Ladders

Abstract

We consider a group G with a normal subgroup N. If σ∈G and τ∈N, then τσ=στ′ for some τ′∈N by normality. Therefore for $f \in \mathbb{F}[V]^{N}$, we have τ⋅(σ⋅f)=τσ⋅f=στ′⋅f=σ⋅f and thus $\sigma \cdot f \in \mathbb{F}[V]^{N}$. This shows that G acts on $\mathbb{F}[V]^{N}$. Clearly $(\mathbb{F}[V]^{N})^{G} = \mathbb{F}[V]^{G}$. Since N acts trivially on $\mathbb{F}[V]^{N}$, in fact, G/N acts on $\mathbb{F}[V]^{N}$ and $(\mathbb{F}[V]^{N})^{G/N} = \mathbb{F}[V]^{G}$. We have seen this in detail in Lemma 1.10.1.