A faithful irreducible is induced from a proper inertia subgroup

Complete reducibility [F1] decomposes $V{\mathrm{\mid }}_A$ into irreducible summands, and [F2] makes every such summand one-dimensional. Thus $V{\mathrm{\mid }}_A$ is the direct sum of its linear weight spaces. The translates of the $\lambda$-weight space are the weight spaces in its $G$-orbit, and their direct sum is $V$ by irreducibility.

If the inertia group $G_{\lambda }$ were $G$, every $a\in A$ would act by a scalar on $V$. Its commutator with every $g\in G$ would then act trivially, so faithfulness would make $A$ central, contrary to hypothesis. Thus $G_{\lambda }<G$.

Put $W=V_{\lambda }$. Distinct left cosets of $G_{\lambda }$ carry $W$ to distinct $A$-weight spaces, whose direct sum is $V$ by step 1.1; the usual transversal model therefore identifies ${\mathrm{Ind}}_{G_{\lambda }}^{G}W$ with $V$. If $0\mathrm{\ne }U\subseteq W$ is a $G_{\lambda }$-submodule, the direct sum of its translates over the left cosets is a nonzero $G$-submodule of $V$. Irreducibility makes this sum all of $V$, and its intersection with $W$ is $U$, so $U=W$. Hence the inducing representation $W$ is irreducible, completing the claim.