# Representation theory of finite groups

(Redirected from Representation of a finite group)

The representation theory of groups is a part of mathematics which examines how groups act on given structures.

Here the focus is in particular on operations of groups on vector spaces. Nevertheless, groups acting on other groups or on sets are also considered. For more details, please refer to the section on permutation representations.

Please note, that except for a few marked exceptions only finite groups will be considered in this article. We will also restrain to vector spaces over fields of characteristic zero. Because the theory of algebraically closed fields of characteristic zero is complete, a theory valid for a special algebraically closed field of characteristic zero is also valid for every other algebraically closed field of characteristic zero. Thus, without loss of generality, we can study vector spaces over ${\displaystyle \mathbb {C} .}$

Representation theory is used in many parts of mathematics, as well as in quantum chemistry and physics. Among other things it is used in algebra to examine the structure of groups. There are also applications in harmonic analysis and number theory. For example, representation theory is used in the modern approach to gain new results about automorphic forms.

## Definition

### Linear representations

Let ${\displaystyle V}$ be a ${\displaystyle K}$–vector space and ${\displaystyle G}$ a finite group. A linear representation of a finite group ${\displaystyle G}$ is a group homomorphism ${\displaystyle \rho :G\to {\text{GL}}(V)={\text{Aut}}(V).}$ That means, a linear representation is a map ${\displaystyle \rho :G\to {\text{GL}}(V)}$ which satisfies ${\displaystyle \rho (st)=\rho (s)\rho (t)}$ for all ${\displaystyle s,t\in G.}$ The vector space ${\displaystyle V}$ is called representation space of ${\displaystyle G.}$ Often the term representation of ${\displaystyle G}$ is also used for the representation space ${\displaystyle V.}$

The representation of a group in a module instead of a vector space is also called a linear representation.

We write ${\displaystyle (\rho ,V_{\rho })}$ for the representation ${\displaystyle \rho :G\to {\text{GL}}(V_{\rho })}$ of ${\displaystyle G.}$ Sometimes we only use ${\displaystyle (\rho ,V),}$ if it is clear to which representation the space ${\displaystyle V}$ belongs.

In this article we will restrain ourselves to the study of finite-dimensional representation spaces, except for the last chapter. As in most cases only a finite number of vectors in ${\displaystyle V}$ is of interest, it is sufficient to study the subrepresentation generated by these vectors. The representation space of this subrepresentation is then finite-dimensional.

The degree of a representation is the dimension of its representation space ${\displaystyle V.}$ The notation ${\displaystyle \dim(\rho )}$ is sometimes used to denote the degree of a representation ${\displaystyle \rho .}$

### Examples

The trivial representation is given by ${\displaystyle \rho (s)={\text{Id}}}$ for all ${\displaystyle s\in G.}$

A representation of degree ${\displaystyle 1}$ of a group ${\displaystyle G}$ is a homomorphism into the multiplicative group ${\displaystyle \rho :G\to {\text{GL}}_{1}(\mathbb {C} )=\mathbb {C} ^{\times }=\mathbb {C} \setminus \{0\}.}$ As every element of ${\displaystyle G}$ is of finite order, the values of ${\displaystyle \rho (s)}$ are roots of unity. For example let ${\displaystyle \rho :G=\mathbb {Z} /4\mathbb {Z} \to \mathbb {C} ^{\times }}$ be a nontrivial linear representation. Since ${\displaystyle \rho }$ is a group homomorphism, it has to satisfy ${\displaystyle \rho ({0})=1.}$ Because ${\displaystyle 1}$ generates ${\displaystyle G,\rho }$ is determined by its value on ${\displaystyle \rho (1).}$ And as ${\displaystyle \rho }$ is nontrivial, ${\displaystyle \rho ({1})\in \{i,-1,-i\}.}$ By this we achieve the result, that the image of ${\displaystyle G}$ under ${\displaystyle \rho }$ has to be a nontrivial subgroup of the group which consists of the fourth roots of unity. This means, that ${\displaystyle \rho }$ has to be one of the following three maps:

${\displaystyle {\begin{cases}\rho _{1}({0})=1\\\rho _{1}({1})=i\\\rho _{1}({2})=-1\\\rho _{1}({3})=-i\end{cases}}\qquad {\begin{cases}\rho _{2}({0})=1\\\rho _{2}({1})=-1\\\rho _{2}({2})=1\\\rho _{2}({3})=-1\end{cases}}\qquad {\begin{cases}\rho _{3}({0})=1\\\rho _{3}({1})=-i\\\rho _{3}({2})=-1\\\rho _{3}({3})=i\end{cases}}}$

Let ${\displaystyle G=\mathbb {Z} /2\mathbb {Z} \times \mathbb {Z} /2\mathbb {Z} }$ and let ${\displaystyle \rho :G\to {\text{GL}}_{2}(\mathbb {C} )}$ be the group homomorphism defined by:

${\displaystyle \rho ({0},{0})={\begin{pmatrix}1&0\\0&1\end{pmatrix}},\quad \rho ({1},{0})={\begin{pmatrix}-1&0\\0&-1\end{pmatrix}},\quad \rho ({0},{1})={\begin{pmatrix}0&1\\1&0\end{pmatrix}},\quad \rho ({1},{1})={\begin{pmatrix}0&-1\\-1&0\end{pmatrix}}.}$

In this case ${\displaystyle \rho }$ is a linear representation of ${\displaystyle G}$ of degree ${\displaystyle 2.}$

#### Permutation representation

Further information: Permutation group

Let ${\displaystyle X}$ be a finite set. Let ${\displaystyle G}$ be a group operating on ${\displaystyle X.}$ The group ${\displaystyle {\text{Aut}}(X)}$ is the group of all permutations on ${\displaystyle X}$ with the composition as operation.

A group acting on a finite set is sometimes considered sufficient for the definition of the permutation representation. However, since we want to construct examples for linear representations, where groups act on vector spaces instead of on arbitrary finite sets, we have to proceed in a different way. In order to construct the permutation representation, we need a vector space ${\displaystyle V}$ with ${\displaystyle \dim(V)=|X|.}$ A basis of ${\displaystyle V}$ can be indexed by the elements of ${\displaystyle X.}$ The permutation representation is the group homomorphism ${\displaystyle \rho :G\to {\text{GL}}(V)}$ given by ${\displaystyle \rho (s)e_{x}=e_{s.x}}$ for all ${\displaystyle s\in G,x\in X.}$ All linear maps ${\displaystyle \rho (s)}$ are uniquely defined by this property.

Example. Let ${\displaystyle X=\{1,2,3\}}$ and ${\displaystyle G={\text{Per}}(3).}$ Then ${\displaystyle G}$ operates on ${\displaystyle X}$ via ${\displaystyle {\text{Aut}}(X)=G.}$ The associated linear representation is ${\displaystyle \rho :G\to {\text{GL}}(V)\cong {\text{GL}}_{3}(\mathbb {C} )}$ with ${\displaystyle \rho (\sigma )e_{x}=e_{\sigma (x)}}$ for ${\displaystyle \sigma \in G,x\in X.}$

#### Left- and right-regular representation

Let ${\displaystyle G}$ be a group and ${\displaystyle V}$ be a vector space of dimension ${\displaystyle |G|}$ with a basis ${\displaystyle (e_{t})_{t\in G}}$ indexed by the elements of ${\displaystyle G.}$ The left-regular representation is a special case of the permutation representation by choosing ${\displaystyle X=G.}$ This means ${\displaystyle \rho (s)e_{t}=e_{st}}$ for all ${\displaystyle s,t\in G.}$ Thus, the family ${\displaystyle (\rho (s)e_{1})_{s\in G}}$ of images of ${\displaystyle e_{1}}$ are a basis of ${\displaystyle V.}$ The degree of the left-regular representation is equal to the order of the group.

The right-regular representation is defined on the same vector space with a similar homomorphism: ${\displaystyle \rho (s)e_{t}=e_{ts^{-1}}.}$ In the same way as before ${\displaystyle (\rho (s)e_{1})_{s\in G}}$ is a basis of ${\displaystyle V.}$ Just as in the case of the left-regular representation, the degree of the right-regular representation is equal to the order of ${\displaystyle G.}$

Both representations are isomorphic via ${\displaystyle e_{s}\mapsto e_{s^{-1}}.}$ For this reason they are not always set apart, and often referred to as the regular representation.

A closer look provides the following result: A given linear representation ${\displaystyle \rho :G\to {\text{GL}}(W)}$ is isomorphic to the left-regular representation, if and only if there exists a ${\displaystyle w\in W,}$ such that ${\displaystyle (\rho (s)w)_{s\in G}}$ is a basis of ${\displaystyle W.}$

Example. Let ${\displaystyle G=\mathbb {Z} /5\mathbb {Z} }$ and ${\displaystyle V=\mathbb {R} ^{5}}$ with the basis ${\displaystyle \{e_{0},\ldots ,e_{4}\}.}$ Then the left-regular representation ${\displaystyle L_{\rho }:G\to {\text{GL}}(V)}$ is defined by ${\displaystyle L_{\rho }(k)e_{l}=e_{l+k}}$ for ${\displaystyle k,l\in \mathbb {Z} /5\mathbb {Z} .}$ The right-regular representation is defined analogously by ${\displaystyle R_{\rho }(k)e_{l}=e_{l-k}}$ for ${\displaystyle k,l\in \mathbb {Z} /5\mathbb {Z} .}$

### Representations, modules and the convolution algebra

Let ${\displaystyle G}$ be a finite group, let ${\displaystyle K}$ be a commutative ring and let ${\displaystyle K[G]}$ be the group algebra of ${\displaystyle G}$ over ${\displaystyle K.}$ This algebra is free and a basis can be indexed by the elements of ${\displaystyle G.}$ Most often the basis is identified with ${\displaystyle G.}$ Every element ${\displaystyle f\in K[G]}$ can then be uniquely expressed as

${\displaystyle f=\sum _{s\in G}a_{s}s}$ with ${\displaystyle a_{s}\in K.}$

The multiplication in ${\displaystyle K[G]}$ extends that in ${\displaystyle G}$ distributively.

Now let ${\displaystyle V}$ be a ${\displaystyle K}$module and let ${\displaystyle \rho :G\to {\text{GL}}(V)}$ be a linear representation of ${\displaystyle G}$ in ${\displaystyle V.}$ We define ${\displaystyle sv=\rho (s)v}$ for all ${\displaystyle s\in G}$ and ${\displaystyle v\in V.}$ By linear extension ${\displaystyle V}$ is endowed with the structure of a left${\displaystyle -K[G]}$–module. Vice versa we obtain a linear representation of ${\displaystyle G}$ starting from a ${\displaystyle K[G]}$–module ${\displaystyle V.}$ Therefore, these terms may be used interchangeably.

Suppose ${\displaystyle K=\mathbb {C} .}$ In this case the left ${\displaystyle \mathbb {C} [G]}$–module given by ${\displaystyle \mathbb {C} [G]}$ itself corresponds to the left-regular representation. In the same way ${\displaystyle \mathbb {C} [G]}$ as a right ${\displaystyle \mathbb {C} [G]}$–module corresponds to the right-regular representation.

In the following we will define the convolution algebra: Let ${\displaystyle G}$ be a group, the set ${\displaystyle L^{1}(G):=\{f:G\to \mathbb {C} \}}$ is a ${\displaystyle \mathbb {C} }$–vector space with the operations addition and scalar multiplication then this vector space is isomorphic to ${\displaystyle \mathbb {C} ^{|G|}.}$ The convolution of two elements ${\displaystyle f,h\in L^{1}(G)}$ defined by

${\displaystyle f*h(s):=\sum _{t\in G}f(t)h(t^{-1}s)}$

makes ${\displaystyle L^{1}(G)}$ an algebra. The algebra ${\displaystyle L^{1}(G)}$ is called the convolution algebra.

The convolution algebra is free and has a basis indexed by the group elements: ${\displaystyle (\delta _{s})_{s\in G},}$ where

${\displaystyle \delta _{s}(t)={\begin{cases}1&t=s\\0&{\text{otherwise}}\end{cases}}.}$

Using the properties of the convolution we obtain: ${\displaystyle \delta _{s}*\delta _{t}=\delta _{st}.}$

We define a map between ${\displaystyle L^{1}(G)}$ and ${\displaystyle \mathbb {C} [G],}$ by defining ${\displaystyle \delta _{s}\mapsto e_{s}}$ on the basis ${\displaystyle (\delta _{s})_{s\in G}}$ and extending it linearly. Obviously the prior map is bijective. A closer inspection of the convolution of two basis elements as shown in the equation above reveals that the multiplication in ${\displaystyle L^{1}(G)}$ corresponds to that in ${\displaystyle \mathbb {C} [G].}$ Thus, the convolution algebra and the group algebra are isomorphic as algebras.

The involution

${\displaystyle f^{*}(s)={f(s^{-1})}}$

turns ${\displaystyle L^{1}(G)}$ into a ${\displaystyle ^{*}}$–algebra. We have ${\displaystyle \delta _{s}^{*}=\delta _{s^{-1}}.}$

A representation ${\displaystyle (\pi ,V_{\pi })}$ of a group ${\displaystyle G}$ extends to a ${\displaystyle ^{*}}$–algebra homomorphism ${\displaystyle \pi :L^{1}(G)\to {\text{End}}(V_{\pi })}$ by ${\displaystyle \pi (\delta _{s})=\pi (s).}$ Since multiplicity is a characteristic property of algebra homomorphisms, ${\displaystyle \pi }$ satisfies ${\displaystyle \pi (f*h)=\pi (f)\pi (h).}$ If ${\displaystyle \pi }$ is unitary, we also obtain ${\displaystyle \pi (f)^{*}=\pi (f^{*}).}$ For the definition of a unitary representation, please refer to the chapter on properties. In that chapter we will see, that, without loss of generality, every linear representation can be assumed to be unitary.

Using the convolution algebra we can implement a Fourier transformation on a group ${\displaystyle G.}$ In the area of harmonic analysis it is shown that the following definition is consistent with the definition of the Fourier transformation on ${\displaystyle \mathbb {R} .}$

Let ${\displaystyle \rho :G\to {\text{GL}}(V_{\rho })}$ be a representation and let ${\displaystyle f\in L^{1}(G)}$ be a ${\displaystyle \mathbb {C} }$-valued function on ${\displaystyle G}$. The Fourier transform ${\displaystyle {\hat {f}}(\rho )\in {\text{End}}(V_{\rho })}$ of ${\displaystyle f}$ is defined as

${\displaystyle {\hat {f}}(\rho )=\sum _{s\in G}f(s)\rho (s).}$

It holds that ${\displaystyle {\widehat {f*g}}(\rho )={\hat {f}}(\rho )\cdot {\hat {g}}(\rho ).}$

### Maps between representations

A map between two representations ${\displaystyle (\rho ,V_{\rho }),\,(\tau ,V_{\tau })}$ of the same group ${\displaystyle G}$ is a linear map ${\displaystyle T:V_{\rho }\to V_{\tau },}$ with the property that ${\displaystyle \tau (s)\circ T=T\circ \rho (s)}$ holds for all ${\displaystyle s\in G.}$

Such a map is also called ${\displaystyle G}$–linear. The kernel, the image and the cokernel of ${\displaystyle T}$ are defined by default. They are again ${\displaystyle G}$–modules. Thus, they provide representations of ${\displaystyle G}$ due to the correlation described in the previous section.

## Properties

Two representations ${\displaystyle (\rho ,V_{\rho }),(\pi ,V_{\pi })}$ are called equivalent or isomorphic, if there exists a ${\displaystyle G}$–linear vector space isomorphism between the representation spaces. In other words, they are isomorphic if there exists a bijective linear map ${\displaystyle T:V_{\rho }\to V_{\pi },}$ such that ${\displaystyle T\circ \rho (s)=\pi (s)\circ T}$ for all ${\displaystyle s\in G.}$ In particular, equivalent representations have the same degree.

A representation ${\displaystyle (\pi ,V_{\pi })}$ is called faithful, if ${\displaystyle \pi }$ is injective. In this case ${\displaystyle \pi }$ induces an isomorphism between ${\displaystyle G}$ and the image ${\displaystyle \pi (G).}$ As the latter is a subgroup of ${\displaystyle {\text{GL}}(V_{\pi }),}$ we can regard ${\displaystyle G}$ via ${\displaystyle \pi }$ as subgroup of ${\displaystyle {\text{Aut}}(V_{\pi }).}$

Let ${\displaystyle \rho :G\to {\text{GL}}(V)}$ be a linear representation of ${\displaystyle G.}$ Let ${\displaystyle W}$ be a ${\displaystyle G}$–invariant subspace of ${\displaystyle V,}$ i.e. ${\displaystyle \rho (s)w\in W}$ for all ${\displaystyle s\in G,w\in W.}$ The restriction ${\displaystyle \rho (s)|_{W}}$ is an isomorphism of ${\displaystyle W}$ onto itself. Because ${\displaystyle \rho (s)|_{W}\circ \rho (t)|_{W}=\rho (st)|_{W}}$ holds for all ${\displaystyle s,t\in G,}$ this construction is a representation of ${\displaystyle G}$ in ${\displaystyle W.}$ It is called subrepresentation of ${\displaystyle V.}$

We can restrict the range as well as the domain:

Let ${\displaystyle H}$ be a subgroup of ${\displaystyle G.}$ Let ${\displaystyle \rho }$ be a linear representation of ${\displaystyle G.}$ We denote by ${\displaystyle {\text{Res}}_{H}(\rho )}$ the restriction of ${\displaystyle \rho }$ to the subgroup ${\displaystyle H.}$

If there is no danger of confusion, we might use only ${\displaystyle {\text{Res}}(\rho )}$ or in short ${\displaystyle {\text{Res}}\rho .}$

The notation ${\displaystyle {\text{Res}}_{H}(V)}$ or in short${\displaystyle {\text{Res}}(V)}$ is also used to denote the restriction of the representation ${\displaystyle V}$ of ${\displaystyle G}$ onto ${\displaystyle H.}$

Let ${\displaystyle f}$ be a function on ${\displaystyle G.}$ We write ${\displaystyle {\text{Res}}_{H}(f)}$ or shortly ${\displaystyle {\text{Res}}(f)}$ for the restriction to the subgroup ${\displaystyle H.}$

A representation ${\displaystyle \rho :G\to {\text{GL}}(V)}$ is called irreducible or simple, if there are no nontrivial ${\displaystyle G}$–invariant vector subspaces of ${\displaystyle V.}$ Here, as well as in the following text, we include the whole vector space as well as the zero-vector space in our definition of trivial vector subspaces. In terms of the group algebra the irreducible representations correspond to the simple ${\displaystyle \mathbb {C} [G]}$–modules.

It can be proved, that the number of irreducible representations of a group ${\displaystyle G}$ (or correspondingly the number of simple ${\displaystyle \mathbb {C} [G]}$–modules) equals the number of conjugacy classes of ${\displaystyle G.}$

A representation is called semisimple or completely reducible, if it can be written as a direct sum of irreducible representations. This is analogue to the definition of the semisimple algebra.

For the definition of the direct sum of representations please refer to the section on direct sums of representations.

A representation is called isotypic, if it is a direct sum of isomorphic, irreducible representations.

Let ${\displaystyle (\rho ,V_{\rho })}$ be a given representation of a group ${\displaystyle G.}$ Let ${\displaystyle \tau }$ be an irreducible representation of ${\displaystyle G.}$ The ${\displaystyle \tau }$isotype ${\displaystyle V_{\rho }(\tau )}$ of ${\displaystyle G}$ is defined as the sum of all irreducible subrepresentations of ${\displaystyle V}$ isomorphic to ${\displaystyle \tau .}$

Every vector space over ${\displaystyle \mathbb {C} }$ can be provided with an inner product. A representation ${\displaystyle \rho }$ of a group ${\displaystyle G}$ in a vector space endowed with an inner product is called unitary, if ${\displaystyle \rho (s)}$ is unitary for every ${\displaystyle s\in G.}$ This means that in particular every ${\displaystyle \rho (s)}$ is diagonalizable. For more details see the article on unitary representations.

A representation is unitary with respect to a given inner product if and only if the inner product is invariant with regard to the induced operation of ${\displaystyle G,}$ that means, if ${\displaystyle (v|u)=(\rho (s)v|\rho (s)u)}$ holds for all ${\displaystyle v,u\in V_{\rho },s\in G.}$

A given inner product ${\displaystyle (\cdot |\cdot )}$ can be replaced by an invariant inner product by exchanging ${\displaystyle (v|u)}$ with

${\displaystyle \sum _{t\in G}(\rho (t)v|\rho (t)u).}$

Thus, without loss of generality, we can assume that every further considered representation is unitary.

Example. Let ${\displaystyle G=D_{6}=\{{\text{id}},\mu ,\mu ^{2},\nu ,\mu \nu ,\mu ^{2}\nu \}}$ be the dihedral group of order ${\displaystyle 6}$ generated by ${\displaystyle \mu ,\nu }$ which fulfil the properties ${\displaystyle {\text{ord}}(\nu )=2,{\text{ord}}(\mu )=3}$ and ${\displaystyle \nu \mu \nu =\mu ^{2}.}$ Let ${\displaystyle \rho :D_{6}\to {\text{GL}}_{3}(\mathbb {C} )}$ be a linear representation of ${\displaystyle D_{6}}$ defined on the generators by:

${\displaystyle \rho (\mu )=\left({\begin{array}{ccc}\cos({\frac {2\pi }{3}})&0&-\sin({\frac {2\pi }{3}})\\0&1&0\\\sin({\frac {2\pi }{3}})&0&\cos({\frac {2\pi }{3}})\end{array}}\right),\,\,\,\,\rho (\nu )=\left({\begin{array}{ccc}-1&0&0\\0&-1&0\\0&0&1\end{array}}\right).}$

This representation is faithful. The subspace ${\displaystyle \mathbb {C} e_{2}}$ is a ${\displaystyle D_{6}}$–invariant subspace. Thus, there exists a nontrivial subrepresentation ${\displaystyle \rho |_{\mathbb {C} e_{2}}:D_{6}\to \mathbb {C} ^{\times }}$ with ${\displaystyle \nu \mapsto -1,\mu \mapsto 1.}$ Therefore, the representation is not irreducible. The mentioned subrepresentation is of degree one and irreducible. The complementary subspace of ${\displaystyle \mathbb {C} e_{2}}$ is ${\displaystyle D_{6}}$–invariant as well. Therefore, we obtain the subrepresentation ${\displaystyle \rho |_{\mathbb {C} e_{1}\oplus \mathbb {C} e_{3}}}$ with

${\displaystyle \nu \mapsto {\begin{pmatrix}-1&0\\0&1\end{pmatrix}},\,\,\,\,\mu \mapsto {\begin{pmatrix}\cos({\frac {2\pi }{3}})&-\sin({\frac {2\pi }{3}})\\\sin({\frac {2\pi }{3}})&\cos({\frac {2\pi }{3}})\end{pmatrix}}.}$

This subrepresentation is also irreducible. That means, the original representation is completely reducible:

${\displaystyle \rho =\rho |_{\mathbb {C} e_{2}}\oplus \rho |_{\mathbb {C} e_{1}\oplus \mathbb {C} e_{3}}.}$

Both subrepresentations are isotypic and are the two only non-zero isotypes of ${\displaystyle \rho .}$

The representation ${\displaystyle \rho }$ is unitary with regard to the standard inner product on ${\displaystyle \mathbb {C} ^{3},}$ because ${\displaystyle \rho (\mu )}$ and ${\displaystyle \rho (\nu )}$ are unitary.

Let ${\displaystyle T:\mathbb {C} ^{3}\to \mathbb {C} ^{3}}$ be any vector space isomorphism. Then ${\displaystyle \eta :D_{6}\to {\text{GL}}_{3}(\mathbb {C} ),}$ which is defined by the equation ${\displaystyle \eta (s):=T\circ \rho (s)\circ T^{-1}}$ for all ${\displaystyle s\in D_{6},}$ is a representation isomorphic to ${\displaystyle \rho .}$

By restricting the domain of the representation to a subgroup, e.g. ${\displaystyle H=\{{\text{id}},\mu ,\mu ^{2}\},}$ we obtain the representation ${\displaystyle {\text{Res}}_{H}(\rho ).}$ This representation is defined by the image ${\displaystyle \rho (\mu ),}$ whose explicit form is shown above.

## Constructions

### The dual representation

Let ${\displaystyle \rho :G\to {\text{GL}}(V)}$ be a given representation. The dual representation or contragredient representation ${\displaystyle \rho ^{*}:G\to {\text{GL}}(V^{*})}$ is a representation of ${\displaystyle G}$ in the dual vector space of ${\displaystyle V.}$ It is defined by the property

${\displaystyle \forall s\in G,v\in V,\alpha \in V^{*}:\qquad \left(\rho ^{*}(s)\alpha \right)(v)=\alpha \left(\rho \left(s^{-1}\right)v\right).}$

With regard to the natural pairing ${\displaystyle \langle \alpha ,v\rangle :=\alpha (v)}$ between ${\displaystyle V^{*}}$ and ${\displaystyle V}$ the definition above provides the equation:

${\displaystyle \forall s\in G,v\in V,\alpha \in V^{*}:\qquad \langle \rho ^{*}(s)(\alpha ),\rho (s)(v)\rangle =\langle \alpha ,v\rangle .}$

Example. Define: ${\displaystyle \rho :\mathbb {Z} /3\mathbb {Z} \to {\text{GL}}_{2}(\mathbb {C} )}$ by:

${\displaystyle \rho (0)={\text{Id}},\quad \rho (1)={\begin{pmatrix}\cos({\tfrac {2\pi }{3}})&-\sin({\tfrac {2\pi }{3}})\\\sin({\tfrac {2\pi }{3}})&\cos({\tfrac {2\pi }{3}})\end{pmatrix}},\quad \rho (2)={\begin{pmatrix}\cos({\frac {4\pi }{3}})&-\sin({\frac {4\pi }{3}})\\\sin({\frac {4\pi }{3}})&\cos({\frac {4\pi }{3}})\end{pmatrix}}.}$

The dual representation ${\displaystyle \rho ^{*}:\mathbb {Z} /3\mathbb {Z} \to {\text{GL}}\left((\mathbb {C} ^{2})^{*}\right)}$ is then given by:

${\displaystyle \rho ^{*}(0)={\text{Id}},\quad \rho ^{*}(1)={\begin{pmatrix}\cos({\frac {4\pi }{3}})&\sin({\frac {4\pi }{3}})\\-\sin({\frac {4\pi }{3}})&\cos({\frac {4\pi }{3}})\end{pmatrix}},\quad \rho ^{*}(2)={\begin{pmatrix}\cos({\frac {2\pi }{3}})&\sin({\frac {2\pi }{3}})\\-\sin({\frac {2\pi }{3}})&\cos({\frac {2\pi }{3}})\end{pmatrix}}.}$

### Direct sum of representations

Let ${\displaystyle (\rho _{1},V_{1})}$ and ${\displaystyle (\rho _{2},V_{2})}$ be a representation of ${\displaystyle G_{1}}$ and ${\displaystyle G_{2},}$ respectively. The direct sum of these representations a linear representation and is defined as

${\displaystyle \forall s_{1}\in G_{1},s_{2}\in G_{2},v_{1}\in V_{1},v_{2}\in V_{2}:\qquad {\begin{cases}\rho _{1}\oplus \rho _{2}:G_{1}\times G_{2}\to {\text{GL}}(V_{1}\oplus V_{2})\\[4pt](\rho _{1}\oplus \rho _{2})(s_{1},s_{2})(v_{1},v_{2}):=\rho _{1}(s_{1})v_{1}\oplus \rho _{2}(s_{2})v_{2}\end{cases}}}$

Let ${\displaystyle \rho _{1},\rho _{2}}$ be representations of the same group ${\displaystyle G.}$ For the sake of simplicity, the direct sum of these representations is defined as a representation of ${\displaystyle G,}$ i.e. it is given as ${\displaystyle \rho _{1}\oplus \rho _{2}:G\to {\text{GL}}(V_{1}\oplus V_{2}),}$ by viewing ${\displaystyle G}$ as the diagonal subgroup of ${\displaystyle G\times G.}$

Example. Let (here ${\displaystyle i}$ and ${\displaystyle \omega }$ are the imaginary unit and the primitive cube root of unity respectively):

${\displaystyle {\begin{cases}\rho _{1}:\mathbb {Z} /2\mathbb {Z} \to {\text{GL}}_{2}(\mathbb {C} )\\[4pt]\rho _{1}(1)={\begin{pmatrix}0&-i\\i&0\end{pmatrix}}\end{cases}}\qquad \qquad {\begin{cases}\rho _{2}:\mathbb {Z} /3\mathbb {Z} \to {\text{GL}}_{3}(\mathbb {C} )\\[6pt]\rho _{2}(1)={\begin{pmatrix}1&0&\omega \\0&\omega &0\\0&0&\omega ^{2}\end{pmatrix}}\end{cases}}}$

Then

${\displaystyle {\begin{cases}\rho _{1}\oplus \rho _{2}:\mathbb {Z} /2\mathbb {Z} \times \mathbb {Z} /3\mathbb {Z} \to {\text{GL}}\left(\mathbb {C} ^{2}\oplus \mathbb {C} ^{3}\right)\\[6pt]\left(\rho _{1}\oplus \rho _{2}\right)(k,l)={\begin{pmatrix}\rho _{1}(k)&0\\0&\rho _{2}(l)\end{pmatrix}}&k\in \mathbb {Z} /2\mathbb {Z} ,l\in \mathbb {Z} /3\mathbb {Z} \end{cases}}}$

As it is sufficient to consider the image of the generating element, we find, that

${\displaystyle (\rho _{1}\oplus \rho _{2})(1,1)={\begin{pmatrix}0&-i&0&0&0\\i&0&0&0&0\\0&0&1&0&\omega \\0&0&0&\omega &0\\0&0&0&0&\omega ^{2}\end{pmatrix}}}$

### Tensor product of representations

Let ${\displaystyle \rho _{1}:G_{1}\to {\text{GL}}(V_{1}),\rho _{2}:G_{2}\to {\text{GL}}(V_{2})}$ be linear representations. We define the linear representation ${\displaystyle \rho _{1}\otimes \rho _{2}:G_{1}\times G_{2}\to {\text{GL}}(V_{1}\otimes V_{2})}$ into the tensor product of ${\displaystyle V_{1}}$ and ${\displaystyle V_{2}}$ by ${\displaystyle \rho _{1}\otimes \rho _{2}(s_{1},s_{2})=\rho _{1}(s_{1})\otimes \rho _{2}(s_{2}),}$ in which ${\displaystyle s_{1}\in G_{1},s_{2}\in G_{2}.}$ This representation is called outer tensor product of the representations ${\displaystyle \rho _{1}}$ and ${\displaystyle \rho _{2}.}$ The existence and uniqueness is a consequence of the properties of the tensor product.

Example. We reexamine the example provided for the direct sum:

${\displaystyle {\begin{cases}\rho _{1}:\mathbb {Z} /2\mathbb {Z} \to {\text{GL}}_{2}(\mathbb {C} )\\[4pt]\rho _{1}(1)={\begin{pmatrix}0&-i\\i&0\end{pmatrix}}\end{cases}}\qquad \qquad {\begin{cases}\rho _{2}:\mathbb {Z} /3\mathbb {Z} \to {\text{GL}}_{3}(\mathbb {C} )\\[6pt]\rho _{2}(1)={\begin{pmatrix}1&0&\omega \\0&\omega &0\\0&0&\omega ^{2}\end{pmatrix}}\end{cases}}}$

The outer tensor product

${\displaystyle {\begin{cases}\rho _{1}\otimes \rho _{2}:\mathbb {Z} /2\mathbb {Z} \times \mathbb {Z} /3\mathbb {Z} \to {\text{GL}}(\mathbb {C} ^{2}\otimes \mathbb {C} ^{3})\\(\rho _{1}\otimes \rho _{2})(k,l)=\rho _{1}(k)\otimes \rho _{2}(l)&k\in \mathbb {Z} /2\mathbb {Z} ,l\in \mathbb {Z} /3\mathbb {Z} \end{cases}}}$

Using the standard basis of ${\displaystyle \mathbb {C} ^{2}\otimes \mathbb {C} ^{3}\cong \mathbb {C} ^{6}}$ we have the following for the generating element we have:

${\displaystyle \rho _{1}\otimes \rho _{2}(1,1)=\rho _{1}(1)\otimes \rho _{2}(1)={\begin{pmatrix}0&0&0&-i&0&-i\omega \\0&0&0&0&-i\omega &0\\0&0&0&0&0&-i\omega ^{2}\\i&0&i\omega &0&0&0\\0&i\omega &0&0&0&0\\0&0&i\omega ^{2}&0&0&0\end{pmatrix}}}$

Remark. Note that the direct sum and the tensor products have different degrees and hence are different representations.

Let ${\displaystyle \rho _{1}:G\to {\text{GL}}(V_{1}),\rho _{2}:G\to {\text{GL}}(V_{2})}$ be two linear representations of the same group. Let ${\displaystyle s}$ be an element of ${\displaystyle G.}$ Then ${\displaystyle \rho (s)\in {\text{GL}}(V_{1}\otimes V_{2})}$ is defined by ${\displaystyle \rho (s)(v_{1}\otimes v_{2})=\rho _{1}(s)v_{1}\otimes \rho _{2}(s)v_{2},}$ for ${\displaystyle v_{1}\in V_{1},v_{2}\in V_{2},}$ and we write ${\displaystyle \rho (s)=\rho _{1}(s)\otimes \rho _{2}(s).}$ Then the map ${\displaystyle s\mapsto \rho (s)}$ defines a linear representation of ${\displaystyle G,}$ which is also called tensor product of the given representations.

These two cases have to be strictly distinguished. The first case is a representation of the group product into the tensor product of the corresponding representation spaces. The second case is a representation of the group ${\displaystyle G}$ into the tensor product of two representation spaces of this one group. But this last case can be viewed as a special case of the first one by focussing on the diagonal subgroup ${\displaystyle G\times G.}$ This definition can be iterated a finite number of times.

Let ${\displaystyle V}$ and ${\displaystyle W}$ be representations of the group ${\displaystyle G.}$ Then ${\displaystyle {\text{Hom}}(V,W)}$ is a representation by virtue of the following identity: ${\displaystyle {\text{Hom}}(V,W)=V^{*}\otimes W}$. Let ${\displaystyle B\in {\text{Hom}}(V,W)}$ and let ${\displaystyle \rho }$ be the representation on ${\displaystyle {\text{Hom}}(V,W).}$ Let ${\displaystyle \rho _{V}}$ be the representation on ${\displaystyle V}$ and ${\displaystyle \rho _{W}}$ the representation on ${\displaystyle W.}$ Then the identity above leads to the following result:

${\displaystyle \rho (s)(B)v=\rho _{W}(s)\circ B\circ \rho _{V}(s)v}$ for all ${\displaystyle s\in G,v\in V.}$
Theorem. The irreducible representations of ${\displaystyle G_{1}\times G_{2}}$ up to isomorphism are exactly the representations ${\displaystyle \rho _{1}\otimes \rho _{2}}$ in which ${\displaystyle \rho _{1}}$ and ${\displaystyle \rho _{2}}$ are irreducible representations of ${\displaystyle G_{1}}$ and ${\displaystyle G_{2},}$ respectively.

#### Symmetric and alternating square

Let ${\displaystyle \rho :G\to V\otimes V}$ be a linear representation of ${\displaystyle G.}$ Let ${\displaystyle (e_{k})}$ be a basis of ${\displaystyle V.}$ Define ${\displaystyle \vartheta :V\otimes V\to V\otimes V}$ by extending ${\displaystyle \vartheta (e_{k}\otimes e_{j})=e_{j}\otimes e_{k}}$ linearly. It holds that ${\displaystyle \vartheta ^{2}=1}$ and therefore ${\displaystyle V\otimes V}$ splits up into ${\displaystyle V\otimes V={\text{Sym}}^{2}(V)\oplus {\text{Alt}}^{2}(V),}$ in which

${\displaystyle {\text{Sym}}^{2}(V)=\{z\in V\otimes V:\vartheta (z)=z\}}$
${\displaystyle {\text{Alt}}^{2}(V)=\bigwedge ^{2}V=\{z\in V\otimes V:\vartheta (z)=-z\}.}$

These subspaces are ${\displaystyle G}$–invariant and by this define subrepresentations which are called the symmetric square and the alternating square, respectively. These subrepresentations are also defined in ${\displaystyle V^{\otimes m},}$ although in this case they are denoted wedge product ${\displaystyle \bigwedge ^{m}V}$ and symmetric product ${\displaystyle {\text{Sym}}^{m}(V).}$ In case that ${\displaystyle m>2,}$ the vector space ${\displaystyle V^{\otimes m}}$ is in general not equal to the direct sum of these two products.

## Decompositions

In order to understand representations more easily, a decomposition of the representation space into the direct sum of simpler subrepresentations would be desirable. This can be achieved for finite groups as we will see in the following results. More detailed explanations and proofs may be found in [1] and [2].

Theorem. (Maschke) Let ${\displaystyle \rho :G\to {\text{GL}}(V)}$ be a linear representation where ${\displaystyle V}$ is a vector space over a field of characteristic zero. Let ${\displaystyle W}$ be a ${\displaystyle G}$-invariant subspace of ${\displaystyle V.}$ Then the complement ${\displaystyle W^{0}}$ of ${\displaystyle W}$ exists in ${\displaystyle V}$ and is ${\displaystyle G}$-invariant.

A subrepresentation and its complement determine a representation uniquely.

The following theorem will be presented in a more general way, as it provides a very beautiful result about representations of compact – and therefore also of finite – groups:

Theorem. Every linear representation of a compact group over a field of characteristic zero is a direct sum of irreducible representations.

Or in the language of ${\displaystyle K[G]}$-modules: If ${\displaystyle {\text{char}}(K)=0,}$ the group algebra ${\displaystyle K[G]}$ is semisimple, i.e. it is the direct sum of simple algebras.

Note that this decomposition is not unique. However, the number of how many times a subrepresentation isomorphic to a given irreducible representation is occurring in this decomposition is independent of the choice of decomposition.

The canonical decomposition

To achieve a unique decomposition, one has to combine all the irreducible subrepresentations that are isomorphic to each other. That means, the representation space is decomposed into a direct sum of its isotypes. This decomposition is uniquely determined. It is called the canonical decomposition.

Let ${\displaystyle (\tau _{j})_{j\in I}}$ be the set of all irreducible representations of a group ${\displaystyle G}$ up to isomorphism. Let ${\displaystyle V}$ be a representation of ${\displaystyle G}$ and let ${\displaystyle \{V(\tau _{j})|j\in I\}}$ be the set of all isotypes of ${\displaystyle V.}$ The projection ${\displaystyle p_{j}:V\to V(\tau _{j})}$ corresponding to the canonical decomposition is given by

${\displaystyle p_{j}={\frac {n_{j}}{g}}\sum _{t\in G}{\overline {\chi _{\tau _{j}}(t)}}\rho (t),}$

where ${\displaystyle n_{j}=\dim(\tau _{j}),}$ ${\displaystyle g={\text{ord}}(G)}$ and ${\displaystyle \chi _{\tau _{j}}}$ is the character belonging to ${\displaystyle \tau _{j}.}$

In the following, we show how to determine the isotype to the trivial representation:

Definition (Projection formula). For every representation ${\displaystyle (\rho ,V)}$ of a group ${\displaystyle G}$ we define

${\displaystyle V^{G}:=\{v\in V:\rho (s)v=v\,\,\,\,\forall \,s\in G\}.}$

In general, ${\displaystyle \rho (s):V\to V}$ is not ${\displaystyle G}$-linear. We define

${\displaystyle P:={\frac {1}{|G|}}\sum _{s\in G}\rho (s)\in {\text{End}}(V).}$

Then ${\displaystyle P}$ is a ${\displaystyle G}$-linear map, because

${\displaystyle \forall t\in G:\qquad \sum _{s\in G}\rho (s)=\sum _{s\in G}\rho (tst^{-1}).}$
Proposition. The map ${\displaystyle P}$ is a projection from ${\displaystyle V}$ to ${\displaystyle V^{G}.}$

This proposition enables us to determine the isotype to the trivial subrepresentation of a given representation explicitly.

How often the trivial representation occurs in ${\displaystyle V}$ is given by ${\displaystyle {\text{Tr}}(P).}$ This result is a consequence of the fact that the eigenvalues of a projection are only ${\displaystyle 0}$ or ${\displaystyle 1}$ and that the eigenspace corresponding to the eigenvalue ${\displaystyle 1}$ is the image of the projection. Since the trace of the projection is the sum of all eigenvalues, we obtain the following result

${\displaystyle \dim(V(1))=\dim(V^{G})=Tr(P)={\frac {1}{|G|}}\sum _{s\in G}\chi _{V}(s),}$

in which ${\displaystyle V(1)}$ denotes the isotype of the trivial representation.

Let ${\displaystyle V_{\pi }}$ be a nontrivial irreducible representation of ${\displaystyle G.}$ Then the isotype to the trivial representation of ${\displaystyle \pi }$ is the null space. That means the following equation holds

${\displaystyle P={\frac {1}{|G|}}\sum _{s\in G}\pi (s)=0.}$

Let ${\displaystyle e_{1},...,e_{n}}$ be a orthonormal basis of ${\displaystyle V_{\pi }.}$ Then we have:

${\displaystyle \sum _{s\in G}{\text{Tr}}(\pi (s))=\sum _{s\in G}\sum _{j=1}^{n}\langle \pi (s)e_{j},e_{j}\rangle =\sum _{j=1}^{n}\left\langle \sum _{s\in G}\pi (s)e_{j},e_{j}\right\rangle =0.}$

Therefore, the following is valid for a nontrivial irreducible representation ${\displaystyle V}$:

${\displaystyle \sum _{s\in G}\chi _{V}(s)=0.}$

Example. Let ${\displaystyle G={\text{Per}}(3)}$ be the permutation groups in three elements. Let ${\displaystyle \rho :{\text{Per}}(3)\to {\text{GL}}_{5}(\mathbb {C} )}$ be a linear representation of ${\displaystyle {\text{Per}}(3)}$ defined on the generating elements as follows:

${\displaystyle \rho (1,2)={\begin{pmatrix}-1&2&0&0&0\\0&1&0&0&0\\0&0&0&1&0\\0&0&1&0&0\\0&0&0&0&1\end{pmatrix}},\quad \rho (1,3)={\begin{pmatrix}{\frac {1}{2}}&{\frac {1}{2}}&0&0&0\\{\frac {1}{2}}&-1&0&0&0\\0&0&0&0&1\\0&0&0&1&0\\0&0&1&0&0\end{pmatrix}},\quad \rho (2,3)={\begin{pmatrix}0&-2&0&0&0\\-{\frac {1}{2}}&0&0&0&0\\0&0&1&0&0\\0&0&0&0&1\\0&0&0&1&0\end{pmatrix}}.}$

This representation can be decomposed on first look into the left-regular representation of ${\displaystyle {\text{Per}}(3),}$which is denoted by ${\displaystyle \pi }$ in the following, and the representation ${\displaystyle \eta :{\text{Per}}(3)\to {\text{GL}}_{2}(\mathbb {C} )}$ with

${\displaystyle \eta (1,2)={\begin{pmatrix}-1&2\\0&1\end{pmatrix}},\quad \eta (1,3)={\begin{pmatrix}{\frac {1}{2}}&{\frac {1}{2}}\\{\frac {1}{2}}&-1\end{pmatrix}},\quad \eta (2,3)={\begin{pmatrix}0&-2\\-{\frac {1}{2}}&0\end{pmatrix}}.}$

With the help of the irreducibility criterion taken from the next chapter, we realize, that ${\displaystyle \eta }$ is irreducible and ${\displaystyle \pi }$ is not. This is, because for the inner product defined in the section ”Inner product and characters” further below, we have ${\displaystyle (\eta |\eta )=1,(\pi |\pi )=2.}$

The subspace ${\displaystyle \mathbb {C} (e_{1}+e_{2}+e_{3})}$ of ${\displaystyle \mathbb {C} ^{3}}$ is invariant with respect to the left-regular representation. Restricted to this subspace we obtain the trivial representation.

The orthogonal complement of ${\displaystyle \mathbb {C} (e_{1}+e_{2}+e_{3})}$ is ${\displaystyle \mathbb {C} (e_{1}-e_{2})\oplus \mathbb {C} (e_{1}+e_{2}-2e_{3}).}$ Restricted to this subspace, which is also ${\displaystyle G}$–invariant as we have seen above, we obtain the representation ${\displaystyle \tau }$ given by

${\displaystyle \tau (1,2)={\begin{pmatrix}-1&0\\0&1\end{pmatrix}},\quad \tau (1,3)={\begin{pmatrix}{\frac {1}{2}}&{\frac {3}{2}}\\{\frac {1}{2}}&-{\frac {1}{2}}\end{pmatrix}},\quad \tau (2,3)={\begin{pmatrix}{\frac {1}{2}}&-{\frac {3}{2}}\\-{\frac {1}{2}}&-{\frac {1}{2}}\end{pmatrix}}.}$

Just like before we can use the irreducibility criterion of the next chapter to prove that ${\displaystyle \tau }$ is irreducible. Now, ${\displaystyle \eta }$ and ${\displaystyle \tau }$ are isomorphic, because ${\displaystyle \eta (s)=B\circ \tau (s)\circ B^{-1}}$ for all ${\displaystyle s\in {\text{Per}}(3),}$ in which ${\displaystyle B:\mathbb {C} ^{2}\to \mathbb {C} ^{2}}$ is given by the matrix

${\displaystyle M_{B}={\begin{pmatrix}2&2\\0&2\end{pmatrix}}.}$

A decomposition of ${\displaystyle (\rho ,\mathbb {C} ^{5})}$ in irreducible subrepresentations is: ${\displaystyle \rho =\tau \oplus \eta \oplus 1}$ where ${\displaystyle 1}$ denotes the trivial representation and

${\displaystyle \mathbb {C} ^{5}=\mathbb {C} (e_{1},e_{2})\oplus \mathbb {C} (e_{3}-e_{4},e_{3}+e_{4}-2e_{5})\oplus \mathbb {C} (e_{3}+e_{4}+e_{5})}$

is the corresponding decomposition of the representation space.

We obtain the canonical decomposition by combining all the isomorphic irreducible subrepresentations: ${\displaystyle \rho _{1}:=\eta \oplus \tau }$ is the ${\displaystyle \tau }$-isotype of ${\displaystyle \rho }$ and consequently the canonical decomposition is given by

${\displaystyle \rho =\rho _{1}\oplus 1,\qquad \mathbb {C} ^{5}=\mathbb {C} (e_{1},e_{2},e_{3}-e_{4},e_{3}+e_{4}-2e_{5})\oplus \mathbb {C} (e_{3}+e_{4}+e_{5}).}$

The theorems above are in general not valid for infinite groups. This will be demonstrated by the following example: let

${\displaystyle G=\{A\in {\text{GL}}_{2}(\mathbb {C} )|\,A\,\,{\text{ is an upper triangular matrix}}\}.}$

Together with the matrix multiplication ${\displaystyle G}$ is an infinite group. ${\displaystyle G}$ acts on ${\displaystyle \mathbb {C} ^{2}}$ by matrix-vector multiplication. We consider the representation ${\displaystyle \rho (A)=A}$ for all ${\displaystyle A\in G.}$ The subspace ${\displaystyle \mathbb {C} e_{1}}$ is a ${\displaystyle G}$-invariant subspace. However, there exists no ${\displaystyle G}$-invariant complement to this subspace. The assumption, that such a complement exists, results in the statement, that every matrix is diagonalizable over ${\displaystyle \mathbb {C} .}$ This is known to be wrong and thus presents the contradiction.

That means, if we consider infinite groups, it is possible that a representation, although being not irreducible, can not be decomposed in a direct sum of irreducible subrepresentations.

## Character theory

### Definitions

Let ${\displaystyle \rho :G\to {\text{GL}}(V)}$ be a linear representation of a finite group ${\displaystyle G}$ into the vector space ${\displaystyle V.}$ We define the map ${\displaystyle \chi _{\rho }}$ by ${\displaystyle \chi _{\rho }(s)={\text{Tr}}(\rho (s)),}$ in which ${\displaystyle {\text{Tr}}(\rho (s))}$ denotes the trace of the linear map ${\displaystyle \rho .}$ The ${\displaystyle \mathbb {C} }$-valued function ${\displaystyle \chi _{\rho }}$ is called character of the representation ${\displaystyle \rho .}$ It is obvious that isomorphic representations have the same character.

Sometimes the character of a representation ${\displaystyle \rho }$ is defined as ${\displaystyle \chi (s)=\dim(\rho ){\text{Tr}}(\rho (s)),}$ in which ${\displaystyle \dim(\rho )}$ denotes the degree of the representation. In this article this definition is not used.

Example 1. If ${\displaystyle \rho }$ is a representation of degree one then its character is given by ${\displaystyle \chi =\rho .}$

Example 2. Let ${\displaystyle V}$ be the permutation representation of ${\displaystyle G}$ corresponding to the left action of ${\displaystyle G}$ on a finite set ${\displaystyle X.}$ Then: ${\displaystyle \chi _{V}(s)=|\{x\in X|s\cdot x=x\}|.}$

Example 3. The character ${\displaystyle \chi _{R}}$ of the regular representation ${\displaystyle R}$ is given by

${\displaystyle \chi _{R}(s)={\begin{cases}0&s\neq e\\|G|&s=e\end{cases}},}$

where ${\displaystyle e}$ denotes the neutral element of ${\displaystyle G.}$ Note, that in this context it is correct to use the notion of regular representation and not to distinguish between left- and right-regular as they are isomorphic and thus have the same character.

Example 4. Let ${\displaystyle \rho :\mathbb {Z} /2\mathbb {Z} \times \mathbb {Z} /2\mathbb {Z} \to {\text{GL}}_{2}(\mathbb {C} )}$ be the representation defined by:

${\displaystyle \rho (0,0)={\begin{pmatrix}1&0\\0&1\end{pmatrix}},\quad \rho (1,0)={\begin{pmatrix}-1&0\\0&-1\end{pmatrix}},\quad \rho (0,1)={\begin{pmatrix}0&1\\1&0\end{pmatrix}},\quad \rho (1,1)={\begin{pmatrix}0&-1\\-1&0\end{pmatrix}}.}$

The character ${\displaystyle \chi _{\rho }}$ is given by

${\displaystyle \chi _{\rho }(0,0)=2,\quad \chi _{\rho }(1,0)=-2,\quad \chi _{\rho }(0,1)=\chi _{\rho }(1,1)=0.}$

This example shows, that the character is in general not a group homomorphism.

### Properties

As shown in the section on properties of linear representations every representation can be assumed to be unitary. A character is called unitary, if it belongs to a unitary representation.

A character is called irreducible, if the corresponding representation is irreducible.

Let ${\displaystyle \chi }$ be the character of a (unitary) representation ${\displaystyle \rho }$ of degree ${\displaystyle n.}$ Then the following holds:[1]

• ${\displaystyle \chi (e)=n,}$ where ${\displaystyle e}$ is the neutral element of ${\displaystyle G.}$
• ${\displaystyle \chi (s^{-1})={\overline {\chi (s)}},\,\,\,\forall \,s\in G.}$
• ${\displaystyle \chi (tst^{-1})=\chi (s),\,\,\forall \,s,t\in G.}$
• ${\displaystyle \chi (s)}$ is the sum of the eigenvalues of ${\displaystyle \rho (s)}$ with multiplicity.
• For ${\displaystyle s\in G}$ of order ${\displaystyle m}$ the following is valid:
• ${\displaystyle \chi (s)}$ is the sum of ${\displaystyle n,m}$–th roots of unity.
• ${\displaystyle |\chi (g)|\leqslant m.}$
• ${\displaystyle \{s\in G|\chi (s)=m\}}$ is a normal subgroup in ${\displaystyle G.}$

Characters of special constructions

Let ${\displaystyle \rho _{1}:G\to {\text{GL}}(V_{1}),\rho _{2}:G\to {\text{GL}}(V_{2})}$ be two linear representations of the same group ${\displaystyle G.}$ Let ${\displaystyle \chi _{1},\chi _{2}}$ be the corresponding characters. Then the following holds:

• The character ${\displaystyle \chi _{1}^{*}}$ of the dual representation ${\displaystyle \rho _{1}^{*}}$ of ${\displaystyle \rho _{1}}$ is given by ${\displaystyle \chi _{1}^{*}={\overline {\chi _{1}}}.}$
• The character ${\displaystyle \chi }$ of the direct sum ${\displaystyle V_{1}\oplus V_{2}}$ is equal to ${\displaystyle \chi _{1}+\chi _{2}.}$
• The character ${\displaystyle \chi }$ of the tensor product of the representations ${\displaystyle V_{1}\otimes V_{2}}$ is given by ${\displaystyle \chi _{1}\chi _{2}.}$
• The character ${\displaystyle \chi }$ of the representation belonging to ${\displaystyle {\text{Hom}}(V_{1},V_{2})}$ is given by ${\displaystyle {\overline {\chi _{1}}}\chi _{2}.}$

Let ${\displaystyle \chi _{1}}$ be the character of ${\displaystyle \rho _{1}:G_{1}\to {\text{GL}}(W_{1})}$ and ${\displaystyle \chi _{2}}$ the character of ${\displaystyle \rho _{2}:G_{2}\to {\text{GL}}(W_{2}).}$ Then the character ${\displaystyle \chi }$ of ${\displaystyle \rho _{1}\otimes \rho _{2}}$ is given by ${\displaystyle \chi (s_{1},s_{2})=\chi _{1}(s_{1})\chi _{2}(s_{2}).}$

Let ${\displaystyle \rho :G\to {\text{GL}}(V)}$ be a linear representation of ${\displaystyle G}$ and let ${\displaystyle \chi }$ be the corresponding character. Let ${\displaystyle \chi _{\sigma }^{(2)}}$ be the character of the symmetric square and let ${\displaystyle \chi _{\alpha }^{(2)}}$ be the character of the alternating square. For every ${\displaystyle s\in G}$ the following holds:

{\displaystyle {\begin{aligned}\chi _{\sigma }^{(2)}(s)&={\tfrac {1}{2}}\left(\chi (s)^{2}+\chi (s^{2})\right)\\\chi _{\alpha }^{(2)}(s)&={\tfrac {1}{2}}\left(\chi (s)^{2}-\chi (s^{2})\right)\\\chi ^{2}&=\chi _{\sigma }^{(2)}+\chi _{\alpha }^{(2)}\end{aligned}}}

### Schur's lemma

Let ${\displaystyle \rho _{1}:G\to {\text{GL}}(V_{1})}$ and ${\displaystyle \rho _{2}:G\to {\text{GL}}(V_{2})}$ be two irreducible representations. Let ${\displaystyle F:V_{1}\to V_{2}}$ be a linear map such that ${\displaystyle \rho _{2}(s)\circ F=F\circ \rho _{1}(s)}$ for all ${\displaystyle s\in G.}$ Then the following is valid:

• If ${\displaystyle \rho _{1}}$ and ${\displaystyle \rho _{2}}$ are not isomorphic, we have ${\displaystyle F=0.}$
• If ${\displaystyle V_{1}=V_{2}}$ and ${\displaystyle \rho _{1}=\rho _{2},}$ ${\displaystyle F}$ is a homothety (i.e. ${\displaystyle F=\lambda {\text{Id}}}$ for a ${\displaystyle \lambda \in \mathbb {C} }$).

Proof. Suppose ${\displaystyle F}$ is nonzero Then ${\displaystyle F\circ \rho _{1}(s)\,(u)=\rho _{2}(s)\circ F(u)=0}$ is valid for all ${\displaystyle u\in \ker(F).}$ Therefore, we obtain ${\displaystyle \rho _{1}(s)u\in \ker(F)}$ for all ${\displaystyle s\in G}$ and ${\displaystyle u\in \ker(F).}$ And we know now, that ${\displaystyle \ker(F)}$ is ${\displaystyle G}$–invariant. Since ${\displaystyle V_{1}}$ is irreducible and ${\displaystyle F\neq 0,}$ we conclude ${\displaystyle \ker(F)=0.}$ Now let ${\displaystyle y\in {\text{Im}}(F).}$ This means, there exists ${\displaystyle v\in V_{1},}$ such that ${\displaystyle Fv=y,}$ and we have ${\displaystyle \rho _{2}(s)y=\rho _{2}(s)Fv=F\rho _{1}(s)v.}$ Thus, we deduce, that ${\displaystyle {\text{Im}}(F)}$ is a ${\displaystyle G}$–invariant subspace. Because ${\displaystyle F}$ is nonzero and ${\displaystyle V_{2}}$ is irreducible, we have ${\displaystyle {\text{Im}}(F)=V_{2}.}$ Therefore, ${\displaystyle F}$ is an isomorphism and the first statement is proven.

Suppose now that ${\displaystyle V_{1}=V_{2},\rho _{1}=\rho _{2}.}$ Since our base field is ${\displaystyle \mathbb {C} ,}$ we know that ${\displaystyle F}$ has at least one eigenvalue ${\displaystyle \lambda .}$ Let ${\displaystyle F'=F-\lambda ,}$ then ${\displaystyle \ker(F')\neq 0}$ and we have ${\displaystyle \rho _{2}(s)\circ F'=F'\circ \rho _{1}(s)}$ for all ${\displaystyle s\in G.}$ According to the considerations above this is only possible, if ${\displaystyle F'=0,}$ i.e. ${\displaystyle F=\lambda .}$

### Inner product and characters

In order to show some particularly interesting results about characters, it is rewarding to consider a more general type of functions on groups:

Definition (Class functions). A function ${\displaystyle \varphi :G\to \mathbb {C} }$ is called a class function if it is constant on conjugacy classes of ${\displaystyle G}$, i.e.

${\displaystyle \forall s,t\in G:\quad \varphi \left(sts^{-1}\right)=\varphi (t).}$

Note that every character is a class function, as the trace of a matrix is preserved under conjugation.

The set of all class functions is a ${\displaystyle \mathbb {C} }$–algebra and is denoted by ${\displaystyle \mathbb {C} _{\text{class}}(G)}$. Its dimension is equal to the number of conjugacy classes of ${\displaystyle G.}$

Proofs of the following results of this chapter may be found in [1], [2] and [3].

An inner product can be defined on the set of all class functions on a finite group:

${\displaystyle (f|h)_{G}={\frac {1}{|G|}}\sum _{t\in G}f(t){\overline {h(t)}}}$

Orthonormal property. If ${\displaystyle \chi _{1},\ldots ,\chi _{k}}$ are the distinct irreducible characters of ${\displaystyle G}$, they form an orthonormal basis for the vector space of all class functions with respect to the inner product defined above, i.e.

• ${\displaystyle (\chi _{i}|\chi _{j})={\begin{cases}1{\text{ if }}i=j\\0{\text{ otherwise }}\end{cases}}.}$
• Every class function ${\displaystyle f}$ may be expressed as a unique linear combination of the irreducible characters ${\displaystyle \chi _{1},\ldots ,\chi _{k}}$.

One might verify that the irreducible characters generate ${\displaystyle \mathbb {C} _{\text{class}}(G)}$ by showing, that there exists no class function unequal to zero which is orthogonal to all the irreducible characters. To see this, denote ${\displaystyle \rho _{f}=\sum _{g}f(g)\rho (g).}$ Then ${\displaystyle \rho _{f}={\frac {|G|}{n}}\langle f,\chi _{V}^{*}\rangle \in End(V)}$ from Schur's lemma. Suppose ${\displaystyle f}$ is a class function which is orthogonal to all the characters. Then by the above we have ${\displaystyle \rho _{f}=0}$ whenever ${\displaystyle \rho }$ is irreducible. But then it follows that ${\displaystyle \rho _{f}=0}$ for all ${\displaystyle \rho }$, by decomposability. Take ${\displaystyle \rho }$ to be the regular representation. Applying ${\displaystyle \rho _{f}}$ to some particular basis element ${\displaystyle g}$, we get ${\displaystyle f(g)=0}$. Since this is true for all ${\displaystyle g}$, we have ${\displaystyle f=0.}$

It follows from the orthonormal property that the number of non-isomorphic irreducible representations of a group ${\displaystyle G}$ is equal to the number of conjugacy classes of ${\displaystyle G.}$

Furthermore, a class function on ${\displaystyle G}$ is a character of ${\displaystyle G}$ if and only if it can be written as a linear combination of the distinct irreducible characters ${\displaystyle \chi _{j}}$ with non-negative integer coefficients: if ${\displaystyle \varphi }$ is a class function on ${\displaystyle G}$ such that ${\displaystyle \varphi =c_{1}\chi _{1}+\cdots +c_{k}\chi _{k}}$ where ${\displaystyle c_{j}}$ non-negative integers, then ${\displaystyle \varphi }$ is the character of the direct sum ${\displaystyle c_{1}\tau _{1}\oplus \cdots \oplus c_{k}\tau _{k}}$ of the representations ${\displaystyle \tau _{j}}$ corresponding to ${\displaystyle \chi _{j}.}$ Conversely, it is always possible to write any character as a sum of irreducible characters.

The inner product defined above can be extended on the set of all ${\displaystyle \mathbb {C} }$-valued functions ${\displaystyle L^{1}(G)}$ on a finite group:

${\displaystyle (f|h)_{G}={\frac {1}{|G|}}\sum _{t\in G}f(t){\overline {h(t)}}}$

Also a symmetric bilinear form can be defined on ${\displaystyle L^{1}(G):}$

${\displaystyle \langle f,h\rangle _{G}={\frac {1}{|G|}}\sum _{t\in G}f(t)h(t^{-1})}$

These two forms match on the set of characters. If there is no danger of confusion the index of both forms ${\displaystyle (\cdot |\cdot )_{G}}$ and ${\displaystyle \langle \cdot |\cdot \rangle _{G}}$ will be omitted.

Let ${\displaystyle V_{1},V_{2}}$ be two ${\displaystyle \mathbb {C} [G]}$–modules. Note that ${\displaystyle \mathbb {C} [G]}$–modules are simply representations of ${\displaystyle G}$. Since the orthonormal property yields the number of irreducible representations of ${\displaystyle G}$ is exactly the number of its conjugacy classes, then there are exactly as many simple ${\displaystyle \mathbb {C} [G]}$–modules (up to isomorphism) as there are conjugacy classes of ${\displaystyle G.}$

We define ${\displaystyle \langle V_{1},V_{2}\rangle _{G}:=\dim({\text{Hom}}^{G}(V_{1},V_{2})),}$ in which ${\displaystyle {\text{Hom}}^{G}(V_{1},V_{2})}$ is the vector space of all ${\displaystyle G}$–linear maps. This form is bilinear with respect to the direct sum.

In the following, these bilinear forms will allow us to obtain some important results with respect to the decomposition and irreducibility of representations.

For instance, let ${\displaystyle \chi _{1}}$ and ${\displaystyle \chi _{2}}$ be the characters of ${\displaystyle V_{1}}$ and ${\displaystyle V_{2},}$ respectively. Then${\displaystyle \langle \chi _{1},\chi _{2}\rangle _{G}=(\chi _{1}|\chi _{2})_{G}=\langle V_{1},V_{2}\rangle _{G}.}$

Once can deduce the following from the results above, of Schur's lemma and of the complete reducibility of representations.

Theorem. Let ${\displaystyle V}$ be a linear representation of ${\displaystyle G}$ with character ${\displaystyle \xi .}$ Let ${\displaystyle V=W_{1}\oplus \cdots \oplus W_{k},}$ where ${\displaystyle W_{j}}$ are irreducible. Let ${\displaystyle (\tau ,W)}$ be an irreducible representation of ${\displaystyle G}$ with character ${\displaystyle \chi .}$ Then the number of subrepresentations ${\displaystyle W_{j}}$ which are isomorphic to ${\displaystyle W}$ is independent of the given decomposition and is equal to the inner product ${\displaystyle (\xi |\chi ),}$ i.e. the ${\displaystyle \tau }$–isotype ${\displaystyle V(\tau )}$ of ${\displaystyle V}$ is independent of the choice of decomposition. We also get:
${\displaystyle (\xi |\chi )={\frac {\dim(V(\tau ))}{\dim(\tau )}}=\langle V,W\rangle }$
and thus
${\displaystyle \dim(V(\tau ))=\dim(\tau )(\xi |\chi ).}$
Corollary. Two representations with the same character are isomorphic. That means, that every representation is determined by its character.

With this we obtain a very handsome result to analyse representations:

Irreducibility criterion. Let ${\displaystyle \chi }$ be the character of the representation ${\displaystyle V,}$ then we have ${\displaystyle (\chi |\chi )\in \mathbb {N} _{0}.}$ And it holds ${\displaystyle (\chi |\chi )=1}$ if and only if ${\displaystyle V}$ is irreducible.

Therefore, using the first theorem, the characters of irreducible representations of ${\displaystyle G}$ form an orthonormal set on ${\displaystyle \mathbb {C} _{\text{class}}(G)}$ with respect to this inner product.

Corollary. Let ${\displaystyle V}$ be a vector space with ${\displaystyle \dim(V)=n.}$ A given irreducible representation ${\displaystyle V}$ of ${\displaystyle G}$ is contained ${\displaystyle n}$–times in the regular representation. That means, that if ${\displaystyle R}$ denotes the regular representation of ${\displaystyle G}$ we have: ${\displaystyle R\cong \oplus (W_{j})^{\oplus \dim(W_{j})},}$ in which ${\displaystyle \{W_{j}|j\in I\}}$ is the set of all irreducible representations of ${\displaystyle G}$ that are pairwise not isomorphic to each other.

In terms of the group algebra this means, that ${\displaystyle \mathbb {C} [G]\cong \oplus _{j}{\text{End}}(W_{j})}$ as algebras.

As a numerical result we get:

${\displaystyle |G|=\chi _{R}(e)=\dim(R)=\sum _{j}\dim \left((W_{j})^{\oplus (\chi _{W_{j}}|\chi _{R})}\right)=\sum _{j}(\chi _{W_{j}}|\chi _{R})\cdot \dim(W_{j})=\sum _{j}\dim(W_{j})^{2},}$

in which ${\displaystyle R}$ is the regular representation and ${\displaystyle \chi _{W_{j}}}$ and ${\displaystyle \chi _{R}}$ are corresponding characters to ${\displaystyle W_{j}}$ and ${\displaystyle R,}$ respectively. It should also be mentioned, that ${\displaystyle e}$ denotes the neutral element of the group.

This formula is a necessary and sufficient condition for all irreducible representations of a group up to isomorphism. It provides us with the means to check whether we found all irreducible representations of a group up to isomorphism.

Similarly, by using the character of the regular representation evaluated at ${\displaystyle s\neq e,}$ we get the equation:

${\displaystyle 0=\chi _{R}(s)=\sum _{j}\dim(W_{j})\cdot \chi _{W_{j}}(s).}$

Using the description of representations via the convolution algebra we achieve an equivalent formulation of these equations:

The Fourier inversion formula:

${\displaystyle f(s)={\frac {1}{|G|}}\sum _{\rho {\text{ irr. rep. of }}G}\dim(V_{\rho })\cdot {\text{Tr}}(\rho (s^{-1})\cdot {\hat {f}}(\rho )).}$

In addition the Plancherel formula holds:

${\displaystyle \sum _{s\in G}f(s^{-1})\psi (s)={\frac {1}{|G|}}\sum _{\rho \,\,{\text{ irred.}}{\text{ rep.}}{\text{ of }}G}\dim(V_{\rho })\cdot {\text{Tr}}({\hat {f}}(\rho ){\hat {h}}(\rho )).}$

In both formulas ${\displaystyle (\rho ,V_{\rho })}$ is a linear representation of a group ${\displaystyle G,s\in G}$ and ${\displaystyle f,h\in L^{1}(G).}$

The corollary above has an additional consequence:

Lemma. Let ${\displaystyle G}$ be a group. Then the following is equivalent:
• ${\displaystyle G}$ is abelian.
• Every function on ${\displaystyle G}$ is a class function.
• All irreducible representations of ${\displaystyle G}$ have degree ${\displaystyle 1.}$

## The induced representation

As was shown in the section on properties of linear representations, we can, by restricting to a subgroup, obtain a representation of a subgroup starting from a representation of a group. Naturally we are interested in the reverse process: Is it possible to obtain the representation of a group starting from a representation of a subgroup? We will see, that the induced representation, which will be defined in the following, provides us with the necessary concept. Admittedly, this construction is not inverse but adjoint to the restriction.

### Definitions

Let ${\displaystyle \rho :G\to {\text{GL}}(V_{\rho })}$ be a linear representation of ${\displaystyle G.}$ Let ${\displaystyle H}$ be a subgroup and ${\displaystyle \rho |_{H}}$ the restriction. Let ${\displaystyle W}$ be a subrepresentation of ${\displaystyle \rho _{H}.}$ We write ${\displaystyle \theta :H\to {\text{GL}}(W)}$ to denote this representation. Let ${\displaystyle s\in G.}$ The vector space ${\displaystyle \rho (s)(W)}$ depends only on the left coset ${\displaystyle sH}$ of ${\displaystyle s.}$ Let ${\displaystyle R}$ be a representative system of ${\displaystyle G/H,}$ then

${\displaystyle \sum _{r\in R}\rho (r)(W)}$

is a subrepresentation of ${\displaystyle V_{\rho }.}$

A representation ${\displaystyle \rho }$ of ${\displaystyle G}$ in ${\displaystyle V_{\rho }}$ is called induced by the representation ${\displaystyle \theta }$ of ${\displaystyle H}$ in ${\displaystyle W,}$ if

${\displaystyle V_{\rho }=\bigoplus _{r\in R}W_{r}.}$

Here ${\displaystyle R}$ denotes a representative system of ${\displaystyle G/H}$ and ${\displaystyle W_{r}=\rho (s)(W)}$ for all ${\displaystyle s\in rH}$ and for all ${\displaystyle r\in R.}$ In other words: the representation ${\displaystyle (\rho ,V_{\rho })}$ is induced by ${\displaystyle (\theta ,W),}$ if every ${\displaystyle v\in V_{\rho }}$ can be written uniquely as

${\displaystyle \sum _{r\in R}w_{r},}$

where ${\displaystyle w_{r}\in W_{r}}$ for every ${\displaystyle r\in R.}$

We denote the representation ${\displaystyle \rho }$ of ${\displaystyle G}$ which is induced by the representation ${\displaystyle \theta }$ of ${\displaystyle H}$ as ${\displaystyle \rho ={\text{Ind}}_{H}^{G}(\theta ),}$ or in short ${\displaystyle \rho ={\text{Ind}}(\theta ),}$ if there is no danger of confusion. The representation space itself is frequently used instead of the representation map, i.e. ${\displaystyle V={\text{Ind}}_{H}^{G}(W),}$ or ${\displaystyle V={\text{Ind}}(W),}$ if the representation ${\displaystyle V}$ is induced by ${\displaystyle W.}$

#### Alternative description of the induced representation

By using the group algebra we obtain an alternative description of the induced representation:

Let ${\displaystyle G}$ be a group, ${\displaystyle V}$ a ${\displaystyle \mathbb {C} [G]}$–module and ${\displaystyle W}$ a ${\displaystyle \mathbb {C} [H]}$–submodule of ${\displaystyle V}$ corresponding to the subgroup ${\displaystyle H}$ of ${\displaystyle G.}$ We say, ${\displaystyle V}$ is induced by ${\displaystyle W,}$ if ${\displaystyle V=\mathbb {C} [G]\otimes _{\mathbb {C} [H]}W,}$ in which ${\displaystyle G}$ acts on the first factor: ${\displaystyle s\cdot (e_{t}\otimes w)=e_{st}\otimes w}$ for all ${\displaystyle s,t\in G,w\in W.}$

### Properties

The results introduced in this section will be presented without proof. These may be found in [1] and [2].

Uniqueness and existence of the induced representation. Let ${\displaystyle (\theta ,W_{\theta })}$ be a linear representation of a subgroup ${\displaystyle H}$ of ${\displaystyle G.}$ Then there exists a linear representation ${\displaystyle (\rho ,V_{\rho })}$ of ${\displaystyle G,}$ which is induced by ${\displaystyle (\theta ,W_{\theta }).}$ Note that this representation is unique up to isomorphism.
Transitivity of induction. Let ${\displaystyle W}$ be a representation of ${\displaystyle H}$ and let ${\displaystyle H\leq G\leq K}$ be an ascending series of groups. Then we have
${\displaystyle {\text{Ind}}_{G}^{K}({\text{Ind}}_{H}^{G}(W))\cong {\text{Ind}}_{H}^{K}(W).}$
Lemma. Let ${\displaystyle (\rho ,V_{\rho })}$ be induced by ${\displaystyle (\theta ,W_{\theta })}$ and let ${\displaystyle \rho ':G\to {\text{GL}}(V')}$ be a linear representation of ${\displaystyle G.}$ Now let ${\displaystyle F:W_{\theta }\to V'}$ be a linear map satisfying the property, that ${\displaystyle F\circ \theta (t)=\rho '(t)\circ F}$ for all ${\displaystyle t\in G.}$ Then there exists a uniquely determined linear map ${\displaystyle F':V_{\rho }\to V',}$ which extends ${\displaystyle F}$ and for which ${\displaystyle F'\circ \rho (s)=\rho '(s)\circ F'}$ is valid for all ${\displaystyle s\in G.}$

This means, that if we interpret ${\displaystyle V'}$ as a ${\displaystyle \mathbb {C} [G]}$–module, we have: ${\displaystyle {\text{Hom}}^{H}(W_{\theta },V')\cong {\text{Hom}}^{G}(V_{\rho },V'),}$ where ${\displaystyle {\text{Hom}}^{G}(V_{\rho },V')}$ is the vector space of all ${\displaystyle \mathbb {C} [G]}$–homomorphisms of ${\displaystyle V_{\rho }}$ to ${\displaystyle V'.}$ The same is valid for ${\displaystyle {\text{Hom}}^{H}(W_{\theta },V').}$

Induction on class functions. In the same way as it was done with representations, we can, using the so-called induction, obtain a class function on the group out of a class function on a subgroup. Let ${\displaystyle \varphi }$ be a class function on ${\displaystyle H.}$ We define the function ${\displaystyle \varphi '}$ on ${\displaystyle G}$ by

${\displaystyle \varphi '(s)={\frac {1}{|H|}}\sum _{t\in G \atop t^{-1}st\in H}^{}\varphi (t^{-1}st).}$

We say ${\displaystyle \varphi '}$ is induced by ${\displaystyle \varphi }$ and write ${\displaystyle {\text{Ind}}_{H}^{G}(\varphi )=\varphi '}$ or ${\displaystyle {\text{Ind}}(\varphi )=\varphi '.}$

Proposition. The function ${\displaystyle {\text{Ind}}(\varphi )}$ is a class function on ${\displaystyle G.}$ If ${\displaystyle \varphi }$ is the character of a representation ${\displaystyle W}$ of ${\displaystyle H,}$ then ${\displaystyle {\text{Ind}}(\varphi )}$ is the character of the induced representation ${\displaystyle {\text{Ind}}(W)}$ of ${\displaystyle G.}$
Lemma. If ${\displaystyle \psi }$ is a class function on ${\displaystyle H}$ and ${\displaystyle \varphi }$ is a class function on ${\displaystyle G,}$ we have: ${\displaystyle {\text{Ind}}(\psi \cdot {\text{Res}}\varphi )=({\text{Ind}}\psi )\cdot \varphi .}$
Theorem. Let ${\displaystyle (\rho ,V_{\rho })}$ be the representation of ${\displaystyle G}$ induced by the representation ${\displaystyle (\theta ,W_{\theta })}$ of the subgroup ${\displaystyle H.}$ Let ${\displaystyle \chi _{\rho }}$ and ${\displaystyle \chi _{\theta }}$ be the corresponding characters. Let ${\displaystyle R}$ be a representative system of ${\displaystyle G/H.}$ The induced character is given by
${\displaystyle \forall t\in G:\qquad \chi _{\rho }(t)=\sum _{r\in R, \atop r^{-1}tr\in H}^{}\chi _{\theta }(r^{-1}tr)={\frac {1}{|H|}}\sum _{s\in G, \atop s^{-1}ts\in H}^{}\chi _{\theta }(s^{-1}ts).}$

### Frobenius reciprocity

The message of the Frobenius reciprocity is, that the maps ${\displaystyle {\text{Res}}}$ and ${\displaystyle {\text{Ind}}}$ are adjoint to each other.

Let ${\displaystyle W}$ be an irreducible representation of ${\displaystyle H}$ and let ${\displaystyle V}$ be an irreducible representation of ${\displaystyle G,}$ then the Frobenius reciprocity tells us, that ${\displaystyle W}$ is contained in ${\displaystyle {\text{Res}}(V)}$ as often as ${\displaystyle {\text{Ind}}(W)}$ is contained in ${\displaystyle V.}$

Frobenius reciprocity. If ${\displaystyle \psi \in \mathbb {C} _{\text{class}}(H)}$ and ${\displaystyle \varphi \in \mathbb {C} _{\text{class}}(G)}$ we have ${\displaystyle \langle \psi ,{\text{Res}}(\varphi )\rangle _{H}=\langle {\text{Ind}}(\psi ),\varphi \rangle _{G}.}$

This statement is also valid for the inner product.

Proof. Every class function can be written as a linear combination of irreducible characters. As ${\displaystyle \langle \cdot ,\cdot \rangle }$ is a bilinear form, we can, without loss of generality, assume ${\displaystyle \psi }$ and ${\displaystyle \varphi }$ to be characters of irreducible representations of ${\displaystyle H}$ in ${\displaystyle W}$ and of ${\displaystyle G}$ in ${\displaystyle V,}$ respectively. We define ${\displaystyle \psi (s)=0}$ for all ${\displaystyle s\in G\setminus H.}$ Then we have

{\displaystyle {\begin{aligned}\langle {\text{Ind}}(\psi ),\varphi \rangle _{G}&={\frac {1}{|G|}}\sum _{t\in G}{\text{Ind}}(\psi )(t)\varphi (t^{-1})\\&={\frac {1}{|G|}}\sum _{t\in G}{\frac {1}{|H|}}\sum _{s\in G \atop s^{-1}ts\in H}\psi (s^{-1}ts)\varphi (t^{-1})\\&={\frac {1}{|G|}}{\frac {1}{|H|}}\sum _{t\in G}\sum _{s\in G}\psi (s^{-1}ts)\varphi ((s^{-1}ts)^{-1})\\&={\frac {1}{|G|}}{\frac {1}{|H|}}\sum _{t\in G}\sum _{s\in G}\psi (t)\varphi (t^{-1})\\&={\frac {1}{|H|}}\sum _{t\in G}\psi (t)\varphi (t^{-1})\\&={\frac {1}{|H|}}\sum _{t\in H}\psi (t)\varphi (t^{-1})\\&={\frac {1}{|H|}}\sum _{t\in H}\psi (t){\text{Res}}(\varphi )(t^{-1})\\&=\langle \psi ,{\text{Res}}(\varphi )\rangle _{H}\end{aligned}}}

In the course of this sequence of equations we used only the definition of induction on class functions and the properties of characters. ${\displaystyle \Box }$

Alternative proof. In terms of the group algebra, i.e. by the alternative description of the induced representation, the Frobenius reciprocity is a special case of a general equation for a change of rings:

${\displaystyle {\text{Hom}}_{\mathbb {C} [H]}(W,U)={\text{Hom}}_{\mathbb {C} [G]}(\mathbb {C} [G]\otimes _{\mathbb {C} [H]}W,U).}$

This equation is by definition equivalent to

${\displaystyle \langle W,{\text{Res}}(U)\rangle _{H}=\langle W,U\rangle _{H}=\langle {\text{Ind}}(W),U\rangle _{G}.}$

As this bilinear form tallies the bilinear form on the corresponding characters, the theorem follows without calculation. ${\displaystyle \Box }$

### Mackey's irreducibility criterion

George Mackey has established a criterion to verify the irreducibility of induced representations. For this we will first need some definitions and some specifications with respect to the notation.

Two representations ${\displaystyle V_{1}}$ and ${\displaystyle V_{2}}$ of a group ${\displaystyle G}$ are called disjoint, if they have no irreducible component in common, i.e. if ${\displaystyle \langle V_{1},V_{2}\rangle _{G}=0.}$

Let ${\displaystyle G}$ be a group and let ${\displaystyle H}$ be a subgroup. We define ${\displaystyle H_{s}=sHs^{-1}\cap H}$ for ${\displaystyle s\in G.}$ Let ${\displaystyle (\rho ,W)}$ be a representation of the subgroup ${\displaystyle H.}$ This defines by restriction a representation ${\displaystyle {\text{Res}}_{s}(\rho )}$ of ${\displaystyle H_{s}.}$ We write ${\displaystyle {\text{Res}}_{s}(\rho )}$ for ${\displaystyle {\text{Res}}_{H_{s}}(\rho ).}$ We also define another representation ${\displaystyle \rho ^{s}}$ of ${\displaystyle H_{s}}$ by ${\displaystyle \rho ^{s}(t)=\rho (s^{-1}ts).}$ These two representations are not to be confused.

Mackey's irreducibility criterion. The induced representation ${\displaystyle V={\text{Ind}}_{H}^{G}(W)}$ is irreducible if and only if the following conditions are satisfied:
• ${\displaystyle W}$ is irreducible
• For each ${\displaystyle s\in G\setminus H}$ the two representations ${\displaystyle \rho ^{s}}$ and ${\displaystyle {\text{Res}}_{s}(\rho )}$ of ${\displaystyle H_{s}}$ are disjoint.[2]

Starting from this theorem we obtain directly the following:

Corollary. Let ${\displaystyle H}$ be a normal subgroup of ${\displaystyle G.}$ Then ${\displaystyle {\text{Ind}}_{H}^{G}(\rho )}$ is irreducible if and only if ${\displaystyle \rho }$ is irreducible and not isomorphic to the conjugates ${\displaystyle \rho ^{s}}$ for ${\displaystyle s\notin H.}$

Proof. As ${\displaystyle H}$ is normal, we have ${\displaystyle H_{s}=H}$ and ${\displaystyle {\text{Res}}_{s}(\rho )=\rho .}$ Thus, the statement follows directly from the criterion of Mackey.${\displaystyle \Box }$

### Applications to special groups

In this chapter we present some applications of the so far presented theory to normal subgroups and to a special group, the semidirect product of a subgroup with an abelian normal subgroup.

Proposition. Let ${\displaystyle A}$ be a normal subgroup of the group ${\displaystyle G}$ and let ${\displaystyle \rho :G\to {\text{GL}}(V)}$ be an irreducible representation of ${\displaystyle G.}$ Then one of the following statements has to be valid:
• either there exists a true subgroup ${\displaystyle H}$ of ${\displaystyle G,}$ which contains ${\displaystyle A,}$ and an irreducible representation ${\displaystyle \eta }$ of ${\displaystyle H,}$ which induces ${\displaystyle \rho ,}$
• or the restriction of ${\displaystyle \rho }$ onto ${\displaystyle A}$ is isotypic.

If ${\displaystyle A}$ is abelian, the second point of the proposition above is equivalent to the statement, that ${\displaystyle \rho (a)}$ is a homothety for every ${\displaystyle a\in A.}$

We obtain also the following

Corollary. Let ${\displaystyle A}$ be an abelian normal subgroup of ${\displaystyle G}$ and let ${\displaystyle \tau }$ be any irreducible representation of ${\displaystyle G.}$ We denote with ${\displaystyle (G:A)}$ the index of ${\displaystyle A}$ in ${\displaystyle G.}$ Then ${\displaystyle \deg(\tau )|(G:A).}$

If ${\displaystyle A}$ is an abelian subgroup of ${\displaystyle G}$ (not necessarily normal), generally ${\displaystyle \deg(\tau )|(G:A)}$ is not satisfied, but nevertheless ${\displaystyle \deg(\tau )\leq (G:A)}$ is still valid.

Now we show, how all irreducible representations of a group ${\displaystyle G,}$ which is the semidirect product of an abelian normal subgroup ${\displaystyle A\vartriangleleft G}$ and a subgroup ${\displaystyle H\leq G,}$ can be classified.

In the following, let ${\displaystyle A}$ and ${\displaystyle H}$ be subgroups of the group ${\displaystyle G,}$ where ${\displaystyle A}$ is assumed to be normal and abelian. Additionally, assume that ${\displaystyle G}$ is the semidirect product of ${\displaystyle H}$ and