Isomorphism of relative holomorphs and matrix similarity
Volker Gebhardt, Alberto J. Hernandez Alvarado, Fernando Szechtman
TL;DR
The paper studies when two relative holomorphs $Hol(V,α)$ and $Hol(V,β)$ are isomorphic, with $V$ an elementary abelian $p$-group and $α,β∈GL(V)$. It proves a sharp criterion: ${\rm Hol}(V,α)\cong{\rm Hol}(V,β)$ if and only if the complements $⟨α⟩$ and $⟨β⟩$ are conjugate in $GL(V)$, i.e., $α$ is similar to $β^i$ for some $i$ with $\,\gcd(i, o(β))=1$. In the unipotent case this reduces to $α$ being similar to $β$, while in general non-conjugate subgroups can still yield isomorphic relative holomorphs, particularly for abelian or cyclic subgroups. The authors develop linear-algebra and module-theoretic tools to analyze similarity, and provide numerous examples demonstrating the sharpness and limits of the main theorem, including extensions to modules over ${\bf Z}/p^m{f Z}$. These results contribute to understanding when the holomorph structure determines the acting complements and have implications for classifying relative holomorphs up to isomorphism.
Abstract
Let $V$ be a finite-dimensional vector space over the field with $p$ elements, where $p$ is a prime number. Given arbitrary $α,β\in \mathrm{GL}(V)$, we consider the semidirect products $V\rtimes\langle α\rangle$ and $V\rtimes\langle β\rangle$, and show that if $V\rtimes\langle α\rangle$ and $V\rtimes\langle β\rangle$ are isomorphic, then $α$ must be similar to a power of $β$ that generates the same subgroup as $β$; 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.