# 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?

• Actually, algebraic integers can be considered beasts from (Galois) group theory. – Hagen von Eitzen Nov 23 '12 at 15:23
• Is the fulton and harris proof the character theoretic one which uses $\langle \chi,\chi \rangle$ to show $|G|/\chi(1)$ is an algebraic integer? – Alexander Gruber Nov 23 '12 at 20:12
• @Alexander More or less. Fulton and Harris presents this theorem as an exercise (Problem 2.38) and works it out at the back of the book. – JHF Nov 23 '12 at 22:14
• “The proof of Kopp and Wiltshire-Gordon seems to use more complicated ideas” ? Their abstract claims otherwise : “the only ingredients are Schur’s lemma, basic counting, and a divisibility argument.” – Ewan Delanoy Dec 1 '12 at 6:20
• @Ewan: While the calculations might not be very sophisticated, Remark 6.4 of the Kopp and Wilton-Gordon indicates that the proof depends on corollary 4.2. Certainly the involvement of surfaces defined by words or equation (52) is a surprising idea, no? So I'm wondering if there's a more group-theoretic proof. – JHF Dec 1 '12 at 23:24

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 $|G|$. $\square$

• Thanks for the detailed explanation @DavidSpeyer, I think this is very nice. – Alexander Gruber Dec 8 '12 at 18:32
• Nice survey! But I'm afraid you got Etingof's proof slightly wrong: What do you mean by "if you expand out $Q_V$ in $\mathbb C[G]$, you'll see that the coefficient of every group element is an algebraic integer"? I don't think there is a "sum of algebraic integers is an algebraic integer" result for noncommutative rings. I think this is the reason Etingof works with numbers rather than elements of $\mathbb C[G]$ or $\mathrm{End}V$. – darij grinberg Apr 10 '13 at 2:37
• @darijgrinberg I think you're right. I'm probably not going to get to this soon; feel free to fix it for me if you want to. You're right that sum of algebraic integers need not be an algebraic integer in a noncommutative ring; consider $\left( \begin{smallmatrix} 0 & p \\ 1/p & 0 \end{smallmatrix} \right) + \left( \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right)$ in $\mathrm{Mat_2}(\mathbb{Q})$. – DES-SupportsMonicaAndTransfolk Apr 10 '13 at 3:06
• @DavidESpeyer I'm wondering whether one could go anywhere toward an elementary proof with the following naive idea: Take a nonzero vector $v \in V$ with $V$ an irrep. Consider the finite set of $\{gv\}$ for $g\in G$. The span of these vectors is apparently $G$-invariant and not the empty set so must be $V$. By the Orbit-Stabilizer theorem $|G|=|Stab(v)||Orb(v)|$. It seems plausible that the finite set $Orb(v)$ could now be a basis of $V$ and thus prove the result, but I was curious whether someone with more rep theory experience had an example to contradict this before I try to proceed. – Brian Klatt Mar 6 '18 at 16:34
• @BrianKlatt : Note that the only representations of $G$ for which $Orb(v)$ is a basis for some $v$ are the permutation representations (linear extensions of group actions). These representations are never irreducible, since the sum of the elements of $Orb(v)$ is a non-zero element which is $G$-invariant. – Matthias Klupsch Sep 20 '18 at 12:00

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.

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.

You should probably use Lagrange Theorem. The basic assumptions over here is G is a finite group over C. Consider any irreducible representation of G to be the subgroup of G. Now any subgroup H (in this case irreducible representation) of a finite group G divides the order of the group by the Lagrange theorem.

• I don't understand what you mean by "consider any irreducible representation of $G$ to be the subgroup of $G$". A representation is not a subgroup. – DES-SupportsMonicaAndTransfolk Dec 2 '12 at 13:08