Let $\chi$ be an irreducible (complex) character of a finite group, $G$. The Schur index $m_{K}(\chi)$ of $\chi$ over the field $K$ is the smallest positive integer $m$ such that $m\chi$ is afforded by a representation over the field $K(\chi)$. The most interesting case is $K=\mathbb{Q}$. Given the character table, or only the particular character one is interested in, one can usually derive bounds for $m(\chi)=m_{\mathbb{Q}}(\chi)$. For example, $m(\chi)$ divides $\chi(1)$ and $n[\chi^n,1_G]$ for all $n\in \mathbb{N}$ (Fein), and the Benard-Schacher Theorem tells us that $\mathbb{Q}(\chi)$ contains a primitive $m(\chi)$-th root of unity.

On the other hand, the example of the quaternion group $Q_8$ and the dihedral group $D_8$ shows that two groups might have identical character tables, but corresponding characters with different Schur indices. I am curious wether there are examples that are even worse than this.

**Notation:** To state this more precisely, I'll make the following assumptions: We are given two finite groups $G$ and $H$, such that there is a bijection $\tau\colon {\rm Cl}(G) \to {\rm Cl}(H)$ from the classes of $G$ to the classes of $H$, and such that $\psi \mapsto \psi \circ \tau$ is a bijection ${\rm Irr}(H)\to {\rm Irr}(G)$. Now:

Is there an example with $m(\chi)/m(\chi\circ\tau)\notin \{1,2,1/2\}$ for some $\chi\in {\rm Irr}(H)$?

Is there an example with $G$ of odd order and $m(\chi) / m(\chi\circ\tau)\neq 1$ for some $\chi \in {\rm Irr}(H)$?

Now let us assume that we know the power maps of the character table. These are the maps $\pi_n^G\colon {\rm Cl}(G)\to {\rm Cl}(G)$ induced by $g\mapsto g^n$. (These maps are stored in the tables of the character table library of GAP.) Given these maps, one can compute $[\chi_C, 1_C]$ for cyclic subgroups $C\leq G$, for example. Also we can compute the Frobenius-Schur Indicator and thus the Schur index over $\mathbb{R}$.

Now assume that $\tau\circ \pi_n^G = \pi_n^H\circ \tau$ in the above situation (then $(G,H)$ is called a Brauer pair).

Is there a Brauer pair $(G,H)$ such that $m(\chi)/m(\chi\circ\tau)\neq 1$ for some $\chi\in {\rm Irr}(H)$?

I would appreciate any examples or (pointers to) results that show the impossibility of such examples.

Thanks

viaa bijection between their conjugacy classes. For a 'bare' such bijection, not guaranteed to preserve any interesting information (such as the size of the class), I'm not sure why the result should be another irreducible representation. – L Spice Oct 13 '15 at 20:10andthat the bijection $\tau$ says how these character tables are "the same". – Frieder Ladisch Oct 16 '15 at 11:40