Abstract Nonsense

Crushing one theorem at a time

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.

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April 16, 2011 - Posted by | Algebra, Representation Theory | , , , ,

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