{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe use the [[action of $G$ on itself by left multiplication|0K5TWJAGPD]]. Let $e$ be the identity of $G$, and for $g \\in G$ let $\\ell_g \\colon G \\to G$, $\\ell_g(x) = g x$.\n\n1. For $g, h, x \\in G$, associativity gives $\\ell_g(\\ell_h(x)) = g(hx) = (gh)x = \\ell_{gh}(x)$. So $\\ell_g \\circ \\ell_h = \\ell_{gh}$.\n2. Also $\\ell_e(x) = ex = x$, so $\\ell_e$ is the identity map of $G$.\n3. By steps 1 and 2, $\\ell_g \\circ \\ell_{g^{-1}} = \\ell_e$ and $\\ell_{g^{-1}} \\circ \\ell_g = \\ell_e$. So $\\ell_g$ is a bijection with inverse $\\ell_{g^{-1}}$, and $\\ell_g \\in \\operatorname{Sym}(G)$, the [[symmetric group|0RJMNEN3BJ]] of $G$.\n4. Let $\\varphi \\colon G \\to \\operatorname{Sym}(G)$, $\\varphi(g) = \\ell_g$. The operation of $\\operatorname{Sym}(G)$ is composition, so step 1 says $\\varphi(gh) = \\varphi(g) \\circ \\varphi(h)$. Thus $\\varphi$ is a [[group homomorphism|0HV9Y7WG5R]].\n5. If $\\varphi(g) = \\varphi(h)$, then $g = \\ell_g(e) = \\ell_h(e) = h$. So $\\varphi$ is injective.\n6. Let $K = \\set{\\ell_g : g \\in G}$ be the image of $\\varphi$. By steps 1 to 3, $K$ contains the identity $\\ell_e$, is closed under composition, and contains the inverse $\\ell_{g^{-1}}$ of each $\\ell_g$. So $K$ is a [[subgroup|0N0PXCTYA7]] of $\\operatorname{Sym}(G)$.\n7. As a map $G \\to K$, $\\varphi$ is a homomorphism, injective by step 5 and onto by the definition of $K$. So it is an [[isomorphism|085PND4KNT]], and $G$ is isomorphic to the subgroup $K$ of $\\operatorname{Sym}(G)$.\n\n## Idea\n\nEach element $g$ shuffles the elements of the group by $x \\mapsto g x$. Composing two shuffles corresponds to multiplying the elements, and different elements give different shuffles because they send $e$ to different places. So the group sits faithfully inside the permutations of its own elements.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "20B30"
    ],
    "name": "Proof of Cayley's theorem by left multiplication",
    "proves": "0151VYVX27",
    "sources": [
      {
        "locator": "Proof of the theorem",
        "url": "https://en.wikipedia.org/wiki/Cayley's_theorem"
      }
    ],
    "tag": "PZPPAXYD3G",
    "uses": [
      "0K5TWJAGPD",
      "0RJMNEN3BJ",
      "0HV9Y7WG5R",
      "0N0PXCTYA7",
      "085PND4KNT"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/PZPPAXYD3G/",
  "links": [
    {
      "field": "proves",
      "target": "0151VYVX27"
    },
    {
      "field": "uses",
      "target": "0K5TWJAGPD"
    },
    {
      "field": "uses",
      "target": "0RJMNEN3BJ"
    },
    {
      "field": "uses",
      "target": "0HV9Y7WG5R"
    },
    {
      "field": "uses",
      "target": "0N0PXCTYA7"
    },
    {
      "field": "uses",
      "target": "085PND4KNT"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/PZPPAXYD3G.md",
  "msc": [
    "20B30"
  ],
  "name": "Proof of Cayley's theorem by left multiplication",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/PZP/PZPPAXYD3G.md",
  "status": null,
  "tag": "PZPPAXYD3G",
  "uri": "m2://PZPPAXYD3G",
  "url": "https://math2.tamnd.com/wiki/proof-of-cayleys-theorem-by-left-multiplication/",
  "writing": {
    "tag": "PZPPAXYD3G",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}