Abstract Nonsense

Crushing one theorem at a time

Irreducible Characters (Pt. II)

Point of post: This post is a continuation of this one.

Irreducible Characters Span Class Functions

We now show that \displaystyle \text{span}\left\{\chi^{(\alpha)}:\alpha\in\widehat{G}\right\}=\text{Cl}(G). Indeed:


Theorem: Let G be a finite group then \displaystyle \text{span}\left\{\chi^{(\alpha)}:\alpha\in\widehat{G}\right\}=\text{Cl}(G).

Proof: Let f\in\text{Cl}(G). Since in particular f\in\mathcal{A}(G) and \displaystyle \text{span }\mathcal{D}=\mathcal{A}(G) as was already proven. Thus, there exists constants c^{(\alpha)}_{i,j} for every \alpha\in\widehat{G} and i,j\in[d_\alpha] such that


\displaystyle f=\sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}c^{(\alpha)}_{i,j}D^{(\alpha)}_{i,j}


It follows then that for any x\in G one has that


\begin{aligned}f(x) &= \frac{1}{|G|}\sum_{g\in G}f\left(gxg^{-1}\right)\\ &= \frac{1}{|G|}\sum_{g\in G}\sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}c^{(\alpha)}_{i,j}D^{(\alpha)}_{i,j}\left(gxg^{-1}\right)\\ &=\frac{1}{|G|}\sum_{g\in G}\sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}c^{(\alpha)}_{i,j}\sum_{p,q=1}^{d_\alpha}D^{(\alpha)}_{i,p}D^{(\alpha)}_{p,q}(x)\overline{D^{(\alpha)}_{j,q}(g)}\\ &= \sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}c^{(\alpha)}_{i,j}\sum_{p,q=1}^{d_\alpha}D^{(\alpha)}_{p,q}(x)\frac{1}{|G|}\sum_{g\in G}D^{(\alpha)}_{i,p}(g)\overline{D^{(\alpha)}_{j,q}(g)}\end{aligned}


which (since the inner most sum is the conjugate of the inner product) is equal to


\displaystyle \begin{aligned}\sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}c^{(\alpha)}_{i,j}\sum_{p,q=1}^{d_\alpha}D^{(\alpha)}_{p,q}(x)\frac{1}{d_\alpha}\delta_{p,q}\delta_{i,j} &= \sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}\frac{1}{d_\alpha}c^{(\alpha)}_{i,j}\delta_{i,j}\sum_{p,q=1}^{d_\alpha}D^{(\alpha)}_{p,q}(x)\delta_{p,q}\\ &=\sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}\frac{1}{d_\alpha}c^{(\alpha)}_{i,j}\delta_{i,j}\sum_{p=1}^{d_\alpha}D^{(\alpha)}_{p,p}(x)\\ &=\sum_{\alpha\in\widehat{G}}\sum_{i,j=1}^{d_\alpha}\frac{1}{d_\alpha}c^{(\alpha)}_{i,j}\delta_{i,j}\chi^{(\alpha)}(x)\\ &=\sum_{\alpha\in\widehat{G}}\left(\frac{1}{d_\alpha}\sum_{i=1}^{d_\alpha}c^{(\alpha)}_{i,i}\right)\chi^{(\alpha)}(x)\end{aligned}


and since x\in G was arbitrary it follows that


\displaystyle f=\sum_{\alpha\in\widehat{G}}\left(\frac{1}{d_\alpha}\sum_{i=1}^{d_\alpha}c^{(\alpha)}_{i,i}\right)\chi^{(\alpha)}


And, since f\in\text{Cl}(G) was arbitrary the conclusion follows. \blacksquare


From this we get the obvious corollary that \#\left(\widehat{G}\right)=k where k is the number of classes of G. From this we can now prove a theorem which, in the future, will help us prove a few interesting structural results about groups. Namely:


Theorem: Let G be a finite group. Then, G is abelian if and only if every one of it’s irreps has degree one.

Proof: The necessity of this theorem was a previous theorem and so it suffices to show sufficiency. To do this we merely note that if each irrep has degree one then


\displaystyle k=\#\left(\widehat{G}\right)=\sum_{\alpha\in\widehat{G}}1^2=\sum_{\alpha\in\widehat{G}}d_\alpha^2=|G|


and thus the number of conjugacy classes of G is equal to its order, which is true if and only if G is abelian. The conclusion follows. \blacksquare



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 25, 2011 - Posted by | Algebra, Representation Theory | , , , , ,


  1. […] As of now all we know about the degrees of irreps of a finite group is that if we select one from each then . In this post we will extend our knowledge further by showing that the number of such that (recall this is just for any ) such that is equal to and in particular it divides . […]

    Pingback by Representation Theory: The Number of Degree One Irreps « Abstract Nonsense | February 28, 2011 | Reply

  2. […] some interesting information about the degrees of the irreps of a finite group . We’ve proven that the sum of the degrees squared must equal the order of the group. Also, we’ve proven […]

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

  3. […] Let be the number of conjugacy classes in and let be a set of representatives such that for . Recall then that and so the elements of can be labeled . We then consider the matrix where . We then […]

    Pingback by Representation Theory: Second Orthogonality Relation For Irreducible Characters « Abstract Nonsense | March 7, 2011 | Reply

  4. […] being by determining the degrees of the irreps for . To do this, we recall that . But, we have that (as for any group) admits the trivial irrep which clearly has degree . […]

    Pingback by Representation Theory: Character Table of S_3 By Finding the Irreducible Representations « Abstract Nonsense | March 22, 2011 | Reply

  5. […] the second one is fairly easy. Namely, we know that the number of non-equivalent irreducible characters of is . And it’s easily verifiable […]

    Pingback by Representation Theory: The Character Table of S_3xZ_3 « Abstract Nonsense | April 11, 2011 | Reply

  6. […] We know that it suffices to prove that every irrep of is of degree one. So, we know […]

    Pingback by Groups of Order pq (pt. I) « Abstract Nonsense | April 19, 2011 | Reply

  7. […] that and . But, since is odd and we know that we may conclude that for some . Thus, since we know that the cardinality of is (where ) we enumerate the elements of as . Thus, we have by prior […]

    Pingback by Representation Theory: The Number of Conjugacy Classes of a Finite Group of Odd Order is Equivalent to The Order of The Group Modulo Sixten « Abstract Nonsense | April 23, 2011 | Reply

  8. […] a class function) we’ve automatically defined a map by extending the map by linearity (recalling that the irreducible characters form a basis for ). But, what we’ll note that is if we […]

    Pingback by Induced Class Functions and the Space of Integral Class Functions (Pt. I) « Abstract Nonsense | April 27, 2011 | Reply

  9. […] Suppose that there were such a . We know then that there exists such that . But, this implies that and so in particular we know that . […]

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

  10. […] the conjugacy classes of a group gives you much information about the group. In particular, we have seen that the number of irreducible characters of a group is equal to the number of conjugacy classes of […]

    Pingback by Conjugacy Classes on the Symmetric Group « Abstract Nonsense | May 10, 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 )

Google+ photo

You are commenting using your Google+ 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 )


Connecting to %s

%d bloggers like this: