Abstract Nonsense

Crushing one theorem at a time

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.

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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 \left|G^{\text{ab}}\right| where G^{\text{ab}} is the abelianization of G. 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).

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Number of Degree One Irrep Classes

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:

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Lemma: Let G be a group, A an abelian group, and G^\text{ab} the abelianization of G. Then, \#\text{Hom}(G,A)=\#\text{Hom}\left(G^{\text{ab}},A\right).

Proof: By our work with the abelinization to every map \phi\in\text{Hom}\left(G,A\right) (since \phi(G)\leqslant A and since A is abelian of course \phi(G) is abelian) there exists a map F_\phi\in\text{Hom}\left(G^{\text{ab}},A\right) for which \phi=F_\phi\circ\pi where \pi:G\to G^{\text{ab}}=G/[G,G] is the canonical projection. Thus we have the map F:\text{Hom}\left(G,A\right)\to\text{Hom}\left(G^{\text{ab}},A\right) with \phi\mapsto F_\phi. Evidently it’s injective since \phi=F_\phi\circ\pi and it’s surjective since if \vartheta\in\text{Hom}\left(G^{\text{ab}},A\right) we can consider the homomorphism \vartheta\circ\pi and note that by definition F_{\vartheta\circ\pi}\left(g[G,G]\right)=\vartheta(\pi(g))=\vartheta(g[G,G]) and so F_{\vartheta\circ\pi}=\vartheta and so \vartheta\in\text{im}(F). Since \vartheta was arbitrary we may conclude that F is a surjection, and thus a bijection and the conclusion follows. \blacksquare

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Now our desired claim follows immediately:

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Theorem: The number of degree one equivalence classes of the finite group G is equal to \left|G^{\text{ab}}\right|

Proof: We recall our notation for the group \widehat{G_{\mathfrak{L}}} of degree one equivalence classes of G, and so it suffices to show that \left|\widehat{G_{\mathfrak{L}}}\right|=\left|G^{\text{ab}}\right|. That said, let \mathbb{T} denote the circle group, then we know that \#\left(\widehat{G_{\mathfrak{L}}}\right)=\#\text{Hom}\left(G,\mathbb{T}\right). That said, by our lemma that \#\text{Hom}\left(G,\mathbb{T}\right)=\#\text{Hom}\left(G^{\text{ab}},\mathbb{T}\right). But, using the fact that for an abelian group every equivalence class is of degree one (and the fact that G^{\text{ab}} is abelian) we know that \#\text{Hom}\left(G^{\text{ab}},\mathbb{T}\right)=\left|\widehat{G^{\text{ab}}_{\mathfrak{L}}}\right|=\left|G^{\text{ab}}\right| and so we may put this all together to get

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\left|\widehat{G_{\mathfrak{L}}}\right|=\#\text{Hom}\left(G,\mathbb{T}\right)=\#\text{Hom}\left(G^{\text{ab}},\mathbb{T}\right)=\left|\widehat{G^{\text{ab}}_{\mathfrak{L}}}\right|=\left|G^{\text{ab}}\right|

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\blacksquare

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References:

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

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May 6, 2011 - Posted by | Algebra, Representation Theory | , , , , , ,

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