Character theory

Character theory studies representations through the traces of the representing linear transformations.

A group of automorphisms of a finite group permutes both its conjugacy classes and its irreducible characters. The two permutation representations have the same character: each automorphism fixes equally many conjugacy classes and irreducible characters. The invertible character table intertwines the two permutation matrices, which have equal traces. The claim concerns fixed classes, not merely fixed elements.

Every complex character of a representation of a finite group $G$ is a rational linear combination of characters of induced representations from one-dimensional group representations of cyclic subgroups $C_j$. One-dimensionality follows because the irreducible representations of a finite cyclic group over the complex numbers are one-dimensional. The rational coefficients distinguish this statement from Brauer induction, which permits integer coefficients by allowing more general subgroups. For a rational-valued character, another form of Artin induction uses rational combinations of induced representations of trivial group representations of cyclic subgroups.

Every complex character of a representation of a finite group $G$ is an integer linear combination of characters of induced representations from one-dimensional group representations of subgroups $H_j\le G$. Negative coefficients are allowed. Unlike Artin induction, the subgroups need not be cyclic groups. Applied to an Artin L-function, compatibility with induced representations turns this character identity into a product of one-dimensional Artin L-functions with integer exponents, reducing the behavior at $s=1$ to abelian extensions.

The character of a finite-dimensional representation $\rho$ is the class function ${\chi }_{\rho }(g)=\mathrm{tr}(\rho (g))$. Isomorphic representations have the same character.

A symmetry transformation acts on localized atomic orbitals both by moving their centres and by transforming their angular shapes. Only centres left in place can contribute to the trace of the corresponding group representation. At each fixed centre, sum the diagonal coefficients of the orbital transformation: a reversed p orbital contributes $-1$, whereas a fixed s orbital contributes $+1$. For Cartesian displacement vectors at a fixed atom, use the trace of the full polar-vector matrix, not merely the number of fixed components. This rule generates characters of a representation without writing its entire matrix.

An ordinary character is the trace character of a finite-dimensional group representation in characteristic zero, usually realized over the complex numbers. It contrasts with a Brauer character in positive characteristic, which lifts eigenvalues on p-regular elements to characteristic-zero roots of unity.

For a group representation $\rho$ with character $\theta$, its determinant character sends $g$ to $\det \rho (g)$. Multiplicativity makes it a linear character, independent of the equivalent matrix realization of $\rho$.

If a Sylow p-subgroup $P$ is abelian, a hypothetical central subgroup $Q\le G^{\prime }$ of order $p$ has a nontrivial linear character extending to $P$. Inducing this extension to $G$ gives a character of degree prime to $p$, hence an irreducible constituent of such degree. On $Q$ its determinant character is a nontrivial power of the chosen linear character, but every determinant character is trivial on $G^{\prime }$. This contradiction proves the displayed restriction.

For a complex character $\chi$ of a finite group, its kernel is the kernel of an affording group representation. Averaging an inner product makes that representation unitary. If its degree is $d$, all eigenvalues of a group element have modulus one, so their sum equals $d$ only when every eigenvalue is one. Thus the displayed character-value criterion agrees with the representation kernel and defines a normal subgroup.

For any group, take the group-algebra image on the direct sum of two finite-dimensional complex semisimple representations. Its radical annihilates their simple summands and hence the faithful direct sum, so the image algebra is semisimple. Equality of traces on group elements gives equality on their linear span. Traces on primitive central blocks then recover every simple multiplicity, proving that equal characters imply isomorphism.

At an element $g$ of a finite group, a complex character value is the sum of the roots of unity which are the eigenvalues of its representing matrix. It is therefore an algebraic integer, and every algebraic conjugate has modulus at most the degree.

An irreducible character is the character of an irreducible representation. Over the complex numbers, every finite-group character decomposes uniquely as a nonnegative integer combination of irreducible characters.

For an irreducible complex character of a representation, the indicator is $1$ for real type, $-1$ for quaternionic type and $0$ for complex type, the latter meaning $\chi \ne \overline{\chi }$. It detects the invariant bilinear form on an irreducible representation. For an odd-order group, squaring permutes its elements, so every nontrivial irreducible character has indicator zero by character orthogonality.

For a complex irreducible representation $D$ of a finite group $G$, the sum $A={\sum }_{g}D(g^2)$ commutes with every $D(h)$: conjugation sends its summands to $D((hg{h}^{-1}{)}^2)$, merely permuting the summation index. Schur's lemma gives $A=cI$. Taking the trace yields $c\dim D={\sum }_g\chi (g^2)=|G|\nu (D)$, proving the identity. Here $\nu (D)$ is the Frobenius-Schur indicator. For a reducible representation the sum is still central in its image, but need not be one scalar on all irreducible constituents.

An odd-order finite group has only odd irreducible degrees, by irreducible character degree divides the group order. Its nontrivial characters occur in distinct conjugate pairs, by the Frobenius-Schur indicator. Therefore $|G|-k(G)={\sum }_{\chi }(\chi (1{)}^2-1)$ is a sum of pairs $2(d^2-1)$ with $d$ odd. Since $d^2\equiv 1(\mathrm{mod}8)$, each pair is divisible by sixteen. Here $k(G)$ is the number of conjugacy classes.

For a real-valued irreducible character of a finite odd-order group, pairing $g$ with $g^{-1}$ gives $⟨\chi ,1_G⟩=(\chi (1)+2A)/|G|$, with $A$ an algebraic integer. If this inner product were zero, $\chi (1)$ would be even, since a rational algebraic integer is an integer. But an irreducible character degree divides the group order, so it is odd. Therefore the character is trivial.

An irreducible character $\phi$ is a constituent of a character $\chi$ when its multiplicity $⟨\chi ,\phi ⟩$ is positive.

For a subgroup $H\le G$, the restriction of a $G$-character $\chi$ to $H$ is the $H$-character obtained by evaluating the same function only on elements of $H$.

An extension of $\theta \in \mathrm{Irr}(N)$ from $N◃G$ is a character $\chi$ of $G$ whose restriction is exactly $\theta$. It is necessarily irreducible, because a nontrivial invariant subspace for $G$ would also be one for $N$. An ordinary extension can fail to exist even when $\theta$ is invariant under $G$.

Let $N◃G$ and let $\theta \in \mathrm{Irr}(N)$ be invariant under $G$. If $\theta (1)$ is coprime to $[G:N]$, then $\theta$ extends to $G$ if and only if its determinant character extends to a linear character of $G$. In particular, if the determinant order is also coprime to the index, there is a unique extension whose determinant order is coprime to the index. When $G/N$ is a p-group and $N=O^p(N)$, every invariant irreducible character of degree prime to $p$ satisfies this latter determinant condition.

Under the invariant, coprime-degree and determinant hypotheses of the coprime-degree determinant extension theorem, without assuming $G/N$ solvable, each cyclic-quotient subgroup $N⟨g⟩$ has a unique determinant-normalized extension. Use its value at $g$ to define $f$. Uniqueness makes $f$ conjugacy-invariant. On an elementary subgroup $E$, it agrees with the restriction of the extension to $NE$, since $NE/N$ is a solvable quotient of $E$. Brauer's characterization of characters therefore makes $f$ a generalized character, with $f_N=\theta$.

Suppose $N◃G$, $G/N$ is solvable, $\theta \in \mathrm{Irr}(N)$ is invariant under $G$, and $\theta (1)$ is coprime to $[G:N]$. If a linear character $\mu$ of $G$ restricts to the determinant character $\det \theta$, then there is a unique character extension $\chi \in \mathrm{Irr}(G)$ with determinant $\mu$. This is the solvable-quotient extension result supplied for the exam's application.

For $N◃G$ and $\theta \in \mathrm{Irr}(N)$, set ${\theta }^g(n)=\theta (g^{-1}ng)$. Its inertia group is its stabilizer under this conjugation action and contains $N$. Write $\mathrm{Irr}(G\mid \theta )$ for the irreducible characters of $G$ whose restriction to $N$ contains $\theta$.

For $N◃G$, $\theta \in \mathrm{Irr}(N)$ and $T=I_G(\theta )$, induction gives the displayed bijection. The inverse takes the $\theta$-isotypic component of an irreducible $G$-module, viewed as a $T$-module. Induction produces distinct $N$-isotypic components indexed by $G/T$. Any nonzero $G$-submodule meets one of them and hence, by translation and irreducibility over $T$, contains them all. The same argument shows that the inverse component is irreducible and determines the original inducing module uniquely.

If $\chi$ is an irreducible character of a finite group $G$ and $H\le G$, thenEquality holds exactly when $\chi$ vanishes on $G\setminus H$.

Let $H◃G$ have index two, and let $\lambda$ be the nontrivial linear character of $G/H$ inflated to $G$. For an irreducible character $\chi$ of $G$, either ${\mathrm{Res}}_H^G\chi$ is irreducible and $\chi \ne \chi \lambda$, or the restriction is a sum of two distinct irreducible characters and $\chi =\chi \lambda$.

A virtual character is an integer linear combination of characters, equivalently an element of the Grothendieck group of finite-dimensional representations.

A complex class function $f$ on a finite group is a generalized character if and only if $f_E$ is a generalized character of every elementary subgroup. Equivalently, $⟨f_E,\alpha {⟩}_E\in \mathbb{Z}$ for every such $E$ and $\alpha \in \mathrm{Irr}(E)$. Ordinary characters additionally have nonnegative global irreducible multiplicities; integrality of the virtual-character lattice is the conclusion of this criterion.

Suppose every nonidentity element of a finite group $K$ is either a pi-element or an element for complementary primes. An elementary subgroup cannot have nontrivial factors of both kinds, since the product of nonidentity commuting coprime-order elements would have mixed prime order. Choose an integer $d$ congruent to $1$ modulo $|K{|}_{\pi }$ and to $0$ modulo $|K{|}_{{\pi }^{\prime }}$. Set $f(1)=d$, $f=1$ on nonidentity pi-elements and $f=0$ on the other nonidentity elements. On each elementary subgroup this is an integer combination of the trivial and regular characters. Brauer's characterization of characters supplies the asserted generalized character.

Write the virtual character as $\psi ={\sum }_{\chi }{a}_{\chi }\chi$ with integer coefficients and irreducible characters $\chi$. Character orthogonality gives $⟨\psi ,\psi ⟩={\sum }_{\chi }{a}_{\chi }^2$. Norm one forces exactly one coefficient to be $1$ or $-1$, and positive degree selects the positive sign.