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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.

For finite-dimensional group representations, the character of their tensor product is the pointwise product of their characters. In a tensor-product basis, the diagonal entries of $A\otimes B$ are $A_{ii}{B}_{jj}$, so their sum is $(\mathrm{tr}A)(\mathrm{tr}B)$. This gives a direct route from a character table to tensor-product decompositions.

A group of order thirty-six with a normal $C_3^2$ and a cyclic quotient of order four acting freely on its eight nonzero elements has six conjugacy classes of sizes $1,4,4,9,9,9$. Its four linear characters are lifted from the cyclic quotient. The remaining irreducible degrees are four and four, by the sum of squared degrees. Column orthogonality forces the last three values of both degree-four rows to vanish. The square of the order-four action is inversion on $C_3^2$, making the first two nontrivial class values real. Orthogonality then gives sum minus one and product minus two, so the rows are $(4,1,-2,0,0,0)$ and $(4,-2,1,0,0,0)$. This illustrates why reality must be justified when reconstructing a character table from orthogonality.

Write the Klein four-group as $\{1,a,b,ab\}$. Its four one-dimensional irreducible representations independently assign signs to $a,b$, so the displayed characters have $p,q\in \{0,1\}$. The character table isEvery conjugacy class is a singleton because the group is abelian. The four rows exhaust the complex irreducibles by the equality between their number and the number of classes. If $H$ is this table, $H{H}^T=4I$ proves character orthogonality. The diagonal three-dimensional representation by half-turns about the coordinate axes is the direct sum of the three nontrivial characters, with total character $(3,-1,-1,-1)$.

In the order of the conjugacy classes of the symmetry group of a square displayed there, its irreducible complex character table isThe four one-dimensional representations assign independent signs to $r$ and $s$. The remaining row is the character of the ordinary planar representation: a rotation has trace $2\cos \theta$ and a reflection has trace zero. Weighted by class sizes, these rows satisfy character orthogonality. Their squared dimensions sum to eight, so they exhaust the irreducible representations.

The class sizes sum to $|G|=36$. A degree-one character is a homomorphism into ${\mathbb{C}}^{\times }$. The supplied ${\chi }_2$ has order four and ${\chi }_2^2={\chi }_1$, so $1,{\chi }_2,{\chi }_1,{\bar{\chi }}_2$ are four distinct irreducible characters. A finite group has as many complex irreducible characters as conjugacy classes, so there are exactly two more. Their degrees $d,e$ satisfy $d^2+e^2=36-4=32$, whose only positive-integer solution is $d=e=4$.

The column orthogonality give ${\sum }_{\chi }|\chi (g){|}^2=|C_G(g)|$. Each of the last three classes has centralizer order $36/9=4$, already exhausted by the four degree-one rows. Thus both new rows vanish on $C_4,C_5,C_6$.

To justify that their remaining values are real, put $P=\ker {\chi }_2$. The supplied values show $P=C_1\cup C_2\cup C_3$, a normal subgroup of order nine, with $G/P\cong C_4$. A group of order nine is Abelian, by the standard p-group center argument and the fact that a cyclic quotient by the center forces the group to be Abelian. If $P$ were cyclic of order nine, conjugation through $G/P$ would act by an automorphism of order dividing both four and $|\mathrm{Aut}(C_9)|=6$, hence dividing two. Its nontrivial conjugacy classes could not have size four. Therefore $P\cong C_3^2$.

Let $h$ map to a generator of $G/P$, and let $A$ be its conjugation action on the two-dimensional vector space $P$ over ${\mathbb{F}}_3$. The two nonzero orbits have length four, so $A^2$ fixes no nonzero vector, while $A^4=I$. Since $t^2-1$ has distinct roots in characteristic three, the involution $A^2$ is diagonalizable; the absence of a $+1$ eigenvector gives $A^2=-I$. Thus conjugation by $h^2$ inverts every element of $P$. Every element in $C_2,C_3$ is therefore conjugate to its inverse, so every character value there is real.

Write one new row as $(4,a,b,0,0,0)$. Its orthogonality with the trivial row gives $4+4a+4b=0$, so $a+b=-1$. Its squared character inner product one gives $16+4a^2+4b^2=36$, so $a^2+b^2=5$ and $ab=-2$. Thus $a,b$ are $1,-2$ in some order. Column orthogonality with the identity column forces the other row's entries to be interchanged. Up to relabelling these two new characters, the completed character table isThe reality argument prevents an unjustified assumption when solving the orthogonality equations. This is character table from a cyclic quotient and two elementary-abelian orbits.

The class equation implies that a nontrivial finite $p$-group has nontrivial center of a group. Here a center of order $9$ would make $G/Z(G)$ cyclic; a group with cyclic quotient by its center of a group is abelian, contradicting the hypothesis. ThusThe quotient of order nine must be noncyclic for the same reason. For $x\notin Z(G)$, $⟨x,Z(G)⟩$ has order at least nine and lies in the proper centralizer $C_G(x)$. Lagrange's theorem therefore forces $\boxed{|C_G(x)|=9}$. Also $1\ne G^{\prime }\le Z(G)$, so $G^{\prime }=Z(G)$ and all commutators are central of order dividing three.

Classification and splitting. First suppose $a$ has order nine. Then $N=⟨a⟩$ is normal of index three and $Z(G)=⟨a^3⟩$. For $b\notin N$, conjugation on $N$ is nontrivial, since otherwise $a,b$ would generate an abelian group. Its order divides three, so, after replacing $b$ by its inverse if needed, $ba{b}^{-1}=a^4$. Write $b^3=a^{3k}$. Central commutators and $3\mid \left(\genfrac{}{}{0ex}{}{3}{2}\right)$ give $(a^{i}b{)}^3=a^{3i}{b}^3$; choose $i=-k$ modulo three to obtain an element $c=a^{i}b$ of order three outside $N$. This provesThere is a unique resulting isomorphism type, since the two nontrivial actions of order three are interchanged by reversing the complement's generator.

If there is no element of order nine, every nonidentity element has order three. Choose $a,b$ whose images form a basis of $G/Z(G)$ and put $c=[a,b]$. These elements generate $G$, and $c$ generates $Z(G)$. The subgroup $N=⟨a,c⟩\cong C_3^2$ has index three and is normal. The line $⟨b⟩$ meets $N$ trivially and complements it. Conjugation by $b$ fixes $c$ and sends $a$ to $ac$ or $a{c}^{-1}$; choosing the central generator removes this sign. Thus there is one isomorphism type,with the nontrivial unipotent action on ${\mathbb{F}}_3^2$. These presentations are realized by the indicated semidirect products and have order 27. Their exponents are nine and three respectively, so they are not isomorphic. This proves both existence and completeness of the two cases.

Their ordinary character tables are the same, up to relabelling. In either group each of the 24 noncentral elements has conjugacy class of size three, giving eight noncentral classes and three central singleton classes. Each noncentral class is $xZ(G)$, since conjugates of $x$ differ from $x$ by central commutators and its class has size three. There are nine linear characters, arising from $G/G^{\prime }\cong C_3^2$, and eleven irreducible characters altogether. The two remaining degrees satisfy $d_1^2+d_2^2=27-9=18$, so both are three.

Put $Z(G)=⟨z⟩$ and $\omega =e^{2\pi i/3}$. Label the noncentral classes by nonzero $(i,j)\in {\mathbb{F}}_3^2$. For $(r,s)\in {\mathbb{F}}_3^2$ the nine linear characters have valuesAt a noncentral element, column orthogonality gives a squared column norm of nine. The nine linear characters already contribute nine, so both degree-three characters vanish there. By the Schur lemma, a central element has value $3{\omega }^k$ in an irreducible representation of degree three. The central character cannot be trivial, since that would factor the representation through the abelian quotient. Orthogonality of the identity and $z$ columns then forces the two nontrivial central characters to occur once each. The final two rows, on $1,z,z^2$ and the eight other columns, areThis description gives the same entire character table in both cases. The differing element orders are additional power-map data, which an ordinary character table need not determine.

For operations ordered as $(E,C_2(z),{\sigma }_{xz},{\sigma }_{yz})$, the characters of the twelve Cartesian displacements areOnly atoms left fixed contribute: all four for $E$ and ${\sigma }_{xz}$, but only C and O for $C_2$ and ${\sigma }_{yz}$. Multiplying their counts by the Cartesian traces $3,-1,1,1$ gives these numbers. Subtractto obtain ${\chi }_{\mathrm{vib}}=(6,0,4,2)$. The character table reduction givesThe dimensions sum to $6=3N-6$, as required. The $B_2$ mode is the out-of-plane wag with this axis convention.

Use the D3h molecular point group for phosphorus pentafluoride, with $z$ along the two axial P–F bonds. For each symmetry operation, only atoms left in place contribute to the character of the Cartesian displacement representation; each contributes the trace of the corresponding $3\times 3$ spatial matrix. In the class order $E,2C_3,3C_2^{\prime },{\sigma }_h,2S_3,3{\sigma }_v$ this givesFor example a $C_2^{\prime }$ fixes P and one equatorial F, and each fixed atom has Cartesian trace $-1$, giving $-2$. reflection in the equatorial plane fixes four atoms, each with trace $1$, giving $4$. Reduction with the character table givesSubtract translations $E^{\prime }+A_2^{\prime \prime }$ and rotations $A_2^{\prime }+E^{\prime \prime }$ to obtainIts dimension is $2+6+2+2=12=3(6)-6$, as required for a nonlinear six-atom molecule. Thus there are eight symmetry-distinct fundamental frequencies in the harmonic picture, with each E mode doubly degenerate. This analysis concerns vibrations about one trigonal-bipyramidal minimum; it does not treat large-amplitude rearrangements between minima.

The C2v molecular point group has four one-dimensional irreducible representations. In the operation order used above, the character table rows are $A_1=(1,1,1,1)$, $A_2=(1,1,-1,-1)$, $B_1=(1,-1,1,-1)$ and $B_2=(1,-1,-1,1)$. Their character inner products with ${\chi }_{\Gamma }=(6,0,2,0)$ giveHence $\Gamma =2A_1\oplus A_2\oplus 2B_1\oplus B_2$, whose dimensions sum to six. The labels $B_1,B_2$ refer to the source axes; swapping $x$ and $y$ swaps these names.

Fix the positive lobes of the real atomic orbitals along the printed positive axes, and write ${\varphi }_1=p_x(1)$, ${\varphi }_4=p_x(4)$, ${\varphi }_5=p_y(5)$ and ${\varphi }_8=p_y(8)$. A coordinate model consistent with the figure is $r_1=(0,a,-b)$, $r_4=(0,-a,-b)$, $r_5=(a,0,b)$ and $r_8=(-a,0,b)$, with $a,b>0$. For the D2d molecular point group, take the two transverse twofold axes along $x+y$ and $x-y$, with vertical mirrors $xz$ and $yz$. The $A_2$ character table row on $(E,2S_4,C_2,2C_2^{\prime },2{\sigma }_d)$ is $(1,1,1,-1,-1)$. The character projection operator isThe eight signed images needed for this projection areso $P_{A_2}{\varphi }_1=({\varphi }_1-{\varphi }_4-{\varphi }_5+{\varphi }_8)/4$. Under the standard orthonormal atomic orbital approximation its squared norm is $4/16=1/4$. Therefore the normalized symmetry-adapted linear combination isAn overall minus sign describes the same quantum state; changing one local atomic orbital phase requires changing its coefficient too. If intersite orbital overlap were retained, normalization would instead use the overlap matrix.

Put $\alpha =e^{2\pi i/3}$, so ${\alpha }^3=1$ and $1+\alpha +{\alpha }^2=0$. Since the group is Abelian, every conjugacy class is a singleton. The character table isFor the character inner product, complex conjugation givesFor $j=k$, all three summands are one, giving unit norm. For $k-j\equiv 1(\mathrm{mod}3)$ the sum is $1+\alpha +{\alpha }^2=0$; for $k-j\equiv 2(\mathrm{mod}3)$ it is $1+{\alpha }^2+\alpha =0$. This checks every distinct pair, proving

In a one-dimensional representation, the relation $srs=r^{-1}$ implies $\rho (r)=\rho (r{)}^{-1}$, so $\rho (r)=\delta \in \{\pm 1\}$. Independently $\rho (s)=ϵ\in \{\pm 1\}$. These give four irreducible representationsA two-dimensional representation $E$ is obtained by restricting the given matrices to the $yz$ plane:The eigenvalues of $E(r)$ are $i$ and $-i$. Its two eigenlines are interchanged by $E(s)$, so neither is invariant under the whole group. Thus $E$ is a complex irreducible representation.

Taking traces gives the character table, with class sizes included for the character inner product:For example $⟨{\chi }_E,{\chi }_E⟩=(4+4)/8=1$, and its inner product with each one-dimensional row is $(2-2)/8=0$. The four one-dimensional rows are mutually orthonormal by direct weighted summation. Their squared dimensions sum to $4\cdot 1^2+2^2=8=|G|$, and there are five rows for five conjugacy classes. This proves completeness of the list.

The irreducible representations can all be specified on the generators:The first is the trivial representation and the second is the sign representation. To see directly that these exhaust the one-dimensional representations, their scalar images commute, so the relation $srs=r^{-1}$ forces $r=r^{-1}$. Combining $r^2=1$ with $r^3=1$ gives $r=1$, and then $s$ may be either sign.

Taking traces gives the character table, in the conjugacy class order of solution:Here $\alpha +{\alpha }^2=-1$ gives the middle entry in the last row. The weighted character inner products of distinct rows vanish, and the norm of each row is one. Together with solution, this proves completeness of the table.

There are three complex irreducible representations, because there are three conjugacy classes. Two are the trivial representation $1$ and the sign representation $\epsilon$: both send $r$ to $1$, while they send $s$ to $1$ and $-1$, respectively. These exhaust the one-dimensional representations, since their scalar images must satisfy $r=r^{-1}$ and $r^3=1$, forcing $r=1$, and then $s^2=1$.

The defining two-dimensional matrix representation $D$ has characterIts character inner product with itself isso it is an irreducible representation. Directly, the two coordinate lines are the distinct eigenspaces of $r$, and $s$ exchanges them, leaving no common invariant line. The sum of squares of irreducible representation dimensions is $1^2+1^2+2^2=6$, proving completeness. In the conjugacy class order of solution, the character table isRepresentatives $r,s$ in this table stand for their full conjugacy classes. This lists all irreducible representations up to equivalence of group representations.

Use the conjugacy class order of solution. The defining representation has characterIts character inner product with itself is $(4+4)/8=1$, proving that it is an irreducible representation. There is also a direct invariant subspace argument: $B$ has two distinct coordinate eigenspaces, and $A$ exchanges them, so no line is invariant under both.

Let $L_{st}$ denote the one-dimensional representation with $L_{st}(A)=s$ and $L_{st}(B)=t$. Then $L_{st}(-I)=1$ and $L_{st}(C)=st$. The complete character table isThe four scalar rows obey character orthogonality, and the dimensions exhaust the group order as established in solution.

The squares of $I,-I,\pm A,\pm B$ are $I$, whereas $(\pm C{)}^2=-I$. Therefore ${\chi }_{\rho }(g^2)=(2,2,2,2,-2)$. The symmetric square and exterior square formulas from solution giveComparison with the character table gives the Clebsch-Gordan decomposition of group representationsFor explicit bases, put $e_1=(1,0{)}^T,e_2=(0,1{)}^T$. The symmetric tensors $e_1\otimes e_1+e_2\otimes e_2$, $e_1\otimes e_2+e_2\otimes e_1$, and $e_1\otimes e_1-e_2\otimes e_2$ transform as $L_{++},L_{+-},L_{-+}$ respectively. The antisymmetric tensor $e_1\otimes e_2-e_2\otimes e_1$ transforms as $L_{--}$, also the determinant representation of $\rho$. Adding these direct sums recovers $\rho \otimes \rho ={⨁}_{s,t}{L}_{st}$.

The symmetric group $S_3$ has three conjugacy classes, determined by cycle type: the identity, three transpositions, and two three-cycles. Hence it has three complex irreducible representations. Two are the trivial representation $1$ and the sign representation $\epsilon$. Their dimensions are one, so the sum of squares of irreducible representation dimensions gives $1+1+d^2=6$, whence $d=2$ for the remaining representation $D$.

Write ${\chi }_D=(2,a,b)$ on these classes. Orthogonality to $1$ and $\epsilon$ gives $2+3a+2b=0$ and $2-3a+2b=0$. Thus $a=0,b=-1$, and the character table isThe character of a tensor product representation is the product of the factor characters. Thus ${\epsilon }^2=1$, $\epsilon \otimes D=D$, and ${\chi }_{D\otimes D}=(4,0,1)={\chi }_1+{\chi }_{\epsilon }+{\chi }_D$. These determine all the Clebsch-Gordan decomposition of group representations:

In a one-dimensional representation, the scalar images of $Q$ and $P$ commute. The relation $PQP=Q^{-1}$ gives $R(Q{)}^2=1$, while $R(Q{)}^5=1$, so $R(Q)=1$ and $R(P)=\pm 1$. These give exactly $R_0$ and $R_1$.

For $k=1,2$, define the two-dimensional representationsThe same invariant subspace argument as in solution proves that each is irreducible. Set $R_2=D_1$ and $R_3=D_2$. Their characters differ at $Q$, so they are inequivalent. The sum of squares of irreducible representation dimensions is already $1^2+1^2+2^2+2^2=10$, proving completeness.

Put $s=2\cos (2\pi /5)=(\sqrt{5}-1)/2$ and $t=2\cos (4\pi /5)=-(\sqrt{5}+1)/2$. Taking traces yields the character tableThe identities $s+t=-1$, $st=-1$, and $s^2+t^2=3$ also verify character orthogonality directly.

A tensor product of group representations has character equal to the product of its factor characters. Multiplying the rows of the character table in solution, using $s^2=2+t$, $t^2=2+s$ and $st=-1$, gives all the Clebsch-Gordan decomposition of group representations:For example, the character of $R_2\otimes R_2$ is $(4,s^2,t^2,0)=(4,2+t,2+s,0)$, the sum of the characters of $R_0$, $R_1$ and $R_3$. Likewise $R_2\otimes R_3$ has character $(4,-1,-1,0)={\chi }_2+{\chi }_3$. The trivial representation and $R_1\otimes R_1=R_0$ account for the remaining entries.