0
$\begingroup$

A matrix $ Q=(q_{ij})$ is called multiplicatively antisymmetric over a field $ F $ if $ q_{ii}=1 $ and $ q_{ij}={q_{ji}}^{-1} $.Let $ \mathcal{Q} $ be the set of all $ n \times n $ multiplicatively antisymmetric matrix. Now we can define an action of $ S_{n} $ on $\mathcal{Q} $ by conjugation of the permutation matrix. Now suppose $ Q \in \mathcal{Q} $ then is there any condition such that $ Stab({Q})=\{\sigma\in S_{n}:\sigma.Q= Q\}= id $ ?

$\endgroup$
4
  • $\begingroup$ What/where is the question ? $\endgroup$ Aug 16, 2022 at 12:10
  • $\begingroup$ I have edited the question. $\endgroup$
    – Sky
    Aug 16, 2022 at 12:30
  • 1
    $\begingroup$ I don't think the condition that the stabilizer is trivial can be expressed better than that the stabilizer is trivial. It suffices to check elements of order $p$ by Cauchy's theorem, but I think one may need to check every element of order $p$. $\endgroup$
    – Will Sawin
    Aug 16, 2022 at 14:18
  • $\begingroup$ can you explain? $\endgroup$
    – Sky
    Aug 17, 2022 at 10:33

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.