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.
 A: 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.
