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|.

\text{ }

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

\text{ }


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|.

\text{ }

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

\text{ }

\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}

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

\text{ }


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 - Posted by | Algebra, Representation Theory | , ,


  1. […] proven that the sum of the degrees squared must equal the order of the group. Also, we’ve proven that the number of degree one irreps of is equal to the order of the abelinization of . In this […]

    Pingback by Representation Theory: The Dimension Theorem « Abstract Nonsense | March 3, 2011 | Reply

  2. […] a last note, we know that where is the abelianization of and is the commutator […]

    Pingback by Representation Theory: The Linear Group of a Group G and the Dual Group « Abstract Nonsense | April 12, 2011 | Reply

  3. […] We recall that the dual group  has order equal the the abelianization . But, by inspection we see that […]

    Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (August-2003) « Abstract Nonsense | May 1, 2011 | Reply

  4. […] of Post: In this post I give a different proof than this one for the number of degree one irreps of a […]

    Pingback by Clearer Proof for the Number of Degree One Irrep Classes « Abstract Nonsense | May 6, 2011 | Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: