On CSQ-normal subgroups of finite groups

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.


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 2.2. 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 jGj D p˛qˇ.
(2) G has a normal Sylow q-subgroup Q; if q > 2, then exp.Q/ D q and if q D 2, then exp.Q/ Ä 4; G also has a cyclic Sylow p-subgroup P D hai. ; c a p 2 ; c a p 1 i, namely, Q D hOEc; a; OEc; a a ; ; OEc; a a p 1 i.
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 Ä t Ä 6, where the groups G t are defined in the following way.
(II) G 2 D ha, c 1 i and jG 2 j D p˛r p and p˛ 1 kr 1, where˛ 2. a p˛D c r where the exponent of t .mod r/ is p˛ 1 . (III) G 3 D ha; b; c 1 i and jG 3 j D 8r 2 and 4 j r 1, where the exponent of s .mod r/ is 4. (IV) G 4 D ha, b, c 1 i and jG 4 j D p˛Cˇr p and p maxf˛;ˇg j r 1, whereˇ 2. a p˛D b pˇD c r , i D 1, 2, : : : , p; where the exponents of t and u .mod r/ are p˛ 1 and pˇ, respectively. (V) G 5 D ha; b; c; c 1 i and jG 5 j D p˛CˇC 1 r p and p maxf˛;ˇg j r 1.
where the exponents of t, v and u .mod r/ are p˛ 1 , pˇand p, respectively. (VI) G 6 D ha; b; c 1 i, jG 6 j D p˛qr p and p˛q j r 1, p j q 1,˛ 1.
where the exponents of t, v .mod r/ are p˛ 1 and q, respectively, and the exponent of u .mod q/ is p.
Proof. (a) By the hypothesis, H is S -quasinormal in hH; H g i for all g 2 G. Then for any x 2 G, we have that H x is S -quasinormal in hH x ; H gx i=hH x ; .H x / g x i for all g 2 G. Then one checks easily that W G ! G, defined by 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 hH x ; .H x / g x i for all g x 2 G. Thus H x is a CSQ-normal subgroup of G.
(b) By the hypothesis, H is S-quasinormal in hH; H g i for all g 2 G. By [5, Theorem 1], we know that H is subnormal in hH; H g i for all g 2 G, so H is subnormal in G by Wielandt's theorem.

Main results
Let Z be a complete set of Sylow subgroups of a group G, that is, for each prime p dividing the order of G, Z contains exactly one Sylow p-subgroup of G. Let Z\E D fP \ E j P 2Zg.
Theorem 3.1. Let G be a group and Z be a complete set of Sylow subgroups of G. Suppose that E E G such that G=E is nilpotent and G is G 1 -free. If every cyclic subgroup of a Sylow subgroup of E contained in Z \ 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 .jGj C jEj/. Let Hence all cyclic subgroups of K p contained in Z are CSQ-normal in H , and thus H and its normal subgroup K satisfy the hypothesis. By the minimal choice of jGj C jEj, H is nilpotent. By Lemma 2.2, we may assume that G D 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 C G. We shall prove that .G=N; EN=N / satisfies the hypothesis. Clearly, .G=N /=.EN=N / Š G=EN is nilpotent and G=N is G 1 -free. Let H=N be a cyclic subgroup of a Sylow subgroup of EN=N \ ZN=N . Then we may assume H D hxN i and hxi is a cyclic subgroup of a Sylow subgroup in E \ Z. By the hypothesis, hxi is CSQ-normal in G and by Lemma 2.1 .b/, H=N is CSQ-normal in G=N . Then .G=ˆ.G/; E=ˆ.G// satisfies the hypothesis of the theorem. The minimality of jGjCjEj implies that G=ˆ.G/ is nilpotent and so is G, a contradiction. Thusˆ.G/ D 1 and so G Š G 1 , 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.
Obviously, the pair .G; E/ satisfy the hypothesis. Nevertheless, it is not nilpotent.
Remark 3.3. The condition of "G is G 1 -free" cannot be removed. For example, let G D S 3 and choose E D A 3 . Then the pair .S 3 ; A 3 / satisfy the hypothesis of Theorem 3.1. Nevertheless, S 3 is not nilpotent.
Corollary 3.4. Let G be a group and Z be a complete set of Sylow subgroups of G. If every cyclic subgroup of a Sylow subgroup of G contained in Z 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 Š G 1 . By the hypothesis, we have that a p-Sylow To prove Theorem 3.6, we need the following Lemma 3.5.
Lemma 3.5. Let G be a group and Z be a complete set of Sylow subgroups of G. Suppose that P is a Sylow psubgroup of G contained in Z, where p is a prime divisor of jGj with .jGj; p 1/ D 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. Let P 1 be a maximal subgroup of P . By the hypothesis, P 1 is CSQ-normal subgroup of G. Then P 1 is subnormal in G by Lemma 2.4, and thus P 1 Ä O p .G/ D 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 .jGj; p 1/ D 1. Then N G .P / D 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 Z be a complete set of Sylow subgroups of G. Suppose that G is G t -free with t 2 f1; 2; 6g and every maximal subgroup of any non-cyclic Sylow subgroup of G contained in Z 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 Z 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 Z.
Let p D min .G/ and P be a Sylow p-subgroup of G contained in Z. 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 andˆ.G/ D 1.
Let N be a minimal normal subgroup of G, then ZN=N be a complete set of Sylow subgroups of G=N . Let PN=N 2 Syl p .G=N /, where P 2 Z 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 D T \ PN D .T \ P /N . Suppose that T \ P D P 1 . Then P 1 \ N D T \ P \ N D P \ N . Hence jP W P 1 j D jPN=N W P 1 N=N j D jPN=N W T =N j D p: By the hypothesis, P 1 is CSQ-normal in G, so P 1 N=N D 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 N 1 is another minimal normal subgroup of G. Then G=N 1 is also supersolvable. Now it follows that G Š G=N \ N 1 is supersolvable, a contradiction. Hence, N is the unique minimal normal subgroup of G. If N Äˆ.G/, then the supersolvability of G=N implies the supersolvability of G. Hence,ˆ.G/ D 1. Therefore, we have Step 2.
Step 3. N D O p .G/ D P , C G .N / D N and jGj D p n r˛1 1 r˛2 2 r˛s s , the Sylow r i -subgroup of G is cyclic, where 1 Ä i Ä s,˛i 1.
By Step 1 and Step 2, we know that N is an elementary abelian p-subgroup and N D F .G/ D O p .G/ Ä P , so C G .N / D N . Assume that N < P . Given a maximal subgroup P 1 of P , by the hypothesis, P 1 is a CSQ-normal subgroup of G, then P 1 is subnormal in G by Lemma 2.4, so P 1 Ä O p .G/ D N < P . If N D P 1 C 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 D P . Suppose that R i is a non-cyclic Sylow r i -subgroup of G contained in Z for some natural number i , 1 Ä i Ä s, and jR i j D r˛i i . Then˛i 2, so we can choose 1 6 D R i1 to be a maximal subgroup of R i 2 Syl r i .G/. By the hypothesis, R i1 is CSQ-normal in G, so R i1 is subnormal in G by Lemma 2.4, so 1 6 D R i1 Ä O r i .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 D P . Hence we have the assertion in Step 3.
Step 4. Let E be a maximal subgroup of G. We show that jG W Ej D jP j D p n or rˇi i , whereˇi Ä˛i . Then E satisfies the hypothesis, so E is supersolvable.
Since G is solvable, jG W Ej D p j or rˇi i , where j Ä n,ˇi Ä˛i . Suppose that jG W Ej D p j . By Step 2 and Step 3, it is easy to show G D NE and N \ E D 1, so E D R 1 R 2 R s and j D n, where R i 2 Syl r i .G/ (1 Ä i Ä 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 Z be a complete set of Sylow subgroups of G. Suppose that G is G 1 -free and G 6 0 -free, where G 6 0 . G 6 and jG 6 0 j D pqr p , that is, the case˛D 1. If every maximal subgroup of every Sylow subgroup of G contained in Z 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 Š G 2 and G Š G 6 , where jG 6 j D p˛qr p and p˛q j r 1, p j q 1,˛ 2. Assume that G Š G 2 . Using the same description as in Lemma 2.3, let V 1 D ha p i. Then it is a maximal subgroup of P . By the hypothesis where the exponent of t .mod r/ is p˛ 1 . Thus r divides Therefore, R i normalizes V 1 and, of course, c i normalizes V 1 . Since i was arbitrary, we conclude that V 1 is normalized by P and R, where P 2 Syl p .G/, R 2 Syl r .G/. If˛ 2, then 1 6 D V 1 C G, which is impossible. If˛D 1, then G Š G 1 , a contradiction. Hence G is not isomorphic to G 1 . As in a similar argument above, we also get that G is not isomorphic to G 6 , where jG 6 j D p˛qr p and p˛q j r 1, p j q 1,˛ 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 -quasinormal in G for D .G/. Then G is supersolvable.
Proof. By the proof of Theorem 3.6 and Theorem 3.7, we only need to check that G Š G 1 and G Š G 6 0 , where jG 6 0 j D pqr p and pq j r 1, p j q 1. Assume that G Š G 1 . By Lemma 2.3, we have G 1 D PQ, where jP j D p and jQj D qˇ.ˇ 2/. By Step 2 and Step 3 of Theorem 3.6, Q is a minimal normal subgroup of G 1 . Choosing Q 1 to be a maximal subgroup of Q, by the hypothesis, we obtain that Q 1 is -quasinormal in G 1 . Then O q .G/ Ä N G .Q 1 /, so P normalizes Q 1 , and thus 1 ¤ Q 1 C G, contrary to the minimality of Q. Hence G © G 1 . Using a similar argument as above, we also get that G is not isomorphic to G 6 0 . The proof is completed.  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 j jGj. By Lemma 3.5, we have that G is solvable. For any 1 ¤ N E G, if 2 − jG=N j, then G=N is a QCLT -group of odd order and hence G=N is supersolvable. Suppose that 2 j jG=N j. Without loss of generality, we assume that every maximal subgroup of a Sylow 2-subgroup of G=N is of the form P 1 N=N , where P 1 is a maximal subgroup of a Sylow 2-subgroup of G. Then P 1 is CSQ-normal in G by hypothesis, so P 1 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 D ML, where M is a maximal subgroup of G, M \ L D 1, L is an elementary abelian p-group and is also the unique minimal normal subgroup of G with order p˛,˛> 1, the Sylow p-subgroup of M is an abelian p-group andˆ.G/ D 1.
If jG 2 j Ä 4, where G 2 2 Syl 2 .G/, then G 2 is a cyclic subgroup or an elementary abelian 2-subgroup. It follows that G is S 4 -free, then G is supersolvable by [10, Theorem 4], a contradiction. Hence we may choose 1 ¤ P 1 to be a maximal subgroup of G 2 . By hypothesis, P 1 is a CSQ-normal subgroup of G. Then P 1 is subnormal in G by Lemma 2.4, thus 1 ¤ P 1 Ä O 2 .G/, hence L Ä O 2 .G/, so we get p D 2. By [7, 6.1, Main lemma], we also get O 2 .G/ D F .G/ D L.
Let M 2 be a Sylow 2-subgroup of M . Then G 2 D M 2 L is a Sylow 2-subgroup of G. Assume that P 1 is a maximal subgroup of M 2 N containing M 2 . Then M 2 < P 1 since jLj D 2˛, where˛> 1. Then P 1 is CSQ-normal in G by the hypothesis, so P 1 is subnormal in G by Lemma 2.4. Thus P 1 Ä O 2 .G/ D L, hence G 2 D M 2 L D P 1 L D L is an elementary abelian Sylow 2-subgroup of G. It follows that G is S 4 -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.