---
title: A Glimpse of Representation and Character Theory
module: "Capstone: Where Algebra Goes Next"
moduleNumber: 12
lessonNumber: 2
order: 1202
summary: >
  Representation theory studies a group by the ways it can act linearly on a
  vector space. Representations are equivalent to modules over the group ring;
  Maschke's theorem gives complete reducibility, the Wedderburn consequences
  bound the irreducible degrees, and character theory reduces a representation to
  a trace invariant governed by the orthogonality relations and displayed in the
  character table of a small group.
topics: ["Capstone: Where Algebra Goes Next"]
sources:
  - book: Dummit & Foote
    ref: "Ch. 18 — Representation Theory and Character Theory; §18.1 Linear Actions and Modules over Group Rings"
  - book: Dummit & Foote
    ref: "§18.3 Character Theory and the Orthogonality Relations; §19.1 Characters of Groups of Small Order"
draft: false
---

A finite group can act on a vector space by invertible linear maps, and
representation theory studies those linear actions. They turn questions about group
structure into questions about matrices, modules, and eventually a small table of
complex numbers. The theory is an application of the
[modules-over-a-ring](/abstract-algebra/module-theory/introduction-to-modules)
viewpoint, with one particular ring — the group ring — playing the role that the
polynomial ring played in
[commutative algebra](/abstract-algebra/capstone/commutative-algebra-and-algebraic-geometry).
Throughout, $G$ is a finite group, $F$ is a field, and $V$ is a vector space over
$F$; the deepest results assume $F = \mathbb{C}$.

## Representations

> **Definition (Representation).** A **linear representation** of $G$ on $V$ is a
> group homomorphism $\varphi: G \to GL(V)$ into the group of invertible linear
> maps on $V$. The **degree** is $\dim_F V$. Choosing a basis turns it into a
> **matrix representation** $G \to GL_n(F)$. A representation is **faithful** if it
> is injective.

The homomorphism condition $\varphi(gh) = \varphi(g)\varphi(h)$ is the entire
content: group multiplication is realized as matrix multiplication. Every group has
several standard representations.[^df-lin]

- **Trivial representation.** Degree $1$, with $\varphi(g) = 1$ for all $g$. Never
  faithful when $|G| > 1$.
- **Regular representation.** Let $G$ act on itself by left multiplication; the
  permutation of basis vectors $g \cdot g_i = g g_i$ gives a faithful degree-$|G|$
  representation whose matrices are permutation matrices.
- **Dihedral and quaternion.** The
  [symmetries of the $n$-gon](/abstract-algebra/groups-and-symmetry/dihedral-and-symmetric-groups)
  give a faithful degree-$2$ real representation of $D_{2n}$ by an explicit rotation
  matrix and a reflection matrix; the
  [quaternion group $Q_8$](/abstract-algebra/groups-and-symmetry/matrix-and-quaternion-groups)
  has a faithful degree-$2$ complex representation.

$$
% caption: A representation sends each group element to an invertible matrix and
% respects products, so the group's multiplication becomes matrix multiplication.
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  n/.style={circle, draw, minimum size=7mm}]
\definecolor{acc}{HTML}{4A6FA5}
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\node[font=\small] at (5,2.0) {$GL_n(\mathbf{C})$};
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$$

### The group ring

A single ring built from $G$ and $F$ holds all representations of $G$ at once.

> **Definition (Group ring).** The **group ring** $FG$ is the set of formal sums
> $\sum_{i} a_i g_i$ with $a_i \in F$ and $g_i \in G$, added coordinatewise and
> multiplied by extending $(a g)(b h) = (ab)(gh)$ linearly.

As an $F$-vector space $FG$ has the elements of $G$ as a basis, so $\dim_F FG =
|G|$, and $F$ sits in its center, making $FG$ an **$F$-algebra**.[^df-lin] It is
commutative exactly when $G$ is abelian, and it has zero divisors whenever $|G| >
1$: for a nonidentity $g$ of order $n$, the element $r = 1 + g + \cdots + g^{n-1}$
satisfies $r(1 - g) = 0$. The group ring plays the same structural role here that
$F[x]$ played for a single linear operator in the
[rational canonical form](/abstract-algebra/modules-over-pids/rational-canonical-form).

A representation and an $FG$-module are the same data. Given $\varphi: G \to
GL(V)$, define a module action by $\bigl(\sum a_i g_i\bigr) \cdot v = \sum a_i\,
\varphi(g_i)(v)$; the module axioms hold precisely because $\varphi$ is a
homomorphism. Conversely an $FG$-module $V$ is an $F$-vector space on which each $g$
acts invertibly and $F$-linearly, which is a representation.[^df-lin]

$$
% caption: Giving a representation is the same as making the vector space a module
% over the group ring; the submodules coincide with the invariant subspaces.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  b/.style={draw, minimum width=40mm, minimum height=14mm, align=center}]
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\node[b] (l) at (0,0) {representation\\ of $G$ on $V$};
\node[b] (r) at (6.8,0) {$FG$-module\\ structure on $V$};
\draw[->, acc, thick] (l.north east) .. controls (2.7,1.4) and (4.1,1.4) .. (r.north west)
  node[midway, above, font=\scriptsize]{read $V$ as a module};
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  node[midway, below, font=\scriptsize]{recover the action};
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$$

Under this dictionary, a subspace $U \subseteq V$ is a submodule exactly when it is
**$G$-invariant** ($g \cdot u \in U$ for all $g, u$), and two representations are
**equivalent** when their modules are isomorphic — equivalently, related by a single
change of basis $P$ with $\varphi(g) = P\, \psi(g)\, P^{-1}$ for all $g$.[^df-lin]

## Complete reducibility

We break a representation into indivisible pieces.

> **Definition (Reducibility).** An $FG$-module $V \neq 0$ is **irreducible** (or
> simple) if its only submodules are $0$ and $V$. It is **indecomposable** if it is
> not a direct sum of two nonzero submodules, and **completely reducible** if it is
> a direct sum of irreducible submodules.

In matrix terms, a reducible representation has a basis putting every $\varphi(g)$
in block upper-triangular form, and a decomposable one puts them all in block
diagonal form simultaneously. Degree-$1$ representations are automatically
irreducible. Whether a representation can be fully broken apart depends on the field.

> **Theorem (Maschke).** Let $G$ be finite and $F$ a field whose characteristic
> does not divide $|G|$. If $V$ is an $FG$-module and $U \subseteq V$ a submodule,
> then $V = U \oplus W$ for some submodule $W$: every submodule is a direct summand.

The proof averages a vector-space projection $\pi_0: V \to U$ over the group to make
it $G$-equivariant:
$$
\pi = \frac{1}{|G|} \sum_{g \in G} g\, \pi_0\, g^{-1},
$$
which requires exactly that $|G|$ be invertible in $F$.[^df-lin] The result is an
$FG$-module projection onto $U$ whose kernel is the complement $W$. When $F$ has
characteristic $0$ — in particular $F = \mathbb{C}$ — the hypothesis is automatic.

> **Corollary (Complete reducibility).** If $\operatorname{char} F \nmid |G|$, then
> every finite-dimensional $FG$-module is completely reducible: a direct sum of
> irreducibles.

$$
% caption: Under Maschke's hypothesis every representation splits; one change of
% basis puts every matrix in the same block-diagonal form built from irreducibles.
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  \draw (0,0) rectangle (2.6,2.6);
  \node at (1.3,-0.42) {block diagonal};
\end{scope}
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$$

The hypothesis is essential. Over $\mathbb{F}_p$ with $g$ of order $p$ acting on a
$2$-dimensional space by a single Jordan block, the invariant line has no invariant
complement: the module is reducible but indecomposable, so not completely
reducible.[^df-lin] Modular representation theory, over fields whose characteristic
divides $|G|$, is a separate and harder subject.

## The Wedderburn count

Complete reducibility says $\mathbb{C}G$ is **semisimple**, and Wedderburn's theorem
describes such a ring as a direct product of matrix algebras over $\mathbb{C}$. Character
theory uses only two numerical consequences, both constraining the
irreducibles.[^df-char]

> **Theorem (Wedderburn consequences over $\mathbb{C}$).** A finite group $G$ has
> only finitely many irreducible complex representations, up to equivalence, say of
> degrees $d_1, \dots, d_r$. Then
> $$
> d_1^2 + d_2^2 + \cdots + d_r^2 = |G|,
> $$
> and $r$ equals the number of conjugacy classes of $G$.

The sum-of-squares identity comes from decomposing the regular representation, which
contains each irreducible exactly $d_i$ times, so $\sum d_i \cdot d_i = \dim
\mathbb{C}G = |G|$. Separately, the number of degree-$1$ representations equals the
index $|G : G'|$ of the
[commutator subgroup](/abstract-algebra/subgroups-and-quotients/composition-series-and-the-alternating-group),
since a degree-$1$ representation factors through the abelianization $G/G'$.[^df-char]

$$
% caption: The squared degrees of the irreducibles tile a region of area equal to
% the group order; for the symmetric group on three letters this is $1 + 1 + 4 = 6$.
\begin{tikzpicture}[font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[fill=acc!10] (0,0) rectangle (1,1); \node at (0.5,0.5){$1$};
\draw[fill=acc!10] (1,0) rectangle (2,1); \node at (1.5,0.5){$1$};
\draw[fill=acc!18] (2,0) rectangle (4,2); \node at (3,1){$4$};
\node[anchor=north] at (0.5,-0.15){$d=1$};
\node[anchor=north] at (1.5,-0.15){$d=1$};
\node[anchor=north] at (3,-0.15){$d=2$};
\node[anchor=west] at (4.4,1){total area $= 6$, the group order};
\end{tikzpicture}
$$

These constraints often force a unique list of degrees.

> **Worked example (degrees from the count).** Two numbers — the group order and the
> index $|G : G'|$ — fix the irreducible degrees of many small groups.
> - **$S_3$** ($|G| = 6$). Nonabelian, so $G' \neq 1$; in fact $G' = A_3$ has index
>   $2$, giving two degree-$1$ characters. Completing $1 + 1 + d^2 = 6$ in positive
>   integers forces $d = 2$, so the degrees are $1, 1, 2$.
> - **$D_8$** ($|G| = 8$). The commutator subgroup $\langle r^2 \rangle$ has index
>   $4$, so four degree-$1$ characters, and $4 + d^2 = 8$ leaves one character of
>   degree $2$. The $4 + 1 = 5$ irreducibles match the five conjugacy classes.
> - **$S_4$** ($|G| = 24$). Here $G' = A_4$ has index $2$, so two degree-$1$
>   characters; the remaining squares sum to $22 = 2^2 + 3^2 + 3^2$, and with five
>   conjugacy classes the degrees are $1, 1, 2, 3, 3$.[^df-char]

## Characters

The trace of $\varphi(g)$ compresses a whole matrix into one scalar, and over
$\mathbb{C}$ the resulting function on $G$ already determines the representation up
to equivalence. Writing down matrices of degree $100$ is unnecessary.

> **Definition (Character).** The **character** of a representation $\varphi$
> afforded by $V$ is $\chi(g) = \operatorname{tr}\varphi(g)$. A **class function**
> is any function $G \to \mathbb{C}$ constant on conjugacy classes.

Because trace is conjugation-invariant, $\chi(h^{-1} g h) = \chi(g)$, so **every
character is a class function**. Two basic values are read off immediately:
$\chi(1)$ is the degree of the representation, and equivalent representations have
equal characters, since similar matrices have equal traces.[^df-char2] The character
of the trivial representation is the constant $1$, called the **principal
character**.

Class functions form a vector space of dimension $r$ (one basis function per
conjugacy class), and the number of irreducible characters is also $r$. Putting a Hermitian inner product on this space makes the irreducibles an
orthonormal basis.

> **Definition (Inner product of class functions).** For class functions $\theta,
> \psi$,
> $$
> \langle \theta, \psi \rangle
> = \frac{1}{|G|} \sum_{g \in G} \theta(g)\, \overline{\psi(g)}.
> $$

> **Theorem (First orthogonality relation).** The irreducible complex characters
> $\chi_1, \dots, \chi_r$ satisfy $\langle \chi_i, \chi_j \rangle = \delta_{ij}$ and
> form an orthonormal basis for the space of class functions. Consequently a class
> function $\theta$ has $\theta = \sum_i \langle \theta, \chi_i \rangle \chi_i$.

$$
% caption: The irreducible characters are orthonormal; the inner product of two of
% them is one when they are equal and zero otherwise.
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\node at (1.5,2.5){$0$}; \node at (2.5,2.5){$0$};
\node at (0.5,1.5){$0$}; \node at (2.5,1.5){$0$};
\node at (0.5,0.5){$0$}; \node at (1.5,0.5){$0$};
\node[anchor=east] at (-0.12,2.5){$X_1$};
\node[anchor=east] at (-0.12,1.5){$X_2$};
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\node[anchor=south] at (0.5,3.12){$X_1$};
\node[anchor=south] at (1.5,3.12){$X_2$};
\node[anchor=south] at (2.5,3.12){$X_3$};
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$$

Orthonormality gives an immediate irreducibility test: a character $\chi$ has
$\langle \chi, \chi \rangle = 1$ **if and only if it is irreducible**, and the
multiplicity of $\chi_i$ in any representation with character $\theta$ is $\langle
\theta, \chi_i \rangle$. The companion statement about columns is the second
orthogonality relation: distinct columns of the table below are orthogonal, and the
squared column norm at $g$ is $|C_G(g)|$.[^df-char2]

## The character table

The **character table** of $G$ arranges the values $\chi_i(g_j)$ in a grid: one row
per irreducible character, one column per conjugacy class. It is a square $r \times
r$ array, and it is a near-complete fingerprint of the group.

For the symmetric group $S_3$, there are three conjugacy classes — the identity, the
three transpositions, and the two $3$-cycles — hence three irreducible characters.
The degrees are $1, 1, 2$ from the count above. The two degree-$1$ characters are the
[trivial and sign homomorphisms](/abstract-algebra/groups-and-symmetry/homomorphisms-and-group-actions),
and the degree-$2$ character is found by orthogonality.[^df-table]

| $S_3$ | $1$ | $(1\,2)$ | $(1\,2\,3)$ |
| --- | --- | --- | --- |
| class size | $1$ | $3$ | $2$ |
| $\chi_{\text{triv}}$ | $1$ | $1$ | $1$ |
| $\chi_{\text{sign}}$ | $1$ | $-1$ | $1$ |
| $\chi_{2}$ | $2$ | $0$ | $-1$ |

> **Worked example (character table of $S_3$).** Build the three rows from the
> degree-$1$ homomorphisms and the permutation representation, using the class sizes
> $1, 3, 2$.
>
> **Rows $1$ and $2$ (degree $1$).** The trivial character is $\chi_{\text{triv}} =
> (1, 1, 1)$. The sign character sends even permutations to $+1$ and odd ones to
> $-1$; transpositions are odd and $3$-cycles even, so $\chi_{\text{sign}} = (1, -1,
> 1)$.
>
> **Row $3$ (degree $2$).** Let $S_3$ act on the three symbols. The character $\pi$
> of this degree-$3$ permutation representation counts fixed points — the identity
> fixes three, a transposition fixes one, a $3$-cycle fixes none:
> $$
> \pi(1) = 3, \qquad \pi\bigl((1\,2)\bigr) = 1, \qquad \pi\bigl((1\,2\,3)\bigr) = 0.
> $$
> The all-ones vector spans a trivial subrepresentation, so
> $\pi = \chi_{\text{triv}} + \chi_2$ and
> $$
> \chi_2 = \pi - \chi_{\text{triv}} = (3 - 1,\; 1 - 1,\; 0 - 1) = (2, 0, -1).
> $$
>
> **Orthonormality.** With the class-weighted inner product $\langle \theta, \psi
> \rangle = \tfrac{1}{6} \sum_c |c|\, \theta(c)\, \overline{\psi(c)}$,
> $$
> \langle \chi_2, \chi_2 \rangle = \tfrac{1}{6}\bigl(1\cdot 2^2 + 3\cdot 0^2 +
> 2\cdot(-1)^2\bigr) = \tfrac{1}{6}(4 + 2) = 1,
> $$
> so $\chi_2$ is irreducible, while
> $$
> \langle \chi_{\text{triv}}, \chi_2 \rangle = \tfrac{1}{6}\bigl(1\cdot 1\cdot 2 +
> 3\cdot 1\cdot 0 + 2\cdot 1\cdot(-1)\bigr) = \tfrac{1}{6}(2 - 2) = 0,
> $$
> $$
> \langle \chi_{\text{triv}}, \chi_{\text{sign}} \rangle = \tfrac{1}{6}\bigl(1 - 3 +
> 2\bigr) = 0.
> $$
> The three rows are orthonormal, and $1^2 + 1^2 + 2^2 = 6 = |G|$ confirms there are
> no further irreducibles.[^df-table]

The same method builds the tables of $D_8$ and $Q_8$, each with four degree-$1$
characters and one degree-$2$ character. The two groups share an identical character
table even though they are not isomorphic, so the table does not always determine the
group.[^df-table]

Multiplicities read off the same inner product. The regular representation decomposes
with each irreducible appearing as often as its degree.

> **Worked example (decomposing the regular representation).** The regular
> representation of $S_3$ has character $\rho(g) = |G|$ at $g = 1$ and $0$ otherwise,
> so $\rho = (6, 0, 0)$ on the three classes. Since $\rho$ vanishes off the identity,
> $$
> \langle \rho, \chi_i \rangle = \tfrac{1}{6}\bigl(1\cdot 6\cdot \overline{\chi_i(1)}\bigr)
> = \chi_i(1) = d_i,
> $$
> so each irreducible occurs with multiplicity equal to its degree:
> $$
> \rho = \chi_{\text{triv}} + \chi_{\text{sign}} + 2\chi_2, \qquad
> \dim = 1 + 1 + 2\cdot 2 = 6 = |G|.
> $$
> This is the sum-of-squares identity $\sum d_i^2 = |G|$ seen one representation at a
> time.[^df-char2]

## Further directions

Character theory converts group-theoretic questions into linear algebra over a
finite table. Counting solutions, detecting normal subgroups (a normal subgroup is a
union of conjugacy classes, visible as a set of columns), and Burnside's theorem that
groups of order $p^a q^b$ are solvable all reduce to manipulating characters. Two
generalizations extend the theory: **induced characters**, which build
representations of $G$ from those of a subgroup, and **modular representation
theory** over fields where Maschke's theorem fails and the group ring is no longer
semisimple. Both study a group through the modules over its group ring — the same
[module-theoretic](/abstract-algebra/module-theory/introduction-to-modules) lens the
[commutative-algebra capstone](/abstract-algebra/capstone/commutative-algebra-and-algebraic-geometry)
applied to polynomial rings.

[^df-lin]: Dummit & Foote, §18.1 — linear and matrix representations, the group ring $FG$ as an $F$-algebra, the equivalence between representations and $FG$-modules, invariant subspaces as submodules, and Maschke's theorem with the averaging projection.
[^df-char]: Dummit & Foote, §18.2 — Wedderburn's theorem, semisimplicity of $\mathbb{C}G$, the finiteness of the irreducibles, the identity $\sum d_i^2 = |G|$, and the count $|G:G'|$ of degree-$1$ representations.
[^df-char2]: Dummit & Foote, §18.3 — characters as traces, class functions, the equality of the number of irreducible characters with the number of conjugacy classes, the Hermitian inner product, and the first and second orthogonality relations.
[^df-table]: Dummit & Foote, §19.1 — the character table, the tables of $\mathbb{Z}_n$, $S_3$, $D_8$, and $Q_8$, and the fact that $D_8$ and $Q_8$ share a character table without being isomorphic.