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A complex representation of a finite group $G$ is a homomorphismfor a finite-dimensional complex vector space $V$. It is an irreducible representation when its only invariant subspaces are $0$ and $V$, and its degree is $\dim V$. Two representations are isomorphic when there is an invertible linear map intertwining their $G$-actions.

Write the dihedral group asIn an irreducible representation choose an eigenvector $0\ne v\in V$ of $\rho (r)$, which is possible because $r^n=1$. If its eigenvalue is $\lambda$, thenThus $W=\mathrm{span}\{v,\rho (s)v\}$ is invariant under both $r$ and $s$. Irreducibility forces $W=V$, proving $\dim V\le 2$ as in irreducible complex representations of a finite dihedral group.

Let $\omega =e^{2\pi i/n}$. Besides the one-dimensional representations described below, define for each indicated $j$These matrices satisfy the defining relations. Since ${\omega }^j\ne {\omega }^{-j}$, the only one-dimensional $r$-invariant subspaces are the two coordinate axes, and $s$ interchanges them. Hence ${\rho }_j$ is irreducible. Their characters satisfywhose values are distinct in the ranges below, so the ${\rho }_j$ are pairwise nonisomorphic. This is the family of two-dimensional representations of a finite dihedral group.

If $n$ is odd, a one-dimensional character must send $r$ to a scalar $\epsilon$ satisfying ${\epsilon }^n=1$ and $\epsilon ={\epsilon }^{-1}$. Thus $\epsilon =1$, while $s$ may independently map to $1$ or $-1$. These give two one-dimensional irreducibles. Takinggives $(n-1)/2$ two-dimensional irreducibles, for a total of

If $n$ is even, $r$ may map to either $1$ or $-1$, and again $s$ may map independently to either sign. This gives four pairwise nonisomorphic one-dimensional irreducibles. Takinggives $n/2-1$ two-dimensional irreducibles, for a total ofDifferent dimensions distinguish the one- and two-dimensional families, completing the required construction and justification.

Representations. A complex group representation is a group homomorphism $\rho :G\to \mathrm{GL}(W)$ to the invertible linear maps on a vector space. It is an irreducible representation when $W\ne 0$ and the only invariant subspaces are $0$ and $W$.

Group table. The supplied matrices satisfy $A=-I$, $C=-B$, and $A^2=B^2=C^2=I$. Their multiplication table isThe table proves closure and gives identity $I$ and an inverse for every element; associativity is inherited from matrix multiplication. The table is symmetric, so the group is abelian. It is the Klein four-group $V\cong C_2\times C_2$.

Dimensions and conjugacy. The sum of squares of irreducible representation dimensions is $|V|=4$. A two-dimensional irreducible would alone contribute four, leaving no room for the ever-present one-dimensional trivial representation. Hence the given two-dimensional representation is reducible. Since the group is abelian, every conjugacy class is a singleton. There are therefore four inequivalent complex irreducible representations, each of dimension one:Taking the trace of the given matrices gives their character of a representation:

Irreducible characters. In a one-dimensional representation, the values on $A$ and $B$ must each be $\pm 1$, since both elements square to $I$. Their values determine the value on $C=AB$. The four choices give the complete character table:Each row is a character because it respects the generating relations, and the rows have unit norm and pairwise zero character inner product. The multiplicities of these rows in $(2,-2,0,0)$ are respectively $0,0,1,1$, by character orthogonality. Thus

Diagonal form. The common invariant lines are spanned by $(1,1{)}^T$ and $(1,-1{)}^T$. Use their normalized vectors as columns ofThen simultaneous diagonalization givesReading the first and second diagonal entries across the four matrices recovers ${\chi }_3$ and ${\chi }_4$, respectively, verifying the decomposition explicitly.

Definitions and labels. A complex group representation on a vector space $V$ is a group homomorphism $\rho :G\to \mathrm{GL}(V)$, so $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. It is an irreducible representation if $V\ne 0$ and its only invariant subspaces are $0$ and $V$.

Read permutation matrices by their action on column basis vectors. The labels in the original PDF areIn particular, the original PDF gives the last row of $C$ as $(1,0,0)$; the provided TeX transcription's $(0,0,1)$ is incorrect and would not give an invertible permutation matrix.

Multiplication and conjugacy. Matrix products act on columns with the right factor first. For example $A^2=B$, $AD=C$, and $DA=E$. The complete multiplication table, with row factor multiplied by column factor, isConjugating a permutation relabels its cycles. Thus the conjugacy classes areof sizes one, two and three. These are respectively the identity, the three-cycles and the transpositions in the symmetric group on three letters.

Dimensions. The sum of squares of irreducible representation dimensions equals the group order, which here is six. A three-dimensional irreducible representation would already contribute nine, so the displayed representation is reducible. There are three inequivalent complex irreducible representations, one for each conjugacy class, including the trivial representation of dimension one. If their dimensions are $1,d_2,d_3$, then $d_2^2+d_3^2=5$, whose positive integer solution is $\{d_2,d_3\}=\{1,2\}$. HenceOne can also see reducibility directly: every permutation matrix fixes $(1,1,1{)}^T$.

Characters. The character of a representation is the trace of its representing matrix, constant on each conjugacy class. The given permutation representation has character equal to the number of fixed basis vectors:in class order $(I;A,B;C,D,E)$. A trivial representation has character $(1,1,1)$. For the second one-dimensional character, a transposition has value $\pm 1$ because its square is the identity. The two three-cycles are conjugate and inverse, so their common value $a$ obeys both $a^2=1$ and $a^3=1$, forcing $a=1$. Character orthogonality with the trivial character then requires $1+2+3b=0$, giving $b=-1$. This determines the sign representation without constructing its matrices.

Write the two-dimensional character as $(2,x,y)$. Character orthogonality with the first two rows givesso $x=-1$, $y=0$. Its squared norm is $(4+2)/6=1$, as required. The complete character table is thereforeThe last row belongs to the standard representation of the symmetric group.

To decompose the given representation, use the character inner productIts multiplicities areThusConcretely the invariant line is spanned by $(1,1,1{)}^T$ and its invariant complement is $\{(x,y,z):x+y+z=0\}$. Their characters are exactly the first and third rows, agreeing with the decomposition obtained entirely from character orthogonality.

If $h=ag{a}^{-1}$, the representation law gives $D(h)=D(a)D(g)D(a{)}^{-1}$. Trace invariance under this similarity transformation givesThus the character of a representation is constant on every conjugacy class.

The trace is invariant under a similarity transformation, by cyclicity of the trace:Hence equivalent representations have the same character of a representation.

A group representation is a homomorphism $D:G\to \mathrm{GL}(V)$ into invertible linear maps. It is a faithful representation when it is injective, equivalently when its kernel contains only the identity. A faithful real two-dimensional representation of the cyclic group of order four isThese are rotations of the plane through $0,\pi /2,\pi ,3\pi /2$. Since $J^2=-I$ and $J^4=I$, their multiplication agrees with the group, and all four matrices are distinct.

For any invertible $S$, set $\tilde{D}(g)=SD(g)S^{-1}$. Thenso this also is a group representation. The character of a representation is ${\chi }_D(g)=\mathrm{tr}D(g)$. For the displayed quarter-turn representation, its values on $1,i,-1,-i$ are $2,0,-2,0$.

A group representation is a homomorphism $D:G\to \mathrm{GL}(V)$. It is a reducible representation if there is a nonzero proper invariant subspace of $V$ preserved by every $D(g)$, and an irreducible representation otherwise. With $S$ invertible,so changing basis gives another representation. Its character of a representation is $\chi (g)=\mathrm{tr}D(g)$, and cyclicity of the trace gives $\tilde{\chi }(g)=\mathrm{tr}(SD(g)S^{-1})=\chi (g)$.

A conjugacy class consists of $hg{h}^{-1}$ as $h$ varies. Since $(hg{h}^{-1}{)}^k=h{g}^k{h}^{-1}$, a power equals the identity exactly when $g^k$ does. Thus conjugation preserves element order, including infinite order.

For the square, let $r$ be rotation by $\pi /2$ and $s$ reflection in a horizontal symmetry axis. The dihedral group has the eight elementsThe conjugacy classes of a finite dihedral group specialize toThe two central rotations are singletons. Reflection conjugation reverses rotations, while conjugation by $r$ shifts a reflection exponent by two, giving the two reflection classes.

For finite groups over $\mathbb{C}$, character orthogonality and completeness make the irreducible characters a basis of the class functions. The latter space has one independent value per conjugacy class. Hence the square group has five inequivalent irreducible complex representations. The sum of squares of irreducible representation dimensions is eight, forcing four one-dimensional representations and one two-dimensional representation. The four scalar choices $r\mapsto \pm 1$, $s\mapsto \pm 1$ supply the former; the ordinary action on the plane supplies the latter.

For a group representation $D:G\to \mathrm{GL}(V)$, a faithful representation is injective, equivalently $D(g)=I$ only for $g=e$. Thus no distinct group elements are represented by the same linear map.

Two equivalent representations are related by one fixed invertible change of basis: $E(g)=S^{-1}D(g)S$ for every $g$. The same $S$ must work simultaneously for all group elements.

An irreducible representation acts on a nonzero vector space and has no invariant subspace other than $\{0\}$ and the whole space. An invariant subspace is one carried into itself by every representing matrix.

The character of a representation isThe cyclic property of the trace makes it invariant under equivalence and constant on conjugacy classes. Its value at the identity is the dimension of the representation.

A group representation assigns each $g\in G$ an invertible linear map $D(g)$ on a fixed vector space, preserving multiplication:Choosing a basis turns those maps into matrices. A faithful representation is injective, equivalently its kernel is $\{e\}$, so distinct elements have distinct matrices.

Two group representations on spaces of the same dimension are called equivalent representations if one fixed invertible change of basis conjugates every representing matrix: $E(g)=S^{-1}D(g)S$ for all $g$. Choosing a different $S$ separately for each element does not establish equivalence.

The character of a representation isThe cyclic property of the trace makes it invariant under a change of basis and constant on conjugacy classes. Thus different characters at even one group element prove inequivalence.

The products in (a), together with $(-A)B=A(-B)=-(AB)$ and $(-A)(-B)=AB$, give closure of the eight-element set. Associativity is inherited from matrix multiplication, $W$ is the identity element, and the inverses from (a), with $(-A{)}^{-1}=-A^{-1}$, remain in the set. Thus the group axioms hold. This is the quaternion group $Q_8$.

In the element order specified in the question, the multiplication table isInterpreting “character of the group” as the character of a representation specified by these matrices, take the trace of each matrix:A character belongs to a representation; the underlying abstract group can have other characters.

The character of a representation is the trace of its representing matrix. For the rotation powers,Each $U^{j}V$ has a planar reflection block of trace zero and third diagonal entry $-1$, giving character $-1$. In the chosen element order,Equivalently its values on the conjugacy classes $\{I\}$, $\{U^2\}$, $\{U,U^3\}$, $\{V,U^{2}V\}$, $\{UV,U^{3}V\}$ are $3,-1,1,-1,-1$.

Distinguish the quaternion group generators $\mathbf{i},\mathbf{j},\mathbf{k}$ from the complex scalar $i$. Denote their proposed image matrices byDirect multiplication givesFor example $AB=\mathrm{diag}(i,-i)=C$ and $ABC=C^2=-I_2$. Sending $I$ to $I_2$ and each negative quaternion element to the negative of its positive image now reproduces every quaternion product. The matrices all have determinant one and are invertible, so this is a group representation in $GL(2,\mathbb{C})$.

The eight matrices $\pm I_2,\pm A,\pm B,\pm C$ are distinct. Therefore the group homomorphism is injective, and the representation is faithful; equivalently its kernel is $\{I\}$.

The character of a representation is the trace of its representing matrix. ConsequentlyThese are constant on the five conjugacy classes $\{I\}$, $\{-I\}$, $\{\pm \mathbf{i}\}$, $\{\pm \mathbf{j}\}$, $\{\pm \mathbf{k}\}$.

The complete list of subgroups isTo see completeness, Lagrange's theorem allows only orders $1,2,4,8$. The only element of order two is $-I$, which fixes the order-two subgroup. Every noncentral element has order four and generates the corresponding listed cyclic subgroup. Any order-four subgroup contains such an element, because there are only two central elements; it must equal the cyclic subgroup generated by that element. This accounts for all possibilities.

The Ramanujan sum is $c_n(k)={\sum }_{1\le j\le n,(j,n)=1}{e}^{2\pi ijk/n}$. It is an integer: insert $1_{(j,n)=1}={\sum }_{d\mid (j,n)}\mu (d)$, sum the geometric series, and obtain $c_n(k)={\sum }_{r\mid (n,k)}r\mu (n/r)$. Here $\mu$ is the Möbius function. Such sums connect roots of unity with integer character of a representation values.

Suppose a finite group $G$ acts by Riemannian isometries on a compact Riemannian manifold $M$, and Gassmann equivalent subgroups $H_1,H_2$ act freely. The quotient manifolds $H_i\backslash M$ have equal Laplace-Beltrami operator spectra. For an eigenfunction space $E$, the quotient multiplicity is $|H_i{|}^{-1}{\sum }_{h\in H_i}\mathrm{tr}(h|E)$, and the equality follows by summing its character of a representation over conjugacy classes. Nonconjugacy inside $G$ alone does not guarantee that the quotients are nonisometric.

A transformation that preserves a specified structure, including the chemical identity of sites in a molecule. Rotations, reflections, inversion and improper rotations can occur in a molecular point group. The associated action on atomic orbitals also changes their orientations and signs, which must be included when computing a character of a representation.