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How can one determine whether a finite subgroup of $GL(n,\mathbb{Q})$ is irreducible, over $\mathbb{Q}$?

I understand that over $\mathbb{C}$, irreducibility is characterized by the squared norm of the character (Math Stack Exchange Q 720765).

Similarly, over $\mathbb{R}$, irreducibility is characterized by the squared norm and the Schur index (Math Stack Exchange Q 2938038).

This post (Math Stack Exchange Q 676303) suggests that, if one has (1) all the (irreducible?) complex representations of the group and (2) the Galois group for a relevant finite extension of $\mathbb{Q}$, then one could effectively construct all irreducible rational representations of the group. One could then check whether the original group is isomorphic to the image of any of these.

In my application, all I have at the start is an arbitrary finite subgroup of $GL(n,\mathbb{Q})$ (in fact, of $GL(n,\mathbb{Z})$), given in terms of generator matrices. I want my approach to be able to handle any such subgroup someone hands me. I don't yet know whether it is feasible for me to compute items (1) and (2) above. I have functions that will compute

  1. the squared norm of a finite matrix group (more precisely, of the character of the representation that maps each element of the group to itself),
  2. the Schur index of a finite matrix group (sum of the trace of the square of each group element, normalized by the order of the group), and
  3. the character table of a finite group (i.e. the character of each irreducible complex representation of the group).
If there is a criterion for irreducibility over $\mathbb{Q}$ similar those over $\mathbb{C}$ and $\mathbb{R}$, it would seem attractive.
ev.gal
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    How is your subgroup given? Generator matrices? Can one assume you have already tested that the group is reducible over $\mathbb{C}$? – ahulpke Nov 24 '21 at 02:07
  • This problem is solvable, but not particularly easy in general. In many examples you can tell by looking at the character. If the character is the sum of distinct algebraically conjugate complex characters then it is irreducible over ${\mathbb Q}$. But if, for example, it is equal to $2 \chi$ for some complex character $\chi$, then you need to know the Schur index (which is computable) of the representation to determine its irreducibility. – Derek Holt Nov 24 '21 at 08:22
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    It might make more sense if you asked about your specific example. – Derek Holt Nov 24 '21 at 08:24
  • @ahulpke : In my application, subgroups are given in terms of generator matrices. Yes, one can assume group (ir)reducibility over $\mathbb{C}$ is known: In my application, I use a function to compute the squared norm of the character of the group, following the discussion in the first post (Q 720765) linked in the question.

    I've edited the final paragraph in the original question to reflect these.

     – ev.gal Nov 24 '21 at 16:53
  • @DerekHolt : In my specific example, I want to analyze any finite subgroup of $GL(n,\mathbb{Z})$ that someone hands me. – ev.gal Nov 24 '21 at 16:56
  • @DerekHolt : Would you please elaborate on the approach in your first reply? or point me to a reference? At present, I can compute most of the pieces you mention (character of the given group, characters of all irreducible complex representations of the group, Schur index of the group). I don't yet know whether I can compute which (irreducible) complex characters are algebraically conjugate, or how to put all these pieces together to determine irreducibility of the group over $\mathbb{Q}$. – ev.gal Nov 24 '21 at 17:01
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    If you have the character table of the group, then it is easy to identify the algebraic conjuagtes. The entries in the conjugates have the same minimal polynomials over ${\mathbb Q}$. The the sum of the characters of the conjugates mutliplied by the Schur Index will give you the character of the corresponding irreducible rational representation. You don't need to construct the rational representations explicitly, which is significantly more difficult. – Derek Holt Nov 24 '21 at 17:31

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