Dual group is cyclic for finite abelian cyclic group

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 $$\gamma(mx)\neq \gamma(kx) \forall m\neq k$$ & $$m,k. If $$\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

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?