I only have superficial familiarity with concepts of homological algebra, and couldn't find this written down explicitly anywhere so I wanted to make sure.
Here is my basic argument:
Let $G$ be a finite groups and $ H \subset G $ a subgroup, and suppose we have the following short exact sequence of complex representations of $H$:
$$ 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 $$
so $ A,B,C $ are all finite-dimensional complex vector spaces on which $H$ acts by invertible linear maps, and the arrows corresponds to $H$-equivariant linear maps.
Since all complex representations of $H$ are completely reducible, this short exact sequence splits. Now, the functor $ \mathrm{Ind} $ is additive, so it preserves split short exact sequences, hence we obtain a split short exact sequence of complex representations of $G$:
$$ 0 \rightarrow \mathrm{Ind} (A) \rightarrow \mathrm{Ind} (B) \rightarrow \mathrm{Ind} (C) \rightarrow 0 $$
which implies that the map from $ \mathrm{Ind} (A) $ to $ \mathrm{Ind} (B) $ is injective, so the first left derived functor of $ \mathrm{Ind} $ vanishes on $C$. Since this short exact sequence can be continued to the left by an infinite sequence of zeros, it seems that all higher order derived functors vanish as well, for $A,B,C$ (and presumably any other finite-dimensional complex representation of $H$).
Question: is this argument correct? And if so, how can it be generalized? Is $ \mathrm{Ind} $ always exact for cases where all representations are completely reducible? When is it nontrivial?
Best Answer
A different approach: induction for $k$-linear representations ($k$ a field) can be defined as $\text{Ind}_H^G(M) := kG \otimes_{kH} M$. The group algebra $kG$ is free (hence flat) as a right $kH$-module, so the functor $kG \otimes_{kh} -$ is exact. This is true in general, including the semisimple case and the case where $\text{char}~k$ divides $|G|$.