Abstract
Let G be a finite group. In this paper, we obtain some new conditions for supersolvability of G with three supersolvable subgroups of pairwise relatively prime indices in G. Our main results improve and extend some known results in the literature.
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1 Introduction
All groups considered in this paper will be finite. We use conventional notions and notation (see [7, 13]). G always denotes a finite group, |G| is the order of G and \(\pi (G)\) denotes the set of all primes dividing |G|.
Recall that a class of groups \(\Im \) is a formation if \(\Im \) contains all homomorphic images of a group in \(\Im ,\) and if G/M and G/N are in \(\Im \), then \(G/(M\cap N)\) is in \(\Im \) for any normal subgroups M and N of G. A formation \(\Im \) is said to be saturated if \(G/\Phi (G)\in \Im \) implies that \(G\in \Im .\) Let \(\mathcal {U}\) be the class of all supersolvable groups. Clearly, \(\mathcal {U}\) is a saturated formation. The \(\mathcal {U}{ -}hypercenter\) of G will be denoted by \(Z_{ \mathcal {U}}{ (G)}\), i.e., \(Z_{\mathcal {U}}{ (G)}\) is the product of all normal subgroups N of G such that every chief factor of G below N has a prime order (see [10]).
A subgroup H of G is said to be \(s-permutable\) in G if H permutes with every Sylow subgroup of G (see [15]); H is said to be \( s-semipermutable\) in G if H permutes with every Sylow subgroup L of G with \((|H|,|L|)=1\) (see [16]). We say that two subgroups H and K of a group G are mutually permutable if H permutes with every subgroup of K and K permutes with every subgroup of H (see [7]). It is well known that the product of two normal supersolvable subgroups is not supersolvable in general (see [20, Chapter 1, pages 8/9]). In [5], it is proved that the product of two mutually permutable supersolvable subgroups is supersolvable provided that the commutator subgroup is nilpotent. In [1], it is also proved that if G is a product of two mutually permutable supersolvable subgroups H and K with Core\(_{G}(H\cap K)=1,\) then G is supersolvable. These two results have been generalized in [8] and [11].
There exists a nonsupersolvable group G with three supersolvable subgroups \(H_{i},\) \(i=1,\) 2, 3, such that \((|G:H_{i}|,|G:H_{j}|)=1\) if \(i\ne j\) as the following example shows:
Example 1.1
-
(1)
There exists a nonsupersolvable group G of order \(7^{2}.2.3\) (see [20, Chapter 3, Corollary 4.6]).
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(2)
G is solvable and every proper subgroup of G is supersolvable (see [20, Chapter 3, Lemma 4.3(i)]).
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(3)
G has three supersolvable subgroups \(H_{1},\) \(H_{2}\) and \(H_{3}\) of orders \(7^{2}.3,\) \(7^{2}.2,\) and 3.2, respectively (see [20, Chapter3, Lemma 4.3(ii)]). Now this example raises the following question:
Question. Suppose that G is a group with three supersolvable subgroups \(H_{i},\) \(i=1,2,3,\) such that \((|G:H_{i}|,|G:H_{j}|)=1\) if \(1\le i\ne j\le 3.\) What are the conditions that should be imposed on G to gurantee it is supersolvable?
Many researchers showed interest in addresing this question. In [9], Doerk proved the following theorem:
Theorem 1.2
If a group G possesses four supersolvable subgroups whose indices in G are pairwise relatively prime, then G is supersolvable.
In [2, 3], we proved the following two theorems:
Theorem 1.3
Let G be a group. Let \(H_{i},\) \( i=1,2,3,\) be supersolvable subgroup with \(|G:H_{i}|=p_{i},\) \(i=1,2,3,\) where \(p_{1},\) \(p_{2}\) and \(p_{3}\) are distinct primes. Then G is supersolvable.
Theorem 1.4
Let G be a group. Let \(H_{1}\) be a nilpotent subgroup of G. Let \(H_{i},\) \(i=2,3,\) be supersolvable subgroups of G. If the indices of \(H_{i},\) \(i=1,2,3,\) in G are pairwise relatively prime, then G is supersolvable.
In [12], Flowers and Wakefield proved the following theorem:
Theorem 1.5
Let G be a group and let \(H_{i},\) \( i=1,2,3,\) be supersolvable subgroups of G with pairwise relatively prime indices in G. Suppose that \(G^{^{\prime }}\) is nilpotent. Then G is supersolvable.
In [6], Ballester-Bolinches and Ezquerro proved the following theorem:
Theorem 1.6
Let G be a group with three supersolvable subgroups \(H_{i},\) \(i=1,2,3,\) with pairwise relatively prime indices in G. Suppose that \(H_{1}\) and \(H_{j}\) are mutually permutable subgroups for every \(j\ne 1.\) Then G is supersolvable.
In this paper, we prove the following theorems:
Theorem 1.7
Let \(H_{1}\), \(H_{2}\), \(H_{3}\) be three supersolvable subgroups of a group G. Let \( |G:H_{1}|=p_{1}^{e_{1}}p_{2}^{e_{2}}\ldots p_{s}^{es},\) \( |G:H_{2}|=p_{s+1}^{e_{s+1}}p_{s+2}^{e_{s+2}}\ldots p_{t}^{e_{t}},\) \( |G:H_{3}|=p_{t+1}p_{t+2}\ldots p_{m},\) where \(p_{1},p_{2},\ldots ,p_{s},\) \( p_{s+1},\ldots ,p_{t},p_{t+1},\ldots ,p_{m}\) are distinct primes such that \( p_{1}<p_{2}<\cdots<p_{s}<p_{s+1}<\cdots<p_{t}<p_{t+1}<\cdots <p_{m}\). Then G is supersolvable.
Theorem 1.8
Let \(H_{1}\), \(H_{2}\), \(H_{3}\) be three supersolvable subgroups of a group G. Let \( |G:H_{1}|=p_{1}^{e_{1}}p_{2}^{e_{2}}\ldots p_{s}^{es},\) \( |G:H_{2}|=p_{s+1}^{e_{s+1}}p_{s+2}^{e_{s+2}}\ldots p_{t}^{e_{t}},\) \( |G:H_{3}|=p_{t+1}^{e_{t+1}}p_{t+2}^{e_{t+2}}\ldots p_{m}^{e_{m}},\) where \( p_{1},p_{2},\ldots ,p_{s},\) \(p_{s+1},\ldots ,p_{t},p_{t+1},\ldots ,p_{m}\) are distinct primes such that \(p_{1}<p_{2}<\cdots<p_{s}<p_{s+1}<\cdots<p_{t}<p_{t+1}<\cdots <p_{m}\). Suppose that every subgroup of G of prime order p is s-semipermutable in G for every p dividing \(|G:H_{3}|.\) Then G is supersolvable.
Theorem 1.9
Let \(H_{1}\), \(H_{2}\), \(H_{3}\) be three supersolvable subgroups of a group G. Let \( |G:H_{1}|=p_{1}^{e_{1}}p_{2}^{e_{2}}\ldots p_{s}^{es},\) \( |G:H_{2}|=p_{s+1}^{e_{s+1}}p_{s+2}^{e_{s+2}}\ldots p_{t}^{e_{t}},\) \( |G:H_{3}|=p_{t+1}^{e_{t+1}}p_{t+2}^{e_{t+2}}\ldots p_{m}^{e_{m}},\) where \( p_{1},p_{2},\ldots ,p_{s},\) \(p_{s+1},\ldots ,p_{t},p_{t+1},\ldots ,p_{m}\) are distinct primes such that \(p_{1}<p_{2}<\cdots<p_{s}<p_{s+1}<\cdots<p_{t}<p_{t+1}<\cdots <p_{m}\). Suppose that \(H_{3}\) and \(H_{2}\cap H_{1}\) are mutually permutable subgroups of G. Then G is supersolvable.
Theorem 1.10
Let \(\Im \) be a saturated formation containing \(\mathcal {U}\). Let G be a group and let \(H_{1},\) \( H_{2}\) and \(H_{3}\) be subgroups of G belong to \(\Im \) with pairwise relatively prime indices in G. Suppose that the commutator subgroup \( G^{^{\prime }}\) of G is nilpotent. Then \(G\in \Im .\)
2 Preliminaries
Lemma 2.1
[2, Theorem 11]. Let G be a group. Suppose that \(H_{1},\) \( H_{2}\) and \(H_{3}\) are subgroups of G whose indices in G are pairwise relatively prime. If \(H_{1},\) \(H_{2}\) and \(H_{3}\) have Sylow towers of supersolvable type, then G has a Sylow tower of supersolvable type.
Lemma 2.2
[18, II, Lemma 7.9]. Let P be a nilpotent normal subgroup of a group G. If \(P\ne 1\) and \(P\cap \Phi (G)=1, \) then P is the direct product of some minimal normal subgroups of G.
Lemma 2.3
[16, Lemma 2.1]. Suppose that H is an s−semipermutable subgroup of G. Then
-
(a)
If \(H\le K\le G\), then H is s−semipermutable in K.
-
(b)
If \(H\le O_{p}(G),\) then H is s−permutable in G.
Lemma 2.4
[17, Theorem 3.3]. Let G be a group. Suppose that P is a normal \(p-\)subgroup of G, where p is an odd prime dividing the order of G. If every subgroup of P of order p is s\( - \)permutable in G, then \(P\le Z_{\mathcal {U}}{ (G).}\)
Lemma 2.5
[1, Theorem A]. Let the group \(G=HK\) be the product of the mutually permutable subgroups H and K. If H and K are supersolvable subgroups of G with Core\(_{G}(H\cap K)=1,\) then G is supersolvable.
Lemma 2.6
[7, Theorem 5.2.21]. Let \(G=HK\) be the product of the mutually permutable subgroups H and K. Let \(\Im \) be a saturated formation containing \(\mathcal {U}\) If H and K belong to \(\Im \) and \(G^{{\prime }}\) is nilpotent, then \(G\in \Im .\)
3 Main Results
Theorem 3.1
Let G be a group and let \(H_{1},\) \( H_{2}\) and \(H_{3}\) be supersolvable subgroups of G. Suppose that \( |G:H_{1}|=p_{_{1}}^{e_{1}},\) \(|G:H_{2}|=p_{2}^{e_{2}},\) \(|G:H_{3}|=p_{_{3}},\) where \(p_{1},\) \(p_{2},\) \(p_{3}\) are distinct primes such that \( p_{1}<p_{2}<p_{3}.\) Then G is supersolvable.
Proof
Suppose that the theorem is false and let G be a counterexample with minimal order. Let P be a Sylow \(p-\)subgroup of G, where p is the largest prime dividing the order of G. Then:
-
(1)
\(P\trianglelefteq G.\) By \(H_{1},\) \(H_{2}\), \(H_{3}\) have Sylow towers of supersolvable type. Then by Lemma 2.1, G has a Sylow tower of supersolvable type. Hence by [14, VI, Satz 9.1(c)], \(P\trianglelefteq G.\)
-
(2)
If N is a nontrivial normal \(p-\)subgroup of G, then G/N is supersolvable. If \(G=H_{3}N,\) then G/N is supersolvable. Suppose now that \(N\le H_{3}.\) It is clear that \(H_{3}/N,\) \(H_{2}/N,\) \(H_{1}/N\) are proper supersolvable subgroups of G/N. Then by the minimal chioce of G, G/N is supersolvable.
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(3)
\(p=p_{3}.\) Suppose that \(p>p_{3}.\) By (1), \(P\trianglelefteq G,\) where P is a Sylow \( p-\)subgroup of G and p is the largest prime dividing the order of G. It is clear that \(P\le H_{i},\) \(i=1,2,3.\) Then by (2), G/P is supersolvable. By Schur−Zassenhouse Theorem [13, Theorem 6.2.1], G has a subgroup K which is a complement to P in G. It is clear that \( (|G:H_{i}|,\) \(|G:K|)=1\) for any \(i=1,2,3.\) Hence by Theorem 1.2, G is supersolvable, a contradiction.
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(4)
The final contradiction. By (1), \(P\trianglelefteq G.\) Put \(A=P\cap \Phi (G).\) If \(A\ne 1,\) then by (2), G/A is supersolvable. Hence by [20, Chapter 1, Corollary 3.2], G is supersolvable, a contradiction. Thus \(A=1.\) Then by Lemma 2.2, \( P=N_{1}\times N_{2}\times \cdots \times N_{n},\) where \(N_{i}\) is a minimal normal subgroup of G, \(i=1,2,3,\ldots ,n.\) By (3), P is a Sylow \(p_{3}-\) subgroup of G. Then \(G=H_{3}N_{j}\) for some j, where \(1\le j\le n.\) Since \(N_{j}\) is a minimal normal subgroup of G and \(|G:H_{3}|=p_{3},\) we have \(|N_{j}|=p_{3}.\) Hence G is supersolvable, a contradiction. \(\square \)
Proof of Theorem 1.7
Suppose that the theorem is false and let G be a counterexample with minimal order. By Wielandt’s Theorem [13, Theorem 6.4.4], G is solvable. Let \(M_{i}\) be a maximal subgroup of G such that \(H_{i}\le M_{i},\) \(i=1,2,3.\) Since \( |G:M_{i}|||G:H_{i}|,\) we have \(|G:M_{1}|=p_{r}^{l_{r}},\) \( |G:M_{2}|=p_{j}^{l_{j}},\) \(|G:M_{3}|=p_{k},\) where \(p_{r},\) \(p_{j},\) \(p_{k}\) are distinct primes such that \(p_{r}<p_{j}<p_{k}.\) It is clear that \( G=H_{2}M_{1}=H_{3}M_{1}\) and so \(|G:H_{2}|=|M_{1}:M_{1}\cap H_{2}|\) and \( |G:H_{3}|=|M_{1}:M_{1}\cap H_{3}|\). Then by the minimal choice of G, \( M_{1} \) is supersolvable. Similarly, \(M_{2}\) and \(M_{3}\) are supersolvable. Hence by Theorem 3.1, G is supersolvable, a contradiction. \(\square \)
As a corollary of the proof of Theorem 1.7, we have
Corollary 3.2
[4]. If a group G has three supersolvable subgroups \(H_{1},H_{2},H_{3}\) whose indices in G are pairwise relatively prime and \(|G:H_{i}|=\) square free order, where \( i=1,2,3, \) then G is supersolvable.
The following result follows at once from Theorem 3.1
Corollary 3.3
[19, Theorem 3]. A group G is a supersolvable group whose order has at least three prime divisors if and only if there exist three maximal supersolvable subgroups of G whose indices are three distinct primes.
Theorem 3.4
Let G be a group and, for \(i=1,2,3,\) let \(H_{i}\) be a supersolvable subgroup of G. Suppose that \( |G:H_{1}|=p_{1}^{e_{1}},\) \(|G:H_{2}|=p_{2}^{e_{2}},\) \( |G:H_{3}|=p_{3}^{e_{3}},\) where \(p_{1},\) \(p_{2},\) \(p_{3}\) are three distinct primes such that \(p_{1}<p_{2}<p_{3}.\) Suppose further that every subgroup of G of order \(p_{3}\) is s−semipermutable in G. Then G is supersolvable.
Proof
Suppose that the theorem is false and let G be a counterexample with minimal order. Then:
-
(1)
\(P\trianglelefteq G,\) where P is a Sylow \(p-\)subgroup of G and p is the largest prime dividing the order of G. By a similar argument as in step (1) in the proof of Theorem 3.1, we have \( P\trianglelefteq G.\)
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(2)
\(p>p_{3},\) where p is the largest prime dividing the order of G. Suppose that \(p=p_{3}.\) By (1), \(P\trianglelefteq G.\) Then by Lemma 2.3(b), every subgroup of P of order \(p_{3}\) is s−permutable in G. Hence by Lemma 2.4, \(P\le Z_{\mathcal {U}}(G).\) It is clear that \(G=PH_{3},\) where P is a Sylow \(p_{3}-\)subgroup of G. Since \(G/P\cong H_{3}/P\cap H_{3}\) is supersolvable and \(P\le Z_{\mathcal {U}}(G),\) it follows easily that G is supersolvable, a contradiction.
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(3)
The final contradiction. By Wielandt’s Theorem [13, Theorem 6.4.4], G is solvable. Then by Hall’s Theorem [13, Theorem 6.4.1], G has a Hall subgroup K such that \(\pi (K)=\{p_{1},\) \(p_{2},\) \(p_{3}\}.\) By (2), \(p>p_{3}>p_{2}>p_{1}\) and so \(K<G.\) It is clear that \(G=H_{i}K\) and so \(|G:H_{i}|=|K:K\cap H_{i}|,\) \(i=1,2,3.\) By Lemma 2.3(a), every subgroup of K of order \(p_{3}\) is s− semipermutable in K. Then K satisfies the hypothesis of the theorem. The minimal choice of G yields that K is supersolvable. It is clear that (\( |G:H_{i}|,|G:K|)=1,\) \(i=1,2,3.\) Then by Theorem 1.2, G is supersolvable, a contradiction. \(\square \)
Proof of Theorem 1.8
Suppose that the theorem is false and let G be a counterexample with minimal order. Let \( M_{i}\) be a maximal subgroup of G such that \(H_{i}\le M_{i},\) \(i=1,2,3.\) Then we have \(|G:M_{1}|=p_{r}^{l_{r}},\) \(|G:M_{2}|=p_{j}^{l_{j}},\) \( |G:M_{3}|=p_{k}^{l_{k}},\) where \(p_{r},\) \(p_{j},\) \(p_{k}\) are distinct primes such that \(p_{r}<p_{j}<p_{k}\) (see the proof of Theorem 1.7). It is clear that \(G=H_{2}M_{1}=H_{3}M_{1}\) and so \(|G:H_{2}|=|M_{1}:M_{1}\cap H_{2}|\) and \(|G:H_{3}|=|M_{1}:M_{1}\cap H_{3}|.\) By hypothesis, every subgroup of G of order p is s−semipermutable in G for every prime p dividing \(|G:M_{3}|.\) By Lemma 2.3(a), every subgroup of \(M_{1}\) of order p is s−semipermutable in \(M_{1}\) for every prime p dividing \( |M_{1}:M_{1}\cap H_{3}|=|G:H_{3}|.\) Then \(M_{_{1}}\) satisfies the hypothesis of the theorem. Hence by the minimal choice of G, \(M_{1}\) is supersolvable. Similarly, \(M_{2}\) and \(M_{3}\) are supersolvable. Hence by Theorem 3.4, G is supersolvable, a contradiction. \(\square \)
Proof of Theorem 1.9
Suppose that the theorem is false and let G be a counterexample with minimal order. Let p be the largest prime dividing the order of G. Then:
-
(1)
\(p=p_{m}.\) Suppose that \(p>p_{m}.\) By Wielandt’s Theorem [13, Theorem 6.4.4], G is solvable. Then by Hall’s Theorem [13, Theorem 6.4.1], there exists a Hall subgroup K of G such that \(\pi (K)=\{p_{1},\) \(p_{2}\),..., \(p_{s},\) \(p_{s+1},\)..., \(p_{t},\) \(p_{t+1},\)..., \(p_{m}\}.\) It is clear that \( G=H_{3}(H_{2}\cap H_{1})\) and \(H_{3}\) and \(H_{2}\cap H_{1}\) are mutually permutable. Since \(G=KH_{i},\) we have \(|G:H_{i}|=|K:K\cap H_{i}|,\) \(i=1,2,3.\) By [7, Lemma 4.1.21(1)], \((K\cap H_{3})(K\cap H_{2}\cap H_{1})\) is a subgroup and \((K\cap H_{3})\) and \((K\cap H_{2}\cap H_{1})\) are mutually permutable. Then K satisfies the hypothesis of the theorem. By the minimal choice of G, K is supersolvable. It is clear that \((|G:H_{i}|,|G:K|)=1\) for any \(i=1,2,3.\) Hence by Theorem 1.2, G is supersolvable, a contradiction.
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(2)
\(P\trianglelefteq G,\) where P is a Sylow \(p_{m}-\)subgroup of G. By (1), \(p_{m}\) is the largest prime dividing the order of G. By a similar argument as in Step (1) in the proof of Theorem 3.1, we have \( P\trianglelefteq G.\)
-
(3)
If N is a nontrivial normal \(p_{m}-\)subgroup of G, then G/N is supersolvable. If \(G=H_{3}N,\) then \(G/N\cong H_{3}/H_{3}\cap N\) is supersolvable. Assume now that \(H_{3}N/N<G/N.\) It is clear that \(N\le H_{2}\) and \(N\le H_{1}.\) By [7, Lemma 4.1.10], \(H_{3}N/N\) and \(H_{2}\cap H_{1}/N\) are mutually permutable subgroups of G/N. Then G/N satisfies the hypothesis of the theorem. By the minimal choice of G, G/N is supersolvable.
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(4)
\(P\cap \Phi (G)=1.\) Suppose that \(P\cap \Phi (G)\ne 1.\) Then by (3), \(G/P\cap \Phi (G)\) is supersolvable. Hence by [20, Chapter 1, Corollary 3.2], G is supersolvable, a contradiction.
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(5)
\(Core_{G}(H_{3}\cap H_{2}\cap H_{1})\ne 1\) and \(G/Core_{G}(H_{3}\cap H_{2}\cap H_{1})\) is supersolvable. Suppose that \(Core_{G}(H_{3}\cap H_{2}\cap H_{1})=1.\) Then by Lemma 2.5, G is supersolvable, a contradiction. Thus \(Core_{G}(H_{3}\cap H_{2}\cap H_{1})\ne 1.\) Put \(A=Core_{G}(H_{3}\cap H_{2}\cap H_{1}).\) Now, it is easy to see that G/A satisfies the hypothesis of the theorem. Then by the minimal choice of G, G/A is supersolvable.
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(6)
P is a minimal normal subgroup of G.
By (4), \(P\cap \Phi (G)=1.\)
Then by Lemma 2.2, \(P=N_{1}\times N_{2}\times \cdots \times N_{n},\) where \(N_{i}\) is a minimal normal subgroup of G, \( i=1,2,\ldots ,n.\) By (3), \(G/N_{i}\) is supersolvable, \(i=1,2,3,\ldots ,n.\) Since the class of supersolvable groups is a saturated formation, we have P is a minimal normal subgroup of G.
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(7)
The final contradiction. By (5), G/A is supersolvable, where \(A=Cor_{G}(H_{3}\cap H_{2}\cap H_{1})\ne 1.\) By (2) and (3), G/P is supersolvable. Since G is not supersolvable, we have \(P\cap A\ne 1.\) By (6), P is a minimal normal subgroup of G and so \(P\le A\le H_{3},\) a contradiction. \(\square \)
Proof of Theorem 1.10
Suppose that the theorem is false and let G be a counterexample with minimal order. Since \(G^{^{\prime }}\) is nilpotent, we have G is solvable. Clearly, the order of G is divided by at least three distinct primes. Let L be a minimal normal subgroup of G. Since G is solvable, it follows that \( |L|=p^{n},\) for some prime p. If \(G=H_{i}L\) for some \(i=1,2,3,\) then \( G/L\in \Im .\) Assume now that \(H_{i}L<G,\) \(i=1,2,3.\) Then \(1<H_{i}L/L<G/L,\) \( i=1,2,3.\) Clearly, \(H_{i}L/L\cong H_{i}/H_{i}\cap L\in \Im \) and \( (G/L)^{^{\prime }}=G^{^{\prime }}L/L\cong G^{^{\prime }}/G^{^{\prime }}\cap L \) is nilpotent. By the minimal choice of G, \(G/L\in \Im .\) Since \(\Im \) is a saturated formation, L is a unique minimal normal subgroup of G and \(\Phi (G)=1.\) By Lemma 2.2, \(F(G)=L\) and so \(G^{^{\prime }}=F(G)=L.\) Since \(|L|=p^{n}\) and G possesses three subgroups \(H_{i},\) \(i=1,2,3,\) whose indices in G are pairwise relatively primes, we can assume that \(p\not \mid |G:H_{i}|\), \(i=1,2.\) Now it follows easily that \(G^{^{\prime }}=F(G)\le H_{i},\) \(i=1,2.\) Then \(H_{i}\trianglelefteq G,\) \(i=1,2.\) Hence \(G=H_{1}H_{2}\) is a mutually permutable product of \(H_{1}\) and \(H_{2}.\) Applying Lemma 2.6, \(G\in \Im ,\) a contradiction. \(\square \)
As a corollary of Theorem 1.10, we have
Theorem 1.5 ([12, Theorem 1.1]). Let G be a group and let \(H_{i},\) \(i=1,2,3,\) be supersolvable subgroups of G with pairwise relatively prime indices in G. Suppose that \( G^{^{\prime }}\) is nilpotent. Then G is supersolvable.
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Asaad, M. On Finite Groups with Three Supersolvable Subgroups of Pairwise Relatively Prime Indices. Bull. Malays. Math. Sci. Soc. 47, 112 (2024). https://doi.org/10.1007/s40840-024-01692-6
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DOI: https://doi.org/10.1007/s40840-024-01692-6