Abstract Nonsense

Crushing one theorem at a time

The Dual Group of an Abelian Group


Point of Post: In this post we show that any abelian group is isomorphic to its dual group.

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Motivation

We now know theorems about the relationship between cyclic groups and their duals, and products of groups and their duals. But, we also know that abelian groups and products of cyclic groups are intimately related. Thus, putting these three bits of information together we shall gather from this post the interesting fact that every abelian group is isomorphic to its dual.

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The Dual Group of an Abelian Group

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We waste no time and head straight for the theorem:

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Theorem: Let G be a finite abelian group. Then, G\cong\widehat{G_\mathfrak{L}}.

Proof: By the structure theorem for finite abelian groups there exists a_1,\cdots,a_n\in\mathbb{N} such that

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\displaystyle G\cong\prod_{j=1}^{n}\mathbb{Z}_{a_j}

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Then, by our theory about the dual of the product of groups we may conclude that

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\displaystyle \widehat{G_\mathfrak{L}}\cong\prod_{j=1}^{n}\widehat{\left(\mathbb{Z}_{a_j}\right)_\mathfrak{L}}

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But, considering our theorems about the dual group of a cyclic group we may conclude that

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\displaystyle \widehat{G_\mathfrak{L}}\cong\prod_{j=1}^{n}\widehat{\left(\mathbb{Z}_{a_j}\right)_\mathfrak{L}}\cong\prod_{j=1}^{n}\mathbb{Z}_{a_j}\cong G

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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 | , , , , ,

1 Comment »

  1. […] 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 […]

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


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