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  "body": "\n## Proof\n\nWe show two inclusions. Let $n \\geq 3$, let $T$ be the set of $3$-cycles in $S_n$, and let $H = \\langle T \\rangle$ be the [[subgroup generated|0RG9SPQSWM]] by $T$. Products are composites of maps, and the right factor acts first.\n\n1. By the [[sign of a cycle|0V0E04DWX2]], each $3$-cycle has sign $(-1)^{2} = 1$. So $T \\subseteq A_n$, by the [[definition of the alternating group|0T13HM377F]].\n2. The set $A_n$ is the kernel of the [[sign homomorphism|0JEJJCB0EX]], so it is a [[subgroup|0C9ZKYSPYR]] of $S_n$. Since $H$ is the smallest subgroup that contains $T$, step 1 gives $H \\subseteq A_n$.\n3. For the other inclusion, we first show that each product of two [[transpositions|02DTRYKGE8]] lies in $H$. Let $\\alpha$ and $\\beta$ be transpositions.\n4. If $\\alpha = \\beta$, then $\\alpha \\beta = e$, and $e \\in H$.\n5. If $\\alpha$ and $\\beta$ move exactly one common point $a$, write $\\alpha = (a\\ b)$ and $\\beta = (a\\ c)$ with $a, b, c$ distinct. Then $\\alpha\\beta$ sends $a \\mapsto c$, $c \\mapsto b$ and $b \\mapsto a$, and fixes the other points.\n6. So in [[cycle notation|0ZKPH2BD1T]] $\\alpha \\beta = (a\\ c\\ b)$, which lies in $T \\subseteq H$.\n7. If $\\alpha$ and $\\beta$ move no common point, write $\\alpha = (a\\ b)$ and $\\beta = (c\\ d)$ with $a, b, c, d$ distinct. Let $\\gamma = (a\\ c\\ b)(a\\ c\\ d)$.\n8. Compute $\\gamma$ point by point: $a \\mapsto c \\mapsto b$, $b \\mapsto b \\mapsto a$, $c \\mapsto d \\mapsto d$, $d \\mapsto a \\mapsto c$. Other points are fixed.\n9. So $\\gamma = (a\\ b)(c\\ d) = \\alpha \\beta$. It is a product of two elements of $T$, so it lies in $H$.\n10. Steps 4 to 9 cover all cases, because two transpositions move two points each and share zero, one or two points. Sharing two points means $\\alpha = \\beta$.\n11. Now let $\\sigma \\in A_n$. By the [[decomposition into transpositions|0XK6FECMW5]], $\\sigma = \\tau_1 \\tau_2 \\cdots \\tau_m$ with transpositions $\\tau_i$.\n12. By the [[definition of the sign|0SK90238BH]], $(-1)^m = \\operatorname{sgn}(\\sigma) = 1$, so $m$ is even. Write $m = 2r$.\n13. Then $\\sigma = (\\tau_1 \\tau_2)(\\tau_3 \\tau_4) \\cdots (\\tau_{2r-1} \\tau_{2r})$. Each bracket lies in $H$ by steps 4 to 10, and $H$ is closed under products. So $\\sigma \\in H$.\n14. Thus $A_n \\subseteq H$. With step 2, $A_n = H$, so $A_n$ is generated by the $3$-cycles.\n\n## Idea\n\nAn even permutation is a product of an even number of transpositions, so it is a product of pairs of transpositions. Each pair is either trivial, a $3$-cycle, or a product of two $3$-cycles. So the $3$-cycles already produce every even permutation.\n",
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