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Does the trivial representation always induce the permutation representation? Is this true for each field $\mathbb{K}$ or just for representations over $\mathbb{C}$?

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The induction of the trivial representation from a subgroup is a permutation representation. A nontrivial group has many permutation representations, so you cannot speak of "the permutation representation". But yeah, this is true over any field, pretty much by definition of induction. –  Alex B. Apr 19 '12 at 15:48
    
Thanks!Is it also true, that the dimension of a permutation repr. comes from the index? (I mean the index of $G$ and the subgroup you use for induction) –  james Apr 19 '12 at 16:00

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I think about the induced representation $Ind_H^G 1$ as the space of functions $$ f: H\backslash G \rightarrow \mathbb{K},$$ and $$ g \in G : f(x) \mapsto f(xg).$$

Than surely $g$ will permute the cosets $H \backslash G$, hence is a permutation action, and since you can write down a basis of functions supported on a single coset $H \gamma$, the dimesnion equals the cardinality of $H \backslash G$, i.e. the index.

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