Theorem: Every subgroup of a cyclic group is cyclic. Moreover, if $| \langle a \rangle|=n$, then the order of any subgroup of $\langle a \rangle$ is a divisor of $n$; and , for each positive divisor $k$ of $n$, the group $\langle a \rangle$ has exactly one subgroup of order $k$- namely $\langle a^{\frac{n}{k}}\rangle$.
I have confusion, in understanding of uniqueness. For example: in $\mathbb Z_6$ there are two subgroups of order $6$ - namely $\langle 1 \rangle$ and $\langle 5 \rangle$ - and $3$ - namely $\langle 2 \rangle $ and $\langle 4 \rangle$.
But according to theorem $\langle 2 \rangle$ and $\langle 1 \rangle$ must be unique subgroups of order $3$ and $6$ respectively.
I don't know where my understanding lacks. Any help will be appreciated. Thank you.
Edit: Precisely the lack in my understanding was that $\langle 5 \rangle$ and $\langle 1 \rangle$ are different subgroups of order 6. But I now understand that theorem says " In a cyclic group $\langle a \rangle$ there is only one subgroup of order $k$, that is $\{e,a,a^2,...,a^{k-1}\}$. Now if any other subgroup of $\langle a \rangle$ has order k then the subgroup is nothing else but above set(subgroup). Having different generator doesn't make different set (subgroup)". Sorry for my dumb question. Before posting the question I was feeling the notion is like above (which i explained now), but I was not able to precisely explain it to myself so i thought there is lack of my knowledge/understanding thats why i post.
