19.84 (2018)
SolvedLet $\mathbb{P}$ be the set of all primes, and let $\sigma = \{\sigma_i \mid i \in I\}$ be some partition of $\mathbb{P}$ into disjoint subsets. A finite group $G$ is said to be $\sigma$-primary if $G$ is a $\sigma_i$-group for some $i$; $\sigma$-nilpotent if $G$ is a direct product of $\sigma$-primary groups; $\sigma$-soluble if every chief factor of $G$ is $\sigma$-primary. A subgroup $A$ of a finite group $G$ is said to be $\sigma$-subnormal in $G$ if there is a chain $A = A_0 \leqslant A_1 \leqslant \dots \leqslant A_n = G$ such that for every $i$ either $A_{i-1} \trianglelefteq A_i$ or $A_i/(A_{i-1})_{A_i}$ is $\sigma$-primary, where $(A_{i-1})_{A_i}$ is the largest normal subgroup of $A_i$ contained in $A_{i-1}$. Suppose that a subgroup $A$ of a finite group $G$ is $\sigma$-subnormal in $\langle A, A^x \rangle$ for all $x \in G$. Is it true that then $A$ is $\sigma$-subnormal in $G$?
Progress
No, not always. For example, in $G = S_5$ with partition $\sigma = \{2, 3\} \cup \{5\}$ the subgroup $A = \langle (12) \rangle$ is $\sigma$-subnormal in $\langle A, A^x \rangle$ for every $x \in G$, but it is not $\sigma$-subnormal in $G$. (V. N. Tyutyanov, Letter of 28 August 2019.)
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