# Representation theory of finite groups (1/4)

A classical result in a first course in abstract algebra is the following: Let $$G$$ be a group of order $$p^2$$ for a prime $$p$$. Then $$G$$ is abelian. The proof offered often goes by Sylow theory. In this little series of texts we aim to give a different proof via the beautiful representation theory of finite groups. In a nutshell, representation theory tries to transfer problems about algebraic structures to problems in linear algebra, which is often better understood. In the case of a group for example, we represent group elements as invertible linear operators. This let us interpret the group with the machinery of linear algebra, which has proven fruitful over and over again. This series of articles will most likely consist of four entries:

1. Definitions, Maschke’s Theorem, Schur’s Lemma.
2. Character theory.
3. The regular representation.
4. Algebraic numbers, the dimension theorem and groups of order $$p^2$$.

Without further ado, let us jump straight into it.

Throughout this series of texts, we always assume $$G$$ is a finite group. A representation of $$G$$ is a homomorphism $$\rho: G \to GL(V)$$ where $$V$$ is a vector space, here taken to be finite dimensional and over $$\mathbb{C}$$. Hence we can interpret $$GL(V)$$ as $$GL_n(\mathbb{C})$$, the group of invertible $$n \times n$$-matrices over $$\mathbb{C}$$. We often write $$\rho_g(v)$$ or $$\rho_gv$$ for $$\rho(g)(v)$$. The degree of a representation $$\rho: G \to GL(V)$$ is defined to be $$\dim(V)$$. A subspace of $$W$$ of $$V$$ is said to be $$G$$-invariant if $$\rho_g(w) \in W$$ for all $$w \in W, g \in G$$. If we restrict a representation $$\rho_g$$ to a $$G$$-invariant subspace $$W$$ we get a subrepresenation, often written as $$\rho\vert_W$$. Two representations $$\rho: G \to GL(V), \varphi: G \to GL(W)$$ are said to be equivalent if there exists a linear isomorphism $$T: V \to W$$ so that $$T\rho_g = \varphi_g T$$ for all $$g \in G$$. The first result we will prove is Maschke’s theorem – a basic, but important result about how representation of finite groups decompose. For this we need the notion of a direct sum of two representations. There is no surprise in the definition: given two representations $$\varphi^{(1)}: G \to GL(V_1), \varphi^{(2)}: G \to GL(V_2)$$, we define the representation $$\varphi^{(1)} \oplus \varphi^{(2)} : G \to GL(V_1 \oplus V_2)$$ by $$(\varphi^{(1)}\oplus \varphi^{(2)})_g (v_1,v_2) \mapsto (\varphi^{(1)}_g(v_1),\varphi^{(2)}_g(v_2))$$. We will now define three important notions in representation theory. To this end, let $$G$$ be a finite group and $$\rho: G \to GL(V)$$ a represenation of $$G$$.

• $$\rho$$ is irreducible if the only $$G$$-invariant subspace of $$V$$ is $$0$$ and $$V$$,
• $$\rho$$ is completely reducible if $$V = V_1 \oplus \cdots \oplus V_n$$ for $$G$$-invariant subspaces so that $$\rho\vert_{V_1}$$ is irreducible for all $$i=1,\dots,n$$.
• $$\rho$$ is decomposable if $$V=V_1 \oplus V_2$$ for proper non-zero $$G$$-invariant subspaces $$V_1,V_2$$. If $$\rho$$ is not decomposable, it is said to be indecomposable.

Maschke’s theorem is the statement that every representation of a finite group is completely reducible. From the definition of the equivalence of representations, it follows readily that irreducibility, complete reducibility and decomposability is preserved by equivalences. To prove Maschke’s theorem we will take advantage of inner product spaces. To this end, let $$V, \langle \cdot, \cdot \rangle$$ be a (finite-dimensional) inner product space over $$\mathbb{C}$$. In a typical introduction course in linear algebra one studies unitary linear operators $$T: V \to V$$ – surjective operators that satisfy $$\langle Tv, Tw \rangle = \langle v,w \rangle$$ for all $$v, w \in V$$. In the same way a representation $$\rho: G \to GL(V)$$ is said to be unitary if for all $$g \in G$$ $$\langle \rho_gv, \rho_gw \rangle = \langle v, w \rangle$$ for all $$v,w \in W$$ and $$\rho_g$$ is surjective. With this defined we can prove our first lemma towards Mashcke’s theorem.

Lemma 1.1:  A unitary representation of a group $$G$$ is either irreducible or decomposable.

Proof. Let $$\rho: G \to GL(V)$$ be a unitary representation, and suppose that it is reducible. Hence we have some $$G$$-invariant subspace $$W$$ that is non-zero and proper. We have $$V = W \oplus W^{\perp}$$, so if $$W^{\perp}$$ is a $$G$$-invariant subspace, this will show that $$G$$ is decomposable. Let $$v \in W, w \in W^{\perp}$$. Then by $$G$$-invariance of $$W$$ and unitarity we have $$0=\langle w,\rho_{g^{-1}}v \rangle = \langle \rho_{g} w, \rho_{g}\rho_{g^{-1}}v \rangle = \langle \rho_{g}w,v \rangle$$ showing that $$W^{\perp}$$ is $$G$$-invariant and finishing the proof. $$\blacksquare$$

The strength of this lemma comes from the next lemma.

Lemma 1.2: A representation $$\rho:G \to GL(V)$$ of a finite group $$G$$ is equivalent to a unitary representation of $$G$$.

Proof. Let $$\dim(V)=n$$. We construct an inner product on $$\mathbb{C}^n$$ . To do this we introduce the averaging trick, which is a trick often used in representation theory of finite groups. This proof is really more about constructing a space to map into, then the isomorphism itself as we very soon will see. To this end, fix a basis $$\mathcal{B}$$ of $$V$$. Let $$T: V \to \mathbb{C}^n$$ be the canonical isomorphism. For any fixed $$g \in G$$ we let $$\varphi_g = T\rho_g T^{-1}$$. Then this gives a representation $$\varphi:G \to GL_n(\mathbb{C})$$. Let $$\langle \cdot, \cdot \rangle$$ be the the standard inner product on $$\mathbb{C}^n$$. As mentioned in the start, we construct a new inner product: $$(v,w) = \sum_{g \in G} \langle \varphi_g v, \varphi_gw \rangle$$ This is one of the places in the theory where the assumption that $$G$$ is finite is crucial. That this is an actual inner product is a routine calculation, which I leave to the reader. What is more interesting is showing that $$\varphi$$ is actually a unitary representation of $$G$$. For this we introduce another trick that also depends on $$G$$ being finite. We first see that $$( \varphi_h v,\varphi_h w ) = \sum_{g \in G} \langle \varphi_g\varphi_h v, \varphi_g \varphi_h w \rangle = \sum_{g \in G} \langle \varphi_{gh} v, \varphi_{gh} w \rangle$$ Right multiplication by $$h$$ is injective since $$G$$ is a group, and since $$G$$ is finite it is also a bijection. This observation tells us that the change of variables $$x=gh$$, won’t alter the sum – that is $$\sum_{g \in G} \langle \varphi_{gh} v, \varphi_{gh} w \rangle = \sum_{x \in G } \langle \varphi_{x}v, \varphi_{x} w \rangle$$  This trick will be used over and over again. This finishes the proof. $$\blacksquare$$

With this lemma proved, we are ready to prove one of the main theorems of this first article. The proof is very similar to classical proof of the fundamental theorem of arithmetic.

Theorem 1.3: (Maschke’s theorem)
Every representation of a finite group is completely reducible.

Proof. Let $$\rho: G \to GL(V)$$ be a representation of a finite group $$G$$. By lemma 1.2, $$\rho$$ is equivalent to a unitary representation, and by lemma 1.1 $$\rho$$ is hence either decomposable or irreducible. With this general fact in mind, we proceed by (strong) induction on the dimension $$n$$ of $$V$$. When $$n = 1$$, $$V$$ doesn’t have any non-zero proper subspaces. We assume now the statement is true whenever $$V$$ is a vector space so that $$\dim(V) \leq n$$, and let $$\rho: G \to GL(V)$$ be a representation so that $$\dim(V)=n+1$$. If $$\rho$$ is irreducible we are done, so suppose it is not. Then it is decomposable, so $$V= W \oplus Z$$ for some non-zero proper $$G$$-invariant subspaces. By the induction hypothesis we have $$W=W_1 \oplus \cdots \oplus W_t \qquad Z = Z_{1} \oplus \cdots \oplus Z_{s}$$ for $$G$$-invariant subspaces $$W_i,Z_i$$ so that $$\rho\vert_{W_i}, \rho\vert_{Z_i}$$ are irreducible representations. Hence we get the complete reduction $$V=W_1\oplus \cdots \oplus W_t \oplus Z_1 \oplus \cdots \oplus Z_{s}$$ which finishes the proof. $$\blacksquare$$

While this proves existence of a decomposition, it doesn’t say anything about uniqueness. We will answer the question regarding uniqueness quite specifically later by using the elegant theory of characters. We will take our first step towards the theory of characters, by ending this article with proving Schur’s lemma and a corollary of it. From now on, we will write $$\rho \sim \varphi$$ if $$\rho$$ and $$\varphi$$ are equivalent representations of $$G$$. A homomorphism of representations of $$G$$, from $$\rho: G \to GL(V)$$ to $$\varphi:G \to GL(W)$$, is a linear map $$T:V \to W$$ so that $$T\rho_g = \varphi_g T$$. This is also sometimes called an intertwiner. We are interested in the set of all homomorphism from $$\rho$$ to $$\varphi$$, denoted $$\text{Hom}_G(\rho,\varphi)$$. Schur’s lemma gives a very specific answer in the case where both $$\rho$$ and $$\varphi$$ are irreducible representations. We observe that $$\text{Hom}_G(\rho,\varphi) \subseteq \text{Hom}_G(V,W)$$ and furthermore $$\text{Hom}_G(\rho,\varphi)$$ is a subspace of $$\text{Hom}_G(V,W)$$.

Lemma 1.4: Let $$T:V \to W \in \text{Hom}_G(\rho,\varphi)$$. Then $$\ker(T)$$ is a $$G$$-invariant subspace of $$V$$ and $$\text{Im}(T)$$ is a $$G$$-invariant subspace of $$W$$.

Proof. Let $$v \in \ker(T), g \in G$$. We have $$T\rho_gv = \varphi_g T v = 0$$ so $$\rho_gv \in \ker(T)$$ and hence $$\ker(T)$$ is a $$G$$-invariant subspace of $$V$$. Let $$w \in \text{Im}(T), g \in G$$, then $$Tv = w$$ for some $$v \in V$$. Then $$\varphi_g w = \varphi_g T v = T\rho_g v$$ so $$\text{Im}(T)$$ is a $$G$$-invariant subspace of $$W$$. $$\blacksquare$$

With this lemma we are ready to prove Schur’s lemma. Recall that we are working with vector spaces over the complex numbers. This is the first place where we really use this. What we use is actually the fundamental theorem of algebra applied to this setting: every linear operator on a finite-dimensional complex vector space has an eigenvalue.

Theorem 1.5: (Schur’s lemma)
Let $$\rho: G \to GL(V), \varphi:G \to GL(W)$$ be irreducible representations of $$G$$ and let $$T \in \text{Hom}_G(\rho,\varphi)$$. Then either $$T$$ is invertible or $$T = 0$$. Consequently $$\text{Hom}_G(\rho,\varphi) = 0$$ if $$\rho \not\sim \varphi$$, and $$T = \lambda I$$ for some $$\lambda \in \mathbb{C}$$ if $$\rho = \varphi$$.

Proof. From lemma 1.4, $$\ker(T)$$ is a $$G$$-invariant subspace of $$V$$, but $$\rho$$ is irreducible so either $$\ker(T) = 0$$ or $$\ker(T)=V$$. If $$\ker(T)=V$$ then $$T = 0$$, so assume $$T \neq 0$$. If $$\ker(T)=0$$, then $$T$$ is injective. From lemma 1.4, $$\text{Im}(T)$$ is a $$G$$-invariant subspace of $$W$$, but $$\varphi$$ is irreducible so $$\text{Im}(T)=0$$ or $$\text{Im}(T)=W$$, but $$\ker(T)=0$$ so $$\text{Im}(T) = W$$. Hence $$T$$ is surjective as well, so invertible. It is clear that $$\text{Hom}_G(\rho,\varphi) = 0$$ if $$\rho \not \sim \varphi$$ from this. Finally, let $$\lambda$$ be an eigenvalue of $$T \in \text{Hom}_G(\rho,\rho)$$. Letting $$I:V \to V$$ be the identity operator, we have $$I \in \text{Hom}_G(\rho,\rho)$$. Since $$\text{Hom}_G(\rho,\rho)$$ is a subspace of $$\text{Hom}(V,V)$$ it follows that $$T-\lambda I \in \text{Hom}_G(\rho,\rho)$$. But by definition $$T-\lambda I$$ is not invertible, and so $$T=\lambda I$$ by the previous parts of the theorem. $$\blacksquare$$

A very interesting corollary of Schur’s lemma is the following.

Corollary 1.6: Let G be an abelian group. Then any irreducible representation $$\rho: G \to GL(V)$$ has degree one.

Proof. Fix some $$h \in G$$. Then for all $$g \in G$$ we have $$\rho_h\rho_g = \rho_{hg} = \rho_{gh} = \rho_g\rho_h$$ that is $$\rho_h \in \text{Hom}_G(\rho,\rho)$$, so by Schur’s lemma we have that $$\rho_h = \lambda_h I$$. Now if $$\alpha \in \mathbb{C}$$ and $$v \in V$$ is a non-zero vector then $$\rho_h(\alpha v) = \lambda_h \alpha v \in \mathbb{C}v$$ so $$\mathbb{C}v$$ is a $$G$$-invariant subspace of $$V$$. Since $$\rho$$ is an irreducible representation, this implies that $$\mathbb{C}v=0$$ or $$\mathbb{C}v = V$$, and since $$v$$ is a non-zero vector, we conclude that $$\mathbb{C}v = V$$. Hence $$\dim(V) = 1$$. $$\blacksquare$$