## Clearer Proof for the Number of Degree One Irrep Classes

**Point of Post: **In this post I give a different proof than this one for the number of degree one irreps of a group.

**Motivation**

In a past post I’d given a proof that he number of degree one equivalence classes of irreps of a finite group is equal to where is the abelianization of . This proof followed pretty closely the proof in Barry Simon’s book, and to be honest I was always a little hazy on the end of it. A while ago when I was thinking about dual groups and trying to develop stuff independent of my books I realized that there was a ‘much simpler’ proof for this fact, one which was much clearer and less ambiguous than Barry Simon’s. I had been meaning to write it up for a while, and this is that post. The only thing is that now that I write it up, I realize that it is actually Barry Simon’s proof secretly, just with much clearer language. Anyways, I like it better because even though it is only different notationally I actually thought that I had come up with a new proof when I first wrote this down–this is how unclear that proof is (no offense Barry Simon).

*Number of Degree One Irrep Clas**ses*

The proof of this theorem shall really just be a small lemma and then a corollary will give the result (the real work is secretly hidden in my discussion of dual groups). Namely:

**Lemma: ***Let be a group, an abelian group, and the abelianization of . Then, .*

**Proof: **By our work with the abelinization to every map (since and since is abelian of course is abelian) there exists a map for which where is the canonical projection. Thus we have the map with . Evidently it’s injective since and it’s surjective since if we can consider the homomorphism and note that by definition and so and so . Since was arbitrary we may conclude that is a surjection, and thus a bijection and the conclusion follows.

Now our desired claim follows immediately:

**Theorem:** *The number of degree one equivalence classes of the finite group is equal to *

**Proof: **We recall our notation for the group of degree one equivalence 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 equivalence class is of degree one (and the fact that is abelian) we know that and so we may put this all together to get

**References:**

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