Abstract Nonsense

Crushing one theorem at a time

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

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February 28, 2011 - Posted by | Algebra, Representation Theory | , ,

4 Comments »

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