Representation Theory IV 1 Introduction 3 Character theory

2 Representation theory of finite groups

Throughout, unless specified otherwise, $G$ denotes a finite group and $V$ denotes a finite dimensional complex vector space (such as $\mathbb{C}^{n}$ ).

2.1 Definitions and first examples

Let $k$ be a field (we will almost always take $k=\mathbb{C}$ ), and let $G$ be a group.

Definition 2.1.

A representation of $G$ over $k$ is a pair $(\rho,V)$ where the

•

$V$ is a vector space over $k,$ and
•

$\rho:G\rightarrow\operatorname{GL}(V)$ is a group homomorphism.

The dimension of the representation is the dimension of $V$ . We will very often say that $V$ is a representation of $G$ , or that $\rho$ is a representation of $G$ , without mentioning the other part of the definition.

There is another way to think of this. Suppose that $(\rho,V)$ is a representation of $G$ . Then we define an action of $G$ on $V$ by $gv=\rho(g)v$ . This is an action because $\rho$ is a homomorphism, and it is linear, meaning that for every $g$ the map taking $v$ to $g v$ is a linear map. Conversely, given a linear action of $G$ on $V$ , we can define $\rho$ by $\rho(g)v=gv$ . In other words:

A representation of $G$ is a linear action on a vector space.

We will often use $\rho(g)v$ and $g v$ interchangeably.

If a basis is given for $V$ , then an (invertible) linear map $V\rightarrow V$ is just the same thing as an (invertible) $n\times n$ matrix, where $n=\dim V$ . So, once you choose a basis, a representation is just the same as a homomorphism $G\rightarrow\operatorname{GL}_{n}(k)$ . In particular:

A one-dimensional representation of $G$ is the same as a homomorphism $G\rightarrow k^{\times}$ .

Example 2.2.

Let $G=S_{n}$ . Recall that there is a homomorphism

\epsilon:S_{n}\rightarrow\{\pm 1\}

taking a permutation to its sign. As $\pm 1\subset\mathbb{C}^{\times},$ this gives a one-dimensional representation of $S_{n}$ called the sign representation.

Example 2.3.

If $V$ is any vector space, then we can always take $\rho:G\rightarrow\operatorname{GL}(V)$ to be the homomorphism sending every element to the identity. We call this the trivial representation on $V$ .

Example 2.4.

Suppose that $G=(\mathbb{Z},+)$ . Then, if $\rho$ is a representation of $G,$ it is completely determined by $V$ and the invertible linear map $\rho(1):V\rightarrow V$ (which can be anything). This is because we then have

\rho(n)=\rho(1+\ldots+1)=\rho(1)^{n}.

Thus a representation of $\mathbb{Z}$ is just a vector space $V$ together with an invertible linear map from $V$ to itself.

We can push this a bit further. Suppose that $G$ is a cyclic group of order $n$ with generator $g,$ so

G=\{e,g,g^{2},\ldots,g^{n-1}\}

and $g^{n}=e$ . Then a representation of $G$ is again determined by $V$ and $\rho(g),$ which can be any linear map $T:V\rightarrow V$ such that $T^{n}=I$ .

One source of more interesting examples is geometry.

Example 2.5.

Let $G=D_{n}$ be the dihedral group of order $2n,$ the group of symmetries (rotations and reflections) of a regular $n$ -gon — see Figure 1. Since each rotation/reflection is an invertible linear map from $\mathbb{R}^{2}\rightarrow\mathbb{R}^{2},$ we get a representation $\rho$ of $G$ on $\mathbb{R}^{2}$ . Letting $r$ be rotation by $2\pi/n$ and $s$ be reflection in the vertical axis, $D_{n}$ has the presentation

\left\langle r,s:r^{n}=s^{2}=e,sr=r^{-1}s\right\rangle.

As an explicit homomorphism $\rho:D_{n}\rightarrow\operatorname{GL}_{2}(\mathbb{R}),$ we have (with $\theta=2\pi/n$ )

	$\displaystyle\rho(r)$	$\displaystyle=\begin{pmatrix}\cos(\theta)&-\sin(\theta)\\ \sin(\theta)&\cos(\theta)\end{pmatrix}$
	$\displaystyle\rho(s)$	$\displaystyle=\begin{pmatrix}-1&0\\ 0&1\end{pmatrix}.$

Refer to caption — Figure 1: The action of the dihedral group on a polygon.

Example 2.6.

Let $G=S_{4}$ . You might remember that this is isomorphic to the group of symmetries (rotations/reflections) of the regular tetrahedron in $\mathbb{R}^{3}$ . We therefore get a representation

\rho:S_{4}\rightarrow\operatorname{GL}_{3}(\mathbb{R}).

It would be a slightly unpleasant exercise to work the matrices out explicitly.

Note that $S_{4}$ is also isomorphic to the group of rotations of the cube, giving another (different!) three-dimensional representation.

Another source of representations comes from actions of groups on (usually finite) sets.

Example 2.7.

Let $G=S_{n},$ and let $V=k^{n}$ . Define a representation of $S_{n}$ on $V$ via

\sigma\cdot\left(x_{1}e_{1}+\ldots+x_{n}e_{n}\right)=x_{1}e_{\sigma(1)}+\ldots% +x_{n}e_{\sigma(n)}

where $e_{1},\ldots,e_{n}$ is the standard basis. This is called the permutation representation (over $k$ ).

There is a warning here! If you write elements of $V$ as $(x_{1},\ldots,x_{n})$ , as well you might, then it is not the case that $g(x_{1},\ldots,x_{n})=(x_{g(1)},\ldots,x_{g(n)})$ . This actually would define a right action, not a left action. The correct formula is

g(x_{1},\ldots,x_{n})=(x_{g^{-1}(1)},\ldots,x_{g^{-1}(n)}).

Definition 2.8.

If $G$ is a group acting on a set $X,$ we consider a $k$ -vector space $V$ with basis $\{e_{x}:x\in X\}$ . It has a representation of $G$ given by $ge_{x}=e_{gx},$ called the permutation representation associated to $X$ .

Remark 2.9.

There is another point of view on this, the functional point of view. We didn’t cover this in lectures; as things stand this is optional. For simplicity, let $X$ be finite, and define

k^{X}=\{f:X\rightarrow k\}.

We give this an action of $G$ by the formula

(gf)(x)=f(g^{-1}(x))

for $g\in G$ , $f\in k^{X}$ , $x\in X$ .

Question 2.10.

Why don’t we define $(gf)(x)=f(gx)$ ?

Let $\delta_{x}\in k^{X}$ be the function sending $x$ to $1$ and everything else to $0$ . Then the $\delta_{x}$ for $x\in X$ are a basis for $k^{X}$ , and for $g\in G$ you can check that

g\delta_{x}=\delta_{gx}.

In other words, the $\delta_{x}$ behave exactly like the $e_{x}$ in the permutation representation; when we have a bit more language, we can say that $k^{X}$ is isomorphic to the permutation representation.

The next example combines geometry and combinatorics.

Example 2.11.

Let $G$ be the group of rotations of the cube and let $X$ be the set of faces. We can think of an element of the permutation representation as being a way of writing a complex number on each face.

This term is all about finite groups, but next term we will consider representations of Lie groups such as the group $\mathrm{SO}(3)$ of rotations of $\mathbb{R}^{3}$ . These have many applications in physics.

Example 2.12.

In a spherically symmetric situation we might be interested in (smooth) solutions $f:\mathbb{R}^{3}\rightarrow\mathbb{C}$ to Laplace’s equation

\nabla f=0

with radial behaviour $f(rx)=r^{l}f(x)$ for scalars $r,$ and some integer $l\geq 0$ . Here $\nabla$ is the Laplacian

\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{% \partial^{2}}{\partial z^{2}}.

These form a representation of $\mathrm{SO}(3)$ of dimension $2l+1,$ and in fact the irreducible finite dimensional representations of $\mathrm{SO}(3)$ are exactly those obtained in this way for integers $l\geq 0$ .

The solutions are called spherical harmonics, and representation theory can be used to find particularly nice bases of these spaces!

Here is another example:

Example 2.13.

Let $G=\operatorname{GL}_{2}(\mathbb{C})$ and let $V=\left\langle X^{2},XY,Y^{2}\right\rangle\subset\mathbb{C}[X,Y]$ . Then $V$ is a representation of $G$ via

g\cdot F(X,Y)=F(aX+cY,bX+dY).

2.2 Formalities

2.2.1 Subrepresentations and irreducibility

Definition 2.14.

A subrepresentation of a representation $(\rho,V)$ of $G$ is a subspace $W\subset V$ such that $\rho(g)w\in W$ for all $w\in W$ .

Then $(\rho_{W},W)$ is also a representation of $G,$ where $\rho_{W}(g)w=\rho(g)w$ .

This has a block matrix interpretation: if we choose a basis for $W$ and extend it to a basis for $V$ , then the matrix of $\rho(g)$ will have the following block matrix form:

\left(\begin{array}[]{c|c}\rho_{W}(g)&\star\\ \hline\cr\mathbf{0}&\rho_{V/W}(g).\end{array}\right).

Here $\rho_{V/W}$ is also a representation of $G$ , the quotient representation.

Definition 2.15.

A representation $V$ of $G$ is irreducible if it is nonzero and has no subrepresentations except for $\{0\}$ and $V$ itself.

Example 2.16.

Let $V$ be the permutation representation of $S_{4}$ acting on the faces of the cube. Remember that we can see a vector in $V$ as a way of writing a complex number on each face. Then we can spot three subrepresentations:

•

the subrepresentation where the numbers on all faces are equal
•

the subrepresentation where the numbers on opposite faces sum to zero
•

the subrepresentation where the numbers on opposite faces are equal and all the labels sum to zero.

Note that in lectures we did the related example of edges of the tetrahedron.

Definition 2.17.

Suppose that $(\rho,V)$ is a representation of $G$ and that $W_{1},W_{2}$ are two subrepresentations such that every element of $V$ can be written uniquely as the sum of an element of $W_{1}$ and an element of $W_{2}$ . Then we say that $V$ is the (internal) direct sum of $W_{1}$ and $W_{2}$ and write $V=W_{1}\oplus W_{2}$ .

This also has a block matrix interpretation: if we choose bases for $W_{1}$ and $W_{2}$ , then the matrix of $\rho(g)$ will have the following block matrix form:

\left(\begin{array}[]{c|c}\rho_{1}(g)&\mathbf{0}\\ \hline\cr\mathbf{0}&\rho_{2}(g).\end{array}\right).

We can generalize this definition to a finite number of subrepresentations $W_{1},\ldots,W_{k}$ . If $V$ is the direct sum of subrepresentations that are irreducible, then we will say that $V$ is decomposable. We will see that complex representations of finite groups are always decomposable.

Example 2.18.

Consider the permutation representation of $S_{3}$ . This is the representation on $V=\mathbb{C}^{3}$ for which $\sigma(e_{i})=e_{\sigma(i)}$ where $e_{1},e_{2},e_{3}$ are the standard basis vectors. It is not irreducible!

Let $W_{0}=\left\langle e_{1}+e_{2}+e_{3}\right\rangle$ . Then $\sigma(e_{1}+e_{2}+e_{3})=e_{1}+e_{2}+e_{3}$ for all $\sigma,$ so $W_{0}$ is a subrepresentation (and the action of $G$ is trivial).

Let $W_{1}=\{(x,y,z):x+y+z=0.\}$ . Then since $S_{3}$ permutes the coordinates of a vector, it doesn’t change their sum; so $\sigma(W_{1})\subset W_{1},$ and $W_{1}$ is a subrepresentation (of dimension two). Since $W_{0}\cap W_{1}=\{0\}$ and their dimensions add to $\dim V=3,$ we have $V=W_{0}\oplus W_{1}$ .

I claim that $W_{1}$ is irreducible. Indeed, suppose $U\subset W_{1}$ is a nonzero subrepresentation; we have to show that $U=W_{1}$ . Let $(x,y,z)\in U$ be nonzero. As $x=y=z$ can’t happen, we can apply an element of $G$ to permute the coordinates so that $x\neq y$ . Then applying $(12),$ we have $(y,x,z)\in U$ . Taking the difference, $(x-y,y-x,0)\in U$ ; scaling, $(1,-1,0)\in U$ . Applying $(23),$ we have $(1,0,-1)\in U$ . But these vectors span $W_{1}$ (e.g. because they are linearly independent and $\dim W_{1}=2$ ), so $U=W_{1}$ as required.

Example 2.19.

If $(\rho,V)$ is a finite-dimensional representation of $G=\mathbb{Z},$ with $T=\rho(1),$ then $T$ has an eigenvector $v$ which spans a one-dimensional subrepresentation of $V$ . Thus $V$ is reducible unless $\dim V=1$ . The irreducible subrepresentations of $V$ are exactly the lines spanned by $T$ -eigenvectors, and $V$ is decomposable if and only if it has a basis of $T$ -eigenvectors. This is equivalent to $T$ being diagonalizable. As you know from Algebra II (or Linear Algebra I?) this is not always the case! For example,

T=\begin{pmatrix}1&1\\ 0&1\end{pmatrix}.

The next definition is optional: we will avoid using quotient representations.

Definition 2.20.

Suppose that $(\rho,V)$ is a representation of $G$ and that $W\subset V$ is a subrepresentation. Then the quotient vector space $V/W$ has a representation $\rho_{V/W}$ of $G$ defined by

\rho_{V/W}(g)\overline{v}=\overline{\rho(g)v}.

Remember that $\overline{v}$ is the coset $v+W$ . You should check that this is well-defined, i.e. that if $\overline{v}=\overline{v}_{1}$ then $\overline{\rho(g)v}=\overline{\rho(g)v_{1}}$ .

2.2.2 Homomorphisms and isomorphisms

Definition 2.21.

Suppose that $(\rho,V)$ and $(\sigma,W)$ are representations of $G$ . Then a $G$ -homomorphism (or homomorphism of representations of $G,$ or map of representations of $G,$ or if we are being lazy just a homomorphism) $V\rightarrow W$ is a linear map $\phi:V\rightarrow W$ such that

\phi(\rho(g)v)=\sigma(g)\phi(v)

for all $v\in V,$ $g\in G$ .

In other words, $\phi$ ’commutes’ with the action of $G$ : $g\phi(v)=\phi(gv)$ . We write $\operatorname{Hom}_{G}(V,W)$ for the (vector space) of $G$ -homomorphisms from $V$ to $W$ .

There is another word that is sometimes used for $G$ -homomorphism: intertwiner, or $G$ -intertwiner.

Definition 2.22.

A $G$ -isomorphism (or just an isomorphism) is a bijective $G$ -homomorphism.

If there is a $G$ -isomorphism $V\rightarrow W$ then we write $V\cong W$ .

Lemma 2.23.

Suppose that $V$ and $W$ are representations of $G$ .

1.

If $T\in\operatorname{Hom}_{G}(V,W)$ is an isomorphism, then $T^{-1}\in\operatorname{Hom}_{G}(W,V)$ .
2.

Suppose $\dim V=\dim W=n,$ and choose bases for them, so that $\rho_{V}(g)$ and $\rho_{W}(g)$ are (invertible) $n\times n$ matrices. Then $\rho_{V}\cong\rho_{W}$ if and only if there exists $T\in\operatorname{GL}_{n}(\mathbb{C})$ such that

$T\rho_{V}(g)T^{-1}=\rho_{W}(g)$

for all $g\in G$ .

Proof.

Exercise. ∎

Lemma 2.24.

Given $\phi\in\operatorname{Hom}_{G}(V,W),$ then $\ker(\phi)\subset V$ and $\operatorname{im}(\phi)\subset W$ are subrepresentations.

Proof.

We know that they are subspaces, so we just have to show that they are preserved by the action of $G$ .

For the kernel: suppose that $v\in\ker(\phi)$ and $g\in G$ . Then $\phi(v)=0,$ and

\phi(g\cdot v)=g\cdot\phi(v)=g\cdot 0=0,

so $g\cdot v\in\ker(\phi)$ . So $\ker(\phi)$ is $G$ -stable as required.

For the image: suppose $w\in\operatorname{im}(\phi)$ . Then $w=\phi(v)$ for some $v\in V$ . Then

g\cdot w=g\cdot\phi(v)=\phi(g\cdot v)

is also in the image of $\phi$ . ∎

Example 2.25.

Let $G=D_{3}\cong S_{3}$ where the isomorphism takes $r\mapsto(123)$ and $s\mapsto(23)$ . Let $(\rho,V)$ be the permutation representation of $G,$ so in the basis $e_{1},e_{2},e_{3}$ we have

\rho(r)=\begin{pmatrix}0&0&1\\ 1&0&0\\ 0&1&0\end{pmatrix},\rho(s)=\begin{pmatrix}1&0&0\\ 0&0&1\\ 0&1&0\end{pmatrix}.

Then let $G$ act on the equilateral triangle centred at 0 with vertex at $(0,1),$ giving a two dimensional representation $\sigma$ on $W=\mathbb{C}^{2}$ such that

\sigma(r)=\begin{pmatrix}1/2&-\sqrt{3}/2\\ \sqrt{3}/2&1/2\end{pmatrix},\sigma(s)=\begin{pmatrix}-1&0\\ 0&1\end{pmatrix}.

Let $v_{1},$ $v_{2},$ $v_{3}$ be the vertices labeled anticlockwise from the top and let $T:V\rightarrow W$ be the linear map taking $e_{i}$ to $v_{i}$ . I claim that $T$ is a $G$ -homomorphism.

The matrix of $T$ is

\begin{pmatrix}0&-\sqrt{3}/2&\sqrt{3}/2\\ 1&-1/2&-1/2\end{pmatrix}

and now we just have to check that

\begin{pmatrix}0&-\sqrt{3}/2&\sqrt{3}/2\\ 1&-1/2&-1/2\end{pmatrix}\begin{pmatrix}0&0&1\\ 1&0&0\\ 0&1&0\end{pmatrix}=\begin{pmatrix}-1/2&-\sqrt{3}/2\\ \sqrt{3}/2&-1/2\end{pmatrix}\begin{pmatrix}0&-\sqrt{3}/2&\sqrt{3}/2\\ 1&-1/2&-1/2\end{pmatrix},

which is true, and a similar equation coming from $s$ .

In fact we could see this without calculation; by the way we defined the isomorphism $D_{3}\cong S_{3},$ we have that

\sigma(g)T(e_{i})=\sigma(g)v_{i}=v_{g\cdot i}=T(e_{g\cdot i})=T(\rho(g)e_{i})

for all $g\in S_{3}$ .

The kernel of the homomorphism $T$ is the subspace spanned by $e_{1}+e_{2}+e_{3},$ and in fact $T$ defines an isomorphism from the subrepresentation

\{(a,b,c)\in V:a+b+c=0\}\subset V

to $W$ .

Remark 2.26.

If $V_{1}$ and $V_{2}$ are representations of $G$ then we may form their (external) direct sum with underlying vector space $V_{1}\oplus V_{2}$ such that

g(v_{1},v_{2})=(gv_{1},gv_{2})

for all $g\in G,$ $v_{1}\in V_{1},$ $v_{2}\in V_{2}$ . If $V_{1}$ and $V_{2}$ happen to be subrepresentations of some other representation $V,$ then saying $V$ is the direct sum of $V_{1}$ and $V_{2}$ is the same as saying that the map

	$\displaystyle V_{1}\oplus V_{2}$	$\displaystyle\rightarrow V$
	$\displaystyle(v_{1},v_{2})$	$\displaystyle\mapsto v_{1}+v_{2}$

is an isomorphism.

2.2.3 Change of group

We haven’t yet covered this section; we will come back to it.

Definition 2.27.

If $H$ is a subgroup of $G$ and $(\rho,V)$ is a representation of $G,$ then we get a representation $(\rho|_{H},V),$ the restriction of $\rho$ to $G,$ which is given by

\rho|_{H}(h)=\rho(h)

for $h\in H$ .

This representation is also written $V|_{H}$ , $\operatorname{Res}^{G}_{H}\rho$ , or $\operatorname{Res}^{G}_{H}V$ .

If $\rho$ is an irreducible representation of $G$ , then $\rho|_{H}$ need not be. As an extreme example, if $H=\{e\}$ then $\rho|_{H}$ is just $\dim(V)$ copies of the trivial representation!

Example 2.28.

Let $\rho$ be the irreducible two-dimensional representation of $S_{3},$ realised as the space

V=\{(x,y,z):x+y+z=0\}

of the permutation representation. Let $H=\left\langle 123\right\rangle\subset S_{3}$ . Then $\rho|_{H}$ is reducible (as it must be: all irreducible representations of $H\cong C_{3}$ are one-dimensional). Indeed, let $v=(1,\omega,\omega^{2})$ and $w=(1,\omega^{2},\omega)$ where $\omega=e^{2\pi i/3}$ . Then $\left\langle v\right\rangle$ and $\left\langle w\right\rangle$ are $H$ -subrepresentations of $V$ with $V=\left\langle v\right\rangle\oplus\left\langle w\right\rangle$ .

The action of $H$ on $\left\langle v\right\rangle$ is through the character taking $(123)$ to $\omega^{-1},$ and the action on $\left\langle w\right\rangle$ is through the character $(123)\mapsto\omega$ .

Exercise: what is the restriction of $\rho$ to $\{e,(12)\}$ ?

If $K$ is a normal subgroup of $G$ , then $G/K$ is a group. If $(\rho,V)$ is a representation of $G/K$ then we define the lift (or inflation) $\tilde{\rho}$ of $\rho$ to be the homomorphism $G\rightarrow\operatorname{GL}(V)$ defined by $\tilde{\rho}(g)=\rho(gK)$ :

\tilde{\rho}:G\twoheadrightarrow G/K\xrightarrow{\rho}\operatorname{GL}(V).

Then $(\tilde{\rho},V)$ is a representation of $G$ .

Example 2.29.

Take $G=S_{4}$ and take $K=\{e,(12)(34),(13)(24),(14)(23)\}$ . Then there is an isomorphism

G/K\cong S_{3}

defined as follows: label the three non-identity elements of $K$ as $\{a,b,c\}$ . Then $G$ acts by conjugation on this set of three elements, giving us a homomorphism $G\to S_{3}$ . Explicitly, we have

(12)\mapsto(bc)

and

(123)\mapsto(acb).

Then the homomorphism is surjective with kernel $K$ (check these statements!) and so defines an isomorphism

G/K\cong S_{3}.

If $(\rho,V)$ is any representation of $G$ , then its kernel is

\ker(\rho)=\{g\in G:\rho(g)=I\}.

This is just the kernel of the homomorphism $\rho:G\rightarrow\operatorname{GL}(V)$ , and so it is a normal subgroup. If the kernel of $\rho$ is trivial, then we say that $\rho$ is faithful. In this case, $\rho$ determines an embedding of $G$ inside $\operatorname{GL}_{n}(\mathbb{C})$ !

Lemma 2.30.

If $K\subset G$ is a normal subgroup, and $\rho$ is a representation of $G,$ then the following are equivalent:

1.

$\ker(\rho)$ contains $K$ ;
2.

$\rho$ is isomorphic to the inflation of a representation of $G/K$ .

Proof.

Exercise! ∎

Corollary 2.31.

If $\rho$ is a representation of $G/K$ then $\tilde{\rho}$ is irreducible if and only if $\rho$ is.

Proof.

We prove the equivalent statement that $\tilde{\rho}$ is reducible if and only if $\rho$ is. If $\rho$ is reducible, so has a nonzero proper subrepresentation $W,$ then so is $\tilde{\rho}$ since $W$ is also a subrepresentation of $\tilde{\rho}$ . Conversely, if $\tilde{\rho}$ has a nonzero proper subrepresentation $\sigma,$ then $K\subset\ker(\rho)\subset\ker(\sigma)$ and so, by the lemma, $\sigma=\tilde{\rho}_{1}$ for a representation $\rho_{1}$ of $G/K$ that is then a nonzero proper subrepresentation of $\rho$ . ∎

2.3 Example: dihedral groups

We list the elements of the dihedral group $D_{n}$ as

\{r^{k},sr^{k}:k=0,\ldots,n-1\}.

We aim to show that Table 1 gives the complete list of representations of $D_{n}$ , for $n$ odd. We leave the case of $n$ even as an exercise (there are two more one-dimensional representations in this case).

Table 1: Representations of

D_{n}

Label	Dimension	$\rho(r)$	$\rho(s)$
$\rho_{k}$ , $1\leq k<n/2$	$2$	$\begin{pmatrix}e^{2\pi ik/n}&0\\ 0&e^{-2\pi ik/n}\end{pmatrix}$	$\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$
$\mathbf{1}$	$1$	$1$	$1$
$\epsilon$	$1$	$1$	$-1$

Proof.

Take $(\rho,V)$ to be an irreducible complex representation of $D_{n}$ . Let $v\in V$ be an eigenvector for $\rho(r)$ with eigenvalue $\lambda$ (which must be an $n$ th root of unity since $\rho(r)^{n}=1$ ). Let $w=\rho(s)v$ . The key calculation is:

	$\displaystyle\rho(r)w$	$\displaystyle=\rho(r)\rho(s)v$
		$\displaystyle=\rho(rs)v$
		$\displaystyle=\rho(sr^{-1})v$
		$\displaystyle=\rho(s)\rho(r)^{-1}v$
		$\displaystyle=\rho(s)(\lambda^{-1}v)$
		$\displaystyle=\lambda^{-1}w.$

We also have $\rho(s)w=\rho(s)^{2}v=v$ and so $\left\langle v,w\right\rangle$ is a subrepresentation of $V$ . As $V$ is irreducible, we see that $V=\left\langle v,w\right\rangle$ .

Case 1

Suppose that $\lambda\neq\lambda^{-1}$ . Then $v$ and $w$ are eigenvectors of $\rho(r)$ with distinct eigenvalues, and so are linearly independent. Thus $\dim V=2$ . In the basis $v, w,$ the representation is

	$\displaystyle\rho(r)$	$\displaystyle=\begin{pmatrix}\lambda&0\\ 0&\lambda^{-1}\end{pmatrix}$
	$\displaystyle\rho(s)$	$\displaystyle=\begin{pmatrix}0&1\\ 1&0\end{pmatrix}.$

If $\lambda=e^{2\pi ik/n}$ for $1\leq k<n/2,$ then we get the representations in the first line of the table. Otherwise, $\lambda=e^{-2\pi ik/n}$ for some $1\leq k<n/2$ and we instead take the basis $w, v$ to get $\rho_{k}$ again.

Case 2

Suppose that $\lambda=\lambda^{-1}$ . Then $\lambda=1$ as $n$ is odd. Since

\rho(r)(v+w)=\rho(s)(v+w)=v+w

we see that $v+w$ spans a subrepresentation of $V$ . If $v+w\neq 0,$ then $V=\left\langle v+w\right\rangle$ is the trivial representation. Otherwise, $\rho(s)v=w=-v$ and we get the representation $\epsilon$ .

Strictly speaking, we have only shown that if $V$ is an irreducible representation then it is given by matrices as in the table. However, it is easy to see that the matrices given in each row of the table do in fact define representations of $D_{n}$ : one only has to check that $\rho(r)^{n}=\rho(s)^{2}=1$ and $\rho(r)\rho(s)=\rho(s)\rho(r)^{-1}$ . ∎

2.4 Schur’s Lemma

We move on to more theoretical considerations. Let $G$ be a group.

Theorem 2.32.

Let $V$ and $W$ be two irreducible finite-dimensional complex representations of $G$ . Let $T:V\to W$ be a $G$ -homomorphism. Then

1.

Either $T$ is an isomorphism or $T=0$ .
2.

If $V=W,$ then

$T=\lambda\mathrm{id}_{V}$

for some scalar $\lambda\in\mathbb{C}$ .
3.

More generally,

$\dim\operatorname{Hom}_{G}(V,W)=\begin{cases}1&\text{ if $V\cong W$}\\ 0&\text{ otherwise}.\end{cases}$

Proof.

Part (1) follows directly from Lemma 2.24: suppose that $T$ is nonzero. Then $\ker(T)\neq V$ and $\operatorname{im}(T)\neq\{0\}$ . Since $V$ and $W$ are irreducible, we must have $\ker(T)=0$ and $\operatorname{im}(T)=V$ .

For (2), we can find an eigenvalue $\lambda$ of the endomorphism $T$ — here we are using that $V$ is a finite-dimensional complex vector space. Then $T-\lambda\mathrm{id}_{V}$ is also a $G$ -homomorphism with non-zero kernel. Since $V$ is irreducible, by part (1) we must have $T-\lambda\mathrm{id}_{V}=0$ . Therefore $T=\lambda\mathrm{id}_{V}$ .

For (3), by part (1) we have $\dim\operatorname{Hom}_{G}(V,W)=0$ if they are not isomorphic. So assume that they are isomorphic, so that there is an isomorphism $S:V\rightarrow W$ . If $T:V\to W$ is a $G$ -homomorphism, then $S^{-1}\circ T:V\to V$ is also a $G$ -homomorphism. By (2), $S^{-1}\circ T=\lambda\mathrm{id}_{V}$ for some $\lambda,$ so $T=\lambda S$ . We have shown that $\operatorname{Hom}_{G}(V,W)$ is one-dimensional, spanned by $S$ . ∎

As a corollary we obtain

Theorem 2.33.

Let $G$ be an abelian group. Then every finite-dimensional irreducible complex representation of $G$ is one-dimensional.

Proof.

Let $(\rho,V)$ be an irreducible representation. For $h\in G,$ set $T_{h}=\rho(h)\in\operatorname{GL}(V)$ . Then $T_{h}$ is actually a $G$ -homomorphism $V\rightarrow V$ : this is because it commutes with $\rho(g)$ for all $g\in G$ . Indeed,

T_{h}\rho(g)=\rho(h)\rho(g)=\rho(hg)=\rho(gh)=\rho(g)\rho(h)=\rho(g)T_{h}.

Hence by Schur’s Lemma, $T_{h}=\rho(h)$ acts by a scalar $\chi(h)$ on $V$ :

\rho(h)v=\chi(h)v

for all $v\in V$ and a non-zero scalar $\chi(h)$ .

But now, any non-zero $v\in V$ spans a $G$ -invariant subspace. Since $V$ is irreducible, this implies that $V$ is one-dimensional, and $\chi=\rho$ is a homomorphism $G\rightarrow\mathbb{C}^{\times}$ . ∎

Remark 2.34.

It is possible to give an alternative proof of this using the fact from linear algebra that any commuting set of linear maps from a finite dimensional vector space to itself has a simultaneous eigenvector.

In Schur’s Lemma and Theorem 2.33, we didn’t assume $G$ was finite but we did assume the irreducible representation $V$ was finite-dimensional. If, in fact, $G$ is finite, then $V$ is automatically finite-dimensional.

Proposition 2.35.

Any irreducible representation of a finite group is finite-dimensional.

Proof.

Let $V$ be an irreducible representation of the finite group $G,$ and choose $v\in V$ nonzero. Then the subspace $V^{\prime}$ spanned by $\{g\cdot v:g\in G\}$ is finite dimensional, and is preserved by $G$ (why?). As $V$ is irreducible, we have $V^{\prime}=V,$ so $V$ is finite dimensional. ∎

A homomorphism $\chi:G\rightarrow\mathbb{C}^{\times}$ is often called a character (though this will later cause an unfortunate clash of notation). If $G$ is abelian, then the group

\hat{G}=\{\text{homomorphisms }\chi:G\rightarrow\mathbb{C}^{\times}\}

is called the character group, or dual group, of $G$ . It is a group under the operation $(\chi_{1}\chi_{2})(g)=\chi_{1}(g)\chi_{2}(g)$ .

Example 2.36.

Let $G=C_{n}$ be a cyclic group of order $n$ . Then $\hat{G}\cong C_{n}$ .

Indeed, pick a generator $g$ of $G$ and let $\omega=e^{2\pi i/n}$ be a primitive $n$ th root of unity. Then a character $\chi$ of $G$ is determined uniquely by $\chi(g),$ which must be an $n$ th root of unity $\omega^{a}$ . If we let $\chi_{a}\in\hat{G}$ be the homorphism such that $\chi_{a}(g)=\omega^{a},$ then the map

a\mapsto\chi_{a}

determines a group isomorphism $\mathbb{Z}/n\mathbb{Z}\rightarrow\hat{G}$ . This is a homomorphism because

\chi_{a+b}(g)=\omega^{a+b}=\omega^{a}\omega^{b}=\chi_{a}(g)\chi_{b}(g).

In fact, if $G$ is any finite abelian group, then $\hat{G}\cong G$ . You can prove this using the cyclic case and the fundamental theorem of finite abelian groups.

For arbitrary groups $G$ the same method of proof gives:

Proposition 2.37.

Let $(\rho,V)$ be an irreducible finite-dimensional representation of $G$ and let

Z=Z(G)=\{z\in G:zg=gz\text{ for all $g\in G$}\}

be the center of $G$ . Then $Z$ acts on $V$ as a character: there is a character $\chi:Z\rightarrow\mathbb{C}^{\times}$ such that

\rho(z)v=\chi(z)v.

for all $z\in Z$ and $v\in V$ .

We call $\chi$ the central character of $\rho$ .

Proof.

Homework; mimic the proof that irreducible representations of abelian groups are one-dimensional. ∎

Finally, we can use our classification of the irreducible representations of abelian groups to get a bound on the dimension of the irreducible representations of any finite group.

Proposition 2.38.

Let $G$ be a finite group, let $A$ be an abelian subgroup of $G$ and let $(\rho,V)$ be an irreducible representation of $G$ . Then

\dim V\leq\frac{|G|}{|A|}=[G:A].

Proof.

Restrict the representation to $A$ and find an irreducible $A$ -subrepresentation $W$ of $V$ . By Theorem 2.33, $W$ is one-dimensional, spanned by a vector $v$ . So there is a character $\chi$ of $A$ such that

\rho(h)v=\chi(h)v

for all $h\in A$ . Now $\{\rho(g)v:g\in G\}$ spans a subrepresentation of $V,$ hence is equal to $V$ by irreducibility. Write $g_{1}A,g_{2}A,\ldots,g_{r}A$ for the left cosets of $A,$ where $r=[G:A]$ . Then for $h\in A,$ we have

\rho(g_{i}h)v=\rho(g_{i})\rho(h)v=\rho(g_{i})\chi(h)v=\chi(h)(\rho(g_{i})v)

But this implies that $V=\left\langle\{\rho(g)v:g\in G\}\right\rangle$ is already spanned by

\{\rho(g_{i})v:i=1,\ldots,r\},

so has dimension at most $r=[G:A]$ . ∎

Example 2.39.

The group $D_{n}$ has an abelian subgroup $C_{n}$ of index two, and so every irreducible representation of $D_{n}$ has dimension at most $2$ .

2.5 Maschke’s theorem and complete reducibility

2.5.1 Maschke’s theorem

Theorem 2.40.

Suppose that $G$ is a finite group and that $V$ is a representation of $G$ over a field of characteristic {not dividing $|G|$ }. Suppose that $W$ is a subrepresentation of $V$ . Then there is a subrepresentation $W^{\prime}$ of $V$ such that

V\cong W\oplus W^{\prime}.

Proof.

First we make a useful definition.

Definition 2.41.

If $W\subset V$ are vector spaces, then a linear map $\pi:V\rightarrow W$ is a projection if

\pi(w)=w

for all $w\in W$ .

Exercise 2.42.

If $\pi:V\rightarrow W$ is a projection, then

V=W\oplus\ker(\pi).

We will construct a projection $\pi:V\rightarrow W$ that is a $G$ -homomorphism (we could call this a $G$ -projection). Given such a $\pi,$ the proof is easy: let $W^{\prime}=\ker(\pi)$ . This is a subrepresentation as $\pi$ is a $G$ -homomorphism, and by exercise we have $V=W\oplus W^{\prime}$ as required.

It remains to construct $\pi$ . Let $\pi_{0}:V\rightarrow W$ be any linear map such that $\pi_{0}|_{W}=\mathrm{id}_{W}$ (to construct it, choose a basis for $W$ and extend it to a basis for $V$ . Then define $\pi$ to be the identity on the basis of $W$ and whatever you like on the other basis vectors). This might not be a $G$ -homomorphism, but we turn it into one using an ‘averaging trick’: define

\pi(v)=\frac{1}{|G|}\sum_{g\in G}g^{-1}\pi_{0}(gv).

Then this is a $G$ -homomorphism: for any $h\in G,$

$\displaystyle h^{-1}\pi(hv)$	$\displaystyle=\frac{1}{\|G\|}\sum_{g\in G}h^{-1}g^{-1}\pi_{0}(ghv)$
	$\displaystyle=\frac{1}{\|G\|}\sum_{k\in G}k^{-1}\pi_{0}(kv)$	writing $k=gh$
	$\displaystyle=\pi(v),$

whence $\pi(hv)=h\pi(v)$ .

Finally, if $v\in W$ then $\pi(v)=\frac{1}{|G|}\sum_{g\in G}g^{-1}gv=v,$ so $\pi$ is a projection. ∎

Maschke’s Theorem and Schur’s Lemma both hold in the situation that $k=\mathbb{C}$ and $G$ is finite. From now on, we assume that this is the case, and that all representations are finite-dimensional.

2.5.2 Complete reducibility

The following corollary of Maschke’s theorem says that any (finite-dimensional) representation of a (finite) group can be written as a direct sum of irreducible representations, in an essentially unique way. So irreducible representations are the ’prime numbers’ of representation theory.

Corollary 2.43.

Let $V$ be a representation of $G$ . Then

V\cong W_{1}\oplus W_{2}\oplus\ldots\oplus W_{r}

for some irreducible representations $W_{1},\ldots,W_{r}$ .

Moreover, the number of times each isomorphism class of irreducible representation shows up in the above decomposition is independent of the exact choice of decomposition.

Proof.

The existence of such a decomposition follows from Maschke’s theorem and induction: let $W_{1}$ be an irreducible subrepresentation, write $V=W_{1}\oplus W_{1}^{\prime}$ by Maschke, repeat starting with $W^{\prime}_{1}$ .

If $W$ is an irreducible representation and $V\cong W_{1}\oplus\ldots\oplus W_{r},$ then

\dim\operatorname{Hom}_{G}(W,V)=\sum\dim\operatorname{Hom}_{G}(W,W_{i})=\#\{i:% W\cong W_{i}\}

by Schur’s Lemma (specifically, part (3) of 2.32). This only depends on $V$ and $W,$ not on the choice of decomposition of $V$ . ∎

In the previous proof, we used an easy, but important, property of $\operatorname{Hom}$ :

Lemma 2.44.

If $V,V^{\prime},W,W^{\prime}$ are representations of $G,$ then

	$\displaystyle\operatorname{Hom}_{G}(V,W\oplus W^{\prime})$	$\displaystyle\cong\operatorname{Hom}_{G}(V,W)\oplus\operatorname{Hom}_{G}(V,W^% {\prime})$
	$\displaystyle\operatorname{Hom}_{G}(V\oplus V^{\prime},W)$	$\displaystyle\cong\operatorname{Hom}_{G}(V,W)\oplus\operatorname{Hom}_{G}(V^{% \prime},W).$

We isolate the following part of the proof of 2.43 for later use.

Lemma 2.45.

If $\rho$ is an irreducible representation of $G$ and $\sigma$ is some other representation of $V,$ then the number of times $\rho$ appears in the irreducible decomposition of $\sigma$ is exactly

\dim\operatorname{Hom}_{G}(\rho,\sigma).

Example 2.46.

We work out the projections constructed in the proof of Maschke’s theorem when $V$ is the permutation representation of $S_{3}$ on $\mathbb{C}^{3}$ . Then $V$ has a subspace $V_{0}=\{(x,y,z):x=y=z\}$ . The map

	$\displaystyle V$	$\displaystyle\rightarrow V_{0}$
	$\displaystyle(x,y,z)$	$\displaystyle\mapsto\frac{1}{3}(x+y+z)(e_{1}+e_{2}+e_{3})$

is a $G$ -equivariant projection. As in the proof of Maschke’s theorem, its kernel $V_{1}=\{(x,y,z):x+y+z=0\}$ is a complement:

V=V_{0}\oplus V_{1}.

The representation $V_{1}$ is irreducible (under the isomorphism $S_{3}\cong D_{3},$ it is $\rho_{1}$ ). We can also write down a $G$ -equivariant projection $V\rightarrow V_{1}$ :

(x,y,z)\mapsto\frac{1}{3}(2x-y-z,2y-x-z,2z-x-y).

The kernel of this projection is $V_{0}$ .

Remark 2.47.

The decomposition of Corollary 2.43 is not unique: if $G$ is the trivial group, then $\mathbb{C}^{2}$ can be written as $\mathbb{C}\oplus\mathbb{C}$ in infinitely many ways, simply by choosing any two distinct lines. However, if we have an irreducible representation $(\rho,W)$ of $G,$ then for any representation $V$ the subspace $V(\rho)$ spanned by all the subrepresentations of $V$ isomorphic to $\rho$ is uniquely determined: it is called the $\rho$ -isotypic component of $V$ . The key example to keep in mind is, if $G\cong C_{n}$ is generated by $g,$ and $\chi$ is a character with $\chi(g)=\omega,$ then $V(\omega)$ is just the $\omega$ -eigenspace of $g$ acting on $V$ .

2.5.3 Unitarizability

We didn’t cover this section, but it is recommended reading.

Recall that a (Hermitian) inner product on a complex vector space $V$ is a map $(v,w):V\times V\rightarrow\mathbb{C}$ such that

	$\displaystyle(v+v^{\prime},w)$	$\displaystyle=(v,w)+(v^{\prime},w),$
	$\displaystyle(v,w+w^{\prime})$	$\displaystyle=(v,w)+(v,w^{\prime}),$
	$\displaystyle(\lambda v,\mu w)$	$\displaystyle=\overline{\lambda}\mu(v,w),\qquad\text{and}$
	$\displaystyle(w,v)$	$\displaystyle=\overline{(v,w)}$

for all $v,v^{\prime},w,w^{\prime}\in V$ , $\lambda,\mu\in\mathbb{C}$ , and that is positive definite, meaning that $(v,v)>0$ for all nonzero $v\in V$ . The standard example is

((w_{1},\ldots,w_{n}),(z_{1},\ldots,z_{n}))=\overline{w}_{1}z_{1}+\ldots+% \overline{w}_{n}z_{n}

on $V=\mathbb{C}^{n}$ . A (Hermitian) inner product space is a vector space together with a chosen Hermitian inner product.

Remark 2.48.

If $(,)$ is a Hermitian inner product on $\mathbb{C}^{n},$ and $\mathbf{w}$ and $\mathbf{z}\in\mathbb{C}^{n}$ are written as column vectors, then we can write

(\mathbf{w},\mathbf{z})=\mathbf{w}^{\dagger}H\mathbf{z}

where $\mathbf{w}^{\dagger}$ is the complex conjugate of the transpose of $\mathbf{w}$ . The matrix $H$ will satisfy

H^{\dagger}=H

and be diagonalizable with positive real eigenvalues.

If $(,)$ is a Hermitian inner product on $V$ and $W$ is a subspace, then the orthogonal complement is

W^{\perp}=\{v\in V:(v,w)=0\text{ for all $w\in W$}\}.

As vector spaces, we have $V=W\oplus W^{\perp}$ .

Definition 2.49.

A representation $V$ of $G$ is unitarizable if there is a $G$ -invariant inner product on $V$ ; that is, a Hermitian inner product such that

(gv,gw)=(v,w)

for all $g\in G$ .

If such an inner product is chosen, then the representation $V$ is said to be unitary.

Exercise 2.50.

Suppose that $V$ is a unitary representation of $G$ and that $W$ is a subrepresentation. Then $W^{\perp}$ is a subrepresentation of $V$ .

Theorem 2.51.

If $V$ is a complex representation of $G,$ then $V$ is unitarizable.

Proof.

Start with any Hermitian inner product, $(,),$ not necessarily $G$ -invariant. Then average it over $G$ . ∎

Exercise 2.52.

Fill in the details of the proof of Theorem 2.51, and use it together with exercise 2.50 to give a second proof of Maschke’s theorem.

2.6 The group ring and the regular representation

Definition 2.53.

Let $G$ be a finite group. The group ring $\mathbb{C}[G]$ has as elements formal linear combinations

\sum_{g\in G}a_{g}[g]

with $a_{g}\in\mathbb{C},$ multiplied according to the ‘rule forced by $[g][h]=[gh]$ ’, that is,

\sum_{g\in G}a_{g}[g]\sum_{g\in G}b_{g}[g]=\sum_{g,h\in G}a_{g}b_{h}[gh].

This is a noncommutative ring (if $G$ is not abelian).

To elaborate, the symbols $[g]$ for $g\in G$ are basis vectors for a vector space of dimension $\dim G$ , which we call the group ring.

Example 2.54.

Let $x=[e]-[(12)]$ and $y=2[(23)]+[(123)]$ be elements of $\mathbb{C}[S_{3}]$ . Then

	$\displaystyle xy$	$\displaystyle=2[e][(23)]-2[(12)][(23)]+[e][(123)]-[(12)][(123)]$
		$\displaystyle=2[(23)]-2[(123)]+[(123)]-[(23)]$
		$\displaystyle=[(23)]-[(123)].$

If $(\rho,V)$ is a representation of $G$ , then we can ’multiply’ any element of $V$ by any element of $\mathbb{C}[G]$ :

(\sum_{g\in G}a_{g}[g])v=\sum_{g\in G}a_{g}\rho(g)v.

The group ring is a vector space of dimension $\dim(G)$ . It has a linear left action of $G$ :

g\cdot\sum a_{h}[h]=\sum a_{h}[gh].

Therefore we have a representation of $G$ on $\mathbb{C}[G]$ called the left regular representation.

In fact, the regular representation is just the permutation representation for the action of $G$ on itself by left multiplication. If we adopt the ’functional point of view’ mentioned before, we get another way of thinking about this.

Definition 2.55.

Let

\mathbb{C}^{G}=\{f:G\rightarrow\mathbb{C}\}

be the space of functions from $G\rightarrow\mathbb{C}$ . Define a representation $\rho$ of $G$ on $\mathbb{C}^{G}$ by

(\rho(g)f)(h)=f(g^{-1}h).

Lemma 2.56.

The representation $\mathbb{C}^{G}$ is isomorphic to the regular representation $\mathbb{C}[G]$ .

This is simply the discussion in Remark 2.9 applied to this partiuclar case.

When we refer to the ’regular representation’, then, we mean either $\mathbb{C}[G]$ or $\mathbb{C}^{G}$ according to which is most convenient.

Theorem 2.57.

Let $V$ be any representation of $G$ . Then there is an isomorphism of vector spaces

\operatorname{Hom}_{G}(\mathbb{C}[G],V)\rightarrow V.

Equivalently, $\dim\operatorname{Hom}_{G}(\mathbb{C}[G],V)=\dim V$ .

Proof.

This is short, but difficult to wrap your head around. The idea is to provide a recipe to turn a $G$ -homomorphism $\mathbb{C}[G]\rightarrow V$ into an element of $V,$ and a recipe to turn an element of $V$ into a $G$ -homomorphism $\mathbb{C}[G]\rightarrow V,$ and check that these recipes are inverse to each other.

If $f:\mathbb{C}[G]\rightarrow V$ is a $G$ -homomorphism, let

\phi(f)=f([e])\in V

Conversely, if $v\in V,$ let $\psi(v)\in\operatorname{Hom}_{G}(\mathbb{C}[G],V)$ be given by

\psi(v)(\sum a_{g}[g])=\sum a_{g}[g]v;

you can check that $\psi(v)$ is indeed a $G$ -homomorphism.

Then I claim that the maps $\phi$ and $\psi$ are linear maps between $V$ and $\operatorname{Hom}_{G}(\mathbb{C}[G],V)$ that are two-sided inverses of each other, so these vector spaces are isomorphic. It is clear that they are linear maps. We must check that

\phi(\psi(v))=v

and

\psi(\phi(f))=f

for all $v\in V,$ $f\in\operatorname{Hom}_{G}(\mathbb{C}[G],V)$ .

•

Let $v\in V$ . Then

$\phi(\psi(v))=\psi(v)([e])=[e]v=v$

as required.
•

Let $f\in\operatorname{Hom}_{G}(\mathbb{C}[G],V)$ and let $f^{\prime}=\psi(\phi(f))$ . Then, for $g\in G,$

$f^{\prime}([g])=[g]\phi(f)=[g]f([e])=f([g][e])=f([g])$

since $f$ is a $G$ -homomorphism. As $f,f^{\prime}$ are linear, this implies $f=f^{\prime}$ as required. ∎

This has a beautiful consequence: the sum of the squares of the dimensions of the irreducible representations is equal to the order of the group. We write $\mathrm{Irr}(G)$ for the set of isomorphism classes of irreducible representations of $G$ .

Theorem 2.58.

1.

Every irreducible representation $\rho$ of $G$ is a constituent of the regular representation with multiplicity $\dim\rho$ . In other words,

$V\cong\bigoplus_{\rho\in\mathrm{Irr}(G)}\rho^{\dim\rho}.$
2.

(Sum of squares formula.) We have

$\sum_{\rho\in\mathrm{Irr}(G)}\dim(\rho)^{2}=|G|$

where the sum runs over the isomorphism classes of irreducible representations of $G$ . In particular, $\mathrm{Irr}(G)$ is finite.

Proof.

1.

By Maschke’s theorem, we can decompose $\mathbb{C}[G]$ as a direct sum of irreducibles in which each isomorphism class $\rho$ of irreducible representations appears $\dim\operatorname{Hom}_{G}(\mathbb{C}[G],\rho)=\dim(\rho)$ times.
2.

Immediate from equating dimensions on both sides of the first part and noting that $\dim\mathbb{C}[G]=|G|$ . ∎

Exercise 2.59.

Verify the sum of squares formula for dihedral groups.

Note that this gives another approach to the classification of representations of dihedral groups. First, write down all the irreducible representations and check that they are non-isomorphic. Then, by the sum of squares formula, you have found everything!