Let $(G,+)$ be a finite abelian group, $\widehat{G}$ the dual group of $G$ (i.e. the set of homomorphisms from $G$ to $\mathbb{C}^*$). If $X$ is a subset of $G$, we define $X^{\bot}$ as the subgroup of $\widehat{G}$ whose elements are the trivial characters on $X$, that is: \begin{equation} X^{\bot}=\{\chi \in \widehat{G};\: X \subset \ker(\chi)\} \end{equation} Let $K$ be any subgroup of $\widehat{G}$ and if $Y$ is a subset of $\widehat{G}$, let’s write $Y^°$ for the following subgroup of $G$: \begin{equation} Y^°= \bigcap_{\chi \in Y} \ker(\chi) \end{equation}
Show that the following applications \begin{equation} \displaystyle H \longmapsto H^{\bot}\: \: \text{and} \: \: K \longmapsto K^° \end{equation} are two reciprocal bijections between the subgroups of $G$ and the subgroups of $\widehat{G}$.
My idea (but it was apparently a bad idea) was to consider the following bijective map , call it $f$, that takes a subgroup $H$ of $G$ to the set of characters of $\widehat{G}$ whose kernel contains $H$ (i.e. $H^{\bot}$): \begin{equation} f(H)=\{\chi \in \widehat{G};\: \chi(h)=1, \: \forall h \in H \}. \end{equation} Then consider the inverse bijective map, say $g$, defined as: \begin{equation} g\big(f(H)\big)=\{g \in G; \: \chi(g)=1,\: \forall \chi \in f(H)\} \end{equation} We first show that this set contains $H$ and by an inclusion reversing argument, that this set is $H$ itself ! We conclude that $g \circ f$ is the identity map. We proceed the same way for $f \circ g$. I thought it could work but it doesn’t because the map $K \longmapsto K^°$ defined in the exercise is totally different from mine: it is defined on any subgroup $K$ in the dual group $\widehat{G}$ and sends $K$ to $K^°$. I’m really confused about that. Any help would be appreciated. Many thanks.