Salvato in:
| Autori principali: | , , |
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| Natura: | Preprint |
| Pubblicazione: |
2024
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| Soggetti: | |
| Accesso online: | https://arxiv.org/abs/2405.10147 |
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Sommario:
- 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.