2023/10/05 by Maria Ferrara, Ferrara, Maria, Marco Trombetti +1
Computer Science · Mathematics · #20E15 #20F50 #Advanced Algebra and Logic #FOS: Mathematics #Finite Group Theory Research #Group Theory (math.GR) #Rings, Modules, and Algebras
paper · pdf · doi:10.48550/arxiv.2310.03391
openalex publication_date 2023/10/05 · openalex created_date 2023/10/09 · openalex updated_date 2026/07/28
Let σ=\σj : j∈ J\ be a partition of the set ℙ of all prime numbers. A subgroup X of a finite group G is~σ-subnormal in G if there exists a chain of subgroups X=X0≤ X1≤…≤ Xn=G such that, for each 1≤ i≤ n-1, Xi-1\trianglelefteq Xi or Xi/(Xi-1)Xi is a σji-group for some ji∈ J. Skiba~[12] studied the main properties of σ-subnormal subgroups in finite groups and showed that the set of all σ-subnormal subgroups plays a relevant role in the structure of a finite soluble group. In [5], we laid the foundation of a general theory of σ-subnormal subgroups (and σ-series) in locally finite groups. It turns out that the main difference between the finite and the locally finite case concerns the behaviour of the join of σ-subnormal subgroups: in finite groups, σ-subnormal subgroups form a sublattice of the lattice of all subgroups [3], but this is no longer true for arbitrary locally finite groups. This is similar to what happens with subnormal subgroups, so it makes sense to study the class \mathfrakSσ^∞ (resp. \mathfrakSσ) of locally finite groups in which the join of (resp. of finitely many) σ-subnormal subgroups is σ-subnormal. Our aim is to study how much one can extend a group in one of these classes before going outside the same class (see for example Theorems~3.6, 3.8, 5.5 and 5.7). Also, σ-subnormality criteria for the join of σ-subnormal subgroups are obtained: similarly to a celebrated theorem of Williams (see [15]), we give a necessary and sufficient conditions for a join of two σ-subnormal subgroups to always be σ-subnormal; consequently, we show that the join of two orthogonal σ-subnormal subgroups is σ-subnormal (extending a result of Roseblade [11]).