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  "body": "\n## Proof\n\nWe write each cycle as a product of transpositions by induction on its length, and then use the disjoint cycle decomposition.\n\n1. Let $n \\ge 2$, and let $S_n$ be the [[symmetric group|0RJMNEN3BJ]], with products taken as compositions: $(\\alpha\\beta)(x) = \\alpha(\\beta(x))$.\n2. Let $k \\ge 3$ and let $c = (a_1\\ a_2\\ \\dots\\ a_k)$ be a [[cycle|0ZKPH2BD1T]]. We claim $c = (a_1\\ a_k)\\, c'$, where $c' = (a_1\\ a_2\\ \\dots\\ a_{k-1})$ and $(a_1\\ a_k)$ is a [[transposition|02DTRYKGE8]].\n3. For $1 \\le i \\le k-2$, $c'(a_i) = a_{i+1}$, which differs from $a_1$ and $a_k$. So $(a_1\\ a_k) c'(a_i) = a_{i+1} = c(a_i)$.\n4. Next, $c'(a_{k-1}) = a_1$, and $(a_1\\ a_k)$ sends $a_1$ to $a_k$. So $(a_1\\ a_k) c'(a_{k-1}) = a_k = c(a_{k-1})$.\n5. Next, $c'$ fixes $a_k$, and $(a_1\\ a_k)$ sends $a_k$ to $a_1$. So $(a_1\\ a_k) c'(a_k) = a_1 = c(a_k)$.\n6. Every other point is fixed by $c'$, by $(a_1\\ a_k)$ and by $c$. With steps 3 to 5, this proves the claim of step 2.\n7. A cycle $(a_1\\ a_2)$ of length $2$ is itself a transposition. By step 2 and induction on $k \\ge 2$, each cycle of length $k$ is a product of $k - 1$ transpositions.\n8. Let $\\sigma \\in S_n$. By the [[decomposition into disjoint cycles|0YWDAZPRPQ]], $\\sigma = c_1 \\cdots c_r$ with cycles $c_i$ of length at least $2$.\n9. If $r \\ge 1$, replace each $c_i$ by a product of transpositions, using step 7. This writes $\\sigma$ as a product of transpositions.\n10. If $r = 0$, then $\\sigma$ is the identity. Since $n \\ge 2$, the transposition $(1\\ 2)$ exists, and $(1\\ 2)(1\\ 2)$ is the identity.\n11. By steps 9 and 10, each permutation in $S_n$ is a product of transpositions.\n\n## Idea\n\nA cycle can be peeled off one point at a time: $(a_1\\ \\dots\\ a_k) = (a_1\\ a_k)(a_1\\ \\dots\\ a_{k-1})$. Repeating this writes a cycle of length $k$ as $k - 1$ transpositions. Every permutation is a product of cycles, and the identity is a transposition applied twice.\n",
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