Isomorphism of the annihilator of a subgroup in the context of group characters.

I am trying to learn about characters of finite abelian groups. A character is a homomorphism from a finite abelian group $G$ into the multiplicative group of complex numbers of absolute value 1. In my textbook (Finite Fields by Lidl and Niederreiter) there is the following question to which I am stuck:

Let $H$ be a subgroup of the finite abelian group $G$. Prove that the annihilator $A$ of H in $\widehat{G}$ (where $\widehat{G}$ is the group of characters of $G$) is isomorphic to $G/H$ and that $\widehat{G}/A$ is isomorphic to $H$.

This looks like something where the 1st isomorphism theorem for groups could be used, but I don't see how. Any ideas would be greatly appreciated.

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First question

Let ${\text{Ann}_{G}}(H)$ denote the annihilator of $H$ in $G$, i.e., $${\text{Ann}_{G}}(H) = \left\{ \phi \in \widehat{G} ~ \middle| ~ \forall h \in H: ~ \phi(h) = 1_{\mathbb{C}} \right\}.$$ Then ${\text{Ann}_{G}}(H) \cong \widehat{G / H}$. In order to prove this, let $q: G \to G / H$ denote the obvious quotient group homomorphism, and define a group homomorphism $\Phi: \widehat{G / H} \to {\text{Ann}_{G}}(H)$ by $$\forall \phi \in \widehat{G / H}: \quad \Phi(\phi) = \phi \circ q.$$

$\Phi$ is injective: Suppose that $\phi \in \widehat{G / H}$ and $\phi \circ q = \mathbf{1}_{G}$. Then $\phi(g + H) = 1_{\mathbb{C}}$ for all $g \in G$, so $\phi = \mathbf{1}_{G / H}$.

$\Phi$ is surjective: Let $\phi \in {\text{Ann}_{G}}(H)$. We can define a map $\dot{\phi}: G / H \to \mathbb{C}$ by $$\forall g \in G: \quad \dot{\phi}(g + H) \stackrel{\text{df}}{=} \phi(g).$$ This is clearly a well-defined map and is a character on $G / H$. As $\phi = \dot{\phi} \circ q$, we are done.

Note: Up to this point, all of our arguments are valid for an arbitrary locally compact Hausdorff abelian group $G$ with $H$ a closed subgroup and all maps involved continuous.

We now turn to the special case when $G$ is finite and abelian with the discrete topology.

Question. How should we use the assumption that $G$ is finite and abelian?

It turns out that any finite and abelian group is isomorphic to its own dual. By assumption, $G$ is finite and abelian, so $G / H$ is also finite and abelian. Hence, $G / H \cong \widehat{G / H}$, which yields $${\text{Ann}_{G}}(H) \cong G / H$$ as desired. Note, however, that the isomorphism between $G / H$ and $\widehat{G / H}$ is not natural.

Second question

Let us first show that $\widehat{G} / {\text{Ann}_{G}}(H) \cong \widehat{H}$. By the answer to the first question, we have $$(\spadesuit) \qquad \left( \widehat{G} / {\text{Ann}_{G}}(H) \right)^{\land} \cong {\text{Ann}_{\widehat{G}}}({\text{Ann}_{G}}(H)).$$

Claim: ${\text{Ann}_{\widehat{G}}}({\text{Ann}_{G}}(H)) \cong H$.

Proof of Claim

Observe that \begin{align} {\text{Ann}_{\widehat{G}}}({\text{Ann}_{G}}(H)) & = \left\{ \Psi \in \widehat{\widehat{G}} ~ \middle| ~ \forall \phi \in {\text{Ann}_{G}}(H): ~ \Psi(\phi) = 1_{\mathbb{C}} \right\} \\ & \cong \{ g \in G \mid \forall \phi \in {\text{Ann}_{G}}(H): ~ \phi(g) = 1_{\mathbb{C}} \} \quad (\text{By Pontryagin Duality.}) \\ & \supseteq H. \quad (\text{By the definition of ${\text{Ann}_{G}}(H)$.}) \end{align} It then remains to prove that we actually have equality in the last line. As ${\text{Ann}_{G}}(H) \cong \widehat{G / H}$, we get $\left( {\text{Ann}_{G}}(H) \right)^{\land} \cong G / H$ by Pontryagin Duality. The isomorphism is explicitly implemented by the map $\Theta: G / H \to \left( {\text{Ann}_{G}}(H) \right)^{\land}$ defined by $$\forall g \in G, ~ \forall \phi \in {\text{Ann}_{G}}(H): \quad [\Theta(g + H)](\phi) \stackrel{\text{df}}{=} \phi(g).$$ If $g \notin H$, then $g + H \neq e_{G / H}$, so there exists a $\phi \in {\text{Ann}_{G}}(H)$ such that $$\phi(g) = [\Theta(g + H)](\phi) \neq 1_{\mathbb{C}}.$$ This readily implies that $$\{ g \in G \mid \forall \phi \in {\text{Ann}_{G}}(H): ~ \phi(g) = 1_{\mathbb{C}} \} = H$$ as desired. $\quad \blacksquare$

It now follows from $(\spadesuit)$ that $\left( \widehat{G} / {\text{Ann}_{G}}(H) \right)^{\land} \cong H$. By Pontryagin Duality yet again, $$\widehat{G} / {\text{Ann}_{G}}(H) \cong \widehat{H}.$$ Finally, as $H$ is finite and abelian, we obtain $\widehat{H} \cong H$, which concludes the argument.

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