A group with H-Frobenius element of even order

Popov, A. A.; Sozutov, A. I.

doi:10.1007/s10469-005-0005-0

Cited by 4 publications

(3 citation statements)

References 11 publications

Supporting

Mentioning

Contrasting

Order By: Relevance

“…Proposition 8 [9]. If G contains an H-Frobenius element a of even order > 2 then G = F C G (i), where F is a periodic abelian subgroup and i is an involution in a .…”

Section: Definitions and Prerequisitesmentioning

confidence: 99%

“…Let G be a group with an H-Frobenius element a of order > 2. The main task of this article was formulated in Question 10.61 in [8]: Is the union of all kernels of Frobenius subgroups with the complement a a subgroup of G?The affirmative answer to this question was known for |a| = 2n [9] and also in the case when |a| / ∈ {3, 5} and a is finite in G [10]. Here, in the first case, the proof substantially used the abelianity of the kernels of the Frobenius subgroups L g and the uniqueness of an involution in the complement; and in the second case, the finiteness of all subgroups L g (g ∈ G).…”

mentioning

confidence: 99%

“…The affirmative answer to this question was known for |a| = 2n [9] and also in the case when |a| / ∈ {3, 5} and a is finite in G [10]. Here, in the first case, the proof substantially used the abelianity of the kernels of the Frobenius subgroups L g and the uniqueness of an involution in the complement; and in the second case, the finiteness of all subgroups L g (g ∈ G).…”

mentioning

confidence: 99%

See 2 more Smart Citations

On groups with Frobenius elements

Popov

Sozutov

2015

Sib Math J

Self Cite

View full text Add to dashboard Cite

We find the conditions under which the set-theoretic union of the kernels of two-generated Frobenius subgroups of a group G with fixed cyclic complement of order 3n is a normal subgroup in G.The article deals with groups with an H-Frobenius element a of order divisible by 3.An element a of a group G is called H-Frobenius if H is a proper subgroup of G and all subgroups of the form L g = a, a g ,Frobenius groups with complements containing a. If, moreover, the complements of all groups L g coincide with a then a is called cyclically H-Frobenius; and if H = N G ( a ) then a is called Frobenius [1, 2]. These systems of Frobenius subgroups naturally arise in the study of groups with various conditions of finiteness, factorizability, exact 2-transitivity, etc. Finite groups with Frobenius element of order > 2 were studied by Fischer [3, 4] and Aschbacher [5]. Arbitrary groups G with a cyclically H-Frobenius element a of simple odd order were considered by Shunkov [6], and in the case of |a| > 2, by Sozutov [7]. It was proved in these articles that the normal closures a G and a H of a cyclically H-Frobenius element a are Frobenius groups with complement a . Let G be a group with an H-Frobenius element a of order > 2. The main task of this article was formulated in Question 10.61 in [8]: Is the union of all kernels of Frobenius subgroups with the complement a a subgroup of G?The affirmative answer to this question was known for |a| = 2n [9] and also in the case when |a| / ∈ {3, 5} and a is finite in G [10]. Here, in the first case, the proof substantially used the abelianity of the kernels of the Frobenius subgroups L g and the uniqueness of an involution in the complement; and in the second case, the finiteness of all subgroups L g (g ∈ G).As is known, the kernel of a finite Frobenius group is nilpotent, and the subgroup in the complement generated by all elements of prime orders is either cyclic or isomorphic to the direct product of a Hall cyclic subgroup and one of the groups SL 2 (3) and SL 2 (5) (this explains why the case |a| ∈ {3, 5} is exceptional). Every subgroup of a finite Frobenius group generated by a pair of conjugate elements is again Frobenius. These properties need not be fulfilled in an infinite Frobenius group. As was proved in [11], each group is isomorphically embeddable in the kernel of a suitable Frobenius group with cyclic complement, and by [12,13], a periodic Frobenius group with abelian kernel need not be locally finite. Therefore, at present, the complete solution of Question 10.61 of [8] seems problematic to the authors. But, for the cases when the order of a divides by 3 or 5, the prognosis is more favorable due to evident progress in the study of Frobenius groups and complements generated by elements of small orders (see, for example, [14][15][16][17][18][19][20][21]). For instance, Zhurtov, Mazurov, and Churkin found a number of conditions promising for the above problem, under which a Frobenius group generated by elements of small orders is finite. The present article relies upon these ...

show abstract

“…Proposition 8 [9]. If G contains an H-Frobenius element a of even order > 2 then G = F C G (i), where F is a periodic abelian subgroup and i is an involution in a .…”

Section: Definitions and Prerequisitesmentioning

confidence: 99%

mentioning

confidence: 99%

mentioning

confidence: 99%

See 1 more Smart Citation