Dear Gabor, Willem, Peter, all,
Indeed I took the Peter's question as if he didn't mind the field to
be extended.

But even if he did, there are tools available for dealing with Q-irreducible
representations, e.g. the Artin's theorem that basically says that the
character of a rational
representation is a difference of permutation characters, each of the
latter a direct sum of representations induced from a cyclic subgroup,
and the related counting result that says that the number of
irreducible Q-representations equals the number of conjugate cyclic
subgroups. So it does not look completely hopeless to construct all
the Q-irreducibles, if needed.

Regards,
Dima

2008/9/27 Gabor Ivanyos <[EMAIL PROTECTED]>:
> Dear Dima, Willem, Peter, all,
>> Formulae for this can be found in Serre's book "Linear representations
>> of finite groups." Springer GTM, Vol. 42.
> I think Peter was asking for decomposition over Q. Let us stress "over Q"`.
> I agree Willem in that it is hard in general. For finite groups it mght be ber
> somewhat easier than the general case, but I am not aware of any serious
> result supportimg this.
> Gabor
>

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