Does the trivial representation always induce the permutation representation? Is this true for each field $\mathbb{K}$ or just for representations over $\mathbb{C}$?
does the trivial representation always induce the permutation representation?
0
$\begingroup$
group-theory
representation-theory
finite-groups
-
2The 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. – 2012-04-19
-
0Thanks!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) – 2012-04-19
1 Answers
1
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.