# Representation theory of finite groups (4/4)

After spending some time in the realm of characters, we make a little detour to the beautiful world of algebraic number theory. Algebraic number theory is, roughly speaking, the study of finite extensions $$K$$ of $$\mathbb{Q}$$ and moreover the ring of integers of K, $$\mathcal{O}_K$$, which is all the elements in $$K$$ that satisifes a monic polynomial with integer coefficients. In such a ring, the amazing property that a non-zero ideal factorize uniquely into prime ideals of $$\mathcal{O}_K$$ holds. I will most definitely try to write on algebraic number theory later, so look out for that. In this article we will prove one of the first things one prove in algebraic number theory, namely that the algebraic integers form a ring.

$$\alpha \in \mathbb{C}$$ is said to be an algebraic integer if $$\alpha$$ satisfies a monic polynomial with integer coefficients. We start by proving a lemma we need:

Lemma 4.1: $$\alpha \in \mathbb{C}$$ is an algebraic integer if and only if $$(1,\alpha,\alpha^2,\alpha^3, \dots)$$ is finitely generated. Here $$(1,\alpha,\alpha^2,\alpha^3, \dots)$$ is the ideal generated by all powers of $$\alpha$$.

Proof. If $$(1,\alpha,\alpha^2,\alpha^3, \dots)$$ is finitely generated, then for some $$n \in \mathbb{N}_{\geq 1}$$ we have for $$a_i \in \mathbb{Z}$$ that $$\alpha^n = a_0+a_1\alpha+a_2\alpha^2+\dots+a_{n-1}\alpha^{n-1}$$ which is the same as saying that $$\alpha$$ satisfy a monic polynomial with integer coefficients. Conversely, if $$\alpha$$ is an algebraic number we can find $$a_i \in \mathbb{Z}$$, not all zero, so that for some $$n \in \mathbb{N}_{\geq 1}$$:  $$\alpha^n = a_0+a_1\alpha+\dots+a_{n-1}\alpha^{n-1}$$ Now we proceed by (strong) induction to prove that $$\alpha^{n+i} \in (1,\alpha,\dots,\alpha^{n-1} )$$. We just proved the base case, so assume $$\alpha^{n+i}\in (1,\alpha,\dots,\alpha^{n-1} )$$ for $$i=0,1,\dots,k$$. Then we have by the induction hypothesis that \begin{alignat*}{2} \alpha^{n+k+1} &=\alpha^{n+k}\alpha = (b_0+b_1\alpha+\dots+b_{n-1}\alpha^{n-1} )\alpha \\ &= b_0\alpha+b_1\alpha^2+\dots+b_{n-1}\alpha^n \\ &= b_0\alpha+b_1\alpha^2+\dots+b_{n-1}(a_0+a_1\alpha+\dots+a_{n-1}\alpha^{n-1}) \\ &= b_{n-1}a_0+(b_0+b_{n-1}a_1)\alpha+ \dots + b_{n-1}a_{n-1}\alpha^{n-1} \end{alignat*} which finishes the proof by induction. $$\blacksquare$$

Using this lemma we can prove that the algebraic integers forms a ring. We will denote this ring by $$\mathbb{A}_{\text{int}}$$.

Theorem 4.2: $$\mathbb{A}_{\text{int}}$$ is a subring of $$\mathbb{C}$$.

Proof. Clearly $$1 \in \mathbb{A}_{\text{int}}$$. What is left to prove is that if $$\alpha,\beta \in \mathbb{A}_{\text{int}}$$ then $$\alpha-\beta, \alpha\beta \in \mathbb{A}_{\text{int}}$$. Let $$\Gamma_{\alpha-\beta} = (1,\alpha-\beta,(\alpha-\beta)^2,(\alpha-\beta)^3)$$ and $$\Gamma_{\alpha\beta} = (1,\alpha\beta,(\alpha\beta)^2,\dots )$$. Defining $$\Gamma_{\alpha},\Gamma_{\beta}$$ similarly, we have by lemma 4.1 that both those are finitely generated, and hence we see that $$\Gamma_{\alpha-\beta}, \Gamma_{\alpha\beta}$$ also has to be finitely generated because these are the subgroups of a finitely generated abelian group. Again by lemma 4.1 that implies $$\alpha-\beta,\alpha\beta \in \mathbb{A}_{\text{int}}$$. $$\blacksquare$$

As one would expect by the name algebraic integers, $$\mathbb{A}_{\text{int}} \cap \mathbb{Q} = \mathbb{Z}$$, an exercise we leave to the reader. Our next goal is to prove that $$\chi(g)$$ is an algebraic integer. For this we use the fact that an invertible matrix of finite order is diagonalizable ( over $$\mathbb{C}$$ ). In the spirit of this article, the proof we present is representation theoretic, moreover it is a corollary of Schur’s lemma.

Lemma 4.3: Let $$A \in GL_n(\mathbb{C})$$ so that $$A^m = I$$ for some $$m \in \mathbb{N}_{\geq 1}$$. Then $$A$$ is diagonalizable with eigenvalues being $$m$$-th roots of unity.

Proof. We study the map $$\rho: \mathbb{Z}/m\mathbb{Z} \to GL_n(\mathbb{C})$$ defined by $$(k \pmod{m}) \mapsto A^{k \pmod{m}}$$. Since $$A^m = I$$, this is a well-defined map, and it is also seen to be a homomorphism, hence a representation. By Maschke’s theorem, we have a complete decomposition: $$\rho \sim \varphi^{(1)} \oplus \cdots \oplus \varphi^{(r)}$$ Since $$\mathbb{Z}/m\mathbb{Z}$$ is abelian, it follows by corollary 1.6 from the first article in this series that $$\varphi^{(i)}$$ is of degree 1, and hence $$\varphi^{(i)}: \mathbb{Z}/m\mathbb{Z} \to \mathbb{C}^{*}$$ and $$r=n$$. Let $$T: \mathbb{C}^n \to \mathbb{C}^n$$ be the equivalence of $$\rho$$ with $$\varphi^{(1)} \oplus \cdots \oplus \varphi^{(r)}$$. Then $$T^{-1}\rho_gT = \begin{pmatrix} \varphi^{(1)}_g & 0 & \cdots & 0 \\ 0 & \varphi^{(2)}_g & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \varphi^{(n)}_g \end{pmatrix}$$ for all $$g \in \mathbb{Z}/m\mathbb{Z}$$. Letting $$g=1$$ we get that $$T^{-1}AT=\begin{pmatrix} \varphi^{(1)}_1 & 0 & \cdots & 0 \\ 0 & \varphi^{(2)}_1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \varphi^{(n)}_1 \end{pmatrix}=\begin{pmatrix}\lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{pmatrix}$$ for $$\lambda_i \in \mathbb{C}^*$$ which are the eigenvalues of $$A$$, so $$A$$ is diagonalizable. Furthermore $$\begin{pmatrix}\lambda_1^m & 0 & \cdots & 0 \\ 0 & \lambda_2^m & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n^m \end{pmatrix} =(T^{-1}AT)^m = T^{-1}A^mT = I$$ so all eigenvalues are $$m$$-th roots of unity. $$\blacksquare$$

Clearly an $$m$$-th root of unity is an algebraic integer. In the following we prove that an arbitrary character evaluated in $$g$$ is indeed a sum of $$m$$-th roots of unity and hence an algebraic integer since $$\mathbb{A}_{\text{int}}$$ is a ring.

Lemma 4.4: Let $$\chi$$ be a character of a finite group $$G$$, with $$|G|=m$$. Then for any $$g \in G$$, there exists $$\lambda_{g1},\dots,\lambda_{gm}$$, which are $$m$$-th roots of unity so that $$\chi(g)=\lambda_{g1}+\dots+\lambda_{gm}$$.

Proof. Let $$\rho: G \to GL_m(\mathbb{C})$$ be a representation with character $$\chi$$. Fix $$g \in G$$. Then $$g^m=1$$. Hence $$I=\rho_{g^m}=\rho_g^m$$, and so $$\rho_g$$ is diagonalizable by lemma 4.3 – say, with eigenvalues $$m$$-th roots of unity $$\lambda_{g1},\dots,\lambda_{gm}$$. Hence $$\chi(g)=\lambda_{g1}+\dots+\lambda_{gm}$$. $$\blacksquare$$

We need one final lemma before we can prove one of the main goals of this section: the dimension theorem. Without further ado let us jump into it.

Lemma 4.5: Let $$G$$ be a finite group and $$\varphi$$ an irreducible representation of $$G$$ with degree $$d$$. Fix some $$g \in G$$, and let $$h$$ be the order of the conjugacy class of $$g$$ in $$G$$. If $$\chi_{\varphi}$$ is the associated character to $$\varphi$$ then $$\frac{h}{d}\chi_{\varphi}(g)$$ is an algebraic integer.

Proof. Let us first list all the conjugacy classes of $$G$$: $$C_1,\dots,C_k$$. Since characters are class functions we let $$\chi_i$$ be the value that $$\chi_{\varphi}$$ takes on $$C_i$$. If we are able to show that $$\frac{|C_i|}{d}\chi_i$$ is an algebraic integer for each $$i$$ we are done. The proof revolves around one operator: $$T_i = \sum_{x \in C_i} \varphi_x$$ We claim three things:
1) $$T_i = \frac{|C_i|}{d}\chi_i I$$
2) $$T_iT_j = \sum_{\ell=1}^k a_{ij\ell} T_\ell$$ for some $$a_{ij\ell} \in \mathbb{Z}$$.
3) $$\alpha \in \mathbb{C}$$ is an algebraic integer if and only if there exists $$\alpha_1,\dots,\alpha_t$$ not all zeros such that $$\alpha\alpha_i=\sum_{j=1}^t a_{ij}\alpha_j$$ where $$a_{ij} \in \mathbb{Z}$$ for all $$1\leq i \leq t$$.

The last claim is a standard fact from algebraic number theory, and we thus leave it to the reader to take on good faith or prove (for the converse direction, use that eigenvalues of integer matrices are algebraic integers).

If we for a minute assume 1) and 2) is true we get that $$\left ( \frac{|C_i|}{d}\chi_i\right)\left(\frac{|C_j|}{d}\chi_j\right) = \sum_{\ell=1}^k a_{ij\ell} \left( \frac{|C_\ell|}{d}\chi_\ell \right)$$ which by 3) implies that $$\frac{|C_i|}{d}\chi_i$$ is an algebraic integer. 2) is first a routine calculation then checking that the coefficients are independent of the conjugacy class. We omit that here. For 1) we start by observing $$\varphi_g T_i \varphi_{g^{-1}} = \sum_{x \in C_i} \varphi_g\varphi_x\varphi_{g^{-1}} = \sum_{x \in C_i} \varphi_{gxg^{-1}} = T_i$$ where we in the very last line has used that $$C_i$$ is closed under conjugation and that conjugation by $$g$$ induces a bijection of $$C_i$$. Thus $$\varphi_g T_i \varphi_{g^{-1}} = T_i$$ for all $$g \in G$$. We can then conclude by Schur’s lemma that $$T_i = \lambda I$$ for some $$\lambda \in \mathbb{C}$$. Finding $$\lambda$$ is easy: $$d\lambda = \text{Tr}(\lambda I ) = \text{Tr}(T_i) = \text{Tr}\left( \sum_{x \in C_i} \varphi_x \right) = \sum_{x \in C_i} \chi_i = |C_i|\chi_i$$ This finishes the proof. $$\blacksquare$$

Theorem 4.6: (Dimension theorem)
Let $$\varphi$$ be an irreducible representation of $$G$$ with degree $$d$$. Then $$d \mid G$$.

Proof. By the first orthogonality relations (Theorem 2.5) we have $$1 = \frac{1}{|G|} \sum_{ g \in G} \overline{\chi_{\varphi}(g)}\chi_{\varphi}(g) = \frac{d}{|G|} \sum_{ g \in G} \overline{\chi_{\varphi}(g)}\frac{\chi_{\varphi}(g)}{d}$$ Let us list all the conjugacy classes of $$G$$ as before: $$C_1,\dots,C_k$$. Again let $$\chi_i$$ denote the value $$\chi$$ takes on $$C_i$$. From the calculation from the orthogonality relations over we then get $$\frac{|G|}{d} = \sum_{ g \in G} \overline{\chi_{\varphi}(g)}\frac{\chi_{\varphi}(g)}{d} = \sum_{i=1}^k \sum_{ g \in C_i} \overline{\chi_{{\varphi}_i}}\left( \frac{1}{d} \chi_{{\varphi}_i} \right) = \sum_{i=1}^k \overline{\chi_{{\varphi}_i}}\left( \frac{|C_i|}{d} \chi_{{\varphi}_i} \right)$$ Now $$\frac{|C_i|}{d} \chi_{{\varphi}_i}$$ is an algebraic integer by Lemma 4.5 and since the algebraic integers are closed under complex conjugation, so is $$\overline{\chi_{{\varphi}_i}}$$. Since the algebraic integers is a ring it follows that the last sum is an algebraic integer, that is $$\frac{|G|}{d}$$ is an algebraic integer, and hence it an actual integer. In other words $$d \mid |G|$$. $$\blacksquare$$

With the dimension theorem proven we can finally prove what we promised in the start of these blog posts:

Theorem 4.7: Let $$G$$ be a group of order $$p^2$$ for a prime $$p$$. Then $$G$$ is abelian.

Proof. By Corollary 3.3 from the previous article we have $$p^2=|G| = d_1^2+\dots+d_k^2$$ where $$d_i$$ are the degrees of the (equivalence classes of) irreducibles representations of $$G$$. By the dimension theorem $$d_i = 1,p,p^2$$ for any $$i$$. The trivial representation has degree $$1$$ and is irreducible, and hence by cardinality $$d_i = 1$$ for all $$i$$. We are now done by Theorem 3.4 $$\blacksquare$$.