On CSQ-normal subgroups of finite groups

Yong Xu; Xianhua Li

doi:10.1515/math-2016-0073

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On CSQ-normal subgroups of finite groups

Yong Xu und Xianhua Li

Veröffentlicht/Copyright: 3. November 2016

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$Open Mathematics$

Aus der Zeitschrift Open Mathematics Band 14 Heft 1

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 H ≤ G and G ϵ G, then H ≤ 〈H, H^g〉 ≤ 〈H, g〉. It is clear that H = 〈H, H^g〉 for all g ϵ G if and only if H ⊴ G. In [2], H is called abnormal in G if 〈H, H^g〉 = 〈H, g〉 for all g ϵ G. In [3], the famous Wielandt Theorem shows that H ⊲⊲ 〈H, H^g〉 for all g ϵ G if and only if H ⊲⊲ G. In [4], H is called pronormal in G if H is conjugate to H^g in 〈H, H^g〉 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, H^g〉. This leads us to investigate the properties of G from the relationship between the subgroup H of G and the union of 〈H, H^g〉 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 in 〈H, H^g〉 for 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 N ⊴ G. Then

If H ≤ K, then H is CSQ-normal in K.
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

There exist some primes p and q such that |G| = p^αq^β.
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〉.
Let c ϵ Q. Then c is a generator if and only if [c, a] ≠ 1.
If c is a generator of Q, then [c, a] = c⁻¹c^a is also a generator of Q.
If c is a generator of Q, then Q = 〈c, c^a, …, c^{a^p−2}, c^{a^p-1} 〉, namely, Q = 〈[c, a], [c, a]^a, …, [c, a]^{a^p−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 G_t for 1 ≤ t ≤ 6, where the groups G_t are defined in the following way.

G₁is a minimal nonabelian group and |G₁| = pq^β, where p ∤ q − 1, β ≥ 2.
G₂ = 〈a, c₁〉 and |G₂| = p^αr^p and p^α−1 || r − 1, where α ≥ 2. apα=c1r=c2r=…=cpr=1;; c_ic_j = c_j c_i; cia=ci+1, i = 1, 2, …, p − 1; cpa=c1t, where the exponent of t (mod r) is p^α−1.
G₃ = 〈a, b, c₁〉 and |G₃| = 8r²and 4 | r − 1, a4=c1r=c2r=1, a² = b², ba = a⁻¹b, c1a=c2,c2a=c1−1,c1b=c1s,c2b=c2s, where the exponent of s (mod r) is 4.
G₄ = 〈a, b, c₁〉 and |G₄| = p^α+β r^p and p^{max{α, β}} | r − 1, where β ≥ 2. apα=bpβ=c1r=c2r=…=cpr=1; c_i c_j = c_j c_i, ab = b^{1+p^β−1}a; cia=ci+1,i=1,2,…,p−1;cpa=c1t,cib=ciu1+ipβ−1,i=1,2,…,pwhere the exponents of t and u (mod r) are p^α−1and p^β, respectively.
G₅ = 〈a, b, c, c₁〉 and |G₅| = p^α+β+1r^p and p^{max{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,…,p−1; cpa=c1t,cic=ciu,cib=civup−i+1, where the exponents of t, v and u (mod r) are p^α−1, p^β and p, respectively.
G₆ = 〈a, b, c₁〉, |G₆| = p^α qr^p and p^αq | r − 1, p | q − 1, α ≥ 1. apα=bq=c1r=c2r=…=cpr=1; c_i c_j = c_j c_i, i, j = 1, 2, …, p; cia=ci+1,i=1,2,…,p−1; cpa=c1t;ba=bu,cib=civui−1,i=1,2,…,p; where the exponents of t, v (mod r) are p^α−1and 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

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

Proof

(a) By the hypothesis, H is S-quasinormal in 〈H, H^g〉 for all g ϵ G. Then for any x ϵ G, we have that H^x is S-quasinormal in 〈H^x; H^gx〉=〈H^x,(H^x)^gx〉 for all g ϵ G. Then one checks easily that ˝ : G → G, defined by

τ(g)=gx,wherex∈G,

is a bijective map. Since g^x runs over G as g does for fixed x, we get that H^x is S-quasinormal in 〈H^x Z, (H^x)^gx〉 for all G^x ϵ G. Thus H^x is a CSQ-normal subgroup of G.

(b) By the hypothesis, H is S-quasinormal in 〈H, H^g〉 for all g ϵ G. By [5, Theorem 1], we know that H is subnormal in 〈H, H^g〉 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 = {P ∩ E | P ϵ}.

Theorem 3.1

Let G be a group and be a complete set of Sylow subgroups of G. Suppose that E ⊴ G 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/H ∩ E ≅ HE/E is nilpotent. Suppose that K = H ∩ E and K_p is a Sylow p-subgroup of K, so = {K_p|p ϵ β(H ∩ E)} is a complete set of Sylow subgroups of H ∩ E. Assume that T is a cyclic subgroup of K_p. Since K ≤ E, there exists x ϵ E such that Kpx≤p∩E, 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 K_p 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 N ⊲ G. 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/N ∩ N/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 G ≅ G₁, 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 = S₃and choose E = A₃. Then the pair (S₃, A₃) satisfy the hypothesis of Theorem 3.1. Nevertheless, S₃is 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 G ≅ G₁. By the hypothesis, we have that a p-Sylow subgroup G_p is a CSQ-normal subgroup of G. Then G_p ⊲⊲ G by Lemma 2.4(b), thus G_p ⊲ G, 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/O_p(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 O_p(G) = 1. Assume that O_p(G) = P. Then G/O_p(G) is a p'-group and of course it is p-nilpotent, a contradiction. Assume that 1 < O_p(G) < P. Obviously, O_p(G) /O_p(G) is a complete set of Sylow subgroups of G/O_p(G) and G/O_p(G) satisfies the hypothesis by Lemma 2.1(b). The minimal choice implies that G/O_p(G) ≅ (G/O_p(G))/O_p(G/O_p(G)) is p-nilpotent, a contradiction. Thus we have O_p(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 P₁ ≤ O_p(G) = 1. Hence P is a cyclic subgroup of order p. Since N_G(P)/C_G(P) ≲ Aut(P), we get that the order of N_G(P)/C_G(P) must divide (|G|, p − 1) = 1. Then N_G(P) /C_G(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 G_t-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/O_p(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 Syl_p(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 = T ∩ PN = (T ∩ P/N. Suppose that T ∩ P = P₁. Then P₁ ∩ N = T ∩ P ∩ N = P ∩ N. Hence

|P:P1|=|PN/N:P1N/N|=|PN/N:T/N|=p.

By the hypothesis, P₁ is CSQ-normal in G, so P₁N/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 G ≅ G/N ∩ N₁ 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 = O_p(G) = P, C_G(N) = N and |G|=pnr1α1r2α2…rsαs, the Sylow r_i-subgroup of G is cyclic, where 1 ≤ i ≤ s, α_i ≥ 1.

By Step 1 and Step 2, we know that N is an elementary abelian p-subgroup and N = F(G) = O_p(G) ≤ P, so C_G(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 P₁ ≤ O_p(G) = N < P. If N = P₁ ⊲ G, 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 R_i is a non-cyclic Sylow r_i-subgroup of G contained in for some natural number i, 1 ≤ i ≤ s, and |Ri|=riαi. Then α_i ≥ 2, so we can choose 1 ≠ R_i1 to be a maximal subgroup of R_i ϵ Syl_ri(G). By the hypothesis, R_i1 is CSQ-normal in G, so R_i1 is subnormal in G by Lemma 2.4, so 1 ≠ R_i1 ≤ O_ri (G). By the uniqueness of N, this is impossible. Hence R_i 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 β_i ≤ α_i. Then E satisfies the hypothesis, so E is supersolvable.

Since G is solvable, |G : E| = p^j or riβi, where j ≤ n, β_i ≤ α_i. Suppose that |G : E| = p^j. By Step 2 and Step 3, it is easy to show G = NE and N ∩ E = 1, so E = R₁R₂ … R_s and j = n, where R_i ϵ Syl_ri (G) (1 ≤ i ≤ s). 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 G_i 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 G_6'-free, where G_6' ≲ G₆and |G_6' | = pqr^p, 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 G ≅ G₂ and G ≅ G₆, where |G₆| = p^αqr^p and p^αq | r − 1, p | q − 1, α ≥ 2. Assume that G ≅ G₂. Using the same description as in Lemma 2.3, let V₁ = 〈a^p〉. 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 = c_i. Then ((ap)−1)ci=ci−1(ap)−1ci∈〈V1,V1ci〉, so

ci−1(ap)−1ciap=ci−1ciap=ci−1(cia)ap−1=ci−1ci+1ap−1=…=ci−1cit∈〈V1,V1ci〉

where the exponent of t (mod r) is p^α−1. Thus r divides

tpα−1−1=(t−1)(tpα−1−1+tpα−1−2+…+1).

If r | t − 1, then c_i commutes with V₁, of course, c_i normalizes V₁. If r ∤ t − 1, then (t − 1, r) = 1, we get

ci=cim(t−1)+nr=(cit−1)m∈〈V1,V1ci〉.

It follows that 〈V1,V1ci〉=〈ap,ci〉. Since V₁ is S-quasinormal in 〈V1,V1ci〉, we have that V₁R_i = 〈a^p〉R_i is a subgroup of G, where R_i ϵ Syl_r (〈a^p, c_i 〉) . By [5, Theorem 1], V₁ is subnormal in V₁R_i, hence V₁ ⊲ V₁R_i. Therefore, R_i normalizes V₁ and, of course, c_i normalizes V₁. Since i was arbitrary, we conclude that V₁ is normalized by P and R, where P ϵ Syl_p(G), R ϵ Syl_r (G). If α ≥ 2, then 1 ≠ V₁ ⊲ G, which is impossible. If α = 1, then G ≅ G₁, 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| = p^αqr^p and p^αq | 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 G ≅ G₁ and G ≅ G_6', where |G_6' | = pqr^p and pq | r − 1, p | q − 1. Assume that G ≅ G₁. By Lemma 2.3, we have G₁ = PQ, where |P| = p and |Q| = q^β(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 O^q (G) ≤ N_G(Q₁), so P normalizes Q₁, and thus 1 ≠ Q₁ ⊲ G, contrary to the minimality of Q. Hence G ≇ G₁. Using a similar argument as above, we also get that G is not isomorphic to G_6'. 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 ≠ N ⊴ G, 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 P₁N/N, where P₁ is a maximal subgroup of a Sylow 2-subgroup of G. Then P₁ is CSQ-normal in G by hypothesis, so P₁N/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, M ∩ L = 1, L is an elementary abelian p-group and is also the unique minimal normal subgroup of G with order p^α, 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 ≠ P₁ ≤ O₂ (G), hence L ≤ O₂ (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 = M₂L is a Sylow 2-subgroup of G. Assume that P₁ is a maximal subgroup of M₂N containing M₂. Then M₂ < P₁ since |L| = 2^α, where α > 1. Then P₁ is CSQ-normal in G by the hypothesis, so P₁ is subnormal in G by Lemma 2.4. Thus P₁ ≤ O₂ (G) = L, hence G₂ = M₂L =P₁L = 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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Received: 2016-5-15

Accepted: 2016-9-12

Published Online: 2016-11-3

Published in Print: 2016-1-1

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