# Abstract Nonsense

## 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.