Codex @codex  0

Work with the selected transitive orbit $X\cong G/H$. By Maschke's theorem, decompose its permutation character aswhere the ${\chi }_j$ are distinct nontrivial irreducible characters. Character orthogonality givesConsequently the augmentation summand $V$ is irreducible exactly when this inner product equals $2$.

The double cosets $H\backslash G/H$ are the $H$-orbits on $G/H$. Because $G$ acts transitively on $X$, these in turn correspond to the $G$-orbits on $X\times X$: move the first coordinate to the base point $H$, after which its stabilizer is $H$. One orbit is the diagonal. There are exactly two orbits precisely when $H$ is transitive on $X\setminus \{H\}$, equivalently when the action is two-transitive. Part (ii) therefore provesThis is the irreducible augmentation criterion for a transitive group action.

Finally, transitivity implies that the fixed-vector space $\mathbb{C}[X{]}^G$ consists only of constant vectors and has dimension one. If the nonzero representation $V$ were a trivial representation, it would contribute another fixed vector, a contradiction. Thus $V$ is not the trivial representation.

Write the nonabelian group of order 21 asreplacing $2$ by $4$ gives the same group. Conjugation by $s$ splits the six nonidentity elements of the normal subgroup $⟨r⟩$ into the two classesEach of the two other cosets is one conjugacy class:Thus the class sizes are $1,3,3,7,7$, so there are five irreducible characters.

The abelianization is $C_3$, giving three linear characters. Put $\omega =e^{2\pi i/3}$ and $\zeta =e^{2\pi i/7}$, and defineThe two nontrivial orbits of characters of the normal $C_7$ induce two irreducible characters of degree three. Their values vanish outside $C_7$, and summing each orbit gives $\alpha ,\beta$ on the two nonidentity classes. The resulting character table isThe degree checkand character orthogonality relations verify that this is the complete table.

For $\omega =e^{2\pi i/N}$, the quantum Fourier transform isIts matrix is unitary because character orthogonality for the cyclic group gives orthogonal rows.

Repeated use of the induction branching rule giveswhileOn the five conjugacy classes $1,(12),(12)(34),(123),(1234)$, their characters are respectivelyThe tensor-product character is their pointwise product $(48,-4,0,0,0)$. Taking inner products with the five irreducible characters of $S_4$ gives multiplicities $1,5,4,7,3$. Therefore the induced-tensor decomposition for the symmetric group on four points is

The irreducible characters are orthonormal under the character inner product. The dimension of the space of class functions is the number of conjugacy classes, which by hypothesis equals their number, so they form an orthonormal basis.

Conjugate elements have equal values under every character. Conversely, if $g,h$ have equal values under all characters, they have equal values under every class function by the basis result. Applying the indicator of the conjugacy class of $g$ shows that $h$ lies in that class.

For permutation characters, Burnside lemma on a product givesTwo pairs $(A,B)$ and $(A^{\prime },B^{\prime })$ lie in the same $S_n$-orbit exactly whenWhen $0\le l\le k\le n/2$, every value $0,1,\dots ,l$ is possible because $k+l\le n$. Therefore the symmetric-group subset permutation representation satisfies

For $1\le r\le n/2$, the inclusion map between subset permutation modules embeds $\mathbb{C}[X_{r-1}]$ into $\mathbb{C}[X_r]$, sois a character. Its norm isBy character orthogonality, ${\chi }_r$ is the character of an irreducible representation.

For $r>n/2$, complementation is an $S_n$-equivariant bijection $X_r\cong X_{n-r}$, soConsequentlythe negative of one of the irreducible characters just found, and is not itself the character of a representation.

The character of $(\rho ,V)$ isThe stated action is a representation because successive action by $(g_2,h_2)$ and $(g_1,h_1)$ gives

Identifying ${\mathrm{Hom}}_{\mathbb{C}}(V,W)$ with $V^{\ast }\otimes W$ gives the character of a Hom representationHere the final equality follows after unitarizing the finite-group representation.

The permutation character of $\mathbb{C}G$ counts fixed points:Thus

On the other hand, the character of
${⨁}_i\mathrm{Hom}(V_i,V_i)$ isThe column form of character orthogonality says that this has exactly the same two values above. Complex representations of a finite group are semisimple, and semisimple representations with the same character are isomorphic. Hence the two-sided regular representation decomposition isas a $G\times G$ representation.

Let $\varphi =(1+\sqrt{5})/2$ and $\bar{\varphi }=(1-\sqrt{5})/2$. In class order $(1,2A,3A,5A,5B)$, character orthogonality giveswith class sizes $1,15,20,12,12$. The rows are orthonormal and their squared degrees sum to $60$, so the table is complete.

A three-dimensional representation of $S_5$ would decompose into irreducibles. Since $S_5$ has irreducible degrees $1,1,4,4,5,5,6$, only one-dimensional constituents could occur, whose restrictions to $A_5$ are trivial. It cannot restrict to either irreducible $3$.

A nonabelian simple group has no nonidentity conjugacy class of prime-power size. Coprime-degree irreducible characters vanish on such an element: the conjugacy-class sum makes the character-to-degree ratio an algebraic integer, and the Kronecker theorem on algebraic integers in the unit disk makes a nonzero ratio a root of unity, forcing a scalar in a faithful group representation. Column character orthogonality then makes $1/p$ an algebraic integer, impossible. This is the character-theoretic ingredient in Burnside's theorem.

For inequivalent chosen unitary irreducible representations of a finite group, uniform expectation givesThe Kronecker deltas in the first case refer to entries in the same chosen basis. Averaging an intertwiner and applying the Schur lemma proves the formula; summing diagonal entries gives character orthogonality.