Article contents
ON THE LATTICE OF $\unicode[STIX]{x1D6F1}_{\mathfrak{I}}$-SUBNORMAL SUBGROUPS OF A FINITE GROUP
Published online by Cambridge University Press: 02 May 2017
Abstract
Let $\unicode[STIX]{x1D70E}=\{\unicode[STIX]{x1D70E}_{i}\mid i\in I\}$ be a partition of the set of all primes $\mathbb{P}$. Let $\unicode[STIX]{x1D70E}_{0}\in \unicode[STIX]{x1D6F1}\subseteq \unicode[STIX]{x1D70E}$ and let $\mathfrak{I}$ be a class of finite $\unicode[STIX]{x1D70E}_{0}$-groups which is closed under extensions, epimorphic images and subgroups. We say that a finite group $G$ is $\unicode[STIX]{x1D6F1}_{\mathfrak{I}}$-primary provided $G$ is either an $\mathfrak{I}$-group or a $\unicode[STIX]{x1D70E}_{i}$-group for some $\unicode[STIX]{x1D70E}_{i}\in \unicode[STIX]{x1D6F1}\setminus \{\unicode[STIX]{x1D70E}_{0}\}$ and we say that a subgroup $A$ of an arbitrary group $G^{\ast }$ is $\unicode[STIX]{x1D6F1}_{\mathfrak{I}}$-subnormal in $G^{\ast }$ if there is a subgroup chain $A=A_{0}\leq A_{1}\leq \cdots \leq A_{t}=G^{\ast }$ such that either $A_{i-1}\unlhd A_{i}$ or $A_{i}/(A_{i-1})_{A_{i}}$ is $\unicode[STIX]{x1D6F1}_{\mathfrak{I}}$-primary for all $i=1,\ldots ,t$. We prove that the set ${\mathcal{L}}_{\unicode[STIX]{x1D6F1}_{\mathfrak{I}}}(G)$ of all $\unicode[STIX]{x1D6F1}_{\mathfrak{I}}$-subnormal subgroups of $G$ forms a sublattice of the lattice of all subgroups of $G$ and we describe the conditions under which the lattice ${\mathcal{L}}_{\unicode[STIX]{x1D6F1}_{\mathfrak{I}}}(G)$ is modular.
- Type
- Research Article
- Information
- Bulletin of the Australian Mathematical Society , Volume 96 , Issue 2 , October 2017 , pp. 233 - 244
- Copyright
- © 2017 Australian Mathematical Publishing Association Inc.
Footnotes
The research is supported by NNSF of China (grant no. 11371335) and the Wu Wen-Tsun Key Laboratory of Mathematics of the Chinese Academy of Sciences.
References
- 7
- Cited by