A Fourier sample cuts a larger common kernel

Choose $g\in K\setminus H$. Under the canonical evaluation identification [F5], [F1] gives $(H^{\perp }{)}^{\perp }=H$, so there is ${\chi }_0\in H^{\perp }$ with ${\chi }_0(g)\mathrm{\ne }1$. By [F6], $H^{\perp }$ is a group; the evaluation map ${ϵ}_g:H^{\perp }\to {\mathbb{C}}^{\times }$, ${ϵ}_g(\chi )=\chi (g)$, is a homomorphism because pointwise multiplication gives ${ϵ}_g(\chi \psi )=\chi (g)\psi (g)$. Its image is finite by [F2] and is a subgroup because it is the image of a group homomorphism. It contains $1$ and the nonidentity value ${\chi }_0(g)$, so it has at least two elements.

Every nonempty fiber of ${ϵ}_g$ is a coset of its kernel: ${ϵ}_g(\chi )={ϵ}_g(\psi )$ exactly when ${ϵ}_g(\chi {\psi }^{-1})=1$. Therefore at most half of $H^{\perp }$ maps to $1$. A uniform $\chi \in H^{\perp }$ has $\chi (g)\mathrm{\ne }1$ with probability at least $1\mathrm{/}2$, and each such character is nontrivial on $K$ because $g\in K$.

For any $\chi \in H^{\perp }$, every $h\in H$ satisfies $\chi (h)=1$, so $H\subseteq K\cap \ker \chi$. The set $\ker \chi$ is a subgroup: it contains $0$ by [F3], and the homomorphism law gives closure under addition and inverses. Its intersection with $K$ is a subgroup by [F4]. If $\chi {\mathrm{\mid }}_K\mathrm{\ne }1$, the homomorphism $\chi {\mathrm{\mid }}_K$ has at least two values; its fibers are cosets of $K\cap \ker \chi$, so $[K:K\cap \ker \chi ]=\mathrm{\mid }\chi (K)\mathrm{\mid }\ge 2$.