Abstract Nonsense

Crushing one theorem at a time

The Dual Group of a Cyclic Group


Point of post: In this post we show that the dual group of cyclic groups are very simple, in fact they are isomorphic to the original group itself.

Motivation

Often it’s difficult to deal directly with the dual group even with our alternate characterization of it. In this post we show that the lay of the land becomes much nicer. Namely, we shall show that if G is cyclic then G\cong\widehat{G_\mathfrak{L}}.

Dual Group of a Cyclic Group

We now proceed to show that if G is cyclic then G\cong \widehat{G_\mathfrak{L}}. The key fact is that it will suffice to prove the result for \mathbb{Z}_n for a fixed but arbitrary n and really to show that \text{Hom}\left(\mathbb{Z}_n,\mathbb{T}\right)\cong\mathbb{Z}_n from where the conclusion follow from previous theorem.

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

Proof: As previously stated if we prove that \text{Hom}\left(\mathbb{Z}_n,\mathbb{T}\right)\cong U_n where U_n is the group of n^{\text{th}}-roots of unity and n=|G| we’ll be done since

\text{ }

G\cong\mathbb{Z}_n\cong U_n\cong\text{Hom}\left(\mathbb{Z}_n,\mathbb{T}\right)\cong\text{Hom}\left(G,\mathbb{T}\right)\cong\widehat{G_\mathfrak{L}}

\text{ }

So, define F:\text{Hom}\left(\mathbb{Z}_n,\mathbb{T}\right)\to U_n by \chi\mapsto \chi(1). This is evidently a homomorphism since F(\chi\lambda)=(\chi\lambda)(1)=\chi(1)\lambda(1)=F(\chi)F(\lambda). It’s injective since 1 is a generator for \mathbb{Z}_n and thus by basic group theory if F(\chi)=\chi(1)=\lambda(1)=F(\lambda) then \chi=\lambda. Lastly, to see it’s surjective we merely note that for any \zeta\in U_n defining \chi:\mathbb{Z}_n\to U_n by \chi(k)=\zeta^k is a homomorphism and thus \chi\in\text{Hom}\left(\mathbb{Z}_n,\mathbb{T}\right). But, this then implies that \zeta=\chi(1)=F(\chi)\in\text{im}(F) from where surjectivity follows. It follows then that F is an isomorphism and thus the entire theorem follows from previous discussion. \blacksquare

\text{ }

\text{ }

\text{ }

References:

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

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

3 Comments »

  1. […] secretly in our last post was the information to obtain the character table for any cyclic group. But, since it’s such […]

    Pingback by Representation Theory: Classifying, Up To Equivalence, the Irreps of a Cyclic Group « Abstract Nonsense | April 13, 2011 | Reply

  2. […] 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 […]

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

  3. […] 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 […]

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


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