# Abstract Nonsense

## The Dimension Theorem

Point of post: In this post we prove what is called the ‘dimension’ theorem which in essence says that the degree of any irrep of a finite group divides the order of the group.

Motivation

So far we’ve obtained some interesting information about the degrees of the irreps of a finite group $G$. We’ve 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 $G$ is equal to the order of the abelinization of $G$. In this post we’ll prove the supremely interesting result that the degree of any irrep must divide the order of the group. One of many uses for this will be that we will be able to prove some interesting results about finite groups.

The Dimension Theorem

Let as always $G$ be a finite group and suppose that we’ve chosen particular representatives $\displaystyle D^{(\alpha)}$ and thus irreducible characters $\chi^{(\alpha)}$ have the representation $\displaystyle \chi^{(\alpha)}(x)=\sum_{j=1}^{n}D^{(\alpha)}_{j,j}(x)$. Our first theorem will show the connection between irreducible characters and algebraic integers. Namely:

Theorem: Let $G$ be a finite group then for every $\alpha\in\widehat{G}$ and every $g\in G$ one has that $\chi^{(\alpha)}(g)\in\mathbb{A}$.

Proof: We merely note that since $|G|<\infty$ one has that $g^{|G|}=e$ and so $D^{(\alpha)}(g)^|G|=D^{(\alpha)}\left(g^|G|\right)=D^{(\alpha)}(g)=I$ and so by basic matrix analysis every eigenvalue of $D^{(\alpha)}$ is a $|G|$-root of unity and thus trivially an algebraic integer. Thus, $\chi^{(\alpha)}(g)$ being the sum of these algebraic integers (it of course, being the trace of $D^{(\alpha)}(g)$) is an algebraic integers (since they form a ring). $\blacksquare$

We now use this to show that $d_\alpha\mid |G|$ for every $\alpha\in\widehat{G}$.

Theorem: Let $G$ be a finite group and $\alpha\in\widehat{G}$. Then, $d_\alpha\mid |G|$.

Proof: Choose an ordering for $G$ so that we can list $G$ as $(g_1,\cdots,g_{n})$ where $|G|=n$. We then define a matrix $A$ by $A_{i,j}=\chi^{(\alpha)}\left(g_i g_j^{-1}\right)$. We then define $v\in\mathbb{C}^n$ by $v=(\chi^{(\alpha)}(g_1),\cdots,\chi^{(\alpha)}(g_n))^{\top}$. We note then by the convolution relations that

\displaystyle \begin{aligned}Av &=\left(\sum_{k=1}^{n}\chi^{(\alpha)}\left(g_1 g_k^{-1}\right)\chi^{(\alpha)}(g_k),\cdots,\sum_{k=1}^{n}\chi^{(\alpha)}\left(g_n g_k^{-1}\right)\chi^{(\alpha)}\left(g_k\right)\right)^{\top}\\ &= \left(\frac{|G|}{d_\alpha}\chi^{(\alpha)}(g_1),\cdots,\frac{|G|}{d_\alpha}\chi^{(\alpha)}(g_n)\right)^{\top}\\ &= \frac{|G|}{d_\alpha}v\end{aligned}

and thus $\displaystyle \frac{|G|}{d_\alpha}$ is an eigenvalue for $A$. That said, by the previous theorem we know that every entry of $A$ is an algebraic integer and thus by previous theorem we may conclude that $\displaystyle \frac{|G|}{d_\alpha}\in\mathbb{A}$. But, since evidently $\displaystyle \frac{|G|}{d_\alpha}\in\mathbb{Q}$ we may conclude that $\displaystyle \frac{|G|}{d_\alpha}\in\mathbb{Z}$ and thus $d_\alpha\mid |G|$ as desired. $\blacksquare$

References:

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

March 3, 2011 -

1. […] our last post we proved the incredible fact that the degree of any irrep divides the order of the group. We now […]

Pingback by Representation Theory: The Dimension Theorem (Strong Version) « Abstract Nonsense | March 7, 2011 | Reply

2. […] This follows from the fact that the values of every character of a group are algebraic […]

Pingback by Representation Theory: Character Tables « Abstract Nonsense | March 21, 2011 | Reply

3. […] the sum runs over all with . We recall next that for every we must have that so that but since this implies that . Thus, with these […]

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

4. […] the dimension theorem tells us that each and so . But, since there is always the degree of which is we may conclude […]

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

5. […] irrep of is complex. Thus, for every in one has 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 […]

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