# Abstract Nonsense

## The Number of Degree One Irreps

Point of post: In this post we prove that the number of degree one irreps of a finite group $G$ is equal to $\left|G^{\text{ab}}\right|$ and thus, in particular, is a divisor of $|G|$.

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Note: A much better proof of this in much clearer and precise language can be found here.

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Motivation

As of now all we know about the degrees of irreps of a finite group $G$ is that if we select one $\rho^{(\alpha)}$ from each $\alpha\in\widehat{G}$ then $\displaystyle \sum_{\alpha\in\widehat{G}}\left(\deg\rho^{(\alpha)}\right)^2=|G|$. In this post we will extend our knowledge further by showing that the number of $\alpha\in\widehat{G}$ such that $d_\alpha$ (recall this is just $\deg\rho^{(\alpha)}$ for any $\rho^{(\alpha)}\in \alpha$) such that $d_\alpha=1$ is equal to $\left|G^{\text{ab}}\right|$ and in particular it divides $|G|$.

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Number of Degree One Irreps

Before we prove our main result we classify when the range of a representation is abelian. In particular:

Theorem: Let $G$ be a finite group and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ be a representation. Then, $\text{im}(\rho)$ is abelian if and only if $\displaystyle \rho\cong\bigoplus_{j=1}^{m}\psi_j$ where each $\psi_j:G\to\mathcal{U}\left(\mathscr{V}_j\right)$ is a degree one representation.

Proof: Suppose first that $\psi_j:G\to\mathcal{U}\left(\mathscr{V}_j\right)$ for $j=1,\cdots,m$ are degree one representations of $G$ such that $\displaystyle \rho\cong\bigoplus_{j=1}^{m}\psi_j$. Note then that in particular each $\psi_j(g)=c_{j,g}\mathbf{1}$ for each $g\in G$ and $j\in[m]$. So, in particular if $\displaystyle \rho=(g)W\bigoplus_{j=1}^{m}\psi_j(g) W^{-1}$ for every $g\in G$ then for every $g,g'\in G$ one has that

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\displaystyle \begin{aligned}\rho(g)\rho(g') &= \left(W\bigoplus_{j=1}^{m}\psi_j(g) W^{-1}\right)\left(W\bigoplus_{j=1}^{m}\psi_j(g') W^{-1}\right)\\ &= W\left(\bigoplus_{j=1}^{m}c_{j,g}\mathbf{1}\right)\left(\bigoplus_{j=1}^{m}c_{j,g'}\mathbf{1}\right)W^{-1}\\ &= W\bigoplus_{j=1}^{m}\left(c_{j,g}\mathbf{1}c_{j,g'}\mathbf{1}\right)W^{-1}\\ &= W\bigoplus_{j=1}^{m}\left(c_{j,g'}\mathbf{1}c_{j,g}\mathbf{1}\right)W^{-1}\\ &= W\left(\bigoplus_{j=1}^{m}c_{j,g'}\mathbf{1}\right)\left(\bigoplus_{j=1}^{m}c_{j,g}\mathbf{1}\right)W^{-1}\\ &= \left(W\bigoplus_{j=1}^{m}\psi_j(g')W^{-1}\right)\left(W\bigoplus_{j=1}^{m}\psi_j(g)W^{-1}\right)\\ &=\rho(g')\rho(g)\end{aligned}

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and since $g,g'\in G$ were arbitrary it follows that $\text{im}(\rho)$ is abelian as desired.

Conversely, suppose that $\text{im}(\rho)$ is abelian and let $\displaystyle \rho\cong\bigoplus_{j=1}^{m}\psi_j$ for some irreps (not necessarily at this point [or obviously so] one degree). Note then that applying the previous paragraphs proof in reverse shows that the abelianess of $\text{im}(\rho)$ implies the abelianess of $\text{im}(\psi_j)$ for each $j=1,\cdots,m$. Note though that for any $g\in G$ and any $j\in[m]$ one has that $\psi_j(g):\mathscr{V}_j\to\mathscr{V}_j$ is such that for any $g'\in G$ one has that $\psi_j(g)\psi_j(g')=\psi_j(g')\psi_j(g)$ and thus $\psi_j(g)$ is an intertwinor for $\psi$ and itself and so by Schur’s Lemma we may conclude that $\psi_j(g)=c_{j,g}\mathbf{1}$ for some $c_{j,g}\ne 0$. Since $g\in G$ was arbitrary it follows that $\psi_j(g)=c_{j,g}\mathbf{1}$ for every $g\in G$ and thus every subspace of $\mathscr{V}_j$ is invariant under $\psi_j$. But, since $\psi_j$ is irreducible this is true if and only if $\dim_\mathbb{C}\mathscr{V}_j=1$. Since $j\in[m]$ was arbitrary the conclusion follows. $\blacksquare$

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What we now show is that there is an equivalence preserving bijection between irreps of $G^{\text{ab}}$ and one-degree irreps of $G$.

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Theorem: Let $A$ denote the set of all irreps on $G^{\text{ab}}$ and $B$ the set of all irreps on $G$ of degree one. Then, there exists a bijection $f:B\to A$ such that $\rho\cong\psi$ if and only if $f(\rho)\cong f(\psi)$.

Proof: Recall from our discussion of the abelinization of a group $G$ that for every group $H$ and every homomomorphism $\phi:G\to H$ such that $\text{im}(\phi)$ is abelian there exists a unique homomorphism $\theta:G^{\text{ab}}\to H$ such that $\theta\circ\pi=\phi$. It follows then that for any representation $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ there exists a representation $f(\rho):G^{\text{ab}}\to\mathcal{U}\left(\mathscr{V}\right)$ such that $\rho=f(\rho)\circ\pi$. This is clearly a bijection, and moreover if $\rho(g)=W\psi(g) W^{-1}$ for every $g\in G$ then for any $\pi(g)\in G^{\text{ab}}$ one has that $f(\rho)(\pi(g))=Wf(\psi)(\pi(g))W^{-1}$ and so $f(\rho)\cong f(\psi)$. Note then that we can turn this bijection (by our last theorem) into a bijection $b:\left\{\bigoplus_{j}\psi_j:\psi_j\text{ is degree one}\right\}\to\left\{\text{irreps of }G^{\text{ab}}\right\}$ which is also equivalence preserving. Consider then the restriction $\widetilde{b}:B \to b(B)$ this is clearly an equivalence preserving bijection and since evidently $\deg \widetilde{b}(\rho)=\deg\rho$ it’s easy to see that $b(B)=A$. $\blacksquare$

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Corollary: The number of $\alpha\in\widehat{G}$ for which $d_\alpha=1$ is equal to $\left|G^{\text{ab}}\right|$.

Proof: By the previous theorem there is an equivalence preserving bijection between $\left\{\text{irreps of }G^{\text{ab}}\right\}$ and one-dimensional irreps of $G$. It follows then that $\#\left(\widehat{G^{\text{ab}}}\right)=\#\left\{\alpha\in\widehat{G}:d_\alpha=1\right\}$ and since $G^{\text{ab}}$ is abelian we know that $\# \widehat{G^{\text{ab}}}=\left|G^{\text{ab}}\right|$, the conclusion follows. $\blacksquare$

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References:

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

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print

February 28, 2011 -

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