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May 4, 2019 at 9:25 history edited Alex B. CC BY-SA 4.0
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May 4, 2019 at 9:24 comment added Alex B. You are absolutely right, I was too hasty. I will leave this up, so that others don't repeat my mistake.
May 4, 2019 at 2:04 comment added Soluble I think there are examples of groups, in which, if the Schur index of an absolutely irreducible character is $m$, and if we consider fields $\mathbb{Q}(\chi)\subset \mathbb{Q}(\zeta)$ where $\zeta$ is primitive $|G|$-th root of $1$, then in between these two fields, there is no splitting field for $\chi$ whose degree over $\mathbb{Q}(\chi)$ is equal to $m$.
May 3, 2019 at 23:43 history edited Alex B. CC BY-SA 4.0
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May 3, 2019 at 22:52 history answered Alex B. CC BY-SA 4.0