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For a nonzero finite-dimensional complex vector space $V$, a unital matrix algebra acting irreducibly on $V$ is the full endomorphism algebra. Its Jacobson radical annihilates $V$: otherwise the radical times $V$ is all of $V$, contradicting nilpotence. Faithfulness kills the radical, and the Artin–Wedderburn theorem then leaves one simple matrix factor acting on its standard column module. This is a matrix-algebra theorem, distinct from the orbit-counting Burnside lemma and the finite-group Burnside's theorem.

If $\pi$ is the character of $\mathbb{C}[X]$, then Burnside lemma applied to the diagonal action on $X\times X$ givesFor a transitive action $X\simeq G/H$, this also equals the number of double cosets in $H\backslash G/H$.

For a group action, the orbit of a group action of $x$ is $Gx=\{gx:g\in G\}$, and its stabilizer subgroup is $G_x=\{g:gx=x\}$. The orbit-stabilizer theorem gives $|Gx||G_x|=|G|$: the map $g{G}_x\mapsto gx$ is a well-defined bijection from stabilizer cosets to the orbit.

Within one orbit, the stabilizer subgroups are conjugate, since $G_{hx}=h{G}_x{h}^{-1}$. Their sizes are therefore equal. Summing stabilizer sizes over a single orbit gives $|Gx||G_x|=|G|$, and summing over all $n$ orbits yieldsNow count the pairs $(g,x)$ fixed by the action in two ways. Holding $x$ fixed gives $|G_x|$ choices of $g$, whereas holding $g$ fixed gives $|\mathrm{Fix}(g)|$ choices of $x$. ThusCombining the two identities proves Burnside lemma:The finiteness hypotheses justify these cardinality sums. If $X$ is empty, every sum and the orbit count are zero, so the formula still applies.

Centre the cube at the origin with faces having normals $\pm e_1,\pm e_2,\pm e_3$. Every rotational symmetry fixes the centre and permutes these normals, so its rotation matrix is a signed permutation matrix of determinant one. Conversely any such matrix permutes the cube’s coordinates up to signs and preserves the cube. For each of $3!$ coordinate permutations, four of the eight choices of signs give determinant one. Consequently the rotational symmetry group of a cube hasThis counts the rotations directly, without an unverified stabilizer calculation. The TeX’s $2^k$ is damaged transcription: the PDF prints 24.

Let $X$ be the $3^6$ face colourings. A colouring fixed by a rotation must be constant on each permutation cycle of faces, giving $3^c$ fixed colourings if the rotation has $c$ face cycles. The rotation types are:

There are three face axes, each with two quarter-turns and one half-turn; six edge axes, each with one half-turn; and four vertex axes, each with two nonidentity third-turns. These rotations are distinct and, with the identity, total 24, so the table is exhaustive. Face-axis rotations fix their two axial faces; the remaining four faces cycle or pair. Edge half-turns pair all faces, while vertex third-turns form two triples.

Burnside lemma now givesThe three colours are labelled, orthogonal reflections are excluded, and no condition requires every colour to appear.

The singleton orbits are exactly $X^G$. Every other orbit represented by $x$ has size $|G|/|G_x|$ by the orbit-stabilizer theorem. Partitioning $X$ into its orbits of a group action therefore gives

For the second identity, countin two ways. Holding $g$ fixed givesHolding $x$ fixed gives $|Z|={\sum }_{x\in X}|G_x|$. Each orbit contributes $|G|/|G_x|$ points, each with stabilizer size $|G_x|$, and hence contributes $|G|$. Thus $|Z|=|G||X/G|$, proving Burnside lemma:

For the rotation group $C_5$, the identity fixes all $n^5$ colorings. Because five is prime, each of the four nonidentity rotations is a single five-cycle and fixes exactly the $n$ constant colorings. Burnside lemma therefore gives the number of coloring necklaces

For the dihedral group $D_{10}$, the five reflections each fix one vertex and interchange two pairs of vertices, so each fixes $n^3$ colorings. Adding these to the identity and rotation contributions gives the number of coloring bracelets

Burnside lemma says that the number of orbits of a finite group $G$ on a finite set $X$ isIndeed, double-count the set $\{(g,x):gx=x\}$. Counting first by $g$ gives the numerator. Counting first by $x$ gives ${\sum }_x|G_x|$; each orbit $O$ contributes $|O||G_x|=|G|$ by the orbit-stabilizer theorem, proving the formula.

Let $\pi (g)=|X^g|$ be the character of the permutation representation. Thenis, by Burnside's lemma, the number of orbits on $X\times X$. A two-transitive group action has exactly two such orbits: the diagonal and the ordered pairs of distinct points. Hence $⟨\pi ,\pi ⟩=2$. Transitivity also gives $⟨\pi ,1_G⟩=1$. If $\pi =1_G+\sum m_i{\chi }_i$ is its irreducible character decomposition, then $2=1+\sum m_i^2$, so exactly one nontrivial irreducible character occurs, with multiplicity one. Thus the permutation character has precisely two distinct irreducible summands.

The Burnside lemma states that for a finite group $G$ acting on a finite set $X$,To prove it, countCounting by $g$ gives ${\sum }_g|X^g|$. Counting by $x$ givesusing the orbit-stabilizer theorem on each orbit $\mathcal{O}$. Dividing by $|G|$ proves the formula.

Let $\pi$ be the character of the permutation representation $\mathbb{C}[X]$. Then $\pi (g)=|X^g|$, and a second application of Burnside's lemma givesFor a two-transitive action there are exactly two orbits on $X\times X$: the diagonal and the ordered pairs of distinct points. Hence $⟨\pi ,\pi ⟩=2$. Transitivity says the trivial representation occurs once. Since the squared multiplicities of the irreducible constituents sum to two, there is exactly one further constituent, with multiplicity one, and it is irreducible and nontrivial. Thusas in the permutation representation of a two-transitive action.

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.