In this paper we show that a finite group $G$ with Quaternion Sylow $2$-subgroup is $2$-nilpotent if, either $3\nmid |G|$ or $G$ is solvable and the order of its Sylow $2$-subgroup is strictly greater than $16$

Hamid Mousavi

Comptes Rendus Mathématique

Mousavi, Hamid

Numérisation de Documents Anciens Mathématiques

Comptes RendusMathématiqueHamid MousaviFinite groups with Quaternion Sylow subgroupVolume 358, issue 9-10 (2020), p. 1097-1099Published online: 5 January 2021https://doi.org/10.5802/crmath.131This article is licensed under theCreative Commons Attribution 4.0 International License.http://creativecommons.org/licenses/by/4.0/Les Comptes Rendus. Mathématique sont membres duCentre Mersenne pour l’édition scientifique ouvertewww.centre-mersenne.orge-ISSN : 1778-3569Comptes RendusMathématique2020, 358, n 9-10, p. 1097-1099https://doi.org/10.5802/crmath.131Group Theory / Théorie des groupesFinite groups with Quaternion SylowsubgroupHamidMousaviaa Department of Mathematical Sciences, University of Tabriz, P.O.Box 51666-16471,Tabriz, IranE-mail: hmousavi@tabrizu.ac.irAbstract. In this paper we show that a finite group G with Quaternion Sylow 2-subgroup is 2-nilpotent if,either 3 - |G| or G is solvable and the order of its Sylow 2-subgroup is strictly greater than 16.Mathematical subject classification (2010). 20D99, 20E45.Manuscript received 7th October 2020, revised 11th October 2020 and 13th October 2020, accepted 13th Octo-ber 2020.1. IntroductionA 2-group of order 2n , with a cyclic maximal subgroup is isomorphic to one of the followinggroup:(i) Cyclic group Z2n and abelian group Z2n−1 ×Z2.(ii) Dihedral group D2n .(iii) Semi-dihedral group SD2n .(iv) Modular group M(2n).(v) Quaternion group Q2n .Let G be a group with a Sylow 2-subgroup S of type (i). Then G is 2-nilpotent, since Aut(S) isa 2-group and so NG (S) = CG (S). In [5], D. Gorenstein and J. Walter gave the characterizationof finite groups with dihedral Sylow 2-subgroups. They proved that, a finite group G with adihedral Sylow 2-subgroup is 2-nilpotent if G contains a normal subgroup of index 4 (see [5,Lemma 2.1(iii)]). In [8, 9], W. Wong obtained the structure of finite groups whose Sylow 2-subgroups are semi-dihedral or modular 2-group. In the latter case G has a normal 2-complement(see [8, Theorem 1]), so a modular 2-group can not be a Sylow 2-subgroup of a non-solvablegroup. In the semi-dihedral case, G is 2-nilpotent if foc(S) = S′, where S denotes a Sylow 2-subgroup of G and foc(S), the focal subgroup of S [8, Theorem 2(I)].In [1], R. Brauer and M. Suzuki proved that any group with a Quaternion Sylow 2-subgroup isnot simple.ISSN (electronic) : 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/1098 Hamid MousaviNow this question seems to be natural: when a group G with a Quaternion Sylow 2-subgroupis 2-nilpotent?In this paper we obtain sufficient conditions for a group G with a Quaternion (ordinary orgeneralized) Sylow 2-subgroup to be 2-nilpotent.Our notations are standard and can be found in [6].Main Theorem. Let G be a finite group with a Quaternion (ordinary or generalized) Sylow 2-subgroup. Then G is 2-nilpotent if:(i) either 3 - |G|;(ii) or G is solvable with a Sylow 2-subgroup of order strictly greater than 16.Proof. Assume that Q ∼= Q2n is a Sylow 2-subgroup of G . If QEG then G = QN , where N is acomplement of Q. Now N acts trivially on Q, since Aut(Q2n ) is a 2-group unless for n = 3, in thelatter case Aut(Q8) ∼= S4 and 3 - |G|, again N acts trivially on Q. Hence N EG and G ∼= Q × N .Therefore we can assume that Q 5G .(i). By [1], G is not simple. Assume that G is a minimal counterexample and M is a maximalnormal subgroup of G . If |M | is odd, then G/M is a simple group with a Quaternion Sylowsubgroup which is a contradiction, so 2 | |M |. By assumption M 6=Q. Suppose that M 6Q, as G/Mis a non-abelian simple group and 3 - |G/M |, as the Suzuki groups are the only non-abelian simplegroups which 3 does not divides its order [4], then G/M is a Suzuki group with Q/M as its Sylow2-subgroup, which is a contradiction for Q/M is either cyclic or dihedral but a Sylow 2-subgroupof a Suzuki group is of exponent 4 and order greater than or equal 64, (see [7, Lemma 1.6(4)]and [2, Lemma 1 & Proposition 3]).Therefore either Q 6M , then M has a normal 2-complement, for |M | < |G|, or M∩Q is a propersubgroup of Q which is cyclic or Q2n−1 , again M has a normal 2-complement. Thus in either caseM has a normal 2-complement M1 which is normal in G . Now G/M1 has a Quaternion Sylow2-subgroup. By the choice of G , G 6= QM1 so G/M1 has a normal 2-complement N /M1. HenceG = NQ which is a contradiction. Therefore G is 2-nilpotent.(ii). Proof by induction on |G|. Assume that G is solvable and n ≥ 5. Obviously Q 6=G ′. If Q 6G ′,as Q 5 G ′, then by induction G ′ is 2-nilpotent with a normal 2-complement M which is normalin G . Now G/M has a normal Sylow 2-subgroup so is 2-nilpotent with a normal 2-complementN /M . Obviously N is normal a 2-complement of G . So assume that G ′∩Q is a proper subgroupof Q, then either G ′∩Q is cyclic which again implies that G ′ is 2-nilpotent and we are done, orG ′∩Q ∼=Q2n−1 and in this case G ′ is 2-nilpotent unless n = 5, 3 | |G ′| and G ′∩Q is non-cyclic anda non-normal subgroup of G ′. In the latter case, Q1 =G ′∩Q ∼=Q16 and G (`) 6=Q1, for all `≥ 2. Bysolvability for some `, G (`) ∩Q1 is a proper subgroup of Q1. We can assume that ` is the smallestnumber such that G (`) ∩Q1 6=Q1, hence G (`) ∩Q1 is cyclic, since G (`) ∩Q1EQ. If G (`) 6Q1, thenG (`−1)/G (`) is abelian with a normal 2-complement M/G (`), thus MG (`)EG (`−1) and so MEG (`−1),for G (`−1) = MQ1. This implies that MEG and so G/M is 2-nilpotent by induction. Therefore G is2-nilpotent. Otherwise G (`) is 2-nilpotent with a normal 2-complement M which is normal in G ,again G/M is 2-nilpotent by induction and we are done. Corollary 1. Let G be a finite group with Quaternion (ordinary or generalized) Sylow 2-subgroup.If a Sylow 3-subgroup of G is normal, then G is 2-nilpotent.Proof. Assume that P is the Sylow 3-subgroup of G . As 3 - |G/P |, G/P is 2-nilpotent by the mainTheorem. So G is 2-nilpotent. Corollary 2. Let G be a finite group with Quaternion (ordinary or generalized) Sylow 2-subgroupsuch that 3 - |G|, then G is solvable.C. R. Mathématique, 2020, 358, n 9-10, 1097-1099Hamid Mousavi 1099Remark 3. In the above lemma for n = 3 and 4 where 3 | |G|, there exist many examples forwhich G is not 2-nilpotent. The smallest order of such groups is a group of order 48. Assumethat G := SmallGroup(48,28) the small group of GAP library [3], in this group G ′ ∼= SL(2,3) anda Sylow 2-subgroup of G is non-normal, obviously G/G ′ ∼= Z2. For another example we considerG ∼= Z5 ×SL(2,3). In this case G ′ ∼= Q8 and G ∼= (Z7 ×Z7)oSL(2,3), where Q8 acts irreducibly onZ7 ×Z7, in this case G ′ ∼= (Z7 ×Z7)oQ8. For non-solvable case let G := SmallGroup(672,1045), inthis group G ′ ∼= SL(2,7) and a Sylow 2-subgroup of G is a Quaternion group of order 32.AcknowledgmentThe author would like to thank professor Stefan Kohl for providing useful information aboutsimple groups, especially the Suzuki groups.References[1] R. Brauer, M. Suzuki, “On finite groups of even order whose 2-Sylow group is a Quaternion group”, Proc. Natl. Acad.Sci. USA 45 (1959), p. 1757-1759.[2] M. J. Collins, “The characterisation of the Suzuki groups by their Sylow 2-subgroups”, Math. Z. 123 (1971), p. 32-48.[3] The GAP Group, “GAP – Groups, Algorithms, and Programming, Version 4.11.0”, 2020, http://www.gap-system.org.[4] G. Glauberman, Factorizations in local subgroups of finite groups, Regional Conference Series in Mathematics, vol. 33,American Mathematical Society, 1977.[5] D. Gorenstein, J. H. Walter, “The characterization of finite groups with dihedral Sylow 2-subgroups I., II., III”, J. Algebra2 (1965), p. 85-151; 218-270; 334-393.[6] I. M. Isaacs, Finite group theory, Graduate Studies in Mathematics, vol. 92, American Mathematical Society, 2008.[7] S. Kohl, “Counting the orbits on finite simple groups under the action of the automorphism group – Suzuki groups vs.linear groups”, Commun. Algebra 30 (2002), no. 7, p. 3515-3532.[8] W. J. Wong, “On finite groups whose 2-Sylow subgroups have cyclic subgroups of index 2”, J. Aust. Math. Soc. 4 (1964),p. 90-112.[9] ——— , “On Finite Groups with Semi-Dihedral Sylow 2-Subgroups”, J. Algebra 4 (1966), p. 52-63.C. R. Mathématique, 2020, 358, n 9-10, 1097-1099

Finite groups with Quaternion Sylow subgroup

Comptes RendusMathématiqueHamid MousaviFinite groups with Quaternion Sylow subgroupVolume 358, issue 9-10 (2020), p. 1097-1099.<https://doi.org/10.5802/crmath.131>© Académie des sciences, Paris and the authors, 2020.Some rights reserved.This article is licensed under theCreative Commons Attribution 4.0 International License.http://creativecommons.org/licenses/by/4.0/Les Comptes Rendus. Mathématique sont membres duCentre Mersenne pour l’édition scientifique ouvertewww.centre-mersenne.orgComptes RendusMathématique2020, 358, n 9-10, p. 1097-1099https://doi.org/10.5802/crmath.131Group Theory / Théorie des groupesFinite groups with Quaternion SylowsubgroupHamidMousaviaa Department of Mathematical Sciences, University of Tabriz, P.O.Box 51666-16471,Tabriz, IranE-mail: hmousavi@tabrizu.ac.irAbstract. In this paper we show that a finite group G with Quaternion Sylow 2-subgroup is 2-nilpotent if,either 3 - |G| or G is solvable and the order of its Sylow 2-subgroup is strictly greater than 16.Mathematical subject classification (2010). 20D99, 20E45.Manuscript received 7th October 2020, revised 11th October 2020 and 13th October 2020, accepted 13th Octo-ber 2020.1. IntroductionA 2-group of order 2n , with a cyclic maximal subgroup is isomorphic to one of the followinggroup:(i) Cyclic group Z2n and abelian group Z2n−1 ×Z2.(ii) Dihedral group D2n .(iii) Semi-dihedral group SD2n .(iv) Modular group M(2n).(v) Quaternion group Q2n .Let G be a group with a Sylow 2-subgroup S of type (i). Then G is 2-nilpotent, since Aut(S) isa 2-group and so NG (S) = CG (S). In [5], D. Gorenstein and J. Walter gave the characterizationof finite groups with dihedral Sylow 2-subgroups. They proved that, a finite group G with adihedral Sylow 2-subgroup is 2-nilpotent if G contains a normal subgroup of index 4 (see [5,Lemma 2.1(iii)]). In [8, 9], W. Wong obtained the structure of finite groups whose Sylow 2-subgroups are semi-dihedral or modular 2-group. In the latter case G has a normal 2-complement(see [8, Theorem 1]), so a modular 2-group can not be a Sylow 2-subgroup of a non-solvablegroup. In the semi-dihedral case, G is 2-nilpotent if foc(S) = S′, where S denotes a Sylow 2-subgroup of G and foc(S), the focal subgroup of S [8, Theorem 2(I)].In [1], R. Brauer and M. Suzuki proved that any group with a Quaternion Sylow 2-subgroup isnot simple.ISSN (electronic) : 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/1098 Hamid MousaviNow this question seems to be natural: when a group G with a Quaternion Sylow 2-subgroupis 2-nilpotent?In this paper we obtain sufficient conditions for a group G with a Quaternion (ordinary orgeneralized) Sylow 2-subgroup to be 2-nilpotent.Our notations are standard and can be found in [6].Main Theorem. Let G be a finite group with a Quaternion (ordinary or generalized) Sylow 2-subgroup. Then G is 2-nilpotent if:(i) either 3 - |G|;(ii) or G is solvable with a Sylow 2-subgroup of order strictly greater than 16.Proof. Assume that Q ∼= Q2n is a Sylow 2-subgroup of G . If QEG then G = QN , where N is acomplement of Q. Now N acts trivially on Q, since Aut(Q2n ) is a 2-group unless for n = 3, in thelatter case Aut(Q8) ∼= S4 and 3 - |G|, again N acts trivially on Q. Hence N EG and G ∼= Q × N .Therefore we can assume that Q 5G .(i). By [1], G is not simple. Assume that G is a minimal counterexample and M is a maximalnormal subgroup of G . If |M | is odd, then G/M is a simple group with a Quaternion Sylowsubgroup which is a contradiction, so 2 | |M |. By assumption M 6=Q. Suppose that M 6Q, as G/Mis a non-abelian simple group and 3 - |G/M |, as the Suzuki groups are the only non-abelian simplegroups which 3 does not divides its order [4], then G/M is a Suzuki group with Q/M as its Sylow2-subgroup, which is a contradiction for Q/M is either cyclic or dihedral but a Sylow 2-subgroupof a Suzuki group is of exponent 4 and order greater than or equal 64, (see [7, Lemma 1.6(4)]and [2, Lemma 1 & Proposition 3]).Therefore either Q 6M , then M has a normal 2-complement, for |M | < |G|, or M∩Q is a propersubgroup of Q which is cyclic or Q2n−1 , again M has a normal 2-complement. Thus in either caseM has a normal 2-complement M1 which is normal in G . Now G/M1 has a Quaternion Sylow2-subgroup. By the choice of G , G 6= QM1 so G/M1 has a normal 2-complement N /M1. HenceG = NQ which is a contradiction. Therefore G is 2-nilpotent.(ii). Proof by induction on |G|. Assume that G is solvable and n ≥ 5. Obviously Q 6=G ′. If Q 6G ′,as Q 5 G ′, then by induction G ′ is 2-nilpotent with a normal 2-complement M which is normalin G . Now G/M has a normal Sylow 2-subgroup so is 2-nilpotent with a normal 2-complementN /M . Obviously N is normal a 2-complement of G . So assume that G ′∩Q is a proper subgroupof Q, then either G ′∩Q is cyclic which again implies that G ′ is 2-nilpotent and we are done, orG ′∩Q ∼=Q2n−1 and in this case G ′ is 2-nilpotent unless n = 5, 3 | |G ′| and G ′∩Q is non-cyclic anda non-normal subgroup of G ′. In the latter case, Q1 =G ′∩Q ∼=Q16 and G (`) 6=Q1, for all `≥ 2. Bysolvability for some `, G (`) ∩Q1 is a proper subgroup of Q1. We can assume that ` is the smallestnumber such that G (`) ∩Q1 6=Q1, hence G (`) ∩Q1 is cyclic, since G (`) ∩Q1EQ. If G (`) 6Q1, thenG (`−1)/G (`) is abelian with a normal 2-complement M/G (`), thus MG (`)EG (`−1) and so MEG (`−1),for G (`−1) = MQ1. This implies that MEG and so G/M is 2-nilpotent by induction. Therefore G is2-nilpotent. Otherwise G (`) is 2-nilpotent with a normal 2-complement M which is normal in G ,again G/M is 2-nilpotent by induction and we are done. Corollary 1. Let G be a finite group with Quaternion (ordinary or generalized) Sylow 2-subgroup.If a Sylow 3-subgroup of G is normal, then G is 2-nilpotent.Proof. Assume that P is the Sylow 3-subgroup of G . As 3 - |G/P |, G/P is 2-nilpotent by the mainTheorem. So G is 2-nilpotent. Corollary 2. Let G be a finite group with Quaternion (ordinary or generalized) Sylow 2-subgroupsuch that 3 - |G|, then G is solvable.C. R. Mathématique, 2020, 358, n 9-10, 1097-1099Hamid Mousavi 1099Remark 3. In the above lemma for n = 3 and 4 where 3 | |G|, there exist many examples forwhich G is not 2-nilpotent. The smallest order of such groups is a group of order 48. Assumethat G := SmallGroup(48,28) the small group of GAP library [3], in this group G ′ ∼= SL(2,3) anda Sylow 2-subgroup of G is non-normal, obviously G/G ′ ∼= Z2. For another example we considerG ∼= Z5 ×SL(2,3). In this case G ′ ∼= Q8 and G ∼= (Z7 ×Z7)oSL(2,3), where Q8 acts irreducibly onZ7 ×Z7, in this case G ′ ∼= (Z7 ×Z7)oQ8. For non-solvable case let G := SmallGroup(672,1045), inthis group G ′ ∼= SL(2,7) and a Sylow 2-subgroup of G is a Quaternion group of order 32.AcknowledgmentThe author would like to thank professor Stefan Kohl for providing useful information aboutsimple groups, especially the Suzuki groups.References[1] R. Brauer, M. Suzuki, “On finite groups of even order whose 2-Sylow group is a Quaternion group”, Proc. Natl. Acad.Sci. USA 45 (1959), p. 1757-1759.[2] M. J. Collins, “The characterisation of the Suzuki groups by their Sylow 2-subgroups”, Math. Z. 123 (1971), p. 32-48.[3] The GAP Group, “GAP – Groups, Algorithms, and Programming, Version 4.11.0”, 2020, http://www.gap-system.org.[4] G. Glauberman, Factorizations in local subgroups of finite groups, Regional Conference Series in Mathematics, vol. 33,American Mathematical Society, 1977.[5] D. Gorenstein, J. H. Walter, “The characterization of finite groups with dihedral Sylow 2-subgroups I., II., III”, J. Algebra2 (1965), p. 85-151; 218-270; 334-393.[6] I. M. Isaacs, Finite group theory, Graduate Studies in Mathematics, vol. 92, American Mathematical Society, 2008.[7] S. Kohl, “Counting the orbits on finite simple groups under the action of the automorphism group – Suzuki groups vs.linear groups”, Commun. Algebra 30 (2002), no. 7, p. 3515-3532.[8] W. J. Wong, “On finite groups whose 2-Sylow subgroups have cyclic subgroups of index 2”, J. Aust. Math. Soc. 4 (1964),p. 90-112.[9] ——— , “On Finite Groups with Semi-Dihedral Sylow 2-Subgroups”, J. Algebra 4 (1966), p. 52-63.C. R. Mathématique, 2020, 358, n 9-10, 1097-1099

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Finite groups with Quaternion Sylow subgroup

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