An induced representation need not be irreducible. For instance, contains every irreducible representation of . Since is irreducible if and only if , and Frobenius reciprocity gives the question comes down to understanding . We do this when is normal.
Let , let be a representation of given by , and let . The conjugate representation is the space with acting by ; this makes sense because . Its character is . The representations and have the same invariant subspaces, so is irreducible if is. For , the map is an isomorphism of representations, since ; so the isomorphism class of depends only on the coset , and .
Lemma 15.4.1 (Restricting an induced representation to a normal subgroup). Let , let be representatives of the cosets of , and let be a representation of . Then
Proof. Each summand of is -invariant: for , we have with , so . This says exactly that the linear isomorphism , , is an isomorphism of representations of from to . ∎
Proposition 15.4.2 (Induction from a normal subgroup). Let and let be an irreducible representation of . Then is the number of cosets with . In particular, is irreducible if and only if for every .
Proof. By Frobenius reciprocity and Lemma 15.4.1, Each is irreducible, so the th term is if and otherwise. The coset itself always counts, since for ; so the sum equals exactly when no other coset counts. ∎
Note. For a subgroup that is not normal, Mackey’s analysis of leads to the following criterion, which we state without proof (see Serre, Linear Representations of Finite Groups, §7.4). Let be an irreducible representation of , given by , and for let . Then is irreducible if and only if, for every , the two representations and of have no irreducible constituent in common. When is normal, and this is Proposition 15.4.2.
Example 15.4.3 (Inducing from a subgroup of index 2). Let . Then is normal, since its two left cosets and are also its two right cosets. Fix and an irreducible representation of . By Proposition 15.4.2:
- if , then is irreducible;
- if , then , so with and irreducible and not isomorphic.
In the second case . Indeed, by Frobenius reciprocity , so contains and ; likewise ; and .
For and , take . Since , we get , so is irreducible, in agreement with Example 15.2.7. On the other hand the trivial representation of is its own conjugate, and splits as (Exercise 15.5.2(2)).