Proofs that the degree of an irrep divides the order of a group

It is a theorem in basic representation theory that the degree of an irreducible representation on $G$ over $\mathbb{C}$ divides the order of $G$. The usual proof of this fact involves algebraic integers (see for example Fulton & Harris' Representation Theory, Serre's Linear representations of finite groups, or Simon's Representations of finite and compact groups). However I find this proof somewhat unsatisfying precisely because it uses algebraic integers, which don't show up much elsewhere in basic representation theory, and it is not at all evident why algebraic integers should be used. I feel that there has to be another proof of this theorem that uses techniques of group theory and representation theory, but the only other proof I know is one by Kopp and Wiltshire-Gordon, but that proof seems to use even more complicated ideas if not machinery!

What are some other proofs of this theorem?

Solution 1:

It seems to me that Etingof et al (pages 51-52) and Kopp and Wiltshire-Gordon give very similar proofs. I'll write them out in as elementary a way as I can, and also give a third proof of my own along similar lines.

Notation: Let $G$ be our finite group, $V$ an irrep over the complex numbers, $\rho_V$ the map $G \to GL(V)$ and $\chi_V$ the character of $V$. We write $\mathrm{Id}_V$ for the identity map $V \to V$. Let $g_1$, $g_2$, ..., $g_c$ be representatives for the conjugacy classes of $G$, and let $C(g)$ be the conjugacy class of $g$.

Lemma: For any $g \in G$, $$\sum_{h \in C(g)} \rho_V(h) = \frac{|C(g)| \chi_V(g)}{\dim V} \mathrm{Id}_V \quad\quad (\ast)$$

Proof: For any $f$ in $G$, we have $$\rho_V(f) \cdot \left( \sum_{h \in C(g)} \rho_V(h) \right) = \sum_{h \in C(g)} \rho_V(fh) =$$ $$\left( \sum_{h \in C(g)} \rho_V(fhf^{-1}) \right) \cdot \rho_V(f) = \left( \sum_{h \in C(g)} \rho_V(h) \right) \cdot \rho_V(f).$$

So the left hand side of $(\ast)$ commutes with every $\rho_V(f)$. By Schur's lemma, this means that the left hand side of $(\ast)$ is $a \mathrm{Id}_V$ for some scalar $a$. Taking traces, we compute that $|C(g)| \chi_V(g) = a \dim V$, so $a = |C(g)| \chi_V(g)/\dim V$ as required. $\square$

Let's define $$P(g) = \sum_{h \in C(g)} h.$$ This is an element in $\mathbb{Z}[G]$. Let $$P_V(g) = \sum_{h \in C(g)} \rho_V(h).$$ This is how $P(g)$ acts on $V$. So the above Lemma shows that $P_V(g) = |C(g)| \chi_V(g)/ \dim V \cdot \mathrm{Id}_V$.

Both the papers I cite want to sum up $P_V(g_i)$ in some way, use the identity $$\sum_{i} |C(g_i)| \chi_V(g_i) \chi_V(g_i^{-1}) = |G| \quad \quad (\ast \ast)$$ and wind up with $\dim V$ in the denominator of something which they can prove, by other means, is an integer. Note that you may recognize $(\ast \ast)$ better in the form $\sum_{g \in G} \chi_V(g) \overline{\chi_V(g)} = |G|$; we get identity $(\ast \ast)$ by grouping together the terms in the same conjugacy class and using $\chi_V(g^{-1}) = \overline{\chi_V(g)}$.

Etingof et al's proof: Consider $$Q_V = \sum_i P_V(g_i) \chi_V(g_i^{-1}).$$ On the one hand, using the Lemma, $$Q_V = \sum_i |C(g_i)| \chi_V(g_i) \chi_V(g_i^{-1}) \frac{1}{\dim V} \mathrm{Id}_V = \frac{|G|}{\dim V} \mathrm{Id}_V$$ using $(\ast\ast)$.

On the other hand if you expand out $Q_V$ in $\mathbb{C}[G]$, you'll see that the coefficient of every group element is an algebraic integer. So $|G|/\dim V$ is an algebraic integer and, since it is rational, must be an integer. $\square$.

Kopp and Wiltshire-Gordon's proof: Set $$R = \sum_{i} \frac{|G|}{|C(g)|} P(g_i) P(g_i^{-1}).$$ Let $R_V = \rho_V(R)$.

Using the Lemma, $$R_V = \sum_i \frac{|G|}{(\dim V)^2} |C(g_i)| \chi_V(g_i) \chi_V(g_i^{-1}) \mathrm{Id} = \frac{|G|^2}{(\dim V)^2} \mathrm{Id}_V.$$

So, similarly, $$\rho_V(R^k) = \frac{|G|^{2k}}{(\dim V)^{2k}} \mathrm{Id}_V.$$

Let $U$ be the regular representation of $G$. So $\chi_U = \sum (\dim V) \chi_V$. We deduce that $$\chi_U(R^k) = \sum_V (\dim V) \frac{|G|^{2k}}{(\dim V)^{2k}} \mathrm{Tr}(\mathrm{Id}_V) = \sum_V (\dim V)^2 \frac{|G|^{2k}}{(\dim V)^{2k}}.$$

One the other hand, $R$ is clearly in $\mathbb{Z}[G]$. (Recall that $|C(g)|$ divides $|G|$ because $|C(g)|$ has a transitive action of $G$ by conjugation.) So $R^k$ is in $\mathbb{Z}[G]$. And the trace of any element of $\mathbb{Z}[G]$ acting on the regular rep is an integer (in fact, it is $|G|$ times the coefficient of the identity). So $\sum_V (\dim V)^2 \frac{|G|^{2k}}{(\dim V)^{2k}}$ is an integer (in fact, one divisible by $|G|$) for all $k$. In Lemma 6.2, the authors show that this forces each $|G|/\dim V$ to be an integer. $\square$

Remark Other parts of the paper make use of the pleasant identity $R = \sum_{g \in g} \sum_{h \in G} ghg^{-1} h^{-1}$. But this doesn't seem to be important if our goal is solely to get this fact.

Variant on the second proof If one considers the action of $R$ on $\mathbb{Z} G$, it is clearly given by a matrix with integer entries. Let $f(\lambda)$ be the characterisitc polynomial of that matrix, so $f$ is a monic polynomial with rational coefficients.

We showed that $R$ acts on the subspace $V^{\oplus \dim V}$ of $\mathbb{C} G$ by $(|G|/\dim V)^2$. So $(|G|/\dim V)^2$ is an eigenvalue of the above integer matrix. So $(|G|/\dim V)^2$ is a root of $f$. By the rational root theorem, we deduce that $(\dim V)^2$ divides $|G|^2$, so $\dim V$ divides $|G|$. $\square$

Solution 2:

I was thinking about something else the other day, and came up with another proof. It's heavier on the number theory, but lighter on the character theory, then the other proofs so far.

Lemma Let $V$ be an irrep of $G$, with representation map $\rho: G \to GL(V)$. Let $A$ be any element of $\mathrm{End}(V)$. Then $$\sum_{g \in G} \rho(g) A \rho(g^{-1}) = \frac{\mathrm{Tr}(A) \cdot |G|}{\dim V} \mathrm{Id}.$$

Proof We have $$\rho(h) \left( \sum_{g \in G} \rho(g) A \rho(g^{-1}) \right) = \sum_{g \in G} \rho(hg) A \rho(g^{-1}) = \left( \sum_{f \in G} \rho(f) A \rho(f^{-1}) \right) \rho(h)$$ by the change of variable $f=hg$. So the left hand side commutes with the $G$-action and is thus a scalar. Taking traces of both sides shows us which scalar it is. $\square$

Lemma With notation as above, we can choose a basis for $V$ so that all the entries of $\rho(g)$ are algebraic integers.

Proof See here. $\square$

Take such a basis, and let $A$ be a matrix with integer entries and trace $1$. Then the first lemma shows that $|G|/\dim V$ is a sum of algebraic integers, hence an algebraic integer, hence an integer.

Solution 3:

Consider the following Lemma: Let $G$ be a finite group, $g \in G$ and $w$ be a given word on $G$. Let $\gamma_G^g(w) = |\{\vec{g} \in G^n: w(\vec{g})=g\}|$. Then $\gamma_G^g(w)=\sum_{\chi} \frac{\overline{\chi(g)}}{|G|} \sum_{\vec{g} \in G^n} \chi(w(\vec{g}))$.

Proof: Let $f: G \to \mathbb{C}$ be defined in the following way: $f(x)=|G|$, if $x \in g^G$ (that is, if $x$ is in the conjugacy class of $g$) and $f(x)=0$ otherwise. Clearly, $f$ is a class function. Now let $A=\sum_{\vec{g} \in G^n} f(w(\vec{g}))$. Since $f(w(\vec{g}))$ is nonzero if and only if $w(\vec{g}) \in g^G$, and since $f$ is a class function, it follows that $A=|g^G| |G| \gamma_G^g(w)$, so $\gamma_G^g(w)=\frac{A}{|g^G||G|}$.

As $f$ is a class function, we have that $f=\sum_{\chi} \langle f,\chi \rangle \chi$, where the sum is taken over all the irreducible characters of $G$. But $\langle f, \chi \rangle = \frac{1}{|G|} \sum_{x \in G} f(x) \overline{\chi(x)} =\frac{1}{|G|} |g^G| |G| \overline{\chi(g)} = |g^G| \overline{\chi(g)}$ (using only the definition of $f$, so $f=\sum_{\chi} |g^G| \overline{\chi(g)} \chi$) and thus

\begin{equation} A= |g^G|\sum_{\vec{g} \in G^n} \sum_{\chi} \overline{\chi(g)}\chi(w(\vec{g}))=|g^G| \sum_{\chi} \overline{\chi(g)} \sum_{\vec{g} \in G^n} \chi(w(\vec{g})) \end{equation}

Since $\gamma_G^g(w)=\frac{A}{|G||g^G|}$, we get $\gamma_G^g(w)=\sum_{\chi}\frac{\overline{\chi(g)}}{|G|}\sum_{\vec{g} \in G^n}\chi(w(\vec{g}))$.

$\square$

Now apply this lemma to the word which is a product of $k+1$ commutators. Using lemmas 3.2 and 3.3 from Kopp and Wiltshire-Gordon (which rely only on Schur's Lemma and are simple, as the group is finite, so you get sums instead of integrals), you'll recover Kopp's and Wiltshire-Gordon's equation 70, bypassing the topological arguments. From that point, you just need the basic number theory he uses in section 6 to finish the proof.

Solution 4:

Here's my attempt at an "elementary" proof, which is basicially a geodesic path through the proof that Serre gives in Linear representations of finite groups.

Theorem. Let $\rho$ be an irreducible representation of $G$. Then $d=\dim \rho$ divides $n=\lvert G\rvert$.

For any $x\in G$, consider the operator $$\def\Cl{\operatorname{Cl}}T_x := \sum_{g\in \Cl(x)} \rho_g.$$ Note that this only depends on the conjugacy class of $x$. We have $\rho_a T_x\rho_{a^{-1}} = T_x$ for any $a\in G$, so by Schur's lemma, $T_x=\lambda_x I$ for some $\def\C{\mathbb{C}}\lambda_x\in \C$. We can actually compute $\lambda_x$ by taking traces: $$\lambda_x = \frac{\lvert\Cl(x)\rvert}{d} \chi(x)$$ where $\chi$ is the character of $\rho$. In particular, $\lambda_e=1$.

By taking the trace of $\sum_{g\in G} \chi(g^{-1})\rho_g = \sum_{x_i} \chi(x_i^{-1}) T_{x_i}$ (sum over a set $\{x_i\}$ of representatives of conjugacy classes), and using the orthogonality relation $\langle\chi,\,\chi\rangle=1$, we get an identity $$\frac{n}{d} = \sum_{x_i}\chi(x_i^{-1})\lambda_{x_i}.$$

Let $R=$ the abelian subgroup of $\C$ generated under addition by the finite set of elements $$\zeta^k \lambda_x, \qquad 0\leq k<n,\quad x\in G,$$ where $\zeta=e^{2\pi i/n}$. Since $\chi(g)$ is a sum of $n$-th roots of unity, we have from the above identity that $n/d\in R$.

Claim. For any $g,h\in G$, there is a formula $$T_gT_h = \sum_{x_i} m_{x_i} T_{x_i}\qquad \text{and therefore}\qquad \lambda_g\lambda_h = \sum_{x_i} m_{x_i} \lambda_{x_i},$$ where the sum is over $\{x_i\}$ a set of representatives of conjugacy classes, and $m_x=$ size of $\{(u,v)\in \Cl(g)\times \Cl(h)\;|\;uv=x\}$, and thus $\def\Z{\mathbb{Z}}m_x\in \Z$. Proof. Straightforward.

By the above claim, $R$ is in fact a subring of $\C$, so multiplication by any element of $R$ such as $n/d$ defines an endomorphism $\phi\colon R\to R$ of the underlying abelian group.

As an abelian group $R$ is certainly finitely generated, and since $R\subseteq \C$ it is torsion free. So by the classification of finitely generated abelian group it is isomorphic to $\Z^N$ for some $N\geq 1$. Choose a basis for $R$ over $\Z$, and let $A$ be the integer matrix representing $\phi$ in this basis.

Since $\def\Q{\mathbb{Q}}n/d\in \Q$, for any $r\in R$ we must have $d\phi(r)=nr$, so $dA=nI$, i.e., $A=(n/d)I$. But $A$ is an $N\times N$ integer matrix with $N\geq1$, so $n/d\in \Z$ as desired.