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  "body": "\n## Proof\n\nWe let $G$ act on the left cosets of $H$ and show that $H$ is the kernel of this action.\n\n1. Let $G$ be a finite group, let $p$ be the smallest prime that divides $|G|$, and let $H \\le G$ have [[index|04QEBF8S49]] $p$.\n2. Let $X$ be the set of [[left cosets|05FD317V2W]] $xH$ with $x \\in G$. By the definition of the index, $|X| = p$.\n3. For $g \\in G$ and a subset $A$ of $G$, write $gA = \\{ga : a \\in A\\}$. Then $g(xH) = (gx)H$ for each $x \\in G$.\n4. For $g \\in G$, define $\\rho_g \\colon X \\to X$ by $\\rho_g(C) = gC$. By step 3, $\\rho_g(xH) = (gx)H$ lies in $X$, so $\\rho_g$ is a map from $X$ to $X$.\n5. For $g, h \\in G$ and $C \\in X$, we have $\\rho_g(\\rho_h(C)) = g(hC) = (gh)C = \\rho_{gh}(C)$. Also $\\rho_e(C) = C$, where $e$ is the identity of $G$.\n6. By step 5, $\\rho_g \\circ \\rho_{g^{-1}} = \\rho_e$ and $\\rho_{g^{-1}} \\circ \\rho_g = \\rho_e$ are the identity of $X$. So $\\rho_g$ is [[bijective|01VG2A0BHW]].\n7. Thus $\\rho_g$ lies in the [[symmetric group|0RJMNEN3BJ]] $\\operatorname{Sym}(X)$. By step 5, the map $\\rho \\colon G \\to \\operatorname{Sym}(X)$, $g \\mapsto \\rho_g$, is a [[group homomorphism|0HV9Y7WG5R]].\n8. By step 5 and step 6, the image $\\operatorname{im} \\rho$ contains $\\rho_e$ and is closed under products and inverses. So $\\operatorname{im} \\rho$ is a [[subgroup|0N0PXCTYA7]] of $\\operatorname{Sym}(X)$.\n9. Let $K = \\ker \\rho$ be the [[kernel|0Z424HK8CQ]]. By [[the kernel is a normal subgroup|0C9ZKYSPYR]], $K$ is a normal subgroup of $G$.\n10. Let $g \\in K$. Then $\\rho_g(H) = H$, that is, $gH = H$. Since $e \\in H$, we get $g = ge \\in gH = H$. So $K \\subseteq H$, and $K$ is a subgroup of $H$.\n11. By the [[first isomorphism theorem|0CXXJNG5B6]], the [[quotient group|0YYWCNTPWR]] $G/K$ is isomorphic to $\\operatorname{im} \\rho$. An [[isomorphism|085PND4KNT]] is a bijection, so $|G/K| = |\\operatorname{im} \\rho|$.\n12. By [[Lagrange's theorem|3C53TD]] and step 8, $|\\operatorname{im} \\rho|$ divides $|\\operatorname{Sym}(X)|$. By the [[number of permutations|0CEBCS0E0E]], $|\\operatorname{Sym}(X)| = p!$.\n13. The set $G/K$ is the set of left cosets of $K$, so $|G/K| = [G : K]$. Let $d = [G : K]$. By steps 11 and 12, $d$ divides $p!$.\n14. Let $m = [H : K]$. By [[the order of a group is the index times the order of the subgroup|04DC0SRN2K]], $|G| = p \\, |H|$, $|H| = m \\, |K|$ and $|G| = d \\, |K|$.\n15. By step 14, $d \\, |K| = p \\, m \\, |K|$. Since $|K| \\ge 1$, we get $d = pm$.\n16. By steps 13 and 15, $pm$ divides $p! = p \\cdot (p-1)!$. Dividing by $p$, we get that $m$ divides $(p-1)!$.\n17. Suppose that $m \u003e 1$. By the [[fundamental theorem of arithmetic|084HH1NDCB]], some prime $q$ divides $m$.\n18. By step 14, $|G| = p \\, m \\, |K|$, so $q$ divides $|G|$. Since $p$ is the smallest prime that divides $|G|$, we get $q \\ge p$.\n19. If $p = 2$, then $(p-1)! = 1$, and the prime $q$ does not divide $1$. This contradicts step 16.\n20. If $p \\ge 3$, then $q$ divides the product $1 \\cdot 2 \\cdots (p-1)$. Applying [[Euclid's lemma|051VYKA1EX]] repeatedly, $q$ divides some factor $j$ with $1 \\le j \\le p-1$.\n21. A prime that divides the positive integer $j$ is at most $j$. So $q \\le p - 1 \u003c p$, which contradicts step 18.\n22. Hence $m = 1$, so $|H| = |K|$ by step 14. Since $K \\subseteq H$ and $H$ is finite, $K = H$.\n23. By step 9, $H = K$ is [[normal|0W77CVBCHQ]] in $G$.\n\n## Idea\n\nThe group permutes the $p$ cosets of $H$, which gives a homomorphism to a group of order $p!$. Its kernel lies inside $H$, and the index of the kernel divides both $|G|$ and $p!$. The only common room is the prime $p$ itself, because $|G|$ has no prime factor below $p$ and $p!$ has only one factor $p$. So the kernel has index $p$, and it must be all of $H$.\n",
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