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We use the standard facts that finite complex representations are completely reducible, irreducible characters are orthonormal, the regular character is ${\sum }_{\chi }\chi (1)\chi$, and the number of irreducible characters equals the number of conjugacy classes. We also use Schur's lemma, Sylow's theorem and the nontrivial center of a nontrivial finite $p$-group.

For an irreducible character $\chi$ of degree $d$, a conjugacy-class sum $z_C$ acts as the scalar ${\lambda }_C=|C|\chi (g)/d$ by Schur's lemma and taking traces. This scalar is an algebraic integer: class sums form an integral basis of the center of the integral group algebra, their multiplication has integer structure constants, and multiplication by $z_C$ satisfies its monic integer characteristic polynomial. That same polynomial annihilates its image in any irreducible representation. Character values are also algebraic integers, since they are sums of roots of unity. Character orthogonality now givesThis is a rational algebraic integer, hence an integer. Therefore $\chi (1)$ divides $|G|$.

The Burnside's pa qb theorem states that every finite group of order $p^a{q}^b$ is soluble. We prove its character-theoretic ingredient. If a class has size $m$ coprime to $d$, integers $u,v$ with $ud+vm=1$ givean algebraic integer. Every algebraic conjugate of this number is the average of $d$ roots of unity and has modulus at most one. If it is nonzero, its algebraic norm is a nonzero integer of absolute value at least one; since every factor has modulus at most one, all must have modulus one. In particular $|\chi (g)|=d$. Equality in the triangle inequality for the unit-modulus eigenvalues implies that $\rho (g)$ is scalar. Thus if $(m,d)=1$, either $\chi (g)=0$ or $\rho (g)$ is scalar.

Suppose a group of order $p^a{q}^b$ were nonabelian simple, with both primes present. Choose a nonidentity $g$ in the center of a Sylow $p$-subgroup. Its centralizer contains that Sylow subgroup, so its class size is a power of $q$. Every nontrivial irreducible representation is faithful by simplicity. If it sent $g$ to a scalar, faithfulness would make $g$ central in the group, impossible in a nonabelian simple group. The ingredient above therefore gives $\chi (g)=0$ for every nontrivial character with degree coprime to $q$. Evaluating the regular character at $g\ne 1$ givesThe right side is an algebraic integer, but $1/q$ is not, a contradiction. A group of prime-power order cannot be nonabelian simple either, by its nontrivial center. Induct on the order: every nonsimple group has a proper nontrivial normal subgroup, and that subgroup and the quotient have smaller orders with at most these two prime divisors. By induction both are soluble; their derived-series extensions make the original group soluble. Simple abelian groups have prime order and start the induction. This proves Burnside's theorem.

For the final divisibility, let $|G|=n$ be odd and let $k$ be its number of conjugacy classes. All irreducible degrees are odd by the divisibility just proved. We use the standard Frobenius-Schur indicator theorem: $\nu (\chi )=n^{-1}{\sum }_g\chi (g^2)$ is zero precisely for an irreducible character not equal to its conjugate, and is $\pm 1$ otherwise. Squaring is a bijection on an odd-order group: for each $g$, its unique square root is $g^{(n+1)/2}$, since every element order divides $n$. For every nontrivial irreducible character, orthogonality to the trivial character consequently gives $\nu (\chi )=n^{-1}{\sum }_g\chi (g)=0$. The nontrivial characters therefore occur in distinct conjugate pairs of equal odd degree. Since an odd square is one modulo eight,is a sum of pairs $2(d^2-1)$, each divisible by sixteen; the trivial character contributes zero. Hence

Work over $\mathbb{C}$, with the standard inclusions $S_r\subseteq S_n$ fixing the letters $r+1,\dots ,n$. In the group algebra, the Young–Jucys–Murphy elements areThus $X_i$ adds all transpositions involving the newest letter. Conjugating by $S_{i-1}$ merely permutes the summands, so $X_i$ commutes with $\mathbb{C}{S}_{i-1}$. In particular, if $j<i$, then $X_j\in \mathbb{C}{S}_{i-1}$ and $X_i{X}_j=X_j{X}_i$. Hence the whole family is commuting.

Put ${\Omega }_r={\sum }_{1\le a<b\le r}(ab)$, the transposition conjugacy-class sum in $S_r$, with ${\Omega }_1=0$. ThenThis identity will connect the Young–Jucys–Murphy elements with the centres of the successive group algebras.

The relevant centralizer of a subalgebra isEquivalently, it is the fixed subspace of $\mathbb{C}{S}_n$ under conjugation by $S_{n-1}$. We prove the Olshanskii centralizer lemma directly, without assuming multiplicity-free restriction.

A permutation has one cycle containing $n$, of some length $r\ge 1$. Let $\nu$ record the lengths of its other nontrivial cycles, omitting fixed points. Two permutations are conjugate under $S_{n-1}$ exactly when these data $(r,\nu )$ agree: a relabelling of the other letters identifies the distinguished cycles and all remaining cycles. Hence the sums $K_{r,\nu }$ over these conjugation orbits form a basis of $Z_{(n-1,1)}$.

Filter these orbit sums by the number of moved letters other than $n$:If $r=1$, the permutation fixes $n$, and $K_{1,\nu }$ is already a conjugacy-class sum in $Z_{n-1}$. For $r\ge 2$, let $C_{\nu }\in Z_{n-1}$ be the class sum with nontrivial cycle lengths $\nu$. ExpandTerms in which the $r-1$ star transpositions use distinct letters, disjoint from the support of the chosen term of $C_{\nu }$, produce precisely $K_{r,\nu }$, each with coefficient one. Indeed, a cycle through $n$ has a unique expression as an ordered product of star transpositions using each of its other letters once; the remaining cycles uniquely determine the term of $C_{\nu }$.

Every other term moves fewer than $t$ letters outside $n$. A repeated letter uses at most $r-2$ distinct such letters in the star product, and an overlap with the support of $C_{\nu }$ likewise reduces the size of their union. Since the whole product commutes with $S_{n-1}$, these remaining terms are a linear combination of lower-filtered orbit sums. ThusInduction on $t$, starting with the identity, puts every orbit sum in $\mathrm{alg}(Z_{n-1},X_n)$. The reverse inclusion holds because both proposed generators commute with $\mathbb{C}{S}_{n-1}$. ThereforeIn particular this centralizer of a subalgebra is commutative: $Z_{n-1}$ is commutative and $X_n$ commutes with it.

For $k=1$, the product is empty and equals the identity, the only $1$-cycle in $S_1$. Suppose the cycle-sum identity for Young–Jucys–Murphy elements holds at $k-1$:Multiplication by $X_k={\sum }_{a<k}(ak)$ gives terms $\sigma (ak)$. With permutation products acting right to left, this inserts $k$ immediately after $a$ in the cycle of $\sigma$: the arrows become $a\mapsto k\mapsto \sigma (a)$, with the other arrows unchanged.

Conversely, in any $k$-cycle there is a unique predecessor $a$ of $k$. Delete $k$ and join that predecessor to the successor of $k$; this uniquely recovers $\sigma$ and the factor $(ak)$. Hence every $k$-cycle occurs exactly once, and no other permutation occurs. Induction provesThus ${\xi }_k$ is a conjugacy-class sum in $\mathbb{C}{S}_k$, although it need not be central in $\mathbb{C}{S}_n$ when $k<n$.

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.