# Abstract Nonsense

## Irreps of an Abelian Group

Point of Post: In this post we discuss the general methodology for finding all the irreps (up to equivalence) of a finite abelian group.

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Motivation

Just as we were able to combine the structure theorem and our knowledge of the dual group of a cyclic group to gain information about the dual group of an abelian group we so shall use similar knowledge to gain the general knowledge about the irreps of a finite abelian group up to equivalence.

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Irreps of a Finite Abelian Group Up To Equivalence

There is, in a real sense, no ‘theorem’ that can be associated to this post. Just a couple of observations. Namely, if we are given an abelian group $G$ we know from the structure theorem that $G\cong\mathbb{Z}_{a_1}\times\cdots\times\mathbb{Z}_{a_n}$. Once one finds such a deomposition everything becomes easy. Namely, we know from our theorems regarding the irreps of the products of groups that every irrep of $G$, up to equivalence, looks like $\rho_1\boxtimes\cdots\boxtimes\rho_n$ where $\rho_j$ is an irrep of $\mathbb{Z}_{a_j}$. But, since $\mathbb{Z}_{a_j}$ is cyclic, we know that it is of the form $\displaystyle k\mapsto \text{exp}\left(\frac{2\pi i k m_j}{a_j}\right)$ for some $m_j\in\{0,\cdots,m_j\}$. Realistically then, that’s all there is to it. But, since any good math post is incomplete with out one formal theorem we make one last observation in theorem form:

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Theorem: Let $G$ be an abelian group and $\mathcal{C}_G$ be the character table for $G$. Then,

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$G\sim M_1\otimes\cdots\otimes M_n$

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where $\sim$ denotes similarity of matrices and each $M_j$ is an non-normalized DFT matrix, and $\otimes$ the Kronecker product.

Proof: As we noted before there exists $a_1,\cdots,a_n\in\mathbb{N}$ such that

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$G\cong\mathbb{Z}_{a_1}\times\cdots\times\mathbb{Z}_{a_n}$

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It follows then from our previous observation about the character table of the product of groups that up to a rearrangement of the columns $\mathcal{C}_G$ will be equal to the Kronecker product of the character tables of $\mathbb{Z}_{a_1},\cdots,\mathbb{Z}_{a_n}$. But, we know that the character table of each $\mathbb{Z}_{a_m}$ is (up to a permutation of the columns) a DFT matrix. The conclusion follows. $\blacksquare$

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References:

1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Math. Soc., 1996. Print.