## Another Way of Viewing the Dual Group

**Point of post: **In this post we discuss another way to isomorphically view the linear group of a group .

**Motivation**

In our last post we discussed how to naturally turn a subset of into a group, namely the subset of linear characters. That said, the definition we gave is not very conducive to proving things about . In this post we prove that is isomorphic to where is the circle group and the group operation is pointwise multiplication. This shall prove to be just the sort of concrete object we need to do some serious analysis concerning .

*Different Way of Viewing the Linear Group of a Group *

The first interesting result that we have is that is isomorphic to something very simple (or at least simpler to say). Namely, let denote the circle group then what we’ll now show is that , the set of all homomorphisms with point wise multiplication. Indeed:

**Theorem: ***Let be a finite group then .*

** Proof: **Define by . Firstly this map is well-defined since for linear and each , one has that . Indeed, it’s a homomorphism because if then we know that so that for every , for some . Thus, for every ,

and as has been said many times before each is a -root of unity and thus trivially in .

So, to see that is a homomorphism we merely note that for each and

so that . To see that is injective we merely note that if then the characters of and are the same and thus as has been said before we know that . Lastly, to show that is surjective we merely note that if then by definition where is given the usual inner product and thus is a linear representation. Moreover, if then by definition so that and so is surjective. The conclusion follows.

**References:**

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

[…] 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 […]

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[…] classes of , and so it suffices to show that . That said, let denote the circle group, then we know that . That said, by our lemma that . But, using the fact that for an abelian group every […]

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