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  "body": "\n## Proof\n\nThe following proof — using more background than Burnside's — is by contradiction. Let $p^a q^b$ be the smallest product of two prime powers, such that there is a non-solvable group $G$ whose order is equal to this number.\n\n- $G$ **is a [[simple group|0C093D1T4M]] with trivial [[center|0DZ93493VN]] and** $a$ **is not zero.**\n\nIf $G$ had a [[nontrivial|0BNPC9850Y]] proper [[normal subgroup|0CB0DX5BVV]] $H$, then (because of the minimality of *$G$*), $H$ and $G / H$ would be solvable, so $G$ as well, which would contradict our assumption. So $G$ is simple.\n\nIf $a$ were zero, $G$ would be a finite [[q-group|06QYRTX9KQ]], hence [[nilpotent|0DZPD2MG33]], and therefore solvable.\n\nSimilarly, $G$ cannot be abelian, otherwise it would be solvable. As $G$ is simple, its center must therefore be trivial.\n\n- **There is an element $g$ of $G$ which has $q^d$ [[conjugates|0DEBXAJ5SA]], for some $d \u003e 0$.**\n\nBy the first statement of Sylow's theorem, $G$ has a [[subgroup|0X1JXT47GA]] $S$ of order $p^a$. Because $S$ is a nontrivial $p$-group, its center $Z(S)$ is nontrivial. Fix a nontrivial element $g \\in Z(S)$. The number of conjugates of $g$ is equal to the index of its centralizer $G_g$, which divides the [[index|0DTXBQPA2S]] $q^b$ of $S$ (because $S$ is a subgroup of $G_g$). Hence this number is of the form $q^d$. Moreover, the integer $d$ is strictly positive, since $g$ is nontrivial and therefore not central in $G$.\n\n- **There exists a nontrivial irreducible representation $\\rho$ with character $\\chi$, such that its dimension $n$ is not divisible by $q$ and the [[complex number|09FTAET5Z4]] $\\chi(g)$ is not zero.**\n\nLet $(\\chi_i)_{1 \\leq i \\leq h}$ be the family of irreducible characters of $G$ over $\\mathbb{C}$ (here $\\chi_1$ denotes the trivial character). Because $g$ is not in the same conjugacy class as 1, the orthogonality relation for the columns of the group's character table gives:\n\n$$0=\\sum_{i=1}^h \\chi_i(1)\\chi_i(g)= 1 + \\sum_{i=2}^h \\chi_i(1)\\chi_i(g).$$\n\nNow the $\\chi_i(g)$ are algebraic integers, because they are sums of [[roots of unity|0S61N75H45]]. If all the nontrivial irreducible characters which don't vanish at $g$ take a value divisible by $q$ at 1, we deduce that\n\n$$-\\frac1q=\\sum_{i\\ge 2,~\\chi_i(g)\\ne 0}\\frac{\\chi_i(1)}q\\chi_i(g)$$\n\nis an algebraic integer (since it is a sum of integer multiples of algebraic integers), which is absurd. This proves the statement.\n\n- **The complex number $q^d \\chi(g) / n$ is an algebraic integer.**\n\nThe set of integer-valued class functions on $G$, $Z(\\mathbb{Z}[G])$, is a [[commutative ring|0310REJJY8]], [[finitely generated|05QY1VEBYD]] over $\\mathbb{Z}$. All of its elements are thus integral over $\\mathbb{Z}$, in particular the mapping $u$ which takes the value 1 on the conjugacy class of $g$ and 0 elsewhere.\n\nThe mapping $A\\colon Z(\\mathbb{Z}[G]) \\rightarrow \\operatorname{End}(\\mathbb{C}^n)$ which sends a class function $f$ to\n\n$$\\sum_{s\\in G} f(s)\\rho(s)$$\n\nis a ring homomorphism. Because $\\rho(s)^{-1} A(u) \\rho(s) = A(u)$ for all $s$, Schur's lemma implies that $A(u)$ is a homothety $\\lambda I_n$. Its [[trace|0XC1RDBY7A]] $n \\lambda$ is equal to\n\n$$\\sum_{s\\in G} f(s)\\chi(s)=q^d\\chi(g).$$\n\nBecause the homothety $\\lambda I_n$ is the homomorphic image of an integral element, this proves that the complex number $\\lambda = q^d \\chi(g) / n$ is an algebraic integer.\n\n- **The complex number $\\chi(g) / n$ is an algebraic integer.**\n\nSince $q$ is relatively prime to $n$, by [[Bézout's identity|0DQVHMYAGA]] there are two integers $x$ and $y$ such that:\n\n$$xq^d + yn=1\\quad\\text{therefore}\\quad \\frac{\\chi(g)}{n}=x\\frac{q^d\\chi(g)}{n} + y\\chi(g).$$\n\nBecause a linear combination with integer coefficients of algebraic integers is again an algebraic integer, this proves the statement.\n\n- **The image of $g$, under the representation $\\rho$, is a homothety.**\n\nLet $\\zeta$ be the complex number $\\chi(g) / n$. It is an algebraic integer, so its norm $N(\\zeta)$ (i.e. the product of its conjugates, that is the roots of its [[minimal polynomial|0VX9EP2GS8]] over $\\mathbb{Q}$) is a nonzero integer. Now $\\zeta$ is the average of roots of unity (the eigenvalues of $\\rho(g)$), hence so are its conjugates, so they all have an absolute value less than or equal to 1. Because the absolute value of their product $N(\\zeta)$ is greater than or equal to 1, their absolute value must all be 1, in particular $\\zeta$, which means that the eigenvalues of $\\rho(g)$ are all equal, so $\\rho(g)$ is a homothety.\n\n- **Conclusion**\n\nLet $N$ be the kernel of $\\rho$. The homothety $\\rho(g)$ is central in $\\operatorname{Im}(\\rho)$ (which is canonically isomorphic to $G / N$), whereas $g$ is not central in $G$. Consequently, the normal subgroup $N$ of the simple group $G$ is nontrivial, hence it is equal to $G$, which contradicts the fact that $\\rho$ is a nontrivial representation.\n\nThis contradiction proves the theorem.\n",
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