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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.

Darth Geek
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1 Answers1

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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.

Mark
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