Representation theory of $\mathbb{Z}_k$ and complex roots of unity Is there a natural relationship between the (characters?) of irreducible representations of $\mathbb{Z}_k$ and the $k$ complex-roots of unity? 
Can they be like thought of as characters of its 1-dimensional irreducible representations? 
If something like this holds what would be the analogous statement for general groups? 
 A: I assume by $\mathbb Z_k$ you mean $\mathbb Z/k\mathbb Z$, the cyclic group of order $k$.  (Most number theorists don't like this notation.)  In this case $\mathbb Z/k\mathbb Z$ is isomorphic to the group of $k$-th roots of unity under multpilication.  Because it is abelian, all irreducible representations are 1-dimensional, i.e., linear characters.  
For finite abelian groups, there is a non-canonical (meaning one needs to make an arbitrary choice) isomorphism between the group and its group of characters (you can multiply characters).  For finite non-abelian groups, the irreducible representations (or characters) do not form a group, but there is a (again non-canonical) bijection between the conjugacy classes of the group and its irreducible representations.
Edit: To be more explicit, let $\zeta$ be a (not necessarily primitive) $k$-th root of 1 in $\mathbb C$, i.e., $\zeta = e^{2\pi i j/k}$ for some $j = 0, 1, \ldots, k-1$.  Now consider the map $\alpha_\zeta(a) = \zeta^a$ for $a \in \mathbb Z/k \mathbb Z$.  This is an irreducible character, and one gets all of them this way.  So the association $\zeta \mapsto \alpha_\zeta$ defines an isomorphism of the $k$-th roots of unity with the character group of $\mathbb Z/k \mathbb Z$.  
One can do something similar for finite abelian groups, but it requires a choice of basis (it's perhaps not obvious, but the above isomorphism used the basis $1$ for $\mathbb Z/k \mathbb Z$.)  E.g., take the basis $(1,0)$, $(0,1)$ for $G =\mathbb Z/m \mathbb Z \times \mathbb Z / n\mathbb Z$.  Now let $\zeta_m$ and $\zeta_n$ be $m$ and $n$-th roots of unity.  We can define an irreducible character of $G$ by sending $(1,0)$ to $\zeta_m$ and $(0,1)$ to $\zeta_n$, and all characters arise this way.  Precisely, the character is given by $(a, b) \mapsto \zeta_m^a \zeta_n^b$.  So the irreducible characters correspond to pairs of $m$ and $n$-th roots of unity.  This correspondence works for finite abelian groups, which are all isomorphic to a product $\mathbb Z/k_1 \mathbb Z \times \cdots \times \mathbb Z/k_n \mathbb Z$, but not for nonabelian groups.
