Let \(H\le G\) and let representations be modules over a commutative coefficient ring \(R\).
- What can be said about the kernel of an irreducible representation of \(G\)?
- What is the kernel of a direct sum of representations?
- Is induction \(\operatorname{Ind}_H^G\) left exact, right exact, or exact?
Solution
Let \(\rho:G\to\operatorname{Aut}_R(V)\) be a representation.
1. Its kernel\[ \ker\rho=\{g\in G:\rho(g)=1_V\} \]is a normal subgroup of \(G\). If \(\rho\) is irreducible, no stronger group-theoretic condition on the kernel holds in general; equivalently, \(\rho\) factors through a faithful irreducible representation of \(G/\ker\rho\).
2. For representations \(\rho_i\),\[ \ker\Bigl(\bigoplus_i\rho_i\Bigr)=\bigcap_i\ker\rho_i, \]because an element acts trivially on the direct sum exactly when it acts trivially on every summand.
3. Induction is\[ \operatorname{Ind}_H^G(M)=R[G]\otimes_{R[H]}M. \]Choose representatives for the right cosets \(H\backslash G\). Then \(R[G]\) is a free right \(R[H]\)-module, one copy of \(R[H]\) for each right coset. Hence it is flat as a right \(R[H]\)-module. Tensoring with a flat module is exact, so\[ \operatorname{Ind}_H^G:R[H]\text{-Mod}\longrightarrow R[G]\text{-Mod} \]is exact: it is both left and right exact.