For any group $G$ define the dual group of $G$ (denoted $\hat{G}$) to be the set of all homomorphisms from $G$ into the multiplicitive group of roots of unity in $\mathbb{C}$. Define a group operation in $\hat{G}$ by point wise multiplication of functions: if $\chi$ and $\psi$ are homomorphisms from $G$ into the group of roots of unit, then $\chi \psi$ is the homomorphism given by $(\chi \psi)(g)=\chi(g)\psi(g)$ for all $g\in G$, where the latter multiplication takes place in $\mathbb{C}$.
Prove that if $G$ is a finite abelian group, then $\hat{G} \cong G$. [Write $G$ as $\langle x_1 \rangle \times \cdot \cdot \cdot \times \langle x_r \rangle$ and if $n_i=|x_i|$ define $\chi_i$ to be the homomorphism which sends $x_i$ to $e^{\frac{2\pi i}{n_i}}$ and sends $x_j$ to $1$ for all $j \neq i$. Prove that $\chi_i$ has order $n_i$ in $\hat{G}$ and $\hat{G}=\langle \chi_1 \rangle \times \cdot \cdot \cdot \times \langle \chi_r \rangle$].
What I've got:
Suppose $G$ is a finite abelian group (written multiplicitively) with generators $\{x_1,...,x_r\}.$ Write $G$ as $$\langle x_1 \rangle \times \cdot \cdot \cdot \times \langle x_r \rangle.$$ For each $x_i$, let $|x_i|=n_i$ and define $\chi_i$ to be the homomorphism which sends $x_i$ to $e^{\frac{2\pi i}{n_i}}$ and sends $x_j$ to $1$ for all $j \neq i$. If $g \in G$ then there exist $m_1,...,m_r$ such that $g=x_1^{m_1}\cdot \cdot \cdot x_r^{m_r}$. Now we have $$(\chi_i(g))^{n_i}=((e^{\frac{2\pi i}{n_i}})^{m_i})^{n_i} = (e^{\frac{2\pi i}{n_i}\cdot n_i})^{m_i}=1.$$ Since $g$ is arbitrary, we must have $|\chi_i|=n_i$.
By the above result and an already known theorem, we have $$|\langle x_1 \rangle \times \cdot \cdot \cdot \times \langle x_r \rangle|=|\langle \chi_1 \rangle \times \cdot \cdot \cdot \times \langle \chi_r \rangle|$$
Now define $f:\langle x_1 \rangle \times \cdot \cdot \cdot \times \langle x_r \rangle \rightarrow \langle \chi_1 \rangle \times \cdot \cdot \cdot \times \langle \chi_r \rangle$ by $$x_1^{k_1} \cdot \cdot \cdot x_r^{k_r} \mapsto \chi_1(x_1)^{k_1}\cdot \cdot \cdot \chi_r(x_r)^{k_r}.$$ Since $G$ is abelian and the $\chi_i$ are homomorphisms, we have that $f$ is a homomorphism, and thus $$\langle x_1 \rangle \times \cdot \cdot \cdot \times \langle x_r \rangle \cong \langle \chi_1 \rangle \times \cdot \cdot \cdot \times \langle \chi_r \rangle.$$
Now it is enough to show that $\hat{G}=\langle \chi_1 \rangle \times \cdot \cdot \cdot \times \langle \chi_r \rangle$. It is clear that $\langle \chi_1 \rangle \times \cdot \cdot \cdot \times \langle \chi_r \rangle \leq \hat{G}$. This is where I am stuck. I cannot see how to show the other inclusion.