups Gi are non
abelian. As H is a normal subgroup of G, by the proposition (2.5) we have
H = Gi1 × · · · × Gis for some subset J = {i1, . . . , is} of I = {1, . . . , n}.
In this case, it is clear that G = H × K where K is the product of
the Gi’s for i ∈ I\J . For the second case, assume that some but not
all of the subgroups Gi are abelian, say for example that G1, . . . , Gr are
abelian and Gr+1, . . . , Gn are non abelian, where 1 < r < n. If we put
A = G1 × · · · ×Gr and K = Gr+1 × · · · ×Gn, then we have G = A×K.
Since K = Gr+1 × · · · ×Gn, where Gr+1, . . . , Gr are non abelian simple
groups, then by the proposition (2.6) the group K is super-perfect. Hence
H is a normal subgroup of G = A×K with K is super-perfect. Then by
the theorem 1 of [2], page 155, H = B×L for some normal subgroup B of
A and some normal subgroup L of K. By the first case, since L is normal
in K and as K is the direct product of simple non abelian groups, then
K = L ×M for some normal subgroup M of K. Moreover, as A is the
direct product of simple abelian groups we know that every subgroup of A
is a direct factor in A. Then we have A = B × C for some subgroup C of
A. Note that the subgroup M (resp. C) is in fact normal in G since it is
a normal subgroup of a direct factor of G. If we denote H ′ = C ×M then
clearly H ′ is a normal complement of H is G. We deduce then that G is
an ND-group. For the last case, all the simple subgroups Gi, for 1 6 i 6 n,
are assumed to be abelian. In this case, the group G is then an abelian
ND-group and the result in this case is obvious. Conversely, suppose that
G is an ND-group with finite length and let’s prove that the idempotent
monoid N∨(G) is atomic. Let H be a normal subgroup of G. Since G is
an ND-group, then H is an ND-group and as G is of finite length then
H is also of finite length. Therefore, by the proposition (3.9) H is the
direct product of a finite number of simple normal subgroups, which are
normal in G as G is a T-group. Thus H is a product of a finite number of
pairwise simple normal subgroups of G and the fact that the idempotent
monoid N∨(G) is atomic is proved.
Theorem 2. Let G be a group. Then the idempotent monoid N∨(G) is
factorial if and only if the group G is an UND-groups with finite length.
250 Factorial lattice of normal subgroups
Proof. Suppose that G is an UND-group with finite length. By the theorem
(4.1), the monoid N∨(G) is in particular atomic. Let H1, . . . , Hn a pairwise
distinct simple normal subgroups and K1, . . . ,Km a pairwise distinct
simple normal subgroups of G such that H1 . . . Hn = K1 . . .Km. We must
prove that n = m and there exist a permutation σ ∈ Sn such that, for
every i ∈ {1, . . . , n}, Ki = Hσ(i). Since G is an UND-group, then by the
lemma (3.5) we have
H1 . . . Hn = H1 × · · · ×Hn and K1 . . .Km = K1 × · · · ×Km.
Let’s show firstly that we have necessarily n = m. Suppose that n 6= m,
for example take n < m. Let i ∈ {1, . . . , n}, since Hi is a subgroup of the
direct product K1×· · ·×Km then by the corollary (3.4), we have Hi = Kji
for a unique index ji ∈ {1, . . . ,m}. Since the subgroups H1, . . . , Hn are
pairwise distinct, then the indices j1, . . . , jn ∈ {1, . . . ,m} are pairwise
distinct. But as n < m, then there exist an index ℓ ∈ {1, . . . ,m} such
that ℓ 6∈ {j1, . . . , jn}. Now Kℓ is a subgroup of H1 × · · · ×Hn, then by
the corollary (3.4) we have Kℓ = Hk for a unique index k ∈ {1, . . . , n},
but Hk = Kjk . Then we have Kℓ = Kjk which implies that ℓ = jk as the
subgroups K1, . . . ,Km are pairwise distinct. We have then ℓ ∈ {j1, . . . , jn}
which is a contradiction. We deduce then that n = m. We will prove now
that there exist a permutation σ ∈ Sn such that, for every i ∈ {1, . . . , n},
Ki = Hσ(i). By the same way as the previous proof, clearly for every
i ∈ {1, . . . , n} there exist a unique index σ(i) ∈ {1, . . . , n} such that
Ki = Hσ(i). We must show that the map σ is in fact injective, but this
is clearly true since the subgroups K1, . . . ,Kn are pairwise distinct. We
deduce then that the idempotent monoid N∨(G) is factorial. Conversely,
suppose that N∨(G) is factorial and let’s prove that the group G is an
UND-group with finite length. As N∨(G) is factorial, it is in particular
atomic and then by the previous theorem the group G is an ND-group
with finite length. Let H be a normal subgroup of G and let K,L two
normal subgroups of G such that H ×K = G and H × L = G. We will
show that K = L. Since N∨(G) is atomic, then H = H1 . . . Hr for some
pairwise distinct simple normal subgroups H1, . . . , Hr, K = Kr+1 . . .Kr+k
for some pairwise distinct simple normal subgroups Kr+1, . . . ,Kr+k and
L = Lr+1 . . . Lr+ℓ for some pairwise distinct simple normal subgroups
Lr+1, . . . , Lr+ℓ. If we put, for i ∈ {1, . . . , r}, Ki = Li = Hi, then we have
K1 . . .KrKr+1 . . .Kr+k = L1 . . . LrLr+1 . . . Lr+ℓ
Since H ∩K = {1} and H ∩ L = {1}, then the normal simple subgroups
K1, . . . ,Kr+k are in fact pairwise distinct and the same holds for the
A. Rajhi 251
normal simple subgroups L1, . . . , Lr+ℓ. As N∨(G) is factorial, then r+k =
r + ℓ which implies that k = ℓ and there exist a permutation σ ∈ Sr+k
such that, for every i ∈ {1, . . . , r + k}, Li = Kσ(i). Obviously, for every
i ∈ {r + 1, . . . , r + k}, σ(i) ∈ {r + 1, . . . , r + k}, indeed if there exist i0 ∈
{r + 1, . . . , r + k} such that σ(i0) ∈ {1, . . . , r} then Li0 = Kσ(i0) = Hσ(i0)
which is impossible as H ∩ L = {1}. Consequently we have
L = Lr+1 . . . Lr+k = Kσ(r+1) . . .Kσ(r+k) = K.
We deduce then that the group G is an UND-group and we have already
proved that G is of finite length.
Note that in an UND-group G, it is easy to check that a non trivial
and proper normal subgroup H of G is an atom of the idempotent monoid
N∨(G) if and only if H⊥ is an atom of the idempotent monoid N∧(G).
Furthermore, obviously for every normal subgroups H1 and H2 of G, we
have
(H1 ∨H2)
⊥ = H⊥
1 ∧H⊥
2 and (H1 ∧H2)
⊥ = H⊥
1 ∨H⊥
2 .
Using this simple facts, we can show easily that for an arbitrary group G,
if the idempotent monoid N∨(G) is factorial then the idempotent monoid
N∧(G) is factorial.
Recall that a partial commutative monoid, see [3] definition (2.1.1), is
a structure (P,⊕, 0), where P is a set, 0 ∈ P , and ⊕ is a partial binary
operation on P satisfying the following properties, for all x, y, z ∈ P :
(P1) Associativity : x⊕ (y ⊕ z) is defined iff (x⊕ y)⊕ z is defined, and
then the two values are equal.
(P2) Commutativity : x⊕ y is defined iff y ⊕ x is defined, and then the
two values are equal.
(P3) Zero element : x⊕ 0 is defined with values x.
To any bounded lattice (L,∧,∨) we can associate a structure of partial
monoid PL as follows : the underlaying set of PL is L and the partial
binary operation of PL is defined by, for all x, y ∈ L,
x⊕ y is defined only when x ∧ y = 0 in case x⊕ y = x ∨ y.
A partial commutative monoid (P,⊕, 0) is a partial refinement monoid
(or have the refinement property), see [3] definition (2.2.1), if for all
x0, x1, y0, y1 ∈ P with x0 ⊕ x1 = y0 ⊕ y1, there are elements ci,j ∈ P ,
for i, j ∈ {0, 1}, such that xi = ci,0 ⊕ ci,1 and yj = c0,j ⊕ c1,j for every
i, j ∈ {0, 1}. If G is a group, we denote by Npar(G) the partial commutative
252 Factorial lattice of normal subgroups
monoid associated to the bounded lattice of normal subgroup of G. We
have then the following result:
Theorem 3. Let G be an ND-group with finite length. Then G is an
UND-group if and only if the partial commutative monoid Npar(G) has
the refinement property.
Proof. Suppose that the partial commutative monoid Npar(G) has the
refinement property and let’s prove that G is an UND-group. Let H be
a normal subgroup of G and let K0,K1 two normal subgroup of G such
that G = H ×K0 and G = H ×K1, that is K0 and K1 are two normal
complement of H in G. Since Npar(G) has the refinement property and
as H ×K0 = H ×K1, then there exist normal subgroups Ci,j of G, for
i, j ∈ {0, 1}, such that H = C0,0×C0,1, H = C0,0×C1,0, K0 = C1,0×C1,1
and K1 = C0,1 × C1,1. It is clear that the subgroup C1,0 is contained in
H and in K1 which implies that C1,0 is contained in H ∩K1 = {1} and
hence C1,0 = {1}. Likewise, the subgroup C0,1 is contained in H and in
K2 which implies that C0,1 = {1}. Therefore we have
K0 = C1,0 × C1,1 = C1,1 = C0,1 × C1,1 = K1.
Consequently, the group G is an UND-group. Conversely, suppose that
the group G is an UND-group and let’s prove that the partial commu-
tative monoid Npar(G) has the refinement property. Let H0, H1,K0,K1
be normal subgroups of G such that H0 ×H1 = K0 ×K1. Since G is an
UND-group with finite length then by theorem 2, the idempotent monoid
N∨(G) is factorial. Therefore, for i ∈ {0, 1}, there exist a family of pairwise
distinct simple normal subgroups (Hi,j)06j6pi such that
Hi = Hi,0Hi,1 . . . Hi,pi , i ∈ {0, 1}
and, for j ∈ {0, 1}, there exist a family of pairwise distinct simple normal
subgroups (Ki,j)06i6qi such that
Kj = K0,jK1,j . . .Kqj ,j , j ∈ {0, 1}.
By the corollary (4.5), all the previous products are in fact direct products.
Hence we have
H0,0×· · ·×H0,p0×H1,0×· · ·×H1,p1 = K0,0×· · ·×Kq0,0×K0,1×· · ·×Kq1,1.
For i, j ∈ {0, 1}, put Ci,j = Hi∩Kj . The subgroups Ci,j are clearly normal
in G and we will show that we have the refinement matrix
K0 K1
H0 C0,0 C0,1
H1 C1,0 C1,1
,
A. Rajhi 253
that is Hi = Ci,0×Ci,1 for 0 6 i 6 1 and Kj = C0,j×C1,j for 0 6 j 6 1.
We will show only that H0 = C0,0 × C0,1, the other equalities are proved
in the same way. Since C0,0 and C0,1 are normal subgroups of H0, then
their product C0,0C0,1 is a subgroup of H0. For the other inclusion, let
j ∈ {0, . . . , p0}. As the simple normal subgroup H0,j is a subgroup of
H0 ×H1, and in particular a subgroup of K0 ×K1, then by the corollary
(4.4) we have H0,j = Kij ,0 for some index ij ∈ {0, . . . , q0} or H0,j = Kij ,1
for some index ij ∈ {0, . . . , q1}. If H0,j = Kij ,0 then obviously H0,j is a
subgroup of H0 ∩K0 and likewise if H0,j = Kij ,1 then H0,j is a subgroup
of H0 ∩K1. Consequently, for every j ∈ {0, . . . , p0}, H0,j is a subgroup of
(H0∩K0)(H0∩K1) = C0,0C0,1 and so H0 6 C0,0C0,1. We deduce then that
H0 = C0,0C0,1, but as C0,0 6 K0 and C0,1 6 K1, then H0 = C0,0 × C0,1.
Consequently, the partial monoid Npar(G) has the refinement property.
References
[1] J. Weigold, “On direct factor in groups”, J. London Math. Soc. 35 (1960), 310-320.
[2] M. D. Miller, “On the lattice of normal subgroups of a direct product”, Pacific
Jounrnal of mathematics. vol. 60, No 2, (1975).
[3] F. Wehrung, “Refinement monoids, equidecomposability types, and Boolean inverse
semigroups”. Lecture Notes in Mathematics, Springer Verlag, vol 2188 (2017).
[4] Thomas W. Hungerford, Algebra, 1th edn. Springer, New York (1980)
[5] Hans Kurzweil and Bernd Stellmacher, The Theory of Finite Groups An Introduction.
Springer, New York (2004)
[6] B. A. Davey and H. A. Priestley, Introduction to Lattices and Order. Cambridge
university press (1990)
[7] R. Schmidt, Subgroup lattices of groups. de Gruyter Expositions in Mathematics
14 (1994).
Contact information
Anis Rajhi Mathematics Department, Faculty of Sciences and
Humanities in Dawadmi, Shaqra University,
11911, Saudi Arabia;
Quantitative Methods Department, Higher
business School, University of Manouba, Manouba
2010, Tunisia
E-Mail(s): anis.rajhi@math.univ-poitiers.fr
Received by the editors: 09.10.2018
and in final form 08.12.2020.
mailto:anis.rajhi@math.univ-poitiers.fr
A. Rajhi
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