Isomorphism of relative holomorphs and matrix similarity

TL;DR

This paper investigates when semidirect products of vector spaces over finite fields with cyclic subgroups of their automorphism groups are isomorphic, revealing conditions under which the subgroups are conjugate or merely isomorphic.

Contribution

It characterizes the isomorphism classes of relative holomorphs of finite vector spaces over prime fields, especially focusing on cyclic and abelian subgroups, and extends results to modules over ${f Z}/p^m{f Z}$.

Findings

Isomorphism of semidirect products implies conjugacy of cyclic subgroups.

Existence of non-conjugate subgroups with isomorphic semidirect products.

Criteria for non-conjugate cyclic subgroups in automorphism groups of finite groups.

Abstract

Let $V$ be a finite-dimensional vector space over the field with $p$ elements, where $p$ is a prime number. Given arbitrary $\alpha,\beta\in \mathrm{GL}(V)$, we consider the semidirect products $V\rtimes\langle \alpha\rangle$ and $V\rtimes\langle \beta\rangle$, and show that if $V\rtimes\langle \alpha\rangle$ and $V\rtimes\langle \beta\rangle$ are isomorphic, then $\alpha$ must be similar to a power of $\beta$ that generates the same subgroup as $\beta$; that is, if $H$ and $K$ are cyclic subgroups of $\mathrm{GL}(V)$ such that $V\rtimes H\cong V\rtimes K$, then $H$ and $K$ must be conjugate subgroups of $\mathrm{GL}(V)$. If we remove the cyclic condition, there exist examples of non-isomorphic, let alone non-conjugate, subgroups $H$ and $K$ of $\mathrm{GL}(V)$ such that $V\rtimes H\cong V\rtimes K$. Even if we require that non-cyclic subgroups $H$ and $K$ of $\mathrm{GL}(V)$ be abelian, we may still have $V\rtimes H\cong V\rtimes K$ with $H$ and $K$ non-conjugate in $\mathrm{GL}(V)$, but in this case, $H$ and $K$ must at least be isomorphic. If we replace $V$ by a free module $U$ over ${\mathbf Z}/p^m{\mathbf Z}$ of finite rank, with $m>1$, it may happen that $U\rtimes H\cong U\rtimes K$ for non-conjugate cyclic subgroups of $\mathrm{GL}(U)$. If we completely abandon our requirements on $V$, a sufficient criterion is given for a finite group $G$ to admit non-conjugate cyclic subgroups $H$ and $K$ of $\mathrm{Aut}(G)$ such that $G\rtimes H\cong G\rtimes K$. This criterion is satisfied by many groups.