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  "body": "\n## Proof\n\nWe argue by strong induction on $\\abs{G}$. Assume that every group of order less than $\\abs{G}$ has a subgroup of order $p^j$ whenever $p^j$ divides its order. Let $e$ be the identity of $G$.\n\n1. If $k = 0$, the subgroup $\\set{e}$ has order $p^0 = 1$. So assume $k \\geq 1$; then $p$ divides $\\abs{G}$. We split into two cases.\n2. Case 1: $p$ divides $\\abs{Z(G)}$. The [[centre|0S7Y9Q990Q]] $Z(G)$ is a subgroup of $G$, by the [[subgroup criterion|05N4TQKPH0]]. Indeed $e \\in Z(G)$. If $u, w \\in Z(G)$ and $g \\in G$, then $w g = g w$ gives $w^{-1} g = g w^{-1}$, so $u w^{-1} g = u g w^{-1} = g u w^{-1}$. By [[Cauchy's theorem|0JM7PA288B]], it has an element $z$ of order $p$.\n3. Let $N = \\langle z \\rangle$. By the [[order of a cyclic subgroup|031M8HX7DA]], $N = \\set{e, z, \\dots, z^{p-1}}$ and $\\abs{N} = p$.\n4. For $g \\in G$, each $z^i$ commutes with $g$, so $g z^i g^{-1} = z^i$. Hence $g N g^{-1} = N$, and $N$ is a [[normal subgroup|0W77CVBCHQ]] of $G$.\n5. Let $\\bar G = G/N$ be the [[quotient group|0YYWCNTPWR]]. By the [[index formula|04DC0SRN2K]], $\\abs{\\bar G} = [G : N] = \\abs{G}/p$. This is less than $\\abs{G}$ and divisible by $p^{k-1}$.\n6. By the induction hypothesis, $\\bar G$ has a subgroup $\\bar H$ of order $p^{k-1}$.\n7. Let $\\pi \\colon G \\to \\bar G$, $\\pi(g) = gN$. By the definition of the quotient operation, $\\pi(gh) = \\pi(g)\\pi(h)$, and $\\pi(g^{-1}) = \\pi(g)^{-1}$.\n8. Let $H = \\set{g \\in G : \\pi(g) \\in \\bar H}$. It contains $e$. If $g, h \\in H$, then $\\pi(g h^{-1}) = \\pi(g) \\pi(h)^{-1} \\in \\bar H$. So $H$ is a [[subgroup|0N0PXCTYA7]] of $G$.\n9. Each element of $\\bar H$ is a coset $aN$ with $a \\in G$. For $x \\in aN$, write $x = an$ with $n \\in N$; then $xN = a(nN) = aN$, because $m \\mapsto nm$ is a bijection of $N$.\n10. By step 9, $x \\in H$ if and only if $x$ lies in some coset belonging to $\\bar H$. So $H$ is the union of the $p^{k-1}$ cosets in $\\bar H$.\n11. Distinct cosets are disjoint: if $x \\in aN \\cap bN$, step 9 gives $aN = xN = bN$. Each coset has [[$\\abs{N} = p$ elements|0DKMBWTD7P]].\n12. By steps 10 and 11, $\\abs{H} = p^{k-1} \\cdot p = p^k$. This settles Case 1.\n13. Case 2: $p$ does not divide $\\abs{Z(G)}$. By the [[class equation|08BPSC6YTE]], $\\abs{G} = \\abs{Z(G)} + \\sum_{i=1}^{r} [G : C_G(a_i)]$, over the conjugacy classes with more than one element.\n14. If $p$ divided every $[G : C_G(a_i)]$, it would divide $\\abs{G} - \\sum_i [G : C_G(a_i)] = \\abs{Z(G)}$. So some $i$ has $p \\nmid [G : C]$, where $C = C_G(a_i)$.\n15. The [[centraliser|0Q0Z1BEXTF]] $C$ is a subgroup of $G$. Indeed $e \\in C$. If $g, h \\in C$, then $h^{-1} a_i = a_i h^{-1}$, so $g h^{-1} a_i = g a_i h^{-1} = a_i g h^{-1}$. By the [[class size formula|0VK5V9P8G3]], $[G : C] = \\abs{\\operatorname{Cl}(a_i)} \u003e 1$.\n16. By the index formula, $\\abs{G} = [G : C] \\cdot \\abs{C}$. With step 15, $\\abs{C} \u003c \\abs{G}$.\n17. A common prime factor of $p^k$ and $[G : C]$ would divide $p$ by [[Euclid's lemma|051VYKA1EX]], so it would be $p$. This is excluded, so $p^k$ and $[G : C]$ are coprime.\n18. Since $p^k$ divides $[G : C] \\cdot \\abs{C}$, the [[generalised Euclid's lemma|03BAREW1Z5]] gives $p^k \\mid \\abs{C}$.\n19. By steps 16 and 18 and the induction hypothesis, $C$ has a subgroup $K$ of order $p^k$. Then $K$ is also a subgroup of $G$. This settles Case 2.\n20. By induction, the statement holds for every finite group $G$ and every $k \\geq 0$ with $p^k \\mid \\abs{G}$.\n21. Write $\\abs{G} = p^n m$ with $p \\nmid m$. Taking $k = n$ gives a subgroup of order $p^n$, which is a [[Sylow $p$-subgroup|0JWZX6FKSH]].\n\n## Idea\n\nIf $p$ divides the order of the centre, Cauchy gives a central subgroup of order $p$. Divide it out, find a subgroup of order $p^{k-1}$ in the smaller quotient, and pull it back. Otherwise the class equation finds a non-central element whose centraliser has index prime to $p$. That smaller centraliser still has order divisible by $p^k$, and induction applies.\n",
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