Equivalence of the existence of nonconjugate and nonisomorphic Hall $\pi$-subgroups

Guo Wen Bin^a, A. A. Buturlakin^bc, D. O. Revin<sup>bca

a</sup> School of Mathematical Sciences, University of Science and Technology of China
^b Sobolev Institute of Mathematics, Siberian Branch of the Russian Academy of Sciences, Novosibirsk
^c Novosibirsk State University

Abstract: Let $\pi$ be some set of primes. A subgroup $H$ of a finite group $G$ is called a Hall $\pi$-subgroup if any prime divisor of the order $|H|$ of the subgroup $H$ belongs to $\pi$ and the index $|G:H|$ is not a multiple of any number in $\pi$. The famous Hall theorem states that a solvable finite group always contains a Hall $\pi$ subgroup and any two Hall $\pi$-subgroups of such group are conjugate. The converse of the Hall theorem is also true: for any nonsolvable group $G$, there exists a set $\pi$ such that $G$ does not contain Hall $\pi$-subgroups. Nevertheless, Hall $\pi$-subgroups may exist in a nonsolvable group. There are examples of sets $\pi$ such that, in any finite group containing a Hall $\pi$-subgroup, all Hall $\pi$-subgroups are conjugate (and, as a consequence, are isomorphic). In 1987 F. Gross showed that any set $\pi$ of odd primes has this property. In addition, in nonsolvable groups for some sets $\pi$, Hall $\pi$-subgroups can be nonconjugate but isomorphic (say, in $PSL_2(7)$ for $\pi=\{2,3\}$) and even nonisomorphic (in $PSL_2(11)$ for $\pi=\{2,3\}$). We prove that the existence of a finite group with nonconjugate Hall $\pi$-subgroups for a set $\pi$ implies the existence of a group with nonisomorphic Hall $\pi$-subgroups. The converse statement is obvious.

Keywords: Hall $\pi$-subgroup, $\mathscr C_\pi$ condition, conjugate subgroups.

UDC: 512.542

MSC: 20D20

Received: 07.05.2018

DOI: 10.21538/0134-4889-2018-24-3-43-50

 English version: Proceedings of the Steklov Institute of Mathematics (Supplement Issues), 2018, 303, suppl. 1, S94–S99

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