I am trying to prove that given a cyclic gropu $G$ and a subgroup $H$ then $(H^\bot)^\bot = H$, where $H^\bot = \{ \alpha \in G | \chi_\alpha(x) = 1 \forall x \in H\}$ and $\chi_\alpha$ are fourier characters. I think I have a proof for that if first I can prove that if $F, N$ are distinct subgroups of $G$ then $F^\bot \neq N^\bot$.
To prove that lemma, I argue as such. We know that since $G$ is cyclic, every subgroup is normal. Additionally we know that for every normal subgroup $T$ there is a homomorphism $\phi: G \rightarrow G / T$ such that the kernel of the homomorphism is $T$. Moreover we know that since $G/T$ is cyclic, then we can define fourier characters for that group, that is between $G/T$ and the group $R$ of $|G/T|$-th roots of unity. In particular we can define $f: G/T \rightarrow R$ such that $f$ is an isomorphism. Thus we can claim that $f \circ \phi: G \rightarrow R^\prime$ is a homomorphism with kernel $T$, where $R^\prime$ is the group of $|G|$-th roots of unity. Thus there exists a fourier character $f \circ \phi$ that is equal to 1 only for elements in $T$. So now to prove the lemma, we notice that if $F, N$ are distinct groups then (wlog) there exists an element in $N$ that is not in $F$. Since $F^\bot$ will contain the character we defined above, we know that $N^\bot$ cannot contain this character. Thus $F^\bot \neq N^\bot$.
Is this reasoning correct? I am still learning abstract algebra and i am not confident in my proofs yet. Please if you can only point out mistakes in the proof and don't give me a full proof.