# 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?

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.

• But how many irreducible representations are there of Z_k and are they somehow related/parametrized by the k complex roots of unity? Some mapping which can then later be generalized to other groups? Jan 22 '15 at 5:06
• I added an explicit parameterization and some comments. Does this clarify things? Jan 22 '15 at 6:05
• Thanks! Do you have a typo about what is $\xi$? I would have thought that there are $k$ different choices of $\xi$ parametrized as $\xi_m = e^{\frac{2\pi i m }{k}}$ for $0\leq m \leq (k-1)$. Then for every choice of $m$ you have a map, $\alpha_m : \mathbb{Z}_k \rightarrow S^1$ as, $\alpha_m (p) = \xi_m^p$. So if these $\alpha_m$s are called irreducible characters" for $\mathbb{Z}_k$ then there are $k$ such irreducible characters - one for each $m$. Now this set of $\{\alpha_m\}$ forms a group on its own under pointwise multiplication on the elements of $\mathbb{Z}_k$. Jan 22 '15 at 6:52
• But here we have an extra special feature that is group $\{\alpha_m\}$ (the Pontryagin dual?) is also canonically isomorphic to the original group $\mathbb{Z}_k$ via the map, $\alpha_m \rightarrow \xi_m$ - right? Jan 22 '15 at 6:55
• Now lets say you introduce $\phi_n = e^{\frac{2 \pi i n }{p} }$ as the $p$ $p^{th}$ roots of unity one for each $1\leq n \leq (p-1)$. And the corresponding $pk$ number of characters of $\mathbb{Z}_p \times \mathbb{Z}_k$ are given by the map $\alpha_{(m,n)} (i,j) \rightarrow \xi_m^i \xi_n^j$ - right? And again this set of maps $\{ \alpha_{(m,n)} \}$ form the Pontryagin dual of the group $\mathbb{Z}_k \times \mathbb{Z}_p$ - right? Jan 22 '15 at 7:03