# Abstract Nonsense

## Classifying, Up To Equivalence, the Irreps of a Cyclic Group

Point of post: In this post we classify, up to equivalence, the irreps of a finite cyclic group. Of course as a consequence we shall have attained the character table for such a group.

Motiviation

Hidden secretly in our last post was the information to obtain the character table for any cyclic group. But, since it’s such a big statement (not in terms of difficulty, but for what comes later) that it’s probably prudent to put it in a post of its own. The basic idea is simple, namely since all cyclic groups are isomorphic to $\mathbb{Z}_n$ for some $n$ it suffices to find the irreps up to equivalence of $\mathbb{Z}_n$. For that we can just exhibit them and since we know $\#\left(\mathbb{Z}_n\right)=n$ we’ll know we’re done when we’ve exhibited $n$ non-equivalent ones. The reason that this relates to the last post is that we implicitly already defined them there!

Irreps Up to Equivalence of a Finite Cyclic Group

Let $G$ be a finite group with $|G|=n$. We shall now exhibit, for each $\alpha\in\widehat{G}$, an element $\rho^{(\alpha)}\in\alpha$ from where we will have a set of representatives from each equivalence class in $\widehat{G}$. We shall then have the set of all irreducible characters by considering $\chi^{(\alpha)}=\text{tr}(\rho^{(\alpha)})$. But, since $G\cong\mathbb{Z}_n$ it will suffice to do this for $\mathbb{Z}_n$. Indeed:

Theorem: Let $\zeta=e^{\frac{2\pi i}{n}}$, then for each $k=0,\cdots,n-1$ define $\chi^{(\alpha_k)}:\mathbb{Z}_n\to\mathcal{U}\left(\mathbb{C}\right)$ (where $\mathbb{C}$ is given the usual structure) by $m\mapsto \zeta^{km}\mathbf{1}$. Then, $\chi^{(\alpha_k)}$ is an irrep of $\mathbb{Z}_n$ and $\chi^{(\alpha_k)}\not\cong\chi^{(\alpha_j)}$ whenever $j\ne k$. Moreover, for each $\alpha\in\widehat{\mathbb{Z}_n}$ there exists some $k=0,\cdots,n-1$ such that $\alpha=\alpha_k$.

Proof: The fact that each $\chi^{(\alpha_k)}$ is irreducible is evident since they’re linear. To see that $\chi^{(\alpha_k)}\not\cong\chi^{(\alpha_j)}$ for $j\ne k$ we merely note that they induce different characters. The last conclusion follows since $\left\{\alpha_0,\cdots,\alpha_{n-1}\right\}\subseteq\widehat{G}$ and since $\#\left(\widehat{G}\right)=n$ we may conclude that $\widehat{G}=\left\{\alpha_0,\cdots,\alpha_{n-1}\right\}$. $\blacksquare$

$\text{ }$

$\text{ }$

From this we are able to construct the character table for any $\mathbb{Z}_n$. Namely the character table is the matrix $\left[a_{i,j}\right]_{i,j=0}^{n-1}$ where $a_{i,j}=\zeta^{ij}$. Graphically,

$\text{ }$

$\text{ }$

$\begin{array}{c|cccc} G\cong\mathbb{Z}_n & 0 & 1 & \cdots & n-1\\ \hline \chi^{(\alpha_0)} & 1 & 1 & \cdots & 1\\ \chi^{(\alpha_1)} & 1 & \zeta & \cdots & \zeta^{n-1}\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ \chi^{(\alpha_{n-1})} & 1 & \zeta^{n-1} & \cdots & \zeta^{(n-1)^2}\end{array}$