{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe split $X$ into the orbits of the powers of $\\sigma$. Each orbit with at least two points gives one cycle.\n\n1. Let $X$ be a finite set and $\\sigma \\in \\operatorname{Sym}(X)$, the [[symmetric group|0RJMNEN3BJ]]. Products are compositions, and $\\sigma^m$ is defined for $m \\in \\mathbb{Z}$ with the [[laws of exponents|00DRK90Z6K]].\n2. For distinct $a_0, \\dots, a_{k-1} \\in X$, the [[cycle|0ZKPH2BD1T]] $(a_0\\ a_1\\ \\dots\\ a_{k-1})$ sends $a_i$ to $a_{i+1}$ for $i \u003c k-1$, sends $a_{k-1}$ to $a_0$, and fixes every other point.\n3. For $x \\in X$, let $O(x) = \\{\\sigma^m(x) : m \\in \\mathbb{Z}\\}$. The relation \"$y \\in O(x)$\" is reflexive ($m = 0$), symmetric ($x = \\sigma^{-m}(y)$), and transitive ($\\sigma^{m'}(\\sigma^m(x)) = \\sigma^{m+m'}(x)$).\n4. So it is an equivalence relation, and the sets $O(x)$ are its classes. By [[equivalence classes form a partition|01HT2CH8BH]], the distinct sets $O(x)$ are pairwise disjoint and cover $X$.\n5. Fix $x \\in X$. Since $X$ is finite, $\\sigma^i(x) = \\sigma^j(x)$ for some $0 \\le i \u003c j$. Applying $\\sigma^{-i}$ gives $\\sigma^{j-i}(x) = x$. Let $k \\ge 1$ be the least integer with $\\sigma^k(x) = x$.\n6. The points $\\sigma^p(x)$ with $0 \\le p \u003c k$ are distinct. Indeed, $\\sigma^i(x) = \\sigma^j(x)$ with $0 \\le i \u003c j \u003c k$ would give $\\sigma^{j-i}(x) = x$ with $0 \u003c j - i \u003c k$.\n7. Since $\\sigma^k$ fixes $x$, so do its inverse $\\sigma^{-k}$ and every power $\\sigma^{qk}$ with $q \\in \\mathbb{Z}$.\n8. Let $m \\in \\mathbb{Z}$. By the [[division algorithm|0CHQKZFK5S]], $m = qk + r$ with $0 \\le r \u003c k$. By step 7, $\\sigma^m(x) = \\sigma^r(\\sigma^{qk}(x)) = \\sigma^r(x)$.\n9. By steps 6 and 8, $O(x) = \\{x, \\sigma(x), \\dots, \\sigma^{k-1}(x)\\}$ has exactly $k$ elements. If $k = 1$, then $O(x) = \\{x\\}$ and $\\sigma(x) = x$.\n10. Let $k \\ge 2$ and $c = (x\\ \\sigma(x)\\ \\dots\\ \\sigma^{k-1}(x))$. By steps 2 and 5, $c(z) = \\sigma(z)$ for $z \\in O(x)$. Also $c$ fixes each point outside $O(x)$ and moves each point of $O(x)$.\n11. So $c$ is determined by the set $O = O(x)$: it equals $\\sigma$ on $O$ and the identity outside $O$. Write $c_O$ for it.\n12. Let $O_1, \\dots, O_r$ be the distinct orbits with at least two points, and let $c_j = c_{O_j}$. By steps 4 and 10, the cycles $c_j$ have length at least $2$ and pairwise disjoint sets of moved points $O_j$.\n13. Let $y \\in X$ lie in a one-point orbit. By step 9, $\\sigma(y) = y$. Each $c_j$ fixes $y$, since $y \\notin O_j$. So $c_1 c_2 \\cdots c_r(y) = y = \\sigma(y)$.\n14. Let $y \\in O_j$. Then $\\sigma(y) \\in O_j$. For $i \\ne j$, the cycle $c_i$ fixes each point of $O_j$, so it fixes $y$ and $\\sigma(y)$.\n15. In $c_1 \\cdots c_r(y)$, the factors $c_i$ with $i \u003e j$ act first and fix $y$. Then $c_j$ gives $\\sigma(y)$, and the factors with $i \u003c j$ fix it. So $c_1 \\cdots c_r(y) = \\sigma(y)$.\n16. By steps 13 to 15, $\\sigma = c_1 c_2 \\cdots c_r$. When $r = 0$, this is the empty product, the identity. This proves existence.\n17. For uniqueness, let $\\sigma = d_1 \\cdots d_s$ with cycles $d_l$ of length at least $2$ and pairwise disjoint sets of moved points $M_l$.\n18. Let $z \\in M_l$. If $d_l(d_l(z)) = d_l(z)$, then $d_l(z) = z$ because $d_l$ is [[injective|0PVG1CRAJN]]. This is false, so $d_l(z) \\in M_l$.\n19. For $i \\ne l$, $d_i$ fixes each point of $M_l$. As in step 15, step 18 gives $\\sigma(z) = d_l(z)$ for each $z \\in M_l$.\n20. Write $d_l = (b_0\\ b_1\\ \\dots\\ b_{t-1})$ with $t \\ge 2$. By step 19, $\\sigma^p(b_0) = b_p$ for $0 \\le p \u003c t$, and $\\sigma^t(b_0) = b_0$.\n21. Since the $b_p$ are distinct, the least $k \\ge 1$ with $\\sigma^k(b_0) = b_0$ is $t$. By steps 9 and 10, $O(b_0) = M_l$, and $c_{O(b_0)} = (b_0\\ \\sigma(b_0)\\ \\dots\\ \\sigma^{t-1}(b_0)) = d_l$.\n22. So $M_l$ is one of the orbits $O_j$, and $d_l = c_j$. Distinct $l$ give distinct $j$, because the sets $M_l$ are nonempty and pairwise disjoint.\n23. Conversely, let $y \\in O_j$. By step 9, $\\sigma(y) \\ne y$, since $O_j$ has at least two points. If every $d_l$ fixed $y$, their product $\\sigma$ would fix $y$.\n24. So $y \\in M_l$ for some $l$. By steps 4 and 21, $O_j = O(y) = M_l$, so $c_j = d_l$.\n25. By steps 22 and 24, $l \\mapsto j$ is a bijection with $d_l = c_j$. So the factors $d_1, \\dots, d_s$ are $c_1, \\dots, c_r$ in some order.\n\n## Idea\n\nFollowing $x, \\sigma(x), \\sigma^2(x), \\dots$ must return to $x$, because $X$ is finite. The points visited form an orbit, and on it $\\sigma$ acts as one cycle. The orbits do not overlap, so these cycles are disjoint and their product is $\\sigma$. Any disjoint cycle decomposition must have the orbits as its supports, which gives uniqueness.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "20B30"
    ],
    "name": "Proof of the decomposition of a permutation into disjoint cycles",
    "proves": "0YWDAZPRPQ",
    "sources": [
      {
        "locator": "Cycle notation",
        "url": "https://en.wikipedia.org/wiki/Permutation"
      }
    ],
    "tag": "PDPM33XT31",
    "uses": [
      "0RJMNEN3BJ",
      "0ZKPH2BD1T",
      "00DRK90Z6K",
      "01HT2CH8BH",
      "0CHQKZFK5S",
      "0PVG1CRAJN"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/PDPM33XT31/",
  "links": [
    {
      "field": "proves",
      "target": "0YWDAZPRPQ"
    },
    {
      "field": "uses",
      "target": "0RJMNEN3BJ"
    },
    {
      "field": "uses",
      "target": "0ZKPH2BD1T"
    },
    {
      "field": "uses",
      "target": "00DRK90Z6K"
    },
    {
      "field": "uses",
      "target": "01HT2CH8BH"
    },
    {
      "field": "uses",
      "target": "0CHQKZFK5S"
    },
    {
      "field": "uses",
      "target": "0PVG1CRAJN"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/PDPM33XT31.md",
  "msc": [
    "20B30"
  ],
  "name": "Proof of the decomposition of a permutation into disjoint cycles",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/PDP/PDPM33XT31.md",
  "status": null,
  "tag": "PDPM33XT31",
  "uri": "m2://PDPM33XT31",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-decomposition-of-a-permutation-into-disjoint-cycles/",
  "writing": {
    "tag": "PDPM33XT31",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}