= Solution
Work over the complex numbers. By <Maschke's theorem>, restriction to the normal subgroup $A$ is a direct sum of its <isotypic components>. For an irreducible $A$-type $W$, denote its isotypic component in $V$ by $V_W$. Normality makes $G$ permute these components, by conjugating the $A$-types. Any union of orbits spans a $G$-invariant subspace, so irreducibility of $V$ means that all occurring components form a single orbit.
If there is only one component, restriction is isotypic, as claimed. Otherwise take one component $V_W$ and its stabilizer $H=\{g:gV_W=V_W\}$, a proper subgroup. This is an <irreducible representation> of $H$: if $0\ne U\subsetneq V_W$ were $H$-invariant, the direct sum of its translates over coset representatives for $G/H$ would be a nonzero proper $G$-invariant subspace of $V$. It is direct because the translates lie in distinct isotypic components, and proper because $\dim U<\dim V_W$.
The natural map from the <induced representation> sends $g\otimes v$ to $gv$. It identifies the direct-sum coset decomposition with all isotypic components of $V$, so is bijective and $G$-equivariant. Consequently
$$
\boxed{V\cong\operatorname{Ind}_H^G V_W.}
$$
This proves the dichotomy directly, including the irreducibility of the representation being induced.
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