Groups as Unions of Proper Subgroups

Haber, Seymour; Rosenfeld, Azriel

doi:10.2307/2310634

Cited by 38 publications

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“…Since a € C 2 , it follows that w = 1. This is a contradiction, and shows that G^ C 2 U C 3 U C 4 . The result that if b = 3 then c :£ 4 follows as explained above.…”

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confidence: 95%

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Groups of breadth four have class five

Macdonald

1978

Glasgow Math. J.

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“…Since a € C 2 , it follows that w = 1. This is a contradiction, and shows that G^ C 2 U C 3 U C 4 . The result that if b = 3 then c :£ 4 follows as explained above.…”

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confidence: 95%

“…It is a known fact that if D = C 2 C\C 3 DC 4 then G/D is non-cyclic of order 4, and though this is due to Scorza (1926), according to [3], a more convenient reference is [4]. Since a € C 2 , it follows that w = 1.…”

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confidence: 99%

Groups of breadth four have class five

Macdonald

1978

Glasgow Math. J.

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“…Par [7], on a aussi i∈I C i ⊆ D. Il en résulte que i∈I C i = 1. D'où C X (T ) = 1 et |T | n, où T = R i | i ∈ I , et il résulte du Lemme 3.6(ii) de [13] que |X| n − 1.…”

Section: Groupes Résolubles De Type Fini Satisfaisant La Condition E(n)unclassified

Groupes satisfaisant une condition d'Engel

Abdollahi¹

2005

Journal of Algebra

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Reçu le 15 novembre 2000 Disponible sur Internet le 10 décembre 2004 Communiqué par J. Tits RésuméSoit n un entier positif. Nous disons qu'un groupe G satisfait la condition E (n), si chaque ensemble de n + 1 éléments de G contient une paire {x, y} telle que [x, k y] = 1, pour un entier positif k = k(x, y). Dans cet article, nous étudions les groupes G satisfaisant cette condition. En particulier, si G est un groupe résoluble de type fini, alors | G Z * (G) | n 113 √ n+2 , où Z * (G) est l'hypercentre de G. AbstractLet n be a positive integer. We say that a group G satisfies the condition E (n), if every set of n + 1 elements of G contains a pair {x, y} such that [x, k y] =

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“…Hermann Heineken [2] A\J C and in B u C. So xy = yx is contained in C, which shows that C is a commutative subgroup.…”

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confidence: 89%

“…On the second mapping /? we impose n p = n for all n e N. Furthermore we put x^ -xufor some us N. For P to be an antiautomorphism it is necessary, that X 2 = (jc 2 y = (xu) 2 , which is equivalent to x~lux = u~l.…”

Section: C(vw) = C(tu) If Tv = Vt and Uw = Wu C(vw) = C(ut) = C(tu) Imentioning

confidence: 99%

On groups with special antiautomorphisms

Heineken

1970

J. Aust. Math. Soc.

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This note is concerned with the following question: What is the structure of those groups which possess two antiautomorphisms different from identity such that every element of the group is fixed by (at least) one of them?C. Ayoub [1] stated this problem after having proved a statement equivalent to the following: The group G is a non-abelian extension of an abelian group by a group of order two, if, and only if, there is an automorphism α ≠ 1 and an antiautomorphism β ≠ 1 such that every element of G is fixed by a α by β. The Theorem at the end of the paper will show that the class of groups considered by C. Ayoub coincides with the class considered here.

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Groups as Unions of Proper Subgroups

Cited by 38 publications

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Groups of breadth four have class five

Groups of breadth four have class five

Groupes satisfaisant une condition d'Engel

On groups with special antiautomorphisms

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