Document Type : Research Paper
Authors
1
Department of Mathematics, William & Mary, Williamsburg, VA 23187, USA
2
Department of Mathematics, Compueter and Information Science, SUNY at Old Westbury, Old Westbury, NY 11568, USA
10.22108/ijgt.2024.142395.1918
Abstract
Sabatini [L. Sabatini, Products of subgroups, subnormality, and relative orders of elements, Ars Math. Contemp., 24 no. 1 (2024) 9 pp.] defined a subgroup $H$ of $G$ to be an exponential subgroup if $x^{|G:H|} \in H$ for all $x \in G$, in which case we write H ≤exp G. Exponential subgroups are a generalization of normal (and subnormal) subgroups: all subnormal subgroups are exponential, but not conversely. Sabatini proved that all subgroups of a finite group $G$ are exponential if and only if $G$ is nilpotent. The purpose of this paper is to explore what the analogues of a simple group and a solvable group should be in relation to exponential subgroups. We say that an exponential subgroup H ≤exp G is exp-trivial if either $H = G$ or the exponent of $G$, $\exp(G)$, divides $|G:H|$, and we say that a group $G$ is exp-trivial if all exponential subgroups of $G$ are exp-trivial. We classify finite exp-simple groups by proving $G$ is exp-simple if and only if $\exp(G) = \exp(G/N)$ for all proper normal subgroups $N$ of $G$, and we illustrate how the class of exp-simple groups differs from the class of simple groups. Furthermore, in an attempt to overcome the obstacle that prevents all subgroups of a generic solvable group from being exponential, we say that a subgroup $H$ of $G$ is weakly exponential if, for all $x \in G$, there exists $g \in G$ such that $x^{|G:H|} \in H^g$. If all subgroups of $G$ are weakly exponential, then $G$ is wexp-solvable. We prove that all solvable groups are wexp-solvable and almost all symmetric and alternating groups are not wexp-solvable. Finally, we completely classify the groups $PSL(2,q)$ that are wexp-solvable. We show that if $\pi(n)$ denotes the number of primes less than $n$ and $w(n)$ denotes the number of primes $p$ less than $n$ such that $PSL(2,p)$ is wexp-solvable, then
\[ \lim_{n \to \infty} \frac{w(n)}{\pi(n)} = \frac{1}{4}.\]
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