Skip to content
Publicly Available Published by De Gruyter May 7, 2019

# Finite groups in which every cyclic subgroup is self-normalizing in its subnormal closure

Guohua Qian
From the journal Journal of Group Theory

## Abstract

For a given prime p, a finite group G is said to be a 𝒞~p-group if every cyclic p-subgroup of G is self-normalizing in its subnormal closure. In this paper, we get some descriptions of 𝒞~p-groups, show that the class of 𝒞~p-groups is a subgroup-closed formation and that Op(G) is a solvable p-nilpotent group for every 𝒞~p-group G. We also prove that if a finite group G is a 𝒞~p-group for all primes p, then every subgroup of G is self-normalizing in its subnormal closure.

## 1 Introduction

In this paper, G always denotes a finite group. For a subgroup H of G, we denote by SG(H) the subnormal closure of H in G, that is, the unique minimal subnormal subgroup of G containing H.

## Definition 1.1.

Let G be a finite group and let p be a given prime.

1. Following [1], G is said to be a 𝒞-group (respectively, a 𝒞p-group) if every subgroup (respectively, every p-subgroup) H of G is self-normalizing in SG(H).

2. G is said to be a 𝒞~-group (respectively, a 𝒞~p-group) if every cyclic subgroup of prime power order (respectively, every cyclic p-subgroup) of G is self-normalizing in its subnormal closure.

It is shown in [1] that a 𝒞p-group G is p-solvable with lp(G), the p-length of G, at most 1. In particular, all 𝒞-groups are solvable. Clearly, a 𝒞-group is necessarily a 𝒞p group for every prime p. Naturally, one may ask if the converse is true. The answer is positive. Actually, we get the following result.

## Theorem 1.2.

For a finite group G and a given prime p, the following equivalences hold:

1. G is a 𝒞p-group if and only if G is a 𝒞~p-group.

2. G is a 𝒞-group if and only if G is a 𝒞~-group.

For a given prime p, it is shown in [1, Theorem 8] that G is a 𝒞p-group if and only if G is p-solvable with lp(G)1, and COp(SG(H))(H)=1 for all subgroups H of G. The following descriptions of 𝒞p-groups appear more clearly.

## Theorem 1.3.

For a finite group G and a prime p, the following statements are equivalent:

1. G is a 𝒞p-group.

2. G is p-solvable with lp(G)1, and if a p-subgroup P of G acts on a p-subgroup V of G, then V=CV(P)[V,P], that is, C[V,P](P)=1.

3. Every subgroup H of G can be expressed as H=NH(P)B, where P is a Sylow p-subgroup of H and B is a normal p-subgroup of H.

In Theorem 1.3 (3), since P acts coprimely on B, we have ([3, Theorem 8.2.7])

CB(P)NH(P)BandB=[B,P]CB(P).

This implies CB(P)=1 and B=[B,P].

Combining Theorems 1.2 and 1.3, we get some descriptions of 𝒞-groups. As shown in [1, Theorem A], the class of 𝒞-groups is a subgroup-closed formation. Actually, this is a direct corollary of Theorem 1.2 and Theorem 1.4.

## Theorem 1.4.

For a given prime p, the class of Cp-groups is a subgroup-closed formation.

Let 𝐅i(G) be the ith ascending Fitting subgroup of G, that is, 𝐅0(G)=1, 𝐅1(G)=𝐅(G) and 𝐅i+1(G)/𝐅i(G)=𝐅(G/𝐅i(G)). The Fitting height of a solvable group G is the smallest integer k such that 𝐅k(G)=G. For a given prime p, we denote by Ωp(G) the characteristic subgroup of G generated by all elements of order p. Clearly, Ωp(G)Op(G).

## Theorem 1.5.

For a Cp-group G, the following statements hold:

1. Op(G) is solvable and p-nilpotent.

2. Ωp(G) has Fitting height at most 2.

Let G be a 𝒞-group. By Theorem 1.5 (2), we get that Ω(G), the characteristic subgroup of G generated by all elements of prime order, has Fitting height at most 2.

All unexplained notation is standard and taken mainly from [3]. The proofs given here are completely self-contained and independent of [1].

## 2 𝒞~p-groups and 𝒞p-groups

In this section, we will investigate 𝒞~p-groups and 𝒞p-groups. For a subgroup D of G, it is easy to see that SG(D)=SSG(D)(D), and this fact will be used freely.

## Lemma 2.1.

For a C~p-group G, it follows that

1. every subgroup of G is a 𝒞~p-group,

2. every quotient group of G is also a 𝒞~p-group,

3. G is p-solvable with lp(G)1.

## Proof.

(1) Let H be a subgroup of G, and let P be a cyclic p-subgroup of H. Since SG(P) is subnormal in G, we have that SG(P)H is subnormal in H. This implies SH(P)SG(P)HSG(P). Now the self-normalization of P in SG(P) yields NSH(P)(P)=P. Hence H is a 𝒞~p-group.

(2) Let NG, and let U/N be a cyclic p-subgroup of G/N. To see the required result, we may assume by induction that N is a minimal normal subgroup of G, and we only need to show the self-normalization of U/N in SG/N(U/N). Assume that SG/N(U/N):=H/N<G/N. Since H is also a 𝒞~p-group by (1), we get by induction that H/N is again a 𝒞~p-group. Thus NH/N(U/N)=U/N, as required. Therefore, we may assume that SG/N(U/N)=G/N.

Let P be a cyclic p-subgroup of G such that U=PN. Note that

NSG(P)(P)=P.

Suppose that SG(P)=G. Then NG(P)=P, so PSylp(G). Let us investigate NG(PN). Clearly, NG(P)NG(PN), and thus NG(P)=NNG(PN)(P). By the Frattini argument and the normality of PN in NG(PN), we get

NG(PN)=PNNNG(PN)(P)=PNNG(P)=PN.

Clearly, NG/N(PN/N)=NG(PN)/N. This yields the required result

NG/N(U/N)=NG/N(PN/N)=NG(PN)/N=PN/N=U/N.

Suppose that SG(P)<G. Let M be a maximal normal subgroup of G such that SG(P)M. Since U/NMN/NG/N and SG/N(U/N)=G/N, we have G/N=MN/N. Now the minimal normality of N implies G=M×N. Note that M is a 𝒞~p-group by (1), and so is G/N. It follows that U/N is self-normalizing in G/N, and we are done.

(3) Note that all subgroups and all quotient groups of G are also 𝒞~p-groups. Suppose that G is not p-solvable. To see a contradiction, we may assume by induction that G is a nonabelian simple group of order divisible by p. Let P be a subgroup of G of order p. Since NG(P)=P by the hypothesis, we have PSylp(G) and NG(P)=CG(P). It follows by [3, Theorem 7.2.1] that G has a normal p-complement, a contradiction. Consequently, G is p-solvable.

Suppose that lp(G)2. To get a contradiction, we may assume, by somewhat more standard inductive arguments, that the following occur:

1. G has a unique minimal normal subgroup, say E;

2. E=Op(G) and Φ(G)=1;

3. G=ME for a maximal subgroup M of G, and M=PK, where M=Op(M), K=Op(K) and 1<PSylp(M) is cyclic. (For example, we show the cyclicity of P. Assume that P is not cyclic. Then G possesses different maximal normal subgroups, say A1,A2. Since lp(A1)1 and lp(A2)1 by induction, we have

lp(G)=lp(A1A2)=max{lp(A1),lp(A2)}1,

a contradiction.)

Now it is easy to see that G is the subnormal closure of P in G. Then we have NG(P)=P by the hypothesis. However, P<PESylp(G), a contradiction. Thus lp(G)1. ∎

## Lemma 2.2.

Let D and V be subgroups of G with (|D|,|V|)=1. Assume that G=DV and that D is self-normalizing in SG(D). Then

C[V,D](D)=1,V=CV(D)[V,D],G=NG(D)[V,D]𝑎𝑛𝑑NG(D)=D×CV(D).

## Proof.

Since D acts coprimely on V, we have V=[V,D]CV(D) by [3, Theorem 8.2.7]. Clearly, [V,D] and D[V,D] are normal in G. Note that (see [3, Theorem 8.2.7]) [V,D,D]=[V,D]. It follows that SG(D)=D[V,D].

Clearly, C[V,D](D)=N[V,D](D). Since NSG(D)(D)=D by the hypothesis, we get

CV(D)[V,D]=C[V,D](D)=N[V,D](D)=1.

This implies V=[V,D]CV(D)=CV(D)[V,D]. Also, we get

NG(D)=NG(D)G=NG(D)(DCV(D)[V,D])=DCV(D)(NG(D)[V,D])=DCV(D)N[V,D](D)=DCV(D)=D×CV(D),

and G=NG(D)[V,D]. ∎

## Lemma 2.3.

For a given prime p, the group G is a Cp-group if and only if G is a C~p-group.

## Proof.

We need only to show the “if” part. Let G be a 𝒞~p-group, and let P be a p-subgroup of G. To see that P is self-normalizing in SG(P), we may assume that P is not cyclic and that SG(P)=G by induction. Notice that G is p-solvable with lp(G)=1. Now SG(P)=G implies

G=Op(G)andG=PV

for a normal p-subgroup V of G. We only need to show that CV(P)=1 since NV(P)=CV(P) and NG(P)=PNV(P).

Let xCV(P), and let D be a maximal subgroup of P. Clearly, both [V,D] and D[V,D] are normal in G. Note that G/D[V,D]=Op(G/D[V,D]). Hence

G/D[V,D]=SG/D[V,D](P[V,D]/D[V,D]).

Since D>1 by the non-cyclicity of P, we have |G/D[V,D]|<|G|. It follows by induction that G/D[V,D] is a 𝒞p-group. Consequently, P[V,D]/D[V,D] is self-normalizing in G/D[V,D]. This implies

xP[V,D],
xP[V,D]CV(P)=[V,D]CV(P)=C[V,D](P)C[V,D](D).

Observe that DV is also a 𝒞p-group by induction; therefore, by Lemma 2.2, we get C[V,D](D)=1. Hence x=1, CV(P)=1, and the proof is complete. ∎

## Corollary 2.4.

For a finite group G and a given prime p, the following statements hold:

1. Assume that G is a 𝒞p-group. Then all subgroups and all quotient groups of G are 𝒞p-groups; also, G is p-solvable with lp(G)1.

2. Assume that G is a 𝒞~-group. Then all subgroups and all quotient groups of G are 𝒞~-groups, G is solvable with lr(G)1 for all primes r, and all subgroups of prime power order of G are self-normalizing in their subnormal closures.

## Proof.

The results follow by Lemma 2.1 and Lemma 2.3. ∎

## Proof of Theorem 1.3.

(1) (2). Assume that G is a 𝒞p-group. By Corollary 2.4, G is p-solvable with lp(G)1. Let P and V be a p-subgroup and a p-subgroup, respectively, of G such that P acts on V. Since PV is again a 𝒞p-group by Corollary 2.4, the required result follows by Lemma 2.2.

(2) (3). Let HG, and let PSylp(H). We may assume that P>1. Clearly, H is p-solvable with lp(H)=1. Then Op(H)=PB for a normal p-subgroup B of H. Clearly, [B,P]=B. Since B=CB(P)[B,P] by the hypothesis, we have 1=CB(P)=NB(P)=NH(P)B. By the Frattini argument, this implies H=NH(P)Op(H)=NH(P)B=NH(P)B, and we are done.

(3) (1). Let D be a nontrivial p-subgroup of G, and let H=SG(D). By the hypothesis, there exist a Sylow p-subgroup P and a normal p-subgroup B of H such that H=NH(P)B. We assume that DP. Since B and PB are normal in H, DB is necessarily subnormal in H. Observe that H=SG(D)=SH(D), and therefore D=P=NH(P). Consequently, D is self-normalizing in its subnormal closure H, as required. ∎

## Proof of Theorem 1.4.

By Corollary 2.4, it suffices to prove that the class of 𝒞p-groups is closed under taking direct products. Let A and B be 𝒞p-groups, and let G=A×B. Then A and B are p-solvable, and each of them has p-length at most 1. Let P be a p-subgroup of G, and write W=SG(P). We work by induction on |A|+|B| to show that P is self-normalizing in W. Since Op(A) and Op(B) are 𝒞p-groups and PWOp(A)×Op(B), by induction, we may assume that A=Op(A) and B=Op(B). In particular, A, B and G are p-nilpotent. Since W=SW(P) is p-nilpotent, we have PSylp(W). Let

VHallp(NW(P)).

To see the self-normalization of P in W, it suffices to show that V=1.

Suppose that PA=PB=G. Since G=A×B, we conclude that A, B and G are p-groups, and we are done. Consequently, we may assume that PB<G. By the definition of the subnormal closure, we easily get

SG/B(PB/B)=SG(P)B/B.

Observe that

VB/BWB/BNG/B(PB/B)=NWB/B(PB/B)=NSG/B(PB/B)(PB/B).

Since G/BA is a 𝒞p-group, we have VB/BPB/B, that is, VPB. Now the p-nilpotency of PB implies VOp(B). Assume that PA<G. The same arguments as above yield VOp(A), and hence VOp(A)Op(B)=1. Assume that PA=G. Then B is a p-group, and thus V=1, as required. ∎

## Proof of Theorem 1.5.

By Corollary 2.4, all subgroups and all quotient groups of G are 𝒞p-groups; also, G is p-solvable with lp(G)1.

(1) Op(G) must be p-nilpotent since lp(G)1. Then G=PV, where PSylp(G) and V=Op(G). To see the solvability of Op(G), we may assume that P>1 and that G=Op(G) by induction. Let 1P1 be a maximal subgroup of P.

Assume that P1>1. Observe that

P1VG,Op(P1V)G,Op(G/Op(P1V))=G/Op(P1V).

Applying the inductive hypothesis to P1V and G/Op(P1V), we get that

Op(P1V)andG/Op(P1V)

are solvable, and so is G.

Assume that P1=1. Then P has order p. Since G=Op(G), we have

SG(P)=G.

This implies NG(P)=P and CV(P)=1. Now G is a Frobenius group with V as its kernel. As is well known, V is nilpotent in this case [3, Theorem 9.5.1], and the result follows.

(2) Let x be an element of order p, and let H=SG(x). Since H is p-solvable with lp(H)=1 and SH(x)=H, we see that H=xV for a normal p-subgroup V of H. We claim that H has Fitting height at most 2. To prove the claim, we may assume that V>1. Since NH(x)=x, H must be a Frobenius group with kernel V. This implies that V is nilpotent, and the claim holds. Now the subnormality of H in G yields H𝐅2(G). Consequently, Ωp(G)𝐅2(G). ∎

## 3 𝒞~-groups and 𝒞-groups

We denote by G𝒩 the nilpotent residual of G, that is, the smallest normal subgroup of G such that G/G𝒩 is nilpotent.

## Lemma 3.1.

Let H be a subgroup of a finite solvable group G. Then SG(H)=G if and only if G=HGN.

## Proof.

Assume that SG(H)=G. Since HG𝒩/G𝒩 is a subgroup of a nilpotent group G/G𝒩, HG𝒩 is subnormal in G. This implies SG(H)HG𝒩, and hence G=HG𝒩. Assume conversely that G=HG𝒩. If SG(H)<G, then there exists a maximal and normal subgroup M of G such that SG(H)M. By the solvability of G, we have G𝒩M. This implies G=HG𝒩M<G, a contradiction. Hence SG(H)=G. ∎

Note that, for a subgroup H of an arbitrary finite group G, “SG(H)=G” does imply “G=HG𝒩”, but the converse is not true. For example, if G is a direct product of a nonabelian simple group and an abelian group H, then G=HG𝒩, but SG(H)=H<G.

Recall that a Carter subgroup of a finite group is a nilpotent and self-normalizing subgroup of the group. The following results about Carter subgroups are well known [2, Chapter VI, § 12].

## Lemma 3.2.

For a finite solvable group G, the following hold:

1. There exists a Carter subgroup of G, and all Carter subgroups of G are conjugate.

2. If T is a Carter subgroup of G, then G=TG𝒩.

## Lemma 3.3.

Assume that G=HV is a C~-group, where a nilpotent group H acts coprimely on a group V. Then C[V,H](H)=1.

## Proof.

Note that, by Corollary 2.4, all subgroups and all quotient groups of G are also 𝒞~-groups. We may assume that 1<H=A×P, where 1<P=Op(H) and A=Op(H) for some prime p. Since H[V,H] is also a 𝒞~-group, we may assume by induction that V=[V,H]. Clearly, [V,P] is normal in G. Since [V,H]=V and P centralizes V/[V,P], we conclude easily that

[V/[V,P],A[V,P]/[V,P]]=V/[V,P].

Observe that (A[V,P]/[V,P])V/[V,P] also satisfies the hypothesis. Thus it follows by induction that

CV/[V,P](A[V,P]/[V,P])=1.

Let xC[V,H](H). Since x centralizes A, we have

x[V,P]CV/[V,P](A[V,P]/[V,P]).

Then xA[V,P], and

xC[V,H](H)A[V,P]=C[V,H](H)[V,P]=C[V,P](H)C[V,P](P).

Note that G is a 𝒞p-group by Lemma 2.3. Considering the action of P on V and applying Theorem 1.3 (2), we get C[V,P](P)=1. Hence x=1, and we are done. ∎

## Proof of Theorem 1.2.

By Lemma 2.3, we only need to show that a 𝒞~-group must be a 𝒞-group. Assume that G is a 𝒞~-group. By Corollary 2.4, all subgroups and all quotient groups of G are 𝒞~-groups; also, G is solvable with lp(G)=1 for all primes p. Assume that G is not a 𝒞-group, and let H<G be such that H is not self-normalizing in SG(H). Let Δ be the set of elements of prime power order, in SG(H), but not in H. Clearly, Δ. In order to see a contradiction, we may assume that |H|+|G| is as small as possible.

By Corollary 2.4, H is not of prime power order, and hence |π(H)|2. Let U=G𝒩. Since all nilpotent groups are 𝒞-groups, we have U>1. Let U/E be a chief factor of G, and assume that U/E is an elementary abelian q-group for some prime q. Now we work for a contradiction via several steps.

(1) SG(H)=G, G=HU, and H is nilpotent. By the minimality of |G|+|H|, we may assume that SG(H)=G. Then G=HU by Lemma 3.1. Suppose that H is not nilpotent, and let W be a Carter subgroup of H. Then W<H, and also H=WH𝒩 by Lemma 3.2. Note that H/(HU)HU/U=G/U is nilpotent; hence it follows that H𝒩HU. Now

G=HU=WH𝒩U=WU,

and thus G=SG(W) by Lemma 3.1. The minimality of |G|+|H| yields that NG(W)=W. Let xΔ, and take M=xH. By the definition of Carter subgroup, Wm is also a Carter subgroup of H for every mM. By Lemma 3.2, it follows that Wm=Wh for some hH. Now

mNM(W)HandM=NM(W)H.

However, NM(W)NG(W)=W; this leads to the contradiction

M=NM(W)H=WH=H.

Hence H is nilpotent.

(2) H is not maximal in G, and G/E=HE/EU/E, E>1. Assume that H is maximal in G. Since SG(H)=G by (1), H is not normal in G. This implies NG(H)=H, a contradiction. Since G=HU, we have G/E=(HE/E)(U/E). If HE=G, then G/EH/(HE) is nilpotent, against the fact that

U=G𝒩>E.

Hence HE<G. As U/E is a chief factor of G, we have G/E=HE/EU/E. Moreover, since H is not maximal in G, we also get E>1.

(3) For every nontrivial normal subgroup N of G, HN/N is a Carter subgroup of G/N, and ΔHN. Since G/N is also a 𝒞~-group and SG/N(HN/N)=G/N, we get by the minimality of |G|+|H| that HN/N is self-normalizing in G/N. This implies that HN/N is a Carter subgroup of G/N. Note that xNG(H) implies xNNG/N(HN/N), and it follows that ΔHN.

(4) For some prime p, Δ consists only of p-elements, and

𝐅(G):=PSylp(G).

Let N be a minimal normal subgroup of G, and assume that N is a p-group for some prime p. Then ΔHN by (3). Let xΔ, and let D be a Hall p-subgroup of H. Obviously, D is a normal Hall p-subgroup of NHN(D), and this means that all p-elements of NHN(D) are contained in D. Observe that

x(HNNG(H))-H(HNNG(D))-D=NHN(D)-D.

It follows that x is a p-element. Hence Δ consists only of p-elements. The arbitrariness of N also implies that all minimal normal subgroups of G are p-groups. In particular, 𝐅(G) is a p-group. Since lp(G)=1, we see that 𝐅(G) is a Sylow p-subgroup of G.

(5) E is a p-group. Assume this is not true. Write N=Op(E) and M=HN. We have ΔM<G by (3) and (2). Clearly, M𝒩N because H is nilpotent, and N/M𝒩 is a p-group because N=Op(N). Observe that, since 𝐅(G) is a p-group and E is not a p-group, we get

1<Op(N)M𝒩NM.

Write D=Op(N). Since, by (3), HD/D is a Carter subgroup of G/D, HD/D is also a Carter subgroup of M/D. It follows by Lemma 3.2 that

M/D=(HD/D)(M/D)𝒩=(HD/D)(M𝒩/D)=HM𝒩/D.

Thus M=HM𝒩. Now Lemma 3.1 implies SM(H)=M, and the minimality of |G|+|H| leads to the contradiction ΔM=.

(6) Final contradiction. Recall that G=HU, U/E is a q-chief factor of G and E is a p-group. Let A be a Hall {p,q}-subgroup of H and let Q be a Sylow q-subgroup of G. Since G has a normal Sylow p-subgroup P by (4), we conclude easily that QP/E is a normal nilpotent subgroup of G/E. Observe that A=1 would imply that G/E=QP/E is nilpotent. Hence A>1. Write X=QP. Clearly, G=AX and [X,A]U.

We claim that [X,A]=U. To prove this claim, we need only to show that U[X,A].

Assume that q=p. Then G=AP and X=P. Since G/[X,A] is clearly nilpotent, we have [X,A]U, and the claim follows.

Assume that qp. Since E is a p-group, we get by the randomness of U/E that E is the unique maximal G-invariant subgroup of U. Consequently, either [X,A]=U or [X,A]E. Observe that if [X,A]E, then G/E is a direct product of a nilpotent {p,q}-subgroup X/E and a nilpotent {p,q}-subgroup, against the non-nilpotency of G/E. Hence [X,A]=U, as claimed.

Now, considering the action of A on X, we get CU(A)=1 by Lemma 3.3. This implies

NU(H)NU(A)=CU(A)=1.

Since G=HU, we get NG(H)=H(UNG(H))=HNU(H)=H. This is the final contradiction, and the proof is complete. ∎

Communicated by Christopher W. Parker

Award Identifier / Grant number: BK20161265

Award Identifier / Grant number: 11871011

Award Identifier / Grant number: 11671063

Funding statement: Project supported by the NSF of Jiangsu Province (No. BK20161265) and the NSF of China (Nos. 11871011, 11671063).

## Acknowledgements

The author is grateful to the referee for his/her valuable comments.

## References

[1] A. Ballester-Bolinches, J. Cossey and Y. Li, On a class of finite soluble groups, J. Group Theory 21 (2018), no. 5, 839–846. 10.1515/jgth-2018-0015Search in Google Scholar

[2] B. Huppert, Endliche Gruppen I, Springer, Berlin, 1967. 10.1007/978-3-642-64981-3Search in Google Scholar

[3] H. Kurzweil and B. Stellmacher, The Theory of Finite Groups. An Introduction, Springer, Berlin, 2004. 10.1007/b97433Search in Google Scholar

Received: 2019-01-13
Revised: 2019-03-25
Published Online: 2019-05-07
Published in Print: 2019-09-01

© 2020 Walter de Gruyter GmbH, Berlin/Boston

Downloaded on 28.11.2022 from frontend.live.degruyter.dgbricks.com/document/doi/10.1515/jgth-2019-0003/html