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We fix following objects:

(1) $G$ is a finite group.

(2) $\chi$ is complex irreducible character of $G$.

(3) $m$ is the Schur index of $\chi$ w.r.t. the rational field $\mathbb{Q}$.

(4) All the fields appearing below will be of characteristic $0$.

The Schur index of $\chi$ is $m$ w.r.t. $\mathbb{Q}$; whereas it is $1$ w.r.t. any splitting field $F$ of $\chi$. Thus, it is natural to ask the following question.

Q. Given a divisor $m'$ of $m$, can $m'$ occur as Schur index of $\chi$ w.r.t. some extension of $\mathbb{Q}$ in $\mathbb{C}$?

Such question may have been considered in different contexts because the Schur index of an absolutely irreducible character w.r.t. $F$ is also the Schur index of simple component of $F[G]$ corresponding to $\chi$ (in the sense - $\chi$ appears as an irreducible component of an $F$-irreducible charatcer when scalars are extended from $F$ to $\mathbb{C}$). The simple components of $F[G]$ are particular examples of simple algebras, where division algebras are important and Schur index is also defined for division algebras. Thus, it may be the case that the problem of above kind could have been considered in the context of index of simple algebras. But I do not have idea. Can one suggest some way towards solution to above question, for example, literature if it is considered.

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The following is wrong, see comment section:

The Schur index over $K$ is, among other things, the degree of a minimal field extension of $K$ over which the underlying representation can be realised over the field $K(\chi)$ of character values. Since there exists an abelian field of definition, $F$, say (e.g. by Brauer's induction theorem all representations of $G$ can be realised over the field $\mathbb{Q}$ adjoin all $|G|$-th roots of unity), there is, for any divisor $m'$ of $m$, a subfield $K'$ of $F$ over which $F$ has degree $m'$, and therefore $\chi$ has Schur index $m'$ over $K'$.

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    $\begingroup$ 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 $m$. $\endgroup$
    – Soluble
    Commented May 4, 2019 at 2:04
  • $\begingroup$ You are absolutely right, I was too hasty. I will leave this up, so that others don't repeat my mistake. $\endgroup$
    – Alex B.
    Commented May 4, 2019 at 9:24

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