We define the dual of a group $G$, denoted $\widehat{G}$, as the group of homomorphisms $\chi:G\to \mathbb{C}^*$. We say that an element of $\widehat{G}$ is a character. Let $H$ and $K$ be two finite abelian groups. Prove that $$\widehat{H}\times \widehat{K}\cong \widehat{H\times K}.$$
This is an important step in proving that if $G$ is a finite abelian group, then $G \cong \widehat{G}$, but all proofs I've found of this statement either gloss over the proof of this step or use some category theory that goes beyond my understanding.
The proof is simple. If $\chi_1$ is a character of $H$ and $\chi_2$ is a character of $K$ then define $\varphi:H\times K\to\mathbb{C}^{\times}$ by $\varphi(h,k)=\chi_1(h)\chi_2(k)$. This is trivially a character.
Conversely, if $\varphi$ is any character of $H\times K$, define $\chi_1:H\to\mathbb{C}^{\times}$ by $\chi_1(h)=\varphi(h,e_K)$, and similarly define $\chi_2:K\to\mathbb{C}^{\times}$ by $\chi_2(k)=\varphi(e_H,k)$. These are characters of $H$ and $K$ respectively.
From here it should be clear how to define an isomorphism between the two groups. I'll leave it to you.