Abstract Nonsense

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.