Abstract Nonsense

Crushing one theorem at a time

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.



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

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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,

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\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}


April 13, 2011 - Posted by | Algebra, Representation Theory | , , , , , ,


  1. […] every irrep of , up to equivalence, looks like where is an irrep of . But, since is cyclic, we know that it is of the form for some . Realistically then, that’s all there is to it. But, since […]

    Pingback by Representation Theory: Irreps of an Abelian Group « Abstract Nonsense | April 16, 2011 | Reply

  2. […] Consider the character of given by and , we know that this is a character. Moreover, let be the trivial character for . Consider then the character […]

    Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (August-2004) « Abstract Nonsense | May 6, 2011 | Reply

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