The First and Second Orthogonality Relations

Q: Why does the second relation produce centraliser orders rather than class sizes?

Because comparing coefficients of a group element h in the identity for C_g divides by m_g, and |G|/m_g = |C_G(g)| by orbit-stabiliser. Intuitively the column at g is long exactly when g has a large centraliser — the identity column, with centraliser G, has squared length |G|.

Q: Do the relations still hold in characteristic p when p does not divide the group order?

Yes, exactly as stated: the proofs use only that |G| and the degrees n_i are invertible in k and that k splits G. What is lost is every argument that treats the values as complex numbers — positivity, the irreducibility test, and any appeal to complex conjugation.

Q: Can two different groups have the same character table?

Yes. The dihedral group of order 8 and the quaternion group of order 8 have identical character tables under a suitable matching of classes, though they are not isomorphic. So the orthogonality relations, however rigid, do not determine the group; recovering the group needs extra data such as the class multiplication coefficients or the power maps.

Q: What is the analogue for a non-splitting field such as the rationals?

The irreducible k-characters are sums over Galois orbits of absolutely irreducible characters, weighted by Schur indices. Orthogonality survives in the form [,] = m^2 |orbit| rather than 1, and the character table stops being square in the sense used here. The *Splitting Fields for Finite Groups* page describes when the problem disappears.

Q: How do I use the relations to decompose a permutation representation?

The permutation character is (g) = |Fix(g)|. In characteristic zero the multiplicity of M_i is [,_i], and [,] is the number of orbits of G on ordered pairs, that is, the rank of the permutation action. In particular the action is doubly transitive exactly when [,] = 2.

Executive Summary

For a finite group $G$ and a splitting field $k$ with $char k ∤ | G |$ , the character table is a square $r \times r$ matrix, $r$ being the common number of conjugacy classes and of simple $k G$ -modules. The orthogonality relations say that this matrix is essentially its own inverse: its rows are orthonormal for a natural bilinear form, and its columns are orthogonal with prescribed squared lengths.

They are due to Frobenius and are the single most used computational tool in character theory. In Lam's development they are cheap: each is two lines from the formulas of the Central Idempotents and Characters page.

$δ_{i j} | G |$ Row product

$| C_{G} (g) |$ Column square

$r \times r$ Table size

1896Frobenius

Overview

Two structures compete for the same $r$ -dimensional space. The class functions $F_{k} (G)$ carry the irreducible characters; the centre $Z (k G)$ carries the class sums and the block idempotents. The orthogonality relations are the statement that the characters and the class sums are, up to explicit scalars, dual bases of one another.

\sum_{g \in G} χ_{i} (g^{- 1}) χ_{j} (g) = δ_{i j} | G | (first, rows)

(8.16A)

\sum_{i = 1}^{r} χ_{i} (g) χ_{i} (h^{- 1}) = {\begin{cases} | C_{G} (g) | & h conjugate to g, \\ [2 p t] 0 & otherwise \end{cases} (second, columns)

(8.16B)

The asymmetry between the two right-hand sides is only apparent: $| G |$ is the order of the centraliser of the identity, and $m_{g} | C_{G} (g) | = | G |$ turns each relation into the other after transposing. The next section makes that precise as a single matrix identity.

Learning Objectives

State $(8.16)$ (A) and (B) with the hypotheses $G$ finite, $k$ splitting, $char k ∤ | G |$ .
Derive (A) by applying $χ_{j}$ to the formula for $e_{i}$ .
Derive (B) by substituting the formula for $e_{i}$ into the formula for $C_{g}$ .
Define the pairings on $F_{k} (G)$ and on $Z (k G)$ and state $(8.17)$ .
Write both relations as the matrix identity $X^{*} D X^{T} = | G | I$ and its inverse form.
Reconstruct a missing character-table row from column orthogonality.

Definitions

Standing hypotheses: G is a finite group, k is a splitting field for G, and the characteristic of k does not divide the order of G. The simple modules are M_1 to M_r with characters chi_i and degrees n_i.

$F_{k} (G)$: The $k$ -space of class functions $G \to k$ . Its dimension is $r$ , the number of conjugacy classes, with the class indicator functions as an obvious basis.
$[α, β]$ on $F_{k} (G)$: $[α, β] = | G |^{- 1} \sum_{g \in G} α (g) β (g^{- 1})$ . This is a symmetric $k$ -bilinear form; substituting $g \mapsto g^{- 1}$ shows the symmetry.
$[α, β]$ on $Z (k G)$: For $α = \sum α_{g} g$ and $β = \sum β_{g} g$ put $\bar{β} = \sum β_{g} g^{- 1}$ and $[α, β] = | G |^{- 1} χ_{reg} (α \bar{β}) = \sum_{g \in G} α_{g} β_{g}$ .
$m_{g}$ , $C_{G} (g)$: The size of the conjugacy class of $g$ and the centraliser of $g$ ; the orbit-stabiliser theorem gives $m_{g} \cdot | C_{G} (g) | = | G |$ .
$X$ , the character table: With $g_{1}, \dots, g_{r}$ representatives of the classes, $X = (χ_{i} (g_{t}))_{i, t}$ ; write $X^{*} = (χ_{i} (g_{t}^{- 1}))_{i, t}$ and $D = diag (m_{g_{1}}, \dots, m_{g_{r}})$ .

Core Concepts

One matrix identity, two readings

Group the sum in (A) by conjugacy classes. Since $χ_{i}$ is a class function, the $m_{g_{t}}$ elements of the $t$ -th class contribute equally, so (A) becomes

X^{*} D X^{T} = | G | I_{r},

(8.16A′)

$(i, j)$ entry: $\sum_{t} m_{g_{t}} χ_{i} (g_{t}^{- 1}) χ_{j} (g_{t}) = δ_{i j} | G |$ .

A square matrix with a one-sided inverse has that same matrix as a two-sided inverse. Hence $(D X^{T}) X^{*} = | G | I_{r}$ , that is $X^{T} X^{*} = | G | D^{- 1}$ , whose $(s, t)$ entry says

\sum_{i = 1}^{r} χ_{i} (g_{s}) χ_{i} (g_{t}^{- 1}) = δ_{st} \frac{| G |}{m_{g_{t}}} = δ_{st} | C_{G} (g_{t}) | .

(8.16B′)

So the second relation is a formal consequence of the first — the reason they come in pairs. Lam instead derives (B) directly by substitution, which has the advantage of never invoking invertibility of $X$ .

Why the table is square

The identity above presupposes that the number of irreducible characters equals the number of conjugacy classes. That is exactly the split semisimple case: taking centres in the Wedderburn decomposition gives $Z (k G) ≅ Z (D_{1}) \times \dots \times Z (D_{r})$ , and when $k$ splits all $D_{i} = k$ , so $\dim_{k} Z (k G) = r$ on both counts. Drop the splitting hypothesis and the table is no longer square; drop $char k ∤ | G |$ and the count changes again — see the Structure of kG modulo Its Radical page.

Two immediate specialisations

Take $g = h = 1$ in (B): $\sum_{i} n_{i}^{2} = | C_{G} (1) | = | G |$ , the dimension count of the Wedderburn decomposition.
Take $g = 1$ and $h \neq 1$ in (B): $\sum_{i} n_{i} χ_{i} (h^{- 1}) = 0$ , which is the vanishing of the regular character off the identity.
Take $i = j$ in (A) with $χ_{i}$ the trivial character: $| G | = | G |$ , no information — the relations are sharpest on the nonlinear characters.

Key Results

Theorem(8.16)Frobenius: first and second orthogonality relations

Let $G$ be a finite group and $k$ a splitting field for $G$ with $char k ∤ | G |$ . Let $χ_{1}, \dots, χ_{r}$ be the irreducible $k$ -characters of $G$ , of degrees $n_{1}, \dots, n_{r}$ . Then:

(A) $displaystyle \sum_{g \in G} χ_{i} (g^{- 1}) χ_{j} (g) = δ_{i j} | G |$ for all $i, j$ .
(B) $displaystyle \sum_{i = 1}^{r} χ_{i} (g) χ_{i} (h^{- 1}) = ε | C_{G} (g) |$ for all $g, h \in G$ , where $ε = 1$ if $g$ and $h$ are conjugate in $G$ and $ε = 0$ otherwise, and $C_{G} (g)$ is the centraliser of $g$ .

Proof

(A). Apply the character $χ_{j}$ to the identity $e_{i} = | G |^{- 1} n_{i} \sum_{g} χ_{i} (g^{- 1}) g$ of $(8.15) (1)$ . The left-hand side gives $χ_{j} (e_{i}) = δ_{i j} n_{j}$ . The right-hand side is $| G |^{- 1} n_{i} \sum_{g} χ_{i} (g^{- 1}) χ_{j} (g)$ by $k$ -linearity of $χ_{j}$ . Hence

\frac{n_{i}}{| G |} \sum_{g \in G} χ_{i} (g^{- 1}) χ_{j} (g) = δ_{i j} n_{i} .

For $i = j$ divide by $n_{i}$ , which is legitimate because $(8.15) (1)$ guarantees $n_{i} \neq 0$ in $k$ ; for $i \neq j$ the same division gives $0$ . Either way (A) follows.

(B). Substitute $(8.15) (1)$ into $(8.15) (2)$ . Starting from $C_{g} = m_{g} \sum_{i} n_{i}^{- 1} χ_{i} (g) e_{i}$ and replacing each $e_{i}$ ,

C_{g} = m_{g} \sum_{i = 1}^{r} \frac{χ_{i} (g)}{n_{i}} \cdot \frac{n_{i}}{| G |} \sum_{h \in G} χ_{i} (h^{- 1}) h = \frac{m_{g}}{| G |} \sum_{h \in G} (\sum_{i = 1}^{r} χ_{i} (g) χ_{i} (h^{- 1})) h .

Now compare coefficients of $h$ on the two sides. On the left the coefficient is $1$ if $h$ is conjugate to $g$ and $0$ otherwise. So for $h$ conjugate to $g$ , $\sum_{i} χ_{i} (g) χ_{i} (h^{- 1}) = | G | / m_{g}$ , and this is $| C_{G} (g) |$ by orbit-stabiliser; for $h$ not conjugate to $g$ the inner sum is $0$ .

Corollary(8.17)Dual bases

Under the hypotheses of $(8.16)$ :

The $r$ irreducible characters $χ_{1}, \dots, χ_{r}$ form an orthonormal basis of $F_{k} (G)$ for the form $[α, β] = | G |^{- 1} \sum_{g} α (g) β (g^{- 1})$ .
The $r$ class sums $C_{g_{1}}, \dots, C_{g_{r}}$ form an orthogonal basis of $Z (k G)$ , with $[C_{g_{t}}, C_{g_{t}}] = m_{g_{t}}$ .
$F_{k} (G)$ and $Z (k G)$ are dual $k$ -spaces under the pairing $(μ, α) \mapsto μ (α)$ obtained by extending $μ$ linearly to $k G$ ; with respect to it ${n_{i}^{- 1} χ_{i}}$ and ${e_{i}}$ are dual bases.

Proof

(1) Orthonormality is exactly (A) divided by $| G |$ . Since $\dim_{k} F_{k} (G) = r$ and orthonormal vectors are linearly independent, the $χ_{i}$ are a basis.

(2) By definition $[α, β] = \sum_{g} α_{g} β_{g}$ . Distinct class sums have disjoint supports, so their pairing is $0$ ; and $C_{g}$ pairs with itself to give $m_{g}$ terms each equal to $1$ .

(3) The pairing is well defined because $μ$ extends linearly and $α$ has finite support. Duality follows from the computation $⟨ n_{i}^{- 1} χ_{i}, e_{j} ⟩ = n_{i}^{- 1} χ_{i} (e_{j}) = n_{i}^{- 1} δ_{i j} n_{i} = δ_{i j}$ , using again that $n_{i}$ is invertible in $k$ . A pairing with dual bases is non-degenerate.

Corollary(8.17a)Coordinates of a class function

Every class function $f \in F_{k} (G)$ has the expansion $f = \sum_{i = 1}^{r} [f, χ_{i}] χ_{i}$ , and for $f, f' \in F_{k} (G)$ one has the Plancherel identity $[f, f'] = \sum_{i = 1}^{r} [f, χ_{i}] [f', χ_{i}]$ .

Proof

Write $f = \sum_{j} c_{j} χ_{j}$ by $(8.17) (1)$ ; pairing with $χ_{i}$ and using orthonormality gives $c_{i} = [f, χ_{i}]$ . For the second identity expand both arguments and use bilinearity together with $[χ_{i}, χ_{j}] = δ_{i j}$ .

Remark—Multiplicities and the irreducibility test

If in addition $char k = 0$ , then for a finite-dimensional $k G$ -module $M$ the coefficient $[χ_{M}, χ_{i}]$ is the multiplicity of $M_{i}$ in $M$ , so it is a non-negative integer, and $M$ is simple if and only if $[χ_{M}, χ_{M}] = 1$ . The characteristic-zero hypothesis is essential: the criterion counts multiplicities as integers, and in characteristic $p$ the quantity $[χ_{M}, χ_{M}]$ is only known modulo $p$ .

Proof Techniques and Method

How these proofs work, and which move to reuse.

Start from an identity in $Z (k G)$ Both relations begin with one of the two formulas of

(8.15)

; nothing else about

G

is used.

Apply a character, or compare coefficientsApplying

χ_{j}

collapses idempotents to

δ_{i j} n_{j}

. Comparing coefficients of a group element extracts one scalar equation per element of

G

Divide by a degreeEvery proof here divides by some

n_{i}

; the licence is the non-vanishing corollary of

(8.15) (1)

, which must be quoted, not assumed.

Transpose to get the companion relationThe column relation is the row relation for the transposed table, because a one-sided inverse of a square matrix is two-sided.

Note what is not used: no averaging over $G$ , no Maschke-style projection, no Schur's Lemma argument about intertwiners. The classical route to orthogonality runs through Schur's Lemma applied to $\sum_{g} D_{i} (g) A D_{j} (g)^{- 1}$ ; Lam's route replaces it by two coefficient comparisons, at the cost of having established the Wedderburn decomposition first.

Worked Example

Take $G = S_{4}$ , $| G | = 24$ , over $k = ℚ$ — a splitting field for $S_{4}$ . The five classes are represented by $1$ , $(1 2)$ , $(1 2 3)$ , $(1 2 3 4)$ and $(1 2) (3 4)$ , of sizes $1, 6, 8, 6, 3$ ; note $1 + 6 + 8 + 6 + 3 = 24$ . Every element of $S_{4}$ is conjugate to its inverse, so $χ_{i} (g^{- 1}) = χ_{i} (g)$ throughout.

Character table of $S_{4}$
	$1$	$(1 2)$	$(1 2 3)$	$(1 2 3 4)$	$(1 2) (3 4)$
class size	$1$	$6$	$8$	$6$	$3$
$χ_{1}$ trivial	$1$	$1$	$1$	$1$	$1$
$χ_{2}$ sign	$1$	$- 1$	$1$	$- 1$	$1$
$χ_{3}$	$2$	$0$	$- 1$	$0$	$2$
$χ_{4}$ standard	$3$	$1$	$0$	$- 1$	$- 1$
$χ_{5} = χ_{4} χ_{2}$	$3$	$- 1$	$0$	$1$	$- 1$

Checking the first relation

Weight each column by its class size. For $i = j = 4$ :

\sum_{g} χ_{4} (g)^{2} = 1 \cdot 9 + 6 \cdot 1 + 8 \cdot 0 + 6 \cdot 1 + 3 \cdot 1 = 24 = | G | .

For $i = 4$ , $j = 5$ : $1 \cdot 9 + 6 (1) (- 1) + 0 + 6 (- 1) (1) + 3 (- 1) (- 1) = 9 - 6 - 6 + 3 = 0$ . For $i = 1$ , $j = 3$ : $1 \cdot 2 + 0 + 8 (- 1) + 0 + 3 \cdot 2 = 2 - 8 + 6 = 0$ .

Checking the second relation

Column squares should be centraliser orders $| G | / m_{g}$ :

Column sums of squares against centraliser orders
Class $g$	$\sum_{i} χ_{i} (g)^{2}$	$\| G \| / m_{g}$
$1$	$1 + 1 + 4 + 9 + 9 = 24$	$24$
$(1 2)$	$1 + 1 + 0 + 1 + 1 = 4$	$24 / 6 = 4$
$(1 2 3)$	$1 + 1 + 1 + 0 + 0 = 3$	$24 / 8 = 3$
$(1 2 3 4)$	$1 + 1 + 0 + 1 + 1 = 4$	$24 / 6 = 4$
$(1 2) (3 4)$	$1 + 1 + 4 + 1 + 1 = 8$	$24 / 3 = 8$

Cross columns must vanish. For $(1 2)$ against $(1 2 3 4)$ : $1 \cdot 1 + (- 1) (- 1) + 0 + (1) (- 1) + (- 1) (1) = 1 + 1 - 1 - 1 = 0$ . For $1$ against $(1 2) (3 4)$ : $1 + 1 + 4 - 3 - 3 = 0$ .

Recovering a missing row

Suppose only $χ_{1}, χ_{2}, χ_{4}, χ_{5}$ are known. From $\sum_{i} n_{i}^{2} = 24$ we get $n_{3}^{2} = 24 - 1 - 1 - 9 - 9 = 4$ , so $n_{3} = 2$ . Each remaining entry $x = χ_{3} (g)$ is then forced by orthogonality of the column at $g$ against the column at $1$ , that is by $\sum_{i} n_{i} χ_{i} (g) = 0$ :

\begin{aligned} g = (1 2) : & 1 - 1 + 2 x + 3 - 3 = 0 & \Rightarrow x = 0, \\ g = (1 2 3) : & 1 + 1 + 2 x + 0 + 0 = 0 & \Rightarrow x = - 1, \\ g = (1 2 3 4) : & 1 - 1 + 2 x - 3 + 3 = 0 & \Rightarrow x = 0, \\ g = (1 2) (3 4) : & 1 + 1 + 2 x - 3 - 3 = 0 & \Rightarrow x = 2 . \end{aligned}

(E.1)

The whole row $(2, 0, - 1, 0, 2)$ is recovered without constructing the two-dimensional module at all.

Process and Workflow

You have a partial character table. What is the next move?

Degrees missingUse

\sum_{i} n_{i}^{2} = | G |

together with

n_{i} ∣ | G |

from

(8.18)

; for small groups this usually pins the multiset of degrees.

One row missingUse column orthogonality against the identity column:

\sum_{i} n_{i} χ_{i} (g) = 0

for

g \neq 1

determines the missing entry once the degree is known.

Two or more rows missingUse column orthogonality to get quadratic constraints, then row orthogonality between the unknown rows; Galois conjugacy often pairs them, as for the two degree-

3

characters of

A_{5}

Table completeVerify with both relations. Row checks catch arithmetic slips; column checks catch a mislabelled class size.

Fix class representativesRecord class sizes; check they sum to

| G |

and that each divides

| G |

Write the known rowsLinear characters come from

G / [G, G]

; permutation characters come from counting fixed points.

Impose orthogonalityEach unknown entry appears linearly in one column relation against the identity column.

Confirm irreducibilityIn characteristic zero,

[χ, χ] = 1

certifies that a candidate character is irreducible.

Comparison and Classification

The two relations side by side
	First (A)	Second (B)
Sums over	the group $G$	the irreducible characters
Fixes	two characters $χ_{i}, χ_{j}$	two elements $g, h$
Reads	rows of $X$ are orthonormal	columns of $X$ are orthogonal
Right-hand side	$δ_{i j} \| G \|$	$ε \| C_{G} (g) \|$
Matrix form	$X^{*} D X^{T} = \| G \| I$	$X^{T} X^{*} = \| G \| D^{- 1}$
Typical use	multiplicities, irreducibility tests	completing a table, class-size deductions
Proved from	$(8.15) (1)$ plus $χ_{j}$	$(8.15) (1)$ inside $(8.15) (2)$

What survives when a hypothesis is dropped
	Table is square	(A) as stated	(B) as stated	$[χ, χ] = 1$ tests irreducibility
$char k = 0$ , $k$ splits $G$	yes	yes	yes	yes
$char k = p ∤ \| G \|$ , $k$ splits $G$	yes	yes	yes	no
$char k ∤ \| G \|$ , $k$ does not split	no	no	no	no
$char k = p ∣ \| G \|$	no	no	no	no

What survives when a hypothesis is dropped

In the third row the irreducible $k$ -characters are sums of Galois orbits of absolutely irreducible characters, possibly with Schur-index multiplicities, and $[χ, χ]$ is the size of the orbit times the square of the index — never $1$ unless the character was already absolutely irreducible.

Applications and Industry Use

Applications here means where this structure is used — inside mathematics and in the engineering and computing disciplines that consume it.

Computational algebra

Character table engines

Orthogonality is both the constraint that determines a table and the assertion that verifies it. GAP and Magma expose it as ScalarProduct and use it internally when normalising Dixon's eigenvectors.

Molecular spectroscopy

Symmetry-adapted modes

The reduction formula $[χ, χ_{i}]$ decomposes the $3 N$ -dimensional displacement representation of a molecule into irreducible representations of its point group, predicting which vibrational modes are infrared or Raman active.

Probability

Random walks on groups

For a class-supported step distribution the eigenvalues of the transition operator are the central character values $χ_{i} (C_{g}) / n_{i}$ , and orthogonality gives the upper-bound lemma controlling mixing time.

Signal processing

Non-abelian Fourier analysis

$(8.17)$ is the Plancherel theorem for a finite group. Fast Fourier transforms on non-abelian groups, and spectral analysis of ranked or partially ordered data, run on exactly this decomposition.

Coding theory

Group codes

Ideals of a semisimple group algebra are generated by sums of the $e_{i}$ ; orthogonality supplies the weight and dimension bookkeeping for the resulting codes.

Group theory

Structural theorems

Burnside's $p^{a} q^{b}$ solvability theorem and Frobenius's theorem on Frobenius kernels are proved by combining orthogonality with the integrality results of the Degrees of Irreducible Characters page.

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

Verification cost. Checking all row pairs is $O (r^{3})$ field operations; checking all column pairs is likewise $O (r^{3})$ . For $r$ in the hundreds this is negligible compared with computing the classes.
Exact arithmetic. Character values live in $ℚ (ζ_{e})$ with $e = \exp G$ . Implementations store them in a cyclotomic field with a sparse basis rather than as floating-point numbers; orthogonality checks are exact and are the standard regression test.
In Dixon's algorithm. The class-multiplication matrices are simultaneously diagonalised modulo a prime $p \equiv 1 (\mod e)$ ; the resulting eigenvectors are central character rows, and the first orthogonality relation is what fixes their normalisation before lifting to characteristic zero.
Numerical use. If character values are computed in floating point, orthogonality residuals give a usable condition estimate; a residual far above machine epsilon means a class was mis-identified, not that the theory failed.

Failure Modes and Common Mistakes

The irreducibility test $[χ_{M}, χ_{M}] = 1$ needs $char k = 0$ . In characteristic $p ∤ | G |$ the pairing lands in $k$ and cannot distinguish multiplicity $1$ from multiplicity $1 + p$ .
Over a non-splitting field the rows are not orthonormal. Do not apply the relations to a table of $ℚ$ -irreducible characters of a group with irrational character values.
The relations say nothing about which entries are integers; the arithmetic of the entries is the subject of the Degrees of Irreducible Characters page.
Non-isomorphic groups can share a character table — the dihedral and quaternion groups of order $8$ do. Orthogonality is a strong constraint but not a complete invariant.
$[χ_{i}, χ_{j}]$ is a symmetric bilinear form, not a Hermitian one; over a field with an automorphism of order $2$ do not silently insert conjugation.

Best Practices

Always record the class sizes next to the table; both relations are wrong without them and the class equation $\sum_{t} m_{g_{t}} = | G |$ is a free check.
State which relation is in use. Papers that write “by orthogonality” without saying rows or columns are the hardest to verify.
When constructing a table by hand, deduce degrees first, then linear characters, then use column orthogonality against the identity column — it is linear in the unknowns.
Verify a completed table with both relations before quoting anything from it; a single mislabelled class size is invisible to row checks alone.
Keep the inverse in $χ (g^{- 1})$ even when the group has only real characters; the habit prevents errors when the next group does not.

Quick Reference

Hypotheses

G

finite,

k

splits

G

char k ∤ | G |

First (A)

\sum_{g} χ_{i} (g^{- 1}) χ_{j} (g) = δ_{i j} | G |

Second (B)

\sum_{i} χ_{i} (g) χ_{i} (h^{- 1}) = ε | C_{G} (g) |

Class-rep form

[χ_{i}, χ_{j}] = | G |^{- 1} \sum_{t} m_{g_{t}} χ_{i} (g_{t}) χ_{j} (g_{t}^{- 1})

Matrix form

X^{*} D X^{T} = | G | I

and

X^{T} X^{*} = | G | D^{- 1}

Dual bases

{n_{i}^{- 1} χ_{i}}

dual to

{e_{i}}

Class sums

[C_{g}, C_{g}] = m_{g}

, orthogonal for distinct classes

Degree sum

\sum_{i} n_{i}^{2} = | G |

— case

g = h = 1

of (B)

Consequences at a glance
Statement	How obtained	Needs
$\sum_{i} n_{i}^{2} = \| G \|$	(B) at $g = h = 1$	split, $char k ∤ \| G \|$
$\sum_{i} n_{i} χ_{i} (h) = 0$ for $h \neq 1$	(B) at $g = 1$	same
$f = \sum_{i} [f, χ_{i}] χ_{i}$	(8.17)(1)	same
Multiplicity of $M_{i}$ in $M$ is $[χ_{M}, χ_{i}]$	(8.17)(1)	additionally $char k = 0$
$M$ simple $iff [χ_{M}, χ_{M}] = 1$	orthonormal expansion	additionally $char k = 0$

Frequently Asked Questions

Are the two relations independent, or does one imply the other?

Over a splitting field they are equivalent. Grouping the first relation by classes gives $X^{*} D X^{T} = | G | I$ ; since $X$ is square, a one-sided inverse is two-sided, so $X^{T} X^{*} = | G | D^{- 1}$ , which is the second relation. Lam nevertheless proves the second one directly by substituting the idempotent formula into the class-sum formula, which avoids appealing to invertibility.

Why does the second relation produce centraliser orders rather than class sizes?

Because comparing coefficients of a group element $h$ in the identity for $C_{g}$ divides by $m_{g}$ , and $| G | / m_{g} = | C_{G} (g) |$ by orbit-stabiliser. Intuitively the column at $g$ is long exactly when $g$ has a large centraliser — the identity column, with centraliser $G$ , has squared length $| G |$ .

Do the relations still hold in characteristic p when p does not divide the group order?

Yes, exactly as stated: the proofs use only that $| G |$ and the degrees $n_{i}$ are invertible in $k$ and that $k$ splits $G$ . What is lost is every argument that treats the values as complex numbers — positivity, the irreducibility test, and any appeal to complex conjugation.

Can two different groups have the same character table?

Yes. The dihedral group of order $8$ and the quaternion group of order $8$ have identical character tables under a suitable matching of classes, though they are not isomorphic. So the orthogonality relations, however rigid, do not determine the group; recovering the group needs extra data such as the class multiplication coefficients or the power maps.

What is the analogue for a non-splitting field such as the rationals?

The irreducible $k$ -characters are sums over Galois orbits of absolutely irreducible characters, weighted by Schur indices. Orthogonality survives in the form $[χ, χ] = m^{2} \cdot | orbit |$ rather than $1$ , and the character table stops being square in the sense used here. The Splitting Fields for Finite Groups page describes when the problem disappears.

How do I use the relations to decompose a permutation representation?

The permutation character is $π (g) = | Fix (g) |$ . In characteristic zero the multiplicity of $M_{i}$ is $[π, χ_{i}]$ , and $[π, π]$ is the number of orbits of $G$ on ordered pairs, that is, the rank of the permutation action. In particular the action is doubly transitive exactly when $[π, π] = 2$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §8, (8.16)–(8.17).
I. M. Isaacs, Character Theory of Finite Groups, Academic Press, 1976, Chapter 2.
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, Chapter 2.
J.-P. Serre, Linear Representations of Finite Groups, Graduate Texts in Mathematics 42, Springer-Verlag, 1977, Part I.
W. Feit, The Representation Theory of Finite Groups, North-Holland Mathematical Library 25, 1982, Chapter I.
P. Diaconis, Group Representations in Probability and Statistics, IMS Lecture Notes–Monograph Series 11, 1988.

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01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

One matrix identity, two readings

Why the table is square

Two immediate specialisations

06Key Results

07Proof Techniques and Method

08Worked Example

Checking the first relation

Checking the second relation

Recovering a missing row

09Process and Workflow

10Comparison and Classification

11Applications and Industry Use

Character table engines

Symmetry-adapted modes

Random walks on groups

Non-abelian Fourier analysis

Group codes

Structural theorems

12Computational Notes

13Failure Modes and Common Mistakes

14Best Practices

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Process and Workflow

Comparison and Classification

Applications and Industry Use

Computational Notes

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions