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 $ ?
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$\begingroup$ What/where is the question ? $\endgroup$– Denis SerreAug 16, 2022 at 12:10
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$\begingroup$ I have edited the question. $\endgroup$– SkyAug 16, 2022 at 12:30
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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 SawinAug 16, 2022 at 14:18
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$\begingroup$ can you explain? $\endgroup$– SkyAug 17, 2022 at 10:33
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