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If $G$ is a finite abelian cyclic group, and $\hat G$ be it's dual group i.e. the group of all homomorphisms of G into $\mathbb{T}^*$. Prove that the dual group is also cyclic.

My attempt: If $G$ is cyclic and $|G|=n$, then $\exists x \in G \ni \forall y \in G, y=mx$ for some $m\in \mathbb{N}$. Take $\gamma \in \hat G \ni \gamma \neq Id \Rightarrow \gamma(x) \neq 1 \Rightarrow \gamma(mx)=\gamma^m(x)$. Now I claim that $\gamma(mx)\neq \gamma(kx) \forall m\neq k$ & $m,k<n$. If $\gamma(mx)= \gamma(kx) \Rightarrow \gamma^{|m-k|}(x)=Id$, I have to show that $\gamma(x)=Id$, and I am unable to show this.

I appreciate any help on this proof.

Thanks

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There is no reason why $\gamma$ needs to be injective. For example, take the cyclic group of order $4$, and consider the character $C_4\to C_4/C_2\cong C_2\cong\{\pm 1\}\subset\mathbb{C}^\times$.

Instead, where can you map a generator of $C_n$? Why does that tell you $\widehat{C_n}$ is cyclic?

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