The groups whose cyclic subgroups are either ascendant or almost self-normalizing
The main result of this paper shows a description of locally finite groups, whose cyclic subgroups are either almost self-normalizing or ascendant. Also, we obtained some natural corollaries of the above situation.
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Інститут прикладної математики і механіки НАН України
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| Cite this: | The groups whose cyclic subgroups are either ascendant or almost self-normalizing / L.A. Kurdachenko, A.A. Pypka, N.N. Semko // Algebra and Discrete Mathematics. — 2016. — Vol. 21, № 1. — С. 111-127. — Бібліогр.: 21 назв. — англ. |
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| author | Kurdachenko, L.A. Pypka, A.A. Semko, N.N. |
| author_facet | Kurdachenko, L.A. Pypka, A.A. Semko, N.N. |
| citation_txt | The groups whose cyclic subgroups are either ascendant or almost self-normalizing / L.A. Kurdachenko, A.A. Pypka, N.N. Semko // Algebra and Discrete Mathematics. — 2016. — Vol. 21, № 1. — С. 111-127. — Бібліогр.: 21 назв. — англ. |
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| description | The main result of this paper shows a description of locally finite groups, whose cyclic subgroups are either almost self-normalizing or ascendant. Also, we obtained some natural corollaries of the above situation.
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 21 (2016). Number 1, pp. 111–127
© Journal “Algebra and Discrete Mathematics”
The groups whose cyclic subgroups
are either ascendant or almost self-normalizing
Leonid A. Kurdachenko, Aleksandr A. Pypka,
Nikolaj N. Semko
Abstract. The main result of this paper shows a description
of locally finite groups, whose cyclic subgroups are either almost self-
normalizing or ascendant. Also, we obtained some natural corollaries
of the above situation.
Introduction
The subgroups of a group G are connected with some natural families
of subgroups. One of them is the following. Let H be a subgroup of a
group G. We construct an ascending series
〈1〉 = H0 6 H1 6 . . . Hα 6 Hα+1 6 . . . Hγ 6 G,
where H1 = H, H2 = NG(H1) = NG(H), Hα+1 = NG(Hα) for every ordi-
nal α < γ, Hλ =
⋃
µ<λ
Hµ for every limit ordinal λ < γ, and NG(Hγ) = Hγ .
This chain is called the upper normalized chain of H in G. Here the two
natural types of subgroups appear. If Hγ = G, then a subgroup H is
called ascendant in G. If Hγ = H (that is NG(H) = H), then a subgroup
H is called self-normalizing in G. Thus, every subgroup of a group is
naturally connected with the two types of subgroups: an ascendant and a
self-normalizing subgroups. The presence of a large family of ascendant
subgroups has a strong influence on the group structure. For example, if
2010 MSC: 20E15, 20F19, 20F22, 20F50.
Key words and phrases: locally finite group, self-normalizing subgroup, ascen-
dant subgroup, subnormal subgroup, Gruenberg radical, Baer radical.
112 The groups with two types of subgroups
every subgroup of a group G is ascendant, then G is locally nilpotent [16].
Moreover, if every cyclic subgroup of a group G is ascendant, then G is
locally nilpotent [6, Theorem 2]. More precisely, the subgroup Gru(G)
of an arbitrary group G, generated by all ascendant cyclic subgroups
of G, is locally nilpotent. This subgroup is called the Gruenberg radical
of G. Every finitely generated subgroup of Gru(G) is ascendant in G and
nilpotent [6, Theorem 2]. A group G is said to be a Gruenberg group, if
G = Gru(G).
L.A. Kurdachenko and H. Smith [13] have considered the groups,
whose subgroups are either subnormal or self-normalizing. A natural
generalization of this paper was an article [12]. In [12] L.A. Kurdachenko
et al. considered the groups, whose finitely generated subgroups are
either ascendant or self-normalizing. From their results it follows that
locally finite groups, whose cyclic subgroups are either ascendant or self-
normalizing, have the same structure. Here we discuss a more general
situation.
We remark that the groups, in which some family of subgroups divides
into two types of subgroups, which often have the opposite properties,
considered by other authors (see, for example, [15], [17]).
Let H be a subgroup of a group G. Then H is called almost self-
normalizing in G, if H has finite index in NG(H).
In this paper we consider the groups whose cyclic subgroups are either
almost self-normalizing or ascendant. The main result is the following
Theorem A. Let G be an infinite locally finite group whose cyclic sub-
groups are either almost self-normalizing or ascendant. Suppose that
G 6= Gru(G). Then the following assertions hold:
(i) a factor-group G/Gru(G) is finite;
(ii) G = Q ⋋ R, where Q is a normal Sylow σ′-subgroup of G, R is a
Sylow σ-subgroup of G, σ = Π(G/Gru(G));
(iii) R is a Chernikov subgroup;
(iv) Gru(G) = CR(Q) × Q;
(v) if g 6∈ Gru(G), then CG(g) is finite;
(vi) Gru(G) is nilpotent-by-finite.
We obtained the following additional information about the structure
of a factor-group G/Gru(G).
Corollary A1. Let G be an infinite locally finite group whose cyclic sub-
groups are either almost self-normalizing or ascendant. Let G 6= Gru(G),
F = G/Gru(G) and σ = Π(F ). Suppose that the Sylow σ′-subgroup of G
is infinite. Then
L. A. Kurdachenko, A. A. Pypka, N. N. Semko 113
(i) if p ∈ σ and p 6= 2, then Sylow p-subgroup of F is cyclic;
(ii) Sylow 2-subgroup of F is cyclic or a generalized quaternion group;
(iii) every subgroup of order pq of F , p, q ∈ σ, is cyclic.
Let G be a Chernikov group and D be the divisible part of G. Put
Sp(G) = Π(D).
Corollary A2. Let G be an infinite locally finite group whose cyclic sub-
groups are either almost self-normalizing or ascendant. Let G 6= Gru(G),
F = G/Gru(G) and σ = Π(F ). Suppose that the Sylow σ′-subgroup of G
is finite and Sp(G) = {p} for some prime p ∈ σ. Then
(i) if q is a prime and q 6∈ {2, p}, then Sylow q-subgroup of F is cyclic;
(ii) if p 6= 2, then Sylow 2-subgroup of F is cyclic or generalized quater-
nion group.
Corollary A3. Let G be an infinite locally finite group whose cyclic sub-
groups are either almost self-normalizing or ascendant. Let G 6= Gru(G),
F = G/Gru(G) and σ = Π(F ). Suppose that the Sylow σ′-subgroup of G
is finite and |Sp(G)| > 2. Then
(i) if q ∈ σ is a prime and q 6= 2, then Sylow q-subgroup of F is cyclic;
(ii) if 2 ∈ σ, then Sylow 2-subgroup of F is cyclic or generalized quater-
nion group.
An important special case of the ascendant subgroups are the subnor-
mal subgroups. A subnormal subgroup is exactly an ascendant subgroup
having finite upper normalized chain. From Theorem A we can obtain
the description of locally finite groups whose cyclic subgroups are either
almost self-normalizing or subnormal.
The subgroup B(G), generated by all cyclic subnormal subgroups of G,
is called the Baer radical of G. Every finitely generated subgroup of B(G)
is subnormal in G and nilpotent (see, for example, [14, Theorem 2.5.1]),
so that a subgroup B(G) is locally nilpotent. A group G is said to be a
Baer group, if G = B(G).
Let G be a group and A be an abelian normal subgroup of G. Then
A is said to be G-quasifinite if A is infinite, but every proper G-invariant
subgroup of A is finite.
Theorem B. Let G be an infinite locally finite group whose cyclic sub-
groups are either almost self-normalizing or subnormal. Suppose that
G 6= B(G). Then the following assertions hold:
(i) a factor-group G/B(G) is finite;
114 The groups with two types of subgroups
(ii) G = Q ⋋ R, where Q is a normal Sylow σ′-subgroup of G, R is a
Sylow σ-subgroup of G, σ = Π(G/B(G));
(iii) R is a Chernikov subgroup;
(iv) B(G) = CR(Q) × Q;
(v) if g 6∈ B(G), then CG(g) is finite;
(vi) B(G) includes a finite G-invariant σ-subgroup K such that
B(G)/K = QK/K × U1/K × . . . × Uk/K,
where Uj/K is a G-quasifinite divisible Chernikov pj-subgroup,
pj ∈ σ, 1 6 j 6 k;
(vii) B(G) is nilpotent.
Trivially, every normal subgroup is a special case of subnormal sub-
group, and we come to
Corollary B1. Let G be an infinite locally finite group whose cyclic
subgroups are either almost self-normalizing or normal. Suppose that G
is a not Dedekind group. Then the following assertions hold:
(i) a factor-group G/B(G) is finite cyclic;
(ii) if g 6∈ B(G), then CG(g) is finite;
(iii) every subgroup of B(G) is G-invariant, in particular, B(G) is a
Dedekind group;
(iv) Sylow 2-subgroup of B(G) is Chernikov, moreover, if this Sylow 2-
subgroup is infinite, then B(G) is abelian and G/B(G) has order 2.
1. Preliminaries and lemmas
Lemma 1. Let G be a group whose cyclic subgroups are either almost
self-normalizing or ascendant (respectively, subnormal). If H is a subgroup
of G, then every cyclic subgroup of H is either almost self-normalizing or
ascendant (respectively, subnormal).
Proof. Let C be a cyclic subgroup of H and suppose that C is not
ascendant (respectively, subnormal) in H. Then C can not be ascendant
(respectively, subnormal) in G. It follows that the index |NG(C) : C| is
finite. An inclusion NH(C) 6 NG(C) shows that the index |NH(C) : C|
is finite.
Lemma 2. Let G be a group whose cyclic subgroups are either almost
self-normalizing or ascendant (respectively, subnormal). If A is an infinite
periodic abelian subgroup of G, then the Gruenberg radical (respectively,
Baer radical) of G includes A.
L. A. Kurdachenko, A. A. Pypka, N. N. Semko 115
Proof. Indeed, for each element x ∈ A we have A 6 CG(x), which follows
that the index |NG(〈x〉) : 〈x〉| is infinite. Thus, x ∈ Gru(G) (respectively,
x ∈ B(G)). Hence A 6 Gru(G) (respectively, A 6 B(G)).
Lemma 3. Let L be a locally nilpotent periodic subgroup of G. If L is
not Chernikov, then the centralizer of every element of L is infinite.
Proof. Suppose first that the set Π(L) is infinite. Let g be an arbitrary
element of L and |g| = pk1
1 · . . . · pks
s , where p1, . . . , ps are primes, pj 6= pm
whenever j 6= m. Since Π(L) is infinite, the set π = Π(L) \ {p1, . . . , ps} is
infinite. Then the Sylow π-subgroup Lπ of L is infinite. The fact, that L
is locally nilpotent, implies the inclusion Lπ 6 CG(g), which follows that
CG(g) is infinite.
Suppose now that the set Π(L) is finite. Since L is not Chernikov,
there exists a prime p such that the Sylow p-subgroup P of L is not
Chernikov. Let x be an arbitrary element of L. If x is a p′-element, then
P 6 CG(x), which follows again that CG(x) is infinite. Assume that x ∈ P .
Since P is not Chernikov, P includes an 〈x〉-invariant abelian subgroup A,
which is not Chernikov [21]. Then its lower layer H = Ω1(A) is an infinite
elementary abelian subgroup. Clearly H is an 〈x〉-invariant subgroup.
Let 1 6= b1 ∈ H. Put K1 = 〈b1〉〈x〉, then K1 is a finite 〈x〉-invariant
subgroup. Since H is elementary abelian, H includes a subgroup B1 such
that H = K1 × B1. We note that the index |H : B1| is finite. Then the
index |H : By
1 | is also finite for every element y ∈ 〈x〉. Since an element
x has finite order, a family {By
1 |y ∈ 〈x〉} is finite. Then the intersection
C1 =
⋂
y∈〈x〉
By
1 has finite index in H. In particular, C1 is infinite. By
such choice C1 is 〈x〉-invariant and K1 ∩ C1 = 〈1〉. Let 1 6= b2 ∈ C1 and
K2 = 〈b2〉〈x〉, then K2 also is a finite 〈x〉-invariant subgroup such that
K1 ∩ K2 = 〈1〉. Since H is elementary abelian, H includes a subgroup B2
such that H = K1K2 × B2. Using the similar arguments and ordinary
induction, we construct the family {Kn|n ∈ N} of finite 〈x〉-invariant
subgroups such that K1 · . . . ·Km ∩Km+1 = 〈1〉 for every m ∈ N. It follows
that 〈Kn|n ∈ N〉 = Drn∈NKn.
Since 〈x, Kn〉 is a finite p-subgroup, it is nilpotent. Since Kn is its
normal subgroup, Kn∩ζ(〈x, Kn〉) 6= 〈1〉. Let 1 6= zn ∈ Kn∩ζ(〈x, Kn〉) and
put Z = 〈zn|n ∈ N〉. An equality 〈Kn|n ∈ N〉 = Drn∈NKn implies that Z
is an infinite elementary abelian subgroup. By its choice Z 6 CG(〈x〉). It
follows that CG(x) is infinite.
116 The groups with two types of subgroups
Corollary 1. Let L be an infinite periodic nilpotent subgroup of G. Then
the centralizer of every element of L is infinite.
Proof. If L is a not Chernikov subgroup, then result follows from Lemma 3.
Therefore, suppose that L is a Chernikov subgroup. Since L is nilpotent,
every subgroup of L is subnormal. In particular, L is a Baer group. Then
L is central-by-finite [8, Corollary 1 to Lemma 4]. In particular, CL(x) is
infinite for each element x ∈ L.
Corollary 2. Let G be a group whose cyclic subgroups are either almost
self-normalizing or ascendant (respectively, subnormal). Suppose that L
is a locally nilpotent periodic subgroup of G. If L is not Chernikov, then
the Gruenberg radical (respectively, Baer radical) of G includes L.
Proof. In fact, by Lemma 3 CG(g) is infinite for each element g ∈ L, which
follows that the index |NG(〈g〉) : 〈g〉| is infinite and hence g ∈ Gru(G)
(respectively, g ∈ B(G)).
Corollary 3. Let G be a group whose cyclic subgroups are either almost
self-normalizing or ascendant (respectively, subnormal). Suppose that p is
a prime such that the Sylow p-subgroup P of G is not Chernikov. Then
the Gruenberg radical (respectively, Baer radical) of G includes P . In
particular, P is normal in G.
Corollary 4. Let G be a group whose cyclic subgroups are either almost
self-normalizing or ascendant (respectively, subnormal). Suppose that L is
an infinite periodic nilpotent subgroup of G. Then the Gruenberg radical
(respectively, Baer radical) of G includes L.
Proof. In fact, by Corollary 1 CG(g) is infinite for each element g ∈ L,
which follows that the index |NG(〈g〉) :〈g〉| is infinite and hence g ∈Gru(G)
(respectively, g ∈ B(G)).
Let G be a Chernikov group. Denote by D(G) the maximal normal
divisible abelian subgroup of G. A subgroup D(G) is called a divisible
part of G.
Lemma 4. Let G be an infinite locally finite group whose cyclic sub-
groups are either almost self-normalizing or ascendant. Then Sylow p-
subgroup of G/Gru(G) is finite for every prime p. Moreover, for every
p ∈ Π(G/Gru(G)) every Sylow p-subgroup of G is Chernikov.
L. A. Kurdachenko, A. A. Pypka, N. N. Semko 117
Proof. Put B = Gru(G). Let p ∈ Π(G/B) and let P/B be the Sylow
p-subgroup of G/B. Let B 6= xB ∈ P/B. Without loss of generality we
can assume that x is an p-element. Let Bp be the Sylow p-subgroup of B.
Since B is locally nilpotent, Bp is G-invariant. Then a product 〈x〉Bp is
a p-subgroup. Suppose that Bp is not Chernikov. Let C be the Sylow p-
subgroup of G, including 〈x〉Bp. Then C is a not Chernikov subgroup, and
Corollary 3 proves that C 6 B. In this case x ∈ B, what contradicts to
the choice of element x. This contradiction shows that Bp is a Chernikov
subgroup.
Suppose now that P/B is not Chernikov. Then P/B includes an
infinite elementary abelian subgroup A/B [2, Theorem 8]. Without loss
of generality we can assume that A/B is countable. Then A/B has an
ascending series of finite subgroups
A1/B 6 A2/B 6 . . . 6 An/B 6 . . .
such that A/B =
⋃
n∈N
An/B. Since A1/B is finite, A1 includes a finite sub-
group K1 such that A1 = K1B. Choose in K1 Sylow p-subgroup S1. Since
K1 is finite, S1(B ∩ K1)/(B ∩ K1) is a Sylow p-subgroup of K1/(B ∩ K1).
On the other hand, K1/(B ∩ K1) ∼= K1B/B is a p-group. It follows that
S1(B ∩ K1)/(B ∩ K1) = K1/(B ∩ K1), or S1(B ∩ K1) = K1. In turn out
it follows that A1 = S1B. Choose in A2 a finite subgroup K2 such that
K1 6 K2 and A2 = K2B. Let S2 be the Sylow p-subgroup of K2, includ-
ing S1. Using the above arguments, we can prove that A2 = S2B. Using
similarly arguments and ordinary induction, we construct an ascending
series
S1 6 S2 6 . . . 6 Sn 6 . . .
of finite p-subgroups such that A =
(
⋃
n∈N
Sn
)
B. Put S =
⋃
n∈N
Sn, then
S is a p-subgroup and isomorphism S/(S ∩ B) ∼= SB/B = A/B shows
that S is not Chernikov. Since Bp is a normal p-subgroup, then SBp is
a p-subgroup. Let D be the Sylow p-subgroup of G, including SBp. The
fact that S is not Chernikov, implies that D is not Chernikov. Corollary 3
proves that D 6 B. In this case S 6 B and therefore SB = A 6 B, what
contradicts to the choice of A. This contradiction shows that P/B is a
Chernikov subgroup.
Let Q be the Sylow p-subgroup of B, then B = Bp × Q. Since Bp and
P/B are Chernikov, P/Q likewise is a Chernikov group. In particular,
it is countable. Then P includes a p-subgroup R such that P = QR
(see, for example, [3, Theorem 2.4.5]). Denote by W the divisible part
118 The groups with two types of subgroups
of R. Since W is abelian and infinite, Lemma 2 shows that W 6 B. An
inclusion Q 6 B implies that WQ 6 B. In turn out, it follows that P/B
is finite.
Let G be a group. Recall that a subgroup H of a group G is called
abnormal in G if g ∈ 〈H, Hg〉 for each element g of G.
Lemma 5. Let G be an infinite locally finite group whose cyclic subgroups
are either almost self-normalizing or ascendant. Then the factor-group
G/Gru(G) is finite. Moreover, if π = Π(G) \ Π(G/Gru(G)) then G =
Q⋋R, where Q is a Sylow π-subgroup of G and R is a Chernikov subgroup.
Proof. Put B = Gru(G). Suppose that G/B is infinite. Then G/B in-
cludes an infinite abelian subgroup A/B [10]. Since the Sylow p-subgroups
of G/B are finite for each prime p by Lemma 4, the set Π(A/B) is infinite.
Let p ∈ Π(A/B) and let P/B be the Sylow p-subgroup of A/B. By this
choice P/B is non-identity, i.e. B does not include P . Lemma 4 shows
that P/B is finite. Being almost locally nilpotent, P has a Carter sub-
group C, that is maximal locally nilpotent self-normalizing subgroup. We
remark also that all Carter subgroup of P are conjugate and abnormal
[20, Theorem 2.1 and Corollary 2.2]. Since C is abnormal in P , CB is also
abnormal. Then CB/B is abnormal in P/B. On the other hand, P/B is
abelian, which follows that CB/B = P/B.
Let a be an arbitrary element of A. Then Ca is maximal locally nilpo-
tent self-normalizing subgroup of P . As we noted above, the subgroups
C and Ca are conjugate in P , that is there exists an element x of P such
that Ca = Cx. It follows that ax−1 ∈ NA(C), which follows the equality
A = PNA(C). Take into account an equality P = CB, we obtain that
A = BNA(C). If we suppose that a subgroup C is not Chernikov, then
the fact that C is locally nilpotent together with Corollary 1 imply the in-
clusion C 6 B. But this contradicts to the choice of P . This contradiction
shows that C is a Chernikov subgroup. The isomorphism
A/B ∼= BNA(C)/B ∼= NA(C)/(B ∩ NA(C))
shows that NA(C)/(B ∩ NA(C)) has infinite set Π(NA(C)/(B ∩ NA(C)))
and is abelian. Since C is abnormal in P , NP (C) = C. It follows that
B ∩ NA(C) = B ∩ P ∩ NA(C) = B ∩ NP (C) = B ∩ C.
Together with C/(B ∩ C) ∼= CB/B = P/B and the fact, that P/B
is finite, it follows that NA(C)/C is an abelian group with infinite set
L. A. Kurdachenko, A. A. Pypka, N. N. Semko 119
Π(NA(C)/C). Put NA(C) = K. Since K/CK(C) is a Chernikov group
(see, for example, [3, Theorem 1.5.16]), CK(C)C/C is an abelian group
with infinite set Π(CK(C)C/C). Let S/C be the Sylow σ′-subgroup of
CK(C)C/C, where σ = Π(C). Clearly S/C is countable, so that S = C⋋U
(see, for example, [3, Theorem 2.4.5]), where U is a Sylow σ′-subgroup
of S. An inclusion C 6 CK(C) implies that S = C × U . The fact that U
is an infinite abelian subgroup implies that a subgroup CS(y) is infinite
for every element y ∈ C. In turn out, it implies that a cyclic subgroup
〈y〉 is ascendant in G. Hence y ∈ B. In other words, C 6 B. But in this
case the equality P = CB implies that P = B, which contradicts to the
choice of P . This contradiction proves that G/B is finite.
In particular, a set π is finite. Let Q be a Sylow π′-subgroup of G. The
choice of π shows that Q 6 B, so that Q is normal in G. Using Lemma 4
we obtain that for each prime p ∈ π every Sylow p-subgroup of G is
Chernikov. It follows that B/Q is countable. Take in account that G/B
is finite, we obtain that G/Q is countable. Then G = Q ⋋ R, where R is
a Sylow π-subgroup of G (see, for example, [3, Theorem 2.4.5]). Clearly
R is a Chernikov subgroup.
Lemma 6. Let G be a group and P be a normal Chernikov divisible
p-subgroup of G such that G/CG(P ) is finite. Then P includes a finite
G-invariant subgroup F such that P/F = U1/F × . . .×Uk/F , where Uj/F
is a G-quasifinite subgroup, 1 6 j 6 k.
Proof. Let
Q = {Q| Q is an infinite G-invariant subgroup of P}.
Since P is Chernikov, it satisfies the minimal condition on subgroups, and
therefore Q has a minimal element, say V1. Clearly V1 is G-quasifinite.
Then V1 must be divisible and hence V1 has a direct complement in
P (see, for example, [4, Theorem 21.2]). It follows that P includes a
G-invariant subgroup R1 such that P = V1R1 and the intersection V1 ∩
R1 is finite (see, for example, [11, Corollary 5.11]). Put F1 = V1 ∩ R1,
then P/F1 = V1/F1 × R1/F1. Clearly V1/F1 is G-quasifinite. Now we
choose in R1/F1 a G-invariant G-quasifinite subgroup V2/F1. Using the
above arguments, we have found a G-invariant subgroup R2/F1 such that
R1/F1 = (V2/F1)(R2/F1) and the intersection V2/F1 ∩ R2/F1 = F2/F1 is
finite. By such choice we have
P/F2 = (V1F2/F2) × V2/F2 × R2/F2.
120 The groups with two types of subgroups
The both factors V1F2/F2 and V2/F2 are G-quasifinite. Using the similar
arguments, after finitely many steps we prove the required result.
Lemma 7. Let G be a locally finite group whose cyclic subgroups are
either almost self-normalizing or ascendant. Suppose that G = Q ⋋ F ,
where Q = Gru(G), Π(Q) ∩ Π(F ) = ∅ and F is a finite subgroup. If Q
is infinite, then
(i) Q is nilpotent-by-finite;
(ii) CF (Q) = 〈1〉;
(iii) if g 6∈ Q, then CG(g) is finite;
(iv) if p ∈ Π(F ) and p 6= 2, then Sylow p-subgroup of F is cyclic;
(v) Sylow 2-subgroup of F is cyclic or generalized quaternion group;
(vi) every subgroup of order pq of F , p, q ∈ Π(F ), is cyclic.
Proof. Choose an arbitrary element g 6∈ Q and suppose that CG(g) is
infinite. Since g has finite order, the index |CG(g) : 〈g〉| is infinite. In turn
out, it implies 〈g〉 has infinite index in its normalizer, which show that
〈g〉 is ascendant in G. This contradiction shows that CG(g) is finite.
Let y ∈ CF (Q), then Q 6 CG(y), in particular, CG(y) is infinite. As
we have seen above in this case y ∈ Gru(G) = Q, i.e. y ∈ Q ∩ F = 〈1〉.
Let p ∈ Π(F ) and g be an element of F , having order p. By above
proved CQ(g) is finite. Then Q is nilpotent-by-finite [19, Theorem 1.2].
Suppose first that Q is not Chernikov. Then Q includes an F -invariant
abelian subgroup A, which is not Chernikov [21]. Let y be an arbitrary
element of F . The equality Π(Q) ∩ Π(F ) = ∅ implies that A = CA(y) ×
[A, y] [1, Proposition 2.12]. By above proved CA(y) is finite, so that [A, y]
has finite index in A. It is valid for every element y ∈ F . Therefore, the
finiteness of F implies that a subgroup C =
⋂
y∈F
[A, y] has finite index in A.
By its choice CC(y) = 〈1〉 for every element y ∈ F . Put E =
⋂
y∈F
Cy, then
E has finite index in A (in particular, E is infinite), E is F -invariant and
CE(y) = 〈1〉 for every element y ∈ F . Since F is finite, we can choose in
E minimal F -invariant subgroup V . Then CV (y) = 〈1〉 for every element
y ∈ F . Therefore, F satisfies the conditions (iv)-(vi) by [9, Satz V.8.15].
Suppose now that Q is a Chernikov subgroup. Denote by D the
divisible part of Q. The equality Π(Q) ∩ Π(F ) = ∅ implies again that
D = CD(y) × [D, y] [1, Proposition 2.12] for every element y ∈ F . If
we suppose that CD(y) 6= 〈1〉, then CD(y) must be infinite, and we
obtain a contradiction with condition (iii). This contradiction shows that
CD(y) = 〈1〉 for every element y ∈ F . Again choose in D a minimal
L. A. Kurdachenko, A. A. Pypka, N. N. Semko 121
F -invariant subgroup W . Then CW (y) = 〈1〉 for every element y ∈ F .
Therefore, F satisfies the conditions (iv)-(vi) by [9, Satz V.8.15].
Lemma 8. Let G be a Chernikov group and P be a divisible part of G.
Suppose that P is a p-subgroup and CG(g) is finite for each p′-element g.
Then the following assertions hold:
(i) if q is a prime and q 6∈ {2, p}, then Sylow q-subgroup of G/Gru(G)
is cyclic;
(ii) if p 6= 2, then Sylow 2-subgroup of G/Gru(G) is cyclic or generalized
quaternion group.
Proof. Put C = Op(G). Our conditions yields that Gru(G) = C. Let q
be a prime, q 6= p, and Q/C be a Sylow q-subgroup of G/C. Since C is
a p-subgroup, Q = C ⋋ R, where R is a Sylow q-subgroup of Q (see, for
example, [3, Theorem 2.4.5]). Choose in R an arbitrary abelian subgroup
A. Let
S = {S| S is an infinite A-invariant subgroup of P}.
Since P is Chernikov, it satisfies the minimal condition on subgroups, and
therefore S has a minimal element, say V . Clearly V is A-quasifinite. In
particular, it follows that V is a divisible abelian subgroup. Since A is a
p′-subgroup, V = CV (y) × [V, y] for each element y ∈ A [1, Proposition
2.12]. The fact, that A is abelian, implies that the subgroups CA(y) and
[V, y] are A-invariant. Therefore, if we assume that CA(y) 6= 〈1〉, then
CA(y) must be infinite, and we obtain a contradiction. This contradiction
shows that CA(y) = 〈1〉 for each element y ∈ A. Using Lemma 3.1 of
paper [7] we obtain that a subgroup A is cyclic. In particular, ζ(R) is
cyclic. Let 〈d〉 = Ω1(ζ(R)). Suppose that x is an element of R, having
order q. If 〈d, x〉 6= 〈d〉, then 〈d, x〉 is an elementary abelian subgroup of
order q2. But in this case it is not cyclic, and we obtain a contradiction
with above proved. This contradiction shows that 〈d, x〉 = 〈d〉. In other
words, R has only one subgroup of order q. Then R (and hence Q/C)
is cyclic, whenever q 6= 2, and Q/C is cyclic or generalized quaternion
group, whenever q = 2 (see, for example, [9, Satz III.8.2]).
Lemma 9. Let G be an infinite periodic group whose cyclic subgroups are
either almost self-normalizing or ascendant. If g 6∈ Gru(G), then CG(g)
is finite.
Proof. Choose an arbitrary element g 6∈ Gru(G) and suppose that CG(g)
is infinite. Since g has finite order, the index |CG(g) : 〈g〉| is infinite. In
122 The groups with two types of subgroups
turn out, it implies that 〈g〉 has infinite index in its normalizer, which
show that 〈g〉 is ascendant in G. This contradiction shows that CG(g) is
finite.
Lemma 10. Let G be an infinite periodic group whose cyclic subgroups
are either almost self-normalizing or ascendant. If K is a finite normal
subgroup of G, then every cyclic subgroup of G/K is either almost self-
normalizing or ascendant. More precisely, for each element g 6∈ Gru(G)
the centralizer CG/K(gK) is finite.
Proof. At once we note that K 6 Gru(G). In fact, since K is finite, every
element g of K has only finitely many conjugates in G. Then CG(g) has
finite index in G, in particular, CG(g) is infinite. As we have seen above,
in this case the index |NG(〈g〉) : 〈g〉| is infinite, which show that 〈g〉 is
ascendant in G.
Let g be an arbitrary element of G \ Gru(G) and suppose that the
index |NG/K(〈gK〉) : 〈gK〉| is infinite. Put V/K = NG/K(〈gK〉) and X =
〈g, K〉. By its choice X is a normal subgroup of infinite subgroup V . Since
g has finite order, a subgroup X is finite. It follows that CV (X) has finite
index in V , in particular, it is infinite. An inclusion CV (X) 6 CV (g) shows
that CG(g) is infinite and we obtain a contradiction. This contradiction
shows that a cyclic subgroup 〈gK〉 has finite index in its normalizer. It
follows that NG/K(〈gK〉) is finite and hence CG/K(gK) is finite.
If g ∈ Gru(G), then a cyclic subgroup 〈g〉 is ascendant in G. Therefore,
〈gK〉 is ascendant in G/K. In other words, every cyclic subgroup of G/K
is either almost self-normalizing or ascendant.
Now we can describe the general structure of locally finite groups,
whose cyclic subgroups are either almost self-normalizing or ascendant.
2. The proofs of the main results
Proof of Theorem A. Put B = Gru(G) and let Q be a Sylow π-subgroup
of G, π = Π(G) \ σ. Lemma 5 shows that G/B is finite and a group G has
a semidirect decomposition: G = Q ⋋ R, where R is a Chernikov Sylow
σ-subgroup of G.
(v) follows from Lemma 9.
Let y ∈ CR(Q), then Q 6 CG(y), in particular, CG(y) is infinite. As
we have seen above in this case y ∈ Gru(G).
Suppose that Q is infinite and consider a factor-group G/C, where
C = CR(Q) is a Sylow σ-subgroup of B. We have G/C = QC/C ⋋ R/C.
L. A. Kurdachenko, A. A. Pypka, N. N. Semko 123
Every cyclic subgroup of QC/C is ascendant in G/C. Let zC 6∈ QC/C and
suppose that CG/C(zC) is infinite. Then CQC/C(zC) = Z/C is infinite.
We have Z = D × C, where D is a Sylow π-subgroup of Z. The fact, that
Z/C is infinite, implies that D is infinite. For every element d ∈ D we
have [z, d] ∈ C. On the other hand, D 6 Q and Q is a normal subgroup
of G, therefore [z, d] ∈ Q, that is [z, d] ∈ C ∩ Q = 〈1〉. This shows that
D 6 CG(z), in particular CG(z) is infinite. However it contradicts to
(v). This contradiction shows that CG/C(zC) is finite for every element
zC 6∈ QC/C. It is not hard to prove that in this case a subgroup 〈zC〉 has
finite index in its normalizer. In other words, every cyclic subgroup of G/C
is ascendant or almost self-normalizing. Furthermore, QC/C = Gru(G/C)
and we can use Lemma 7. By Lemma 7 QC/C and hence Q is nilpotent-
by-finite.
If Q is finite, then Gru(G) is a Chernikov subgroup, in particular, it
is abelian-by-finite.
Now we can obtain some additional information about the structure
of factor-group G/Gru(G).
Proof of Corollary A1. Put B = Gru(G), then B = P × R, where P is
a Sylow p-subgroup of B and R is a Sylow p′-subgroup of B. We have
G = Q⋋R, where R is a Chernikov Sylow π-subgroup of G, π = Π(G)\σ.
Consider a factor-group G/C, where C = CR(Q) = R ∩ B is a Sylow σ-
subgroup of B. We have G/C = QC/C ⋋R/C. As in a proof of Theorem
A we can show that every cyclic subgroup of QC/C is ascendant in G/C
and every cyclic subgroup of G/C, which does not lie in QC/C, is almost
self-normalizing. Furthermore, QC/C = Gru(G/C), hence we can apply
Lemma 7 and prove this result.
Proof of Corollary A2. Put B = Gru(G). Then B = P × R, where P
is a Sylow p-subgroup of B and R is a Sylow p′-subgroup of B. By our
conditions R is finite. Consider a factor-group G/R. Suppose that the
Sylow p′-subgroup S/R of Gru(G/R) is non-identity. Let R 6= xR ∈ S/R.
Since Sylow p-subgroup of Gru(G/R) includes BR/R, x 6∈ B. Lemma 10
shows that CG/R(xR) is finite. On the other hand, since Gru(G/R) is
locally nilpotent, PR/R 6 CG/R(xR) and we obtain a contradiction. This
contradiction shows that Gru(G/R) is a p-subgroup. If xR 6∈ Gru(G/R)
and xR is a p′-element, then x 6∈ B and CG(x) is finite by Lemma 9.
Using Lemma 10 we obtain that CG/R(xR) is finite. The application of
Lemma 8 yields that Sylow q-subgroup of (G/R)/Gru(G/R) is cyclic
whenever q 6∈ {2, p}, and Sylow 2-subgroup of (G/R)/Gru(G/R) is cyclic
124 The groups with two types of subgroups
or generalized quaternion group, if p 6= 2. Remain to note that the Sylow q-
subgroups of F are isomorphic to Sylow q-subgroups of (G/R)/Gru(G/R),
because Gru(G/R) is a p-subgroup.
Proof of Corollary A3. Let p, r ∈ Sp(G), B = Gru(G). Then B = P ×
R×S, where P is a Sylow p-subgroup of B, R is a Sylow r-subgroup of B,
S is a Sylow {p, r}′-subgroup of B. Consider a factor-group G/D, where
D = R × S. Let D 6= xD and assume that xD is a p′-element. In this
case xD 6∈ PD/D = B/D. Let Z/D = CB/D(xD). We have Z = C × D,
where C is a Sylow p-subgroup of Z. By choice of C we have [x, C] 6 D.
On the other hand, C 6 P and since P is G-invariant, then [x, C] 6 P .
Thus [x, C] 6 D ∩ P = 〈1〉, which shows that CB/D(xD) = CB(x)D/D.
The fact that x 6∈ B together with Lemma 9 shows that CB(x) is finite, so
that CB/D(xD) is finite. Since G/B is finite, CG/D(xD) is likewise finite.
If we suppose now that Gru(G/D) is a not p-subgroup, then it con-
tains some p′-element yD. Since Gru(G/D) is locally nilpotent, B/D 6
CG/D(xD), in particular, CG/D(xD) is infinite, which contradicts to above
proved. This contradiction proves that Gru(G/D) is a p-subgroup.
The application of Lemma 8 yields that the Sylow q-subgroup of
the factor-group (G/D)/Gru(G/D) is cyclic whenever q 6∈ {2, p}, and
Sylow 2-subgroup of (G/D)/Gru(G/D) is cyclic or generalized quaternion
group, if p 6= 2. Since Gru(G/D) is a p-subgroup, the Sylow q-subgroups
of F are isomorphic to Sylow q-subgroups of (G/D)/Gru(G/D).
Consider now a factor-group G/PS. Using the above arguments, we
obtain that the Sylow p-subgroup of F is cyclic whenever p 6= 2, and Sylow
p-subgroup of F is cyclic or generalized quaternion group, if p = 2.
Proof of Theorem B. Put B = B(G). Repeating almost word to word the
proof of Theorem A, we will prove assertions (i)-(v). If Q is infinite, then
using arguments of a proof of Lemma 7, we obtain that Q is nilpotent-by-
finite. Let W be a nilpotent normal subgroup of Q, having finite index.
Then Q = WH for some finite subgroup H. An inclusion H 6 B(G)
implies that H is subnormal in G. Then WH is a nilpotent subgroup
[8, Lemma 4]. A subgroup R ∩ B is Chernikov, and being a Baer group,
it is central-by-finite [8, Corollary 1 to Lemma 4]. In particular, R ∩ B is
nilpotent. Moreover, let D be a divisible part of R∩B. Then D 6 ζ(R∩B).
Let T be a finite subgroup such that R ∩ B = TD. Clearly T is normal
in R ∩ B, and hence in B. Since G/B is finite, T G = U is also finite.
Then (R ∩ B)/U = UD/U is divisible. Therefore, using Lemma 6 we
obtain (vi).
L. A. Kurdachenko, A. A. Pypka, N. N. Semko 125
Proof of Corollary B1. Let L be a locally nilpotent radical of G. Since
B(G) 6 L, G/L is finite by Theorem B. In particular, L is infinite.
Suppose first that L is not Chernikov. Then CL(g) is infinite by
Lemma 3. It follows that every cyclic subgroup of L is G-invariant. Then
every subgroup of L is G-invariant. In particular, L = B(G). Furthermore,
L is a Dedekind group. Then either L is abelian, or L = Q×E ×R, where
Q is a quaternion group, E is elementary abelian 2-subgroup and R is
an abelian 2′-subgroup (see, for example, [14, Theorem 6.1.1]). We note
here, that E must be finite. In fact, every non-identity cyclic subgroup
of E is G-invariant, and being a subgroup of order 2, lies in the center
of G. Hence if we suppose that E is infinite, then ζ(G) is infinite. But
in this case CG(g) is infinite for each element g ∈ G, so that every cyclic
subgroup of G is normal in G, and G must be Dedekind.
Let H be an infinite subgroup of L. If x ∈ CG(H), then H 6 CG(x),
so that CG(x) is infinite and x ∈ B(G). Hence CG(H) 6 L, in particular,
CG(L) 6 L. The fact, that every subgroup of L is G-invariant, implies
that G/CG(L) is abelian (see, for example, [18, Theorem 1.5.1]).
If Sylow 2-subgroup D of L is infinite, then by above proved L is abelian.
If we suppose that Ω1(D) is infinite, then using the above arguments,
we obtain that G is a Dedekind group. This implies that Ω1(D) is finite.
Then D is a Chernikov group. Being infinite, D includes a quasicyclic
2-subgroup W . As we have seen above, W is G-invariant and CG(W ) 6 L.
Furthermore, G/CG(W ) is isomorphic to a periodic subgroup of Aut(W ).
We recall that Aut(W ) is isomorphic to the multiplicative group of a ring
of integer 2-adic numbers (see, for example, [5, Section 113, Example 3]).
Recall also, that a periodic subgroup of the multiplicative group of a ring
of integer 2-adic numbers has order 2 (see, for example, [5, Section 128,
Example 2]). Thus in this case the factor-group G/L has order 2.
Suppose now that there exists an odd prime p such that Sylow p-
subgroup P of L is infinite. By above proved P is abelian. Assume that
Ω1(P ) is infinite. Then again we have an inclusion CG(Ω1(P )) 6 L. Since
every subgroup of Ω1(P ) is G-invariant, G/CG(Ω1(P )) is a cyclic group,
whose order divides p−1. Hence G/L is a cyclic group, whose order divides
p − 1. Suppose now that Ω1(P ) is finite. Then P is a Chernikov group.
In this case the orders of elements of P are not bounded. Since every
subgroup of P is G-invariant, G/P is isomorphic to a periodic subgroup
of the multiplicative group of a ring of integer p-adic numbers (see, for
example, [18, Theorem 1.5.6]). We recall that a periodic subgroup of the
multiplicative group of a ring of integer p-adic numbers is cyclic and its
126 The groups with two types of subgroups
order divides p − 1 (see, for example, [5, Section 128, Example 2]). Thus
in this case the factor-group G/L is cyclic and its order divides p − 1.
Suppose now that the Sylow p-subgroups of L are finite for all primes p.
Since L is not Chernikov subgroup, Π(L) is infinite. Let σ = Π(G/L), then
σ is finite by Theorem B. It follows that Sylow σ-subgroup K of L is finite.
Theorem B shows that CG(g) is finite for each g 6∈ L. Lemma 10 implies
that CG/K(gK) is finite for each g 6∈ L. Obviously L/K is normal Sylow
σ′-subgroup of G/K, so that G/K = L/K ⋋ S/K, where S/K is a finite
Sylow σ-subgroup of G/K. Since Π(L/K) is infinite, we can find in L/K
a finite σ′-subgroup R/K such that R/K ∩ CG/K(gK) = 〈1〉 for every
element gK ∈ S/K (recall that every subgroup of L/K is G-invariant).
Taking into account the fact, that S/K is abelian (S/K ∼= G/L) and
Satz V.8.15 of a book [9], we obtain that every Sylow subgroup of S/K
is cyclic, and therefore S/K is cyclic.
Consider now the case when L is a Chernikov subgroup. By Theorem B
the Baer radical B(G) is nilpotent. Corollary 1 shows that every element
of B(G) has infinite centralizers. Then every cyclic subgroup of B(G)
is G-invariant, and therefore every subgroup of B(G) is G-invariant. As
above we can shows that B(G) includes a centralizer of each its infinite
subgroup. Since G/B(G) is finite, B(G) is infinite. Being Chernikov,
B(G) includes a quasicyclic p-subgroup for some prime p. Using the above
arguments, we obtain that G/B(G) is cyclic and its order divides p − 1.
Furthermore, if p = 2, then B(G) is abelian and G/B(G) has order 2.
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Contact information
L. A. Kurdachenko,
A. A. Pypka
Department of Geometry and Algebra, Oles
Honchar Dnipropetrovsk National University,
Gagarin prospect, 72, Dnipropetrovsk, 49010,
Ukraine
E-Mail(s): lkurdachenko@i.ua,
pypka@ua.fm
N. N. Semko Department of Mathematics, National Univer-
sity of The State Tax Service of Ukraine, Irpen,
Ukraine
E-Mail(s): n_semko@mail.ru
Received by the editors: 10.11.2015.
|
| id | nasplib_isofts_kiev_ua-123456789-155208 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1726-3255 |
| language | English |
| last_indexed | 2025-12-07T15:18:21Z |
| publishDate | 2016 |
| publisher | Інститут прикладної математики і механіки НАН України |
| record_format | dspace |
| spelling | Kurdachenko, L.A. Pypka, A.A. Semko, N.N. 2019-06-16T10:59:41Z 2019-06-16T10:59:41Z 2016 The groups whose cyclic subgroups are either ascendant or almost self-normalizing / L.A. Kurdachenko, A.A. Pypka, N.N. Semko // Algebra and Discrete Mathematics. — 2016. — Vol. 21, № 1. — С. 111-127. — Бібліогр.: 21 назв. — англ. 1726-3255 2010 MSC:20E15, 20F19, 20F22, 20F50. https://nasplib.isofts.kiev.ua/handle/123456789/155208 The main result of this paper shows a description of locally finite groups, whose cyclic subgroups are either almost self-normalizing or ascendant. Also, we obtained some natural corollaries of the above situation. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics The groups whose cyclic subgroups are either ascendant or almost self-normalizing Article published earlier |
| spellingShingle | The groups whose cyclic subgroups are either ascendant or almost self-normalizing Kurdachenko, L.A. Pypka, A.A. Semko, N.N. |
| title | The groups whose cyclic subgroups are either ascendant or almost self-normalizing |
| title_full | The groups whose cyclic subgroups are either ascendant or almost self-normalizing |
| title_fullStr | The groups whose cyclic subgroups are either ascendant or almost self-normalizing |
| title_full_unstemmed | The groups whose cyclic subgroups are either ascendant or almost self-normalizing |
| title_short | The groups whose cyclic subgroups are either ascendant or almost self-normalizing |
| title_sort | groups whose cyclic subgroups are either ascendant or almost self-normalizing |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/155208 |
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