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Open Mathematics

formerly Central European Journal of Mathematics

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Volume 14, Issue 1 (Jan 2016)

Issues

On CSQ-normal subgroups of finite groups

Yong Xu
  • School of Mathematics and Statistics, Henan University of Science and Technology, Luoyang, Henan 471003, China
  • School of Mathematics and Statistics, Southwest University, Chongqing, 400715, China
  • Email:
/ Xianhua Li
  • Corresponding author
  • School of Mathematical Science, Soochow University, Suzhou, Jiangsu 215006, China
Published Online: 2016-11-03 | DOI: https://doi.org/10.1515/math-2016-0073

Abstract

We introduce a new subgroup embedding property of finite groups called CSQ-normality of subgroups. Using this subgroup property, we determine the structure of finite groups with some CSQ-normal subgroups of Sylow subgroups. As an application of our results, some recent results are generalized.

KeyWords: CSQ-normal subgroup; Nilpotent subgroup; Supersolvable subgroup

MSC 2010: 20D10; 20D15

1 Introduction

All groups in this paper are finite. Let π(G) stand for the set of all prime divisors of the order of a group G. The other notations and terminologies in this paper are standard (see [1]).

Let HG and G ϵ G, then H ≤ 〈H, Hg〉 ≤ 〈H, g〉. It is clear that H = 〈H, Hg〉 for all g ϵ G if and only if HG. In [2], H is called abnormal in G if 〈H, Hg〉 = 〈H, g〉 for all g ϵ G. In [3], the famous Wielandt Theorem shows that H ⊲⊲ 〈H, Hg〉 for all g ϵ G if and only if H ⊲⊲ G. In [4], H is called pronormal in G if H is conjugate to Hg in 〈H, Hg〉 for all g ϵ G. These show that the normalities of a subgroup H in G may be determined by the normalities of a subgroup H in 〈H, Hg〉. This leads us to investigate the properties of G from the relationship between the subgroup H of G and the union of 〈H, Hg〉 for all g ϵ G. On the other hand, Kegel in [5] introduced the concept of S-quasinormal subgroups. A subgroup H of a group G is said to be s-permutable, S-quasinormal, or π-quasinormal in G if PH = HP for all Sylow subgroups P of G. In this paper, we introduce a new generalized normality of subgroups, CSQ-normality, and obtain a criterion for nilpotency and supersolvablity of a group by using the CSQ-normality of subgroups. Now we recall the following definitions. Let G be a finite group. For every n ||G|, if G has a subgroup of order n, then G is called a CLT-group. Furthermore, G is called a QCLT-group if the image of G under every homomorphism is a CLT-group. As an application of our results, some recent results are generalized. For example, Humphreys [6] proved that a QCLT-group of odd order is supersolvable, and we will prove that a QCLT-group of even order is also supersolvable if the maximal subgroups of its Sylow 2-subgroup are all CSQ-normal subgroups.

Definition 1.1: Let H be a subgroup of a group G. We say that H is CSQ-normal in G if H is S-quasinormal inH, Hgfor all g ϵ G.

By [5, Lemma 3], we know that all S-quasinormal subgroups are CSQ-normal subgroups. The following example shows that a CSQ-normal subgroup is not necessarily a S-quasinormal subgroup.

Example 1.2: Let G = A4, H = 〈(12)(34)〉. Obviously, H is not S-quasinormal in G but CSQ-normal in G.

2 Basic definitions and preliminary results

The lemma presented below is crucial in the sequel. The proof is a routine check, and we omit its details.

Lemma 2.1: Let H be a CSQ-normal subgroup of a group G and NG. Then

  • (a)

    If HK, then H is CSQ-normal in K.

  • (b)

    HN|N is CSQ-normal in G|N.

Lemma 2.2: Suppose that every proper subgroup of a group G is nilpotent but G itself is not nilpotent. Then

  • (1)

    There exist some primes p and q such that |G| = pαqβ.

  • (2)

    G has a normal Sylow q-subgroup Q; if q > 2, then exp(Q) = q and if q = 2, then exp(Q) ≤ 4; G also has a cyclic Sylow p-subgroup P = 〈a〉.

  • (3)

    Let c ϵ Q. Then c is a generator if and only if [c, a] ≠ 1.

  • (4)

    If c is a generator of Q, then [c, a] = c−1ca is also a generator of Q.

  • (5)

    If c is a generator of Q, then Q = 〈c, ca, …, cap−2, cap-1 〉, namely, Q = 〈[c, a], [c, a]a, …, [c, a]ap−1〉.

Proof: By [7, Theorem 1.1], the result is true.

As in [8], a minimal nonsupersolvable group is a nonsupersolvable group whose proper subgroups and quotients are supersolvable.

Lemma 2.3: Suppose that a group G is minimal nonsupersolvable. Then G is isomorphic to a group of the form Gt for 1 ≤ t ≤ 6, where the groups Gt are defined in the following way.

  • (I)

    G1 is a minimal nonabelian group and |G1| = pqβ, where pq − 1, β ≥ 2.

  • (II)

    G2 = 〈a, c1and |G2| = pαrp and pα−1 || r − 1, where α ≥ 2. apα=c1r=c2r==cpr=1;; ci cj = cj ci; cia=ci+1, i = 1, 2, …, p − 1; cpa=c1t, where the exponent of t (mod r) is pα−1.

  • (III)

    G3 = 〈a, b, c1and |G3| = 8r2 and 4 | r − 1, a4=c1r=c2r=1, a2 = b2, ba = a−1b, c1a=c2,c2a=c11,c1b=c1s,c2b=c2s, where the exponent of s (mod r) is 4.

  • (IV)

    G4 = 〈a, b, c1and |G4| = pα+β rp and pmax{α, β} | r − 1, where β ≥ 2. apα=bpβ=c1r=c2r==cpr=1; ci cj = cj ci, ab = b1+pβ−1 a; cia=ci+1,i=1,2,,p1;cpa=c1t,cib=ciu1+ipβ1,i=1,2,,p where the exponents of t and u (mod r) are pα−1 and pβ, respectively.

  • (V)

    G5 = 〈a, b, c, c1and |G5| = pα+β+1rp and pmax{α, β} | r − 1. apα=bpβ=cp=c1r=c2r==cpr=1; cicj=cjci,ba=abc,ca=ac,cb=bc,cia=ci+1,i=1,2,,p1; cpa=c1t,cic=ciu,cib=civupi+1, where the exponents of t, v and u (mod r) are pα−1, pβ and p, respectively.

  • (VI)

    G6 = 〈a, b, c1〉, |G6| = pα qrp and pαq | r − 1, p | q − 1, α ≥ 1. apα=bq=c1r=c2r==cpr=1; ci cj = cj ci, i, j = 1, 2, …, p; cia=ci+1,i=1,2,,p1; cpa=c1t;ba=bu,cib=civui1,i=1,2,,p; where the exponents of t, v (mod r) are pα−1 and q, respectively, and the exponent of u (mod q) is p.

Proof: See [8, Corollary 2.2].

Lemma 2.4: Let H be a CSQ-normal subgroup of G. Then

  • (a)

    H x is also a CSQ-normal subgroup of G for any x ϵ G.

  • (b)

    H is subnormal in G.

Proof: (a) By the hypothesis, H is S-quasinormal in 〈H, Hg〉 for all g ϵ G. Then for any x ϵ G, we have that Hx is S-quasinormal in 〈Hx; Hgx〉=〈Hx,(Hx)gx〉 for all g ϵ G. Then one checks easily that ˝ : GG, defined by τ(g)=gx,wherexG,is a bijective map. Since gx runs over G as g does for fixed x, we get that Hx is S-quasinormal in 〈Hx Z, (Hx)gx〉 for all Gx ϵ G. Thus Hx is a CSQ-normal subgroup of G.(b) By the hypothesis, H is S-quasinormal in 〈H, Hg〉 for all g ϵ G. By [5, Theorem 1], we know that H is subnormal in 〈H, Hg〉 for all g ϵ G, so H is subnormal in G by Wielandt’s theorem.

3 Main results

Let be a complete set of Sylow subgroups of a group G, that is, for each prime p dividing the order of G, contains exactly one Sylow p-subgroup of G. Let E = {PE | P ϵ}.

Theorem 3.1: Let G be a group and be a complete set of Sylow subgroups of G. Suppose that EG such that G/E is nilpotent and G is G1-free. If every cyclic subgroup of a Sylow subgroup of E contained in E is a CSQ-normal subgroup of G, then G is nilpotent.

Proof: Assume that the result is false, and let G be a counterexample with least (|G| + |E|).Let H < G. Of course, H is G1-free. Obviously, H/HEHE/E is nilpotent. Suppose that K = HE and Kp is a Sylow p-subgroup of K, so = {Kp|p ϵ β(HE)} is a complete set of Sylow subgroups of HE. Assume that T is a cyclic subgroup of Kp. Since KE, there exists x ϵ E such that KpxpE, where P ϵ . By the hypothesis and Lemma 2.4(a), we get that T is CSQ-normal in G. Then T is CSQ-normal in H by Lemma 2.1(a). Hence all cyclic subgroups of Kp contained in are CSQ-normal in H, and thus H and its normal subgroup K satisfy the hypothesis. By the minimal choice of |G| + |E|, H is nilpotent. By Lemma 2.2, we may assume that G = P*Q, where Q is a normal Sylow q-subgroup of G and P* is a cyclic Sylow p-subgroup of G.Suppose that NG. We shall prove that (G/N, EN/N) satisfies the hypothesis. Clearly, (G/N) = (EN/N) ≅ G/EN is nilpotent and G/N is G1-free. Let H/N be a cyclic subgroup of a Sylow subgroup of EN/NN/N. Then we may assume H = 〈xN 〉 and 〈x〉 is a cyclic subgroup of a Sylow subgroup in E. By the hypothesis, 〈x〉 is CSQ-normal in G and by Lemma 2.1(b), H/N is CSQ-normal in G/N. Then (G/Φ(G), E/Φ(G)) satisfies the hypothesis of the theorem. The minimality of |G| + |E| implies that G/Φ(G) is nilpotent and so is G, a contradiction. Thus Φ(G) = 1 and so GG1, again a contradiction. This shows that there exists no counterexample, so the result is true. □

Remark 3.2: We cannot replace the condition “cyclic subgroup of Sylow subgroup” by “minimal subgroup of a Sylow subgroup” in Theorem 3.1. For example, let G = E = (Z2 × Z2 × Z2 × Z2 × Z2 × Z2) ⋊ Z9. Obviously, the pair (G, E) satisfy the hypothesis. Nevertheless, it is not nilpotent.

Remark 3.3: The condition of “G is G1-free” cannot be removed. For example, let G = S3 and choose E = A3. Then the pair (S3, A3) satisfy the hypothesis of Theorem 3.1. Nevertheless, S3 is not nilpotent.

Corollary 3.4: Let G be a group and be a complete set of Sylow subgroups of G. If every cyclic subgroup of a Sylow subgroup of G contained in is a CSQ-normal subgroup of G, then G is nilpotent.

Proof: By the proof of Theorem 3.1, we just need to check that GG1. By the hypothesis, we have that a p-Sylow subgroup Gp is a CSQ-normal subgroup of G. Then Gp ⊲⊲ G by Lemma 2.4(b), thus GpG, so G is nilpotent. The proof is completed. □

To prove Theorem 3.6, we need the following Lemma 3.5.

Lemma 3.5: Let G be a group and be a complete set of Sylow subgroups of G. Suppose that P is a Sylow p-subgroup of G contained in , where p is a prime divisor of |G| with (|G|, p − 1) = 1. If every maximal subgroup of P is CSQ-normal in G, then G/Op(G) is p-nilpotent and hence G is solvable.

Proof: Assume that the result is false and let G be a counterexample of smallest order.First of all, we show that Op(G) = 1. Assume that Op(G) = P. Then G/Op(G) is a p'-group and of course it is p-nilpotent, a contradiction. Assume that 1 < Op(G) < P. Obviously, Op(G) /Op(G) is a complete set of Sylow subgroups of G/Op(G) and G/Op(G) satisfies the hypothesis by Lemma 2.1(b). The minimal choice implies that G/Op(G) ≅ (G/Op(G))/Op(G/Op(G)) is p-nilpotent, a contradiction. Thus we have Op(G) = 1.Let P1 be a maximal subgroup of P. By the hypothesis, P1 is CSQ-normal subgroup of G. Then P1 is subnormal in G by Lemma 2.4, and thus P1Op(G) = 1. Hence P is a cyclic subgroup of order p. Since NG(P)/CG(P) ≲ Aut(P), we get that the order of NG(P)/CG(P) must divide (|G|, p − 1) = 1. Then NG(P) /CG(P) . Thus G is p-nilpotent by [1, Burnside’s theorem], a contradiction. We conclude that there is no counterexample and Lemma 3.5 is proved. □

Theorem 3.6: Let G be a group and be a complete set of Sylow subgroups of G. Suppose that G is Gt-free with t ϵ {1, 2, 6} and every maximal subgroup of any non-cyclic Sylow subgroup of G contained in is CSQ-normal in G. Then G is supersolvable.

Proof: Assume that the theorem is false and let G be a counterexample of smallest order. We proceed in a number of steps.If every Sylow subgroup of G contained in is cyclic, then every Sylow subgroup of G is cyclic, thus G is supersolvable. Next we assume that there is a non-cyclic Sylow p-subgroup contained in .Step 1. G is solvable.Let p = min π(G) and P be a Sylow p-subgroup of G contained in . If P is cyclic, then G is p-nilpotent, so G is solvable. If P is not cyclic, then G/Op(G) is p-nilpotent by Lemma 3.5, thus G is solvable. Hence we have Step 1.Step 2. G has a unique minimal normal subgroup N and Φ(G) = 1.Let N be a minimal normal subgroup of G, then N/N be a complete set of Sylow subgroups of G/N. Let PN/N ϵ Sylp(G/N), where P ϵ and PN/N is non-cyclic. (Of course, P is non-cyclic.) Assume that T/N be a maximal subgroup of PN/N. Then T = TPN = (TP/N. Suppose that TP = P1. Then P1N = TPN = PN. Hence |P:P1|=|PN/N:P1N/N|=|PN/N:T/N|=p.By the hypothesis, P1 is CSQ-normal in G, so P1 N/N = T /N is CSQ-normal in G/N by Lemma 2.1(b). Thus G/N satisfies the hypothesis. By the choice of G, we obtain that G/N is supersolvable. Similarly, if N1 is another minimal normal subgroup of G. Then G/N1 is also supersolvable. Now it follows that GG/NN1 is supersolvable, a contradiction. Hence, N is the unique minimal normal subgroup of G. If N ≤ Φ(G), then the supersolvability of G/N implies the supersolvability of G. Hence, Φ(G) = 1. Therefore, we have Step 2.Step 3. N = Op(G) = P, CG(N) = N and |G|=pnr1α1r2α2rsαs, the Sylow ri-subgroup of G is cyclic, where 1 ≤ is, αi ≥ 1.By Step 1 and Step 2, we know that N is an elementary abelian p-subgroup and N = F(G) = Op(G) ≤ P, so CG(N) = N. Assume that N < P. Given a maximal subgroup P1 of P, by the hypothesis, P1 is a CSQ-normal subgroup of G, then P1 is subnormal in G by Lemma 2.4, so P1Op(G) = N < P. If N = P1G, we get that P has a unique maximal subgroup, so P is cyclic and hence so is N. By Step 2, we obtain that G/N is supersolvable, hence so is G, a contradiction. Therefore, we have N = P. Suppose that Ri is a non-cyclic Sylow ri-subgroup of G contained in for some natural number i, 1 ≤ is, and |Ri|=riαi. Then αi ≥ 2, so we can choose 1 ≠ Ri1 to be a maximal subgroup of Ri ϵ Sylri(G). By the hypothesis, Ri1 is CSQ-normal in G, so Ri1 is subnormal in G by Lemma 2.4, so 1 ≠ Ri1Ori (G). By the uniqueness of N, this is impossible. Hence Ri is cyclic, and thus all Sylow subgroups B of G are cyclic except B = P. Hence we have the assertion in Step 3.Step 4. Let E be a maximal subgroup of G. We show that |G : E| = |P| = pn or riβi, where βiαi. Then E satisfies the hypothesis, so E is supersolvable.Since G is solvable, |G : E| = pj or riβi, where jn, βiαi. Suppose that |G : E| = pj. By Step 2 and Step 3, it is easy to show G = NE and NE = 1, so E = R1 R2Rs and j = n, where Ri ϵ Sylri (G) (1 ≤ is). It is clear that E satisfies the hypothesis by Lemma 2.1(a), so E is supersolvable.Step 5. Final contradiction.By Step 2 and Step 4, we know that G is minimal nonsupersolvable. On the other hand, by Step 4 and the hypothesis, G is not isomorphic to any group Gi in Lemma 2.3. We conclude that there is no minimal counterexample and Theorem 3.6 is proved. □

If we remove “non-cyclic” in the hypothesis of Theorem 3.6, we can get the following Theorem.

Theorem 3.7: Let G be a group and be a complete set of Sylow subgroups of G. Suppose that G is G1 -free and G6'-free, where G6'G6 and |G6' | = pqrp, that is, the case α = 1. If every maximal subgroup of every Sylow subgroup of G contained in is a CSQ-normal subgroup of G, then G is supersolvable.

Proof: By the proof of Theorem 3.6, we only need to check GG2 and GG6, where |G6| = pαqrp and pαq | r − 1, p | q − 1, α ≥ 2. Assume that GG2. Using the same description as in Lemma 2.3, let V1 = 〈ap〉. Then it is a maximal subgroup of P. By the hypothesis V1 is a CSQ-normal subgroup of G, so V1 is S-quasinormal in V1,V1g for all g ϵ G. Choosing g = ci. Then ((ap)1)ci=ci1(ap)1ciV1,V1ci, so ci1(ap)1ciap=ci1ciap=ci1(cia)ap1=ci1ci+1ap1==ci1citV1,V1ciwhere the exponent of t (mod r) is pα−1. Thus r divides tpα11=(t1)(tpα11+tpα12++1).If r | t − 1, then ci commutes with V1, of course, ci normalizes V1. If rt − 1, then (t − 1, r) = 1, we get ci=cim(t1)+nr=(cit1)mV1,V1ci.It follows that V1,V1ci=ap,ci. Since V1 is S-quasinormal in V1,V1ci, we have that V1Ri = 〈apRi is a subgroup of G, where Ri ϵ Sylr (〈ap, ci 〉) . By [5, Theorem 1], V1 is subnormal in V1 Ri, hence V1V1 Ri. Therefore, Ri normalizes V1 and, of course, ci normalizes V1. Since i was arbitrary, we conclude that V1 is normalized by P and R, where P ϵ Sylp(G), R ϵ Sylr (G). If α ≥ 2, then 1 ≠ V1G, which is impossible. If α = 1, then GG1, a contradiction. Hence G is not isomorphic to G1. As in a similar argument above, we also get that G is not isomorphic to G6, where |G6| = pαqrp and pαq | r − 1, p | q − 1, α ≥ 2. The proof is completed. □

Corollary 3.8: [9, Theorem 2] Let G be a group with the property that maximal subgroups of Sylow subgroups are π-quasinormal in G for π = π(G). Then G is supersolvable.

Proof: By the proof of Theorem 3.6 and Theorem 3.7, we only need to check that GG1 and GG6', where |G6' | = pqrp and pq | r − 1, p | q − 1. Assume that GG1. By Lemma 2.3, we have G1 = PQ, where |P| = p and |Q| = qβ(β ≥ 2). By Step 2 and Step 3 of Theorem 3.6, Q is a minimal normal subgroup of G1. Choosing Q1 to be a maximal subgroup of Q, by the hypothesis, we obtain that Q1 is π-quasinormal in G1. Then Oq (G) ≤ NG(Q1), so P normalizes Q1, and thus 1 ≠ Q1G, contrary to the minimality of Q. Hence GG1. Using a similar argument as above, we also get that G is not isomorphic to G6'. The proof is completed. □

Corollary 3.9: ([9, Theorem 1]). Let G be a group with the property that maximal subgroups of Sylow subgroups are normal in G. Then G is supersolvable.

Theorem 3.10: Let G be a QCLT-group. If every maximal subgroup of a Sylow 2-subgroup of G is CSQ-normal in G, then G is supersolvable.

Proof: Assume that the Theorem is false and let G be a counterexample of smallest order.Assume first that G has odd order. Since G is a QCLT-group, by [6], we have that G is supersolvable. Now we assume that 2 ||G|. By Lemma 3.5, we have that G is solvable. For any 1 ≠ NG, if 2 ∤ |G/N|, then G/N is a QCLT-group of odd order and hence G/N is supersolvable. Suppose that 2 ||G/N|. Without loss of generality, we assume that every maximal subgroup of a Sylow 2-subgroup of G/N is of the form P1 N/N, where P1 is a maximal subgroup of a Sylow 2-subgroup of G. Then P1 is CSQ-normal in G by hypothesis, so P1 N/N is CSQ-normal in G/N by Lemma 2.1(b). Hence the quotient group G/N satisfies the hypothesis. By the choice of G, we have that G is a solvable outer-supersolvable group. Then, by [7, Theorem 7.1], G = ML, where M is a maximal subgroup of G, ML = 1, L is an elementary abelian p-group and is also the unique minimal normal subgroup of G with order pα, α > 1, the Sylow p-subgroup of M is an abelian p-group and Φ(G) = 1.If |G2| ≤ 4, where G2 ϵ Syl2 (G), then G2 is a cyclic subgroup or an elementary abelian 2-subgroup. It follows that G is S4-free, then G is supersolvable by [10, Theorem 4], a contradiction. Hence we may choose 1 ≠ P1 to be a maximal subgroup of G2. By hypothesis, P1 is a CSQ-normal subgroup of G. Then P1 is subnormal in G by Lemma 2.4, thus 1 ≠ P1O2 (G), hence LO2 (G), so we get p = 2. By [7, §6.1, Main lemma], we also get O2 (G) = F(G) = L.Let M2 be a Sylow 2-subgroup of M. Then G2 = M2 L is a Sylow 2-subgroup of G. Assume that P1 is a maximal subgroup of M2 N containing M2. Then M2 < P1 since |L| = 2α, where α > 1. Then P1 is CSQ-normal in G by the hypothesis, so P1 is subnormal in G by Lemma 2.4. Thus P1O2 (G) = L, hence G2 = M2 L =P1 L = L is an elementary abelian Sylow 2-subgroup of G. It follows that G is S4-free, so G is supersolvable by [10, Theorem 4], a contradiction. Hence the minimal counterexample does not exist. Therefore G is supersolvable. □

Theorem 3.11: Let G be a QCLT-group. If every 2-maximal subgroup of a Sylow 2-subgroup of G is CSQ-normal in G. Then G is supersolvable.

Proof: The proof is similar to Theorem 3.10 and omitted here. □

Acknowledgement

This work was supported by the National Natural Science Foundation of China (Grant N. 11671324, 11601225, 11171243, 11326056), the China Postdoctoral Science Foundation (N. 2015M582492), the Natural Science Foundation of Jiangsu Province (N. BK20140451) and the Henan University of Science and Technology Science Fund for Innovative Teams (N. 2015XTD010).

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About the article

Received: 2016-05-15

Accepted: 2016-09-12

Published Online: 2016-11-03

Published in Print: 2016-01-01


Citation Information: Open Mathematics, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2016-0073. Export Citation

© Xu and Li, published by De Gruyter Open. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. (CC BY-NC-ND 3.0)

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