Groups all cyclic subgroups of which are BN A-subgroups
Suppose that $G$ is a finite group and $H$ is a subgroup of $G$. We say that $H$ is a BN A-subgroup of $G$ if either $H^x = H$ or $x \in \langle H, H^x\rangle$ for all $x \in G$. The BN A-subgroups of $G$ are between normal and abnormal subgroups of $G$. We obtain some new characterizations for f...
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| author | He, X. Li, S. Wang, Youyu Ге, Х. Лі, С. Ван, Й. |
| author_facet | He, X. Li, S. Wang, Youyu Ге, Х. Лі, С. Ван, Й. |
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| description | Suppose that $G$ is a finite group and $H$ is a subgroup of $G$. We say that $H$ is a BN A-subgroup of $G$ if either $H^x = H$ or $x \in \langle H, H^x\rangle$ for all $x \in G$. The BN A-subgroups of $G$ are between normal and abnormal subgroups of $G$. We obtain some new characterizations for finite groups based on the assumption that all cyclic subgroups are BN A-subgroups. |
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UDC 512.5
X. He, S. Li (Dep. Math., Guangxi Univ., Nanning, Guangxi, China),
Y. Wang (Lingnan College and Dep. Math., Sun Yat-sen Univ., Guangzhou, China)
GROUPS ALL CYCLIC SUBGROUPS OF WHICH ARE \bfitB \bfitN \bfitA -SUBGROUPS*
ГРУПИ, ВСI ЦИКЛIЧНI ПIДГРУПИ ЯКИХ Є \bfitB \bfitN \bfitA -ПIДГРУПАМИ
Suppose that G is a finite group and H is a subgroup of G. We say that H is a BNA-subgroup of G if either Hx = H
or x \in \langle H,Hx\rangle for all x \in G. The BNA-subgroups of G are between normal and abnormal subgroups of G. We obtain
some new characterizations for finite groups based on the assumption that all cyclic subgroups are BNA-subgroups.
Нехай G — скiнченна група, а H — пiдгрупа групи G. Говоримо, що H — BNA-пiдгрупа групи G, якщо Hx = H
або x \in \langle H,Hx\rangle для всiх x \in G. BNA-пiдгрупи групи G знаходяться мiж нормальними та анормальними
пiдгрупами G. Отримано деякi новi характеристики скiнченних груп на основi припущення, що всi циклiчнi
пiдгрупи є BNA-пiдгрупами.
1. Introduction. Throughout this article, all groups are finite. We use conventional notions and
notation. The reader is referred to [1]. G always denotes a finite group, | G| is the order of G, \pi (G)
denotes the set of all primes dividing | G| , Gp is a Sylow p-subgroup of G for some p \in \pi (G), G\prime
is the derived subgroup of G, H \lessdot G means that H is a maximal subgroup of G.
A class \scrF of groups is called a formation provided that (i) if G \in \scrF and H \triangleright G, then G/H \in \scrF ,
and (ii) if G/M and G/N are in \scrF , then G/(M \cap N) is in \scrF for all normal subgroups M,N of
G. A formation \scrF is said to be saturated if G/\Phi (G) \in \scrF implies that G \in \scrF (see [1], Chapter VI).
Throughout this article, we use \scrU to denote the formation of all supersoluble groups. Clearly, \scrU is
saturated [1].
An interesting investigation in finite groups theory is to determine the structure of finite groups
using the generalized normal subgroups. Normal subgroup is a fundamental concept. As it is well
known, groups whose all subgroups are normal are called Dedekind groups. Restricting the system
of groups on which the condition of normality is imposed, one can obtain some generalizations of
Dedekind groups. For instance, groups whose all subgroups are quasinormal are called Iwasawa
groups, and groups all subgroups of which are subnormal are called nilpotent groups.
Recall that a subgroup H of G is said to be an abnormal subgroup if x \in \langle H,Hx\rangle for all x \in G.
There have been several papers on normal subgroups and abnormal subgroups. A. Fattahi classified
the finite groups with only normal and abnormal subgroups [2]. G. Ebert and S. Bauman studied
the finite groups whose subgroups are either subnormal or abnormal [3]. Cuccia showed that if G is
a finite group and, for every minimal subgroup X of G, CG(X) is either subnormal or abnormal,
then G is soluble [4]. Recently, Liu and Li in [5] classified CLT -groups with normal or abnormal
subgroups. The concept of abnormal subgroup is, in a sense, opposite of that of normal subgroup.
Precisely speaking, G has only a subgroup, itself, that is both normal and abnormal in G. Each
maximal subgroup of G is either normal or abnormal. In [6], we analyze and introduce the following
generalization of normal subgroup and abnormal subgroup.
* Supported by the NNSF of China (No. 11401116, No. 11226045, No. 11361006, No. 11461004), NSF of Guangxi
(No. 2014GXNSFBA118003).
c\bigcirc X. HE, S. LI, Y. WANG, 2017
284 ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 2
GROUPS ALL CYCLIC SUBGROUPS OF WHICH ARE BNA-SUBGROUPS 285
Definition 1.1 ([6], Definition 1.1). A subgroup H of G is called a BNA-subgroup (Between
Normal subgroup and Abnormal subgroup) of G if either Hx = H or x \in \langle H,Hx\rangle for all x \in G,
H is also said to be BNA-normal in G.
In this article, we investigate the groups whose all cyclic subgroups of prime power order are
BNA-subgroups and prove the following main result.
Theorem 1.1. All cyclic subgroups of G of prime power order are BNA-subgroups of G if and
only if one of the following statements holds:
(1) G is a Dedekind group.
(2) G = G\prime \rtimes \langle x\rangle , where G\prime is a Hall subgroup of G and is Abelian, x induces a power
automorphism of order p on G\prime , p is the smallest prime of \pi (G).
(3) G is a finite group whose every subgroup is a BNA-subgroup of G.
(4) G is a BNA-subgroup transitive group.
2. Preliminaries. In this section, we collect some known results which are needed in the proof
of our results.
Lemma 2.1 ([6], Lemma 2.1). Let H \leq K \leq G and N \trianglelefteq G. Suppose that H is a BNA-
subgroup of G. Then the following statements hold:
(1) H is a BNA-subgroup of K.
(2) HN is a BNA-subgroup of G.
(3) HN/N is a BNA-subgroup of G/N.
(4) Any maximal subgroup of G is a BNA-subgroup of G.
Lemma 2.2 ([6], Lemma 2.2). Let H be a BNA-subgroup of G. Then the following statements
hold:
(1) The normal closure HG = H or HG = G.
(2) If, in addition, H is subnormal in G, then H is normal in G.
Lemma 2.3. Let H be a BNA-subgroup of G. If there exists a proper normal subgroup M
of G with H \leq M, then H \unlhd G.
Proof. It is easy to follow from (1) of Lemma 2.2.
Lemma 2.4 ([6], Lemma 2.3). Let H be a BNA-subgroup of G. Then:
(1) NG(H) \leq \langle H,Hx\rangle whenever Hx \not = H.
(2) If NG(H) \leq K \leq G, then K is an abnormal subgroup of G.
Lemma 2.5. Let H \leq G. Then H is a BNA-subgroup of G if and only if NG(H) is abnormal
in G and NG(H) \leq \langle H,Hx\rangle whenever x /\in NG(H).
Proof. The necessity is clear by Lemma 2.4. We now prove the sufficiency. Let x be an element
of G such that Hx \not = H. It’s sufficient to show that x \in \langle H,Hx\rangle .
In fact, by hypothesis, NG(H) \leq \langle H,Hx\rangle . On the other hand, (Hx)x
- 1
= H \not = Hx. Applying
again the hypothesis, we have that NG(H
x) \leq \langle Hx, (Hx)x
- 1\rangle = \langle Hx, H\rangle . So \langle NG(H), NG(H
x)\rangle \leq
\leq \langle H,Hx\rangle . Thus it suffices to show that x \in \langle NG(H), NG(H
x)\rangle . Indeed, NG(H) is abnormal in
G by hypothesis, which requires x \in \langle NG(H), NG(H
x)\rangle .
Lemma 2.6 ([7], Proposition 3.11). Let \scrF 1 = LF (F1) and \scrF 2 = LF (F2), where Fi is both an
integrated and full formation function of \scrF i, i = 1, 2. Then the following statements are equivalent:
(1) \scrF 1 \subseteq \scrF 2.
(2) F1(p) \subseteq F2(p) for all p \in P.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 2
286 X. HE, S. LI, Y. WANG
3. Main results.
Theorem 3.1. Suppose that all minimal subgroups and cyclic subgroups of order 4 of G are
BNA-subgroups of G. Then G is supersoluble.
Proof. Let p be the smallest prime of \pi (G) and P be a Sylow p-subgroup of G. If X is a minimal
subgroup or a cyclic subgroup of order 4 in P (if it exists), then X is subnormal in NG(P ). It follows
from Lemma 2.2 that X is normal in NG(P ). Theorem 1 of [8] implies that G is p-nilpotent. Let
H be the complement of P in G. Then G/H \sim = P is supersoluble. It follows from Lemma 2.3 that
all minimal subgroups of H are normal in G. Thus G is supersoluble.
In fact, we have the following stronger result.
Theorem 3.2. Let \scrF be a saturated formation containing \scrU , H a normal subgroup of G such
that G/H \in \scrF . Suppose that all minimal subgroups and cyclic subgroups of order 4 of H are
BNA-subgroups of G. Then G \in \scrF .
Proof. It is easy to see that H is supersoluble by Theorem 3.1. Let q be the largest prime dividing
| H| and Q a Sylow q-subgroup of H. Clearly, Q \unlhd H and so Q \lhd G. Consider the factor groups
(G/Q,H/Q). Since (G/Q)/(H/Q) \sim = G/H \in \scrF , it follows by Lemma 2.1 that (G/Q,H/Q)
satisfy the hypotheses of the theorem. Therefore G/Q \in \scrF by induction on G/Q. Hypotheses of
the theorem implies that arbitrary minimal subgroup L of Q is a BNA-subgroup of G. It follows
by Lemma 2.2 that L\unlhd G. Let Fi, i = 1, 2, be the integrated and full formation function such that
\scrU = LF (F1) and \scrF = LF (F2), respectively. Since L is a normal subgroup of G of prime order,
G/CG(L) \in F1(q). It follows by Lemma 2.6 that G/CG(L) \in F2(q) and so L \leq Z\scrF
\infty (G). Therefore
G \in \scrF by Theorem 4 of [9].
Theorem 3.3. Let G be a nilpotent group. If all cyclic subgroups of G of prime power order
are BNA-subgroups of G, then G is either Abelian or Hamiltonian.
Proof. Let H be an arbitrary cyclic subgroup of G. It is easy to see that H \unlhd G by Lemma 2.2.
Therefore all subgroups of G are normal and the result follows.
Recall that G is called a \scrT -group if H \lhd K \lhd G always implies that H \lhd G [10, p. 388].
We give the following definition.
Definition 3.1 Let H and K be subgroups of G. We call that G is a BNA-subgroup transitive
group if H is a BNA-subgroup of K and K is a BNA-subgroup of G always imply that H is a
BNA-subgroup of G.
Theorem 3.4. All cyclic subgroups of G are BNA-subgroups of G if and only if one of the
following statements holds:
(1) G is a Dedekind group.
(2) G = G\prime \rtimes \langle x\rangle , where G\prime is a Hall subgroup of G and is Abelian, x induces a power
automorphism of order p on G\prime , p is the smallest prime of \pi (G).
(3) G is a finite group whose every subgroup is a BNA-subgroup of G.
(4) G is a BNA-subgroup transitive group.
In fact, we can show the following stronger result.
Theorem 3.5. All cyclic subgroups of G of prime power order are BNA-subgroups of G if and
only if one of the following statements holds:
(1) G is a Dedekind group.
(2) G = G\prime \rtimes \langle x\rangle , where G\prime is a Hall subgroup of G and is Abelian, x induces a power
automorphism of order p on G\prime , p is the smallest prime of \pi (G).
(3) G is a finite group whose every subgroup is a BNA-subgroup of G.
(4) G is a BNA-subgroup transitive group.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 2
GROUPS ALL CYCLIC SUBGROUPS OF WHICH ARE BNA-SUBGROUPS 287
Proof. Necessity. It follows by the hypotheses of the theorem and Theorem 3.1 that G is
supersoluble. If G is nilpotent, then G is a Dedekind group by Theorem 3.3. Therefore we need
only investigate the case that G is nonnilpotent.
Let K be an arbitrary proper subgroup of G which is normal in G. Then every cyclic subgroup
C of K of prime power order satisfies that CG \leq K < G. It is clear that CG = C by Lemma
2.2, that is, C is normal in G. In particular, all subgroups of K are normal in G, therefore K is a
Dedekind group.
Let p denote the smallest prime dividing the order of G. Since G is supersoluble, there is a
normal subgroup M in G of index p. Then
G = MP, (\ast )
where P = \langle x\rangle with xp \in M. By the above argument, every subgroup of M is normal in G.
Consider the following two cases:
Case 1. M is non-Abelian.
In this case, M is an Hamiltonian group. It follows by [10, p. 139] that
M = Q8 \times A\times B,
where Q8 is the quaternion group of order 8, A is an elementary Abelian 2-group and B is an
Abelian group of odd order. In particular, M is of even order. Since p is the smallest prime dividing
the order of G, then p = 2 and | G : M | = 2. Let T be a Sylow 2-subgroup of G which contains P.
Since T is a nilpotent group and satisfies the hypotheses of the theorem, it is clear that all subgroups
of T are normal in T. Of course, T \leq NG(P ) and T is an Hamiltonian group, T = Q8 \times P, where
P = \langle x\rangle with x2 = 1.
Denote K := [B]P. If P is nonnormal in G, since T \leq NG(P ), then there exists b \in B such
that P b \not = P. Lemma 2.4 implies that NG(P ) \leq \langle P, P b\rangle . Moreover, every subgroup of M is normal
in G by the above argument. Hence P \langle b\rangle is a subgroup. Consequently, T \leq \langle P, P b\rangle \leq P \langle b\rangle , which
gives that T = P, contrary to the fact that T is non-Abelian. Therefore, G = M \times P and we
conclude that G is an Hamiltonian group.
Case 2. M is Abelian.
According to (\ast ), we can get that G = MP, where P = \langle x\rangle with xp \in M, p is the smallest
prime dividing the order of G.
Since G is nonnilpotent, it is easy to see that P is nonnormal in G. Thus, there is an element c in
M of prime power order such that [c, P ] \not = 1. Then c is not a p-element. It follows by Lemma 2.4
that NG(P ) \leq \langle P, P c\rangle . Furthermore, every subgroup of M is normal in G by the above argument.
Hence P \langle c\rangle is a subgroup of G. Consequently, NG(P ) \leq P \langle c\rangle and so P is a Sylow p-subgroup
of G. Let X be an arbitrary subgroup of M. It is clear that Xx = X. Therefore x induces an
automorphism with fixed point free, this implies that | P | = p. Obviously, M = G\prime .
Now we have finished the proof of (1) and (2).
By the above proof, it is easy to see that every subgroup of G is a BNA-subgroup of G.
Therefore, (3) holds.
(3) obviously implies (4).
Now we have finished the proof of the necessity.
Sufficiency. Suppose that G is a Dedekind group, then any subgroup of G is normal in G and so
every cyclic subgroup of G of prime power order is a BNA-subgroup of G.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 2
288 X. HE, S. LI, Y. WANG
Suppose that G = G\prime \rtimes \langle x\rangle , where G\prime is a Hall subgroup of G and is Abelian, x induces a
power automorphism of order p on G\prime , p is the smallest prime of \pi (G). Let L be an arbitrary cyclic
subgroup of G of prime power order. Then L has the following two cases:
L \leq G\prime or L \leq \langle xg\rangle for some g \in G\prime .
If L \leq G\prime , then L is normal in G and so is a BNA-subgroup of G.
If L < \langle xg\rangle , then L\times G\prime \leq \langle (xg)p\rangle G\prime \lhd G by Lemma 2.8 of [11]. Since \langle (xg)p\rangle G\prime \lhd G, then
\langle (xg)p\rangle \lhd G. Thus L\lhd G and then L is a BNA-subgroup of G. If L = \langle xg\rangle , then L is pronormal
in G and so NG(L) is abnormal in G by Theorem 7 of [12]. Obviously, NG(L) = L. If not, there is
a subgroup of G\prime which centralizes L and is also centralized by L. This contradicts with x induces a
power automorphism of G\prime of order p. It follows from Lemma 2.5 that L is a BNA-subgroup of G.
Suppose that G is a finite group whose every subgroup is a BNA-subgroup of G. Then the result
is obvious.
Suppose that G is a BNA-subgroup transitive group. Let H be a cyclic subgroup of G of prime
power order. It is easy to know that any maximal subgroup of G is a BNA-subgroup of G by
Lemma 2.1. We know that G has an ascending series of maximal subgroups
H = H0 \lessdot H1 \lessdot H2 \lessdot . . .\lessdot Hr - 1 \lessdot Hr = G,
where Hi - 1 is a maximal subgroup of Hi.
It follows by Lemma 2.1 that maximal subgroup Hi - 1 of Hi is a BNA-subgroup of Hi for
i = 1, 2, . . . , r. Since G is a BNA-subgroup transitive group, then H is a BNA-subgroup of G.
This completes the proof of the theorem.
References
1. Huppert B. Endliche Gruppen I. – Berlin etc.: Springer-Verlag, 1967. – 793 p.
2. Fattahi A. Groups with only normal and abnormal subgroups // J. Algebra. – 1974. – 28, № 1. – P. 15 – 19.
3. Ebert G., Bauman S. A note on subnormal and abnormal chains // J. Algebra. – 1975. – 36, № 2. – P. 287 – 293.
4. Cuccia P., Liotta M. A condition on the minimal subgroups of a finite group // Boll. Unione mat. ital. – 1982. – 1,
№ 6. – P. 303 – 308.
5. Liu J., Li S., He J. CLT -groups with normal or abnormal subgroups // J. Algebra. – 2012. – 362. – P. 99 – 106.
6. He X., Li S., Wang Y. On BNA-normality and solvability of finite groups // Rend. Semin. mat. Univ. Padova. – 2014,
available online at http://rendiconti.math.unipd.it/forthcoming/downloads/HeLiWang - logo.pdf.
7. Doerk K., Hawkes T. O. Finite soluble groups. – Berlin; New York: Walter De Gruyter, 1992.
8. Li S. On minimal subgroups of finite groups // Communs Algebra. – 1994. – 22, № 6. – P. 1913 – 1918.
9. Ballester-Bolinches A., Pedraza-Aguilera M. C. On minimal subgroup of finite groups // Acta Math. Hung. – 1996. –
73, № 4. – P. 335 – 342.
10. Robinson D. J. S. A course in the theory of groups. – Berlin; New York: Springer-Verlag, 1993.
11. Wei H., Wang Y. On C\ast -normality and its properties // J. Group Theory. – 2007. – 10. – P. 211 – 223.
12. Baand M. S., Borevich Z. I. On arrangement of intermediate subgroups // Rings and Linear Groups. – Krasnodar:
Kubansk. Univ., 1988. – P. 14 – 41.
Received 17.12.13,
after revision — 06.08.16
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 2
|
| id | umjimathkievua-article-1695 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:10:47Z |
| publishDate | 2017 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/31/65ba2762bb75f4c469b087d6cbe4b131.pdf |
| spelling | umjimathkievua-article-16952019-12-05T09:23:56Z Groups all cyclic subgroups of which are BN A-subgroups Групи, всi циклiчнi пiдгрупи яких є BN A -пiдгрупами He, X. Li, S. Wang, Youyu Ге, Х. Лі, С. Ван, Й. Suppose that $G$ is a finite group and $H$ is a subgroup of $G$. We say that $H$ is a BN A-subgroup of $G$ if either $H^x = H$ or $x \in \langle H, H^x\rangle$ for all $x \in G$. The BN A-subgroups of $G$ are between normal and abnormal subgroups of $G$. We obtain some new characterizations for finite groups based on the assumption that all cyclic subgroups are BN A-subgroups. Нехай $G$ — скiнченна група, а $H$ — пiдгрупа групи $G$. Говоримо, що $H$ — BN A-пiдгрупа групи $G$, якщо $H^x = H$ або $x \in \langle H, H^x\rangle$ для всiх $x \in G$. BN A-пiдгрупи групи $G$ знаходяться мiж нормальними та анормальними пiдгрупами $G$. Отримано деякi новi характеристики скiнченних груп на основi припущення, що всi циклiчнi пiдгрупи є BN A-пiдгрупами. Institute of Mathematics, NAS of Ukraine 2017-02-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1695 Ukrains’kyi Matematychnyi Zhurnal; Vol. 69 No. 2 (2017); 284-288 Український математичний журнал; Том 69 № 2 (2017); 284-288 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1695/677 Copyright (c) 2017 He X.; Li S.; Wang Youyu |
| spellingShingle | He, X. Li, S. Wang, Youyu Ге, Х. Лі, С. Ван, Й. Groups all cyclic subgroups of which are BN A-subgroups |
| title | Groups all cyclic subgroups of which are BN A-subgroups |
| title_alt | Групи, всi циклiчнi пiдгрупи яких є BN A -пiдгрупами |
| title_full | Groups all cyclic subgroups of which are BN A-subgroups |
| title_fullStr | Groups all cyclic subgroups of which are BN A-subgroups |
| title_full_unstemmed | Groups all cyclic subgroups of which are BN A-subgroups |
| title_short | Groups all cyclic subgroups of which are BN A-subgroups |
| title_sort | groups all cyclic subgroups of which are bn a-subgroups |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1695 |
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