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}.

*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 G*_{1}*-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*.

Assume that the result is false, and let *G* be a counterexample with least (|*G*| + |*E*|).

Let *H* < *G*. Of course, *H* is *G*_{1}-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 ${K}_{p}^{x}\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}p\phantom{\rule{thinmathspace}{0ex}}\cap E\phantom{\rule{thinmathspace}{0ex}}$, 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 *G*_{1}-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* ≅ *G*_{1}, again a contradiction. This shows that there exists no counterexample, so the result is true. □

*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*.

By the proof of Theorem 3.1, we just need to check that *G* ≅ *G*_{1}. 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.

*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*.

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 *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} ≤ *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. □

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* = *P*_{1}. Then *P*_{1} ∩ *N* = *T* ∩ *P* ∩ *N* = *P* ∩ *N*. Hence
$$|P:\phantom{\rule{thinmathspace}{0ex}}{P}_{1}|\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}PN/N\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}{P}_{1}N/N|\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}|PN/N\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}T/N|\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}p.$$

By the hypothesis, *P*_{1} is *CSQ*-normal in *G*, so *P*_{1} *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 *N*_{1} is another minimal normal subgroup of *G*. Then *G/N*_{1} is also supersolvable. Now it follows that *G* ≅ *G/N* ∩ *N*_{1} 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 $|G|\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{p}^{n}{r}_{1}^{\alpha 1}{r}_{2}^{\alpha 2}\dots {r}_{s}^{\alpha s}$, 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 *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} ≤ *O*_{p}(*G*) = *N* < *P*. If *N* = *P*_{1} ⊲ *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 $|{R}_{i}|\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{r}_{i}^{\alpha i}$. 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*| = *p*^{n} or ${r}_{i}^{{\beta}_{i}}$, where *β*_{i} ≤ *α*_{i}. Then *E* satisfies the hypothesis, so *E* is supersolvable.

Since *G* is solvable, |*G* : *E*| = *p*^{j} or ${r}_{i}^{{\beta}_{i}}$, 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* = *R*_{1} *R*_{2} … *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.

*Let G be a group and be a complete set of Sylow subgroups of G. Suppose that G is G*_{1} *-free and G*_{6'}*-free, where G*_{6'} ≲ *G*_{6} *and* |*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*.

By the proof of Theorem 3.6, we only need to check *G* ≅ *G*_{2} and *G* ≅ *G*_{6}, where |*G*_{6}| = *p*^{α}*qr*^{p} and *p*^{α}*q* | *r* − 1, *p* | *q* − 1, *α* ≥ 2. Assume that *G* ≅ *G*_{2}. Using the same description as in Lemma 2.3, let *V*_{1} = 〈*a*^{p}〉. Then it is a maximal subgroup of *P*. By the hypothesis *V*_{1} is a *CSQ*-normal subgroup of *G*, so *V*_{1} is *S*-quasinormal in $\u3008{V}_{1},\phantom{\rule{thinmathspace}{0ex}}{V}_{1}^{g}\u3009$ for all *g* ϵ *G*. Choosing *g* = *c*_{i}. Then $(({a}^{p}{)}^{-1}{)}^{{c}_{i}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{c}_{i}^{-1}({a}^{p}{)}^{-1}\phantom{\rule{thinmathspace}{0ex}}{c}_{i}\phantom{\rule{thinmathspace}{0ex}}\in \phantom{\rule{thinmathspace}{0ex}}\u3008{V}_{1},\phantom{\rule{thinmathspace}{0ex}}{V}_{1}^{{c}_{i}}\u3009$, so
$${c}_{i}^{-1}({a}^{p}{)}^{-1}{c}_{i}{a}^{p}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{c}_{i}^{-1}{c}_{i}^{{a}^{p}}=\phantom{\rule{thinmathspace}{0ex}}{c}_{i}^{-1}({c}_{i}^{a}{)}^{{a}^{p-1}}=\phantom{\rule{thinmathspace}{0ex}}{c}_{i}^{-1}{c}_{i+1}^{{a}^{p-1}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\dots \phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{c}_{i}^{-1}{c}_{i}^{t}\in \u3008{V}_{1},\phantom{\rule{thinmathspace}{0ex}}{V}_{1}^{{c}_{i}}\u3009$$

where the exponent of *t* (mod *r*) is *p*^{α−1}. Thus *r* divides
$${{t}^{p}}^{{}^{\alpha -1}}-\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}(t\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}1)({t}^{{p}^{\alpha -1}-1}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{t}^{{p}^{\alpha -1}-2}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\dots +1).$$

If *r* | *t* − 1, then *c*_{i} commutes with *V*_{1}, of course, *c*_{i} normalizes *V*_{1}. If *r* ∤ *t* − 1, then (*t* − 1, *r*) = 1, we get
$${c}_{i}=\phantom{\rule{thinmathspace}{0ex}}{c}_{i}^{m(t-1)+nr}=\phantom{\rule{thinmathspace}{0ex}}({c}_{i}^{t-1}{)}^{m}\in \u3008{V}_{1},\phantom{\rule{thinmathspace}{0ex}}{V}_{1}^{{c}_{i}}\u3009.$$

It follows that $\u3008{V}_{1},\phantom{\rule{thinmathspace}{0ex}}{V}_{1}^{{c}_{i}}\u3009\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\u3008{a}^{p},\phantom{\rule{thinmathspace}{0ex}}{c}_{i}\u3009$. Since *V*_{1} is *S*-quasinormal in $\u3008{V}_{1},\phantom{\rule{thinmathspace}{0ex}}{V}_{1}^{{c}_{i}}\u3009$, we have that *V*_{1}*R*_{i} = 〈*a*^{p}〉*R*_{i} is a subgroup of *G*, where *R*_{i} ϵ *Syl*_{r} (〈*a*^{p}, c_{i} 〉) . By [5, Theorem 1], *V*_{1} is subnormal in *V*_{1} *R*_{i}, hence *V*_{1} ⊲ *V*_{1} *R*_{i}. 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* ϵ *Syl*_{p}(*G*), *R* ϵ *Syl*_{r} (*G*). If *α* ≥ 2, then 1 ≠ *V*_{1} ⊲ *G*, which is impossible. If *α* = 1, then *G* ≅ *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 |*G*_{6}| = *p*^{α}*qr*^{p} and *p*^{α}*q* | *r* − 1, *p* | *q* − 1, *α* ≥ 2. The proof is completed. □

*[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*.

By the proof of Theorem 3.6 and Theorem 3.7, we only need to check that *G* ≅ *G*_{1} and *G* ≅ *G*_{6'}, where |*G*_{6'} | = *pqr*^{p} and *pq* | *r* − 1, *p* | *q* − 1. Assume that *G* ≅ *G*_{1}. By Lemma 2.3, we have *G*_{1} = *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 *G*_{1}. Choosing *Q*_{1} to be a maximal subgroup of *Q*, by the hypothesis, we obtain that *Q*_{1} is *π*-quasinormal in *G*_{1}. Then *O*^{q} (*G*) ≤ *N*_{G}(*Q*_{1}), so *P* normalizes *Q*_{1}, and thus 1 ≠ *Q*_{1} ⊲ *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'}. The proof is completed. □

([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*.

*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*.

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 *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* = *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 |*G*_{2}| ≤ 4, where *G*_{2} ϵ *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} ≤ *O*_{2} (*G*), hence *L* ≤ *O*_{2} (*G*), so we get *p* = 2. By [7, §6.1, Main lemma], we also get *O*_{2} (*G*) = *F*(*G*) = *L*.

Let *M*_{2} be a Sylow 2-subgroup of *M*. Then *G*_{2} = *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 |*L*| = 2^{α}, where *α* > 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} ≤ *O*_{2} (*G*) = *L*, hence *G*_{2} = *M*_{2} *L* =*P*_{1} *L* = *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. □

*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*.

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