{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe prove three general facts, then the implications $1 \\Rightarrow 2 \\Rightarrow 3 \\Rightarrow 1$.\n\n1. Let $N$ be a [[normal subgroup|0W77CVBCHQ]] of a group $X$ and let $x \\in X$. Then $xN$ lies in the [[centre|0S7Y9Q990Q]] of the [[quotient|0YYWCNTPWR]] $X/N$ if and only if $x^{-1}g^{-1}xg \\in N$ for all $g \\in X$. Indeed, $xgN = gxN$ holds if and only if $(gx)^{-1}xg \\in N$.\n2. Let $H$ and $K$ be normal subgroups of $X$. By the [[second isomorphism theorem|0PPP2J2DJV]], $HK$ is a subgroup. It is normal, because $g(hk)g^{-1} = (ghg^{-1})(gkg^{-1})$.\n3. Let $H$ and $K$ be normal in $X$ with $H \\cap K = \\{e\\}$, and let $h \\in H$, $k \\in K$. The element $c = h^{-1}(k^{-1}hk)$ lies in $H$, and $c = (h^{-1}k^{-1}h)k$ lies in $K$. So $c = e$ and $hk = kh$.\n4. In the situation of step 3, the map $H \\times K \\to HK$, $(h, k) \\mapsto hk$, from the [[direct product|0YP7WYXF9V]] is a homomorphism, since $h_1k_1h_2k_2 = h_1h_2k_1k_2$. It is onto. If $hk = e$, then $h = k^{-1} \\in H \\cap K$, so $h = k = e$. So the map is an [[isomorphism|085PND4KNT]].\n5. ($1 \\Rightarrow 2$.) Let $G$ be [[nilpotent|0DZPD2MG33]] with central series $\\{e\\} = N_0 \\subseteq \\cdots \\subseteq N_n = G$. Let $H \\neq G$ be a subgroup. We show that $H$ is strictly smaller than its [[normaliser|0AT1ZSNM25]] $N_G(H)$.\n6. Since $N_0 \\subseteq H$ and $N_n \\not\\subseteq H$, there is a largest $i \u003c n$ with $N_i \\subseteq H$. Pick $x \\in N_{i+1}$ with $x \\notin H$.\n7. Let $h \\in H$. Since $N_{i+1}/N_i$ lies in the centre of $G/N_i$, step 1 with $x^{-1} \\in N_{i+1}$ and $g = h^{-1}$ gives $xhx^{-1}h^{-1} \\in N_i \\subseteq H$. So $xhx^{-1} \\in H$.\n8. In the same way, step 1 with $x$ and $g = h^{-1}$ gives $x^{-1}hxh^{-1} \\in H$, so $x^{-1}hx \\in H$. With step 7, $xHx^{-1} = H$. So $x \\in N_G(H)$ and $x \\notin H$.\n9. Let $p$ be a prime and let $P$ be a [[Sylow $p$-subgroup|0JWZX6FKSH]] of $G$; it exists by the [[first Sylow theorem|0V98WY94B8]]. Let $M = N_G(P)$ and $K = N_G(M)$. These are subgroups, with $P \\trianglelefteq M$ and $M \\trianglelefteq K$.\n10. The order $|M|$ divides $|G|$ and is divisible by $|P|$, by [[Lagrange's theorem|3C53TD]]. So $P$ is a Sylow $p$-subgroup of $M$.\n11. By the [[Frattini argument|0QTZDKJEDV]] in the group $K$, with the normal subgroup $M$, we get $K = N_K(P)\\,M$. Since $N_K(P) \\subseteq N_G(P) = M$, this gives $K \\subseteq M$. So $N_G(M) = M$.\n12. If $M \\neq G$, then steps 5 to 8 give $M \\subsetneq N_G(M)$, which contradicts step 11. So $N_G(P) = G$, and $P$ is normal in $G$.\n13. By the [[second Sylow theorem|09GCQ5STVJ]], each Sylow $p$-subgroup of $G$ equals $gPg^{-1} = P$ for some $g \\in G$. So $G$ has exactly one Sylow $p$-subgroup. This proves 2.\n14. ($2 \\Rightarrow 3$.) Write $|G| = p_1^{a_1} \\cdots p_r^{a_r}$ with distinct primes $p_k$ and $a_k \\geq 1$. Let $P_k$ be the only Sylow $p_k$-subgroup of $G$.\n15. For $g \\in G$, the subgroup $gP_kg^{-1}$ has order $|P_k|$, so it is a Sylow $p_k$-subgroup. Hence $gP_kg^{-1} = P_k$, and $P_k$ is normal in $G$.\n16. Let $Q_0 = \\{e\\}$ and $Q_k = Q_{k-1} P_k$ for $1 \\leq k \\leq r$. By step 2 and induction on $k$, each $Q_k$ is a normal subgroup of $G$.\n17. We show $|Q_k| = p_1^{a_1} \\cdots p_k^{a_k}$ by induction on $k$. The case $k = 0$ is clear. Let $k \\geq 1$.\n18. By Lagrange's theorem, $|Q_{k-1} \\cap P_k|$ divides both $|Q_{k-1}| = p_1^{a_1} \\cdots p_{k-1}^{a_{k-1}}$ and $p_k^{a_k}$. These numbers are coprime, so $Q_{k-1} \\cap P_k = \\{e\\}$.\n19. By the [[product formula|08PY4GNAPK]], $|Q_k| = |Q_{k-1}|\\,|P_k| = p_1^{a_1} \\cdots p_k^{a_k}$.\n20. So for each $k \\geq 1$, the group $Q_k$ is the [[internal direct product|0Q2MVQ7JMT]] of its normal subgroups $Q_{k-1}$ and $P_k$. Also $|Q_r| = |G|$, so $Q_r = P_1 \\cdots P_r = G$. This is what 3 says.\n21. ($3 \\Rightarrow 1$.) Assume 3, in the sense of step 20. So $G$ has Sylow subgroups $P_1, \\dots, P_r$, and the groups $Q_0 = \\{e\\}$ and $Q_k = Q_{k-1}P_k$ satisfy the following. Each $Q_k$ with $k \\geq 1$ is the internal direct product of its normal subgroups $Q_{k-1}$ and $P_k$, and $Q_r = G$.\n22. Let $X$ and $Y$ be nilpotent with central series $(N_i)$ and $(M_i)$. Adding copies of $X$ or $Y$ at the top keeps a central series, since a trivial factor lies in any centre. So both series have the same length $c$.\n23. Let $L_i = N_i \\times M_i$. Each $L_i$ is normal in $X \\times Y$, since conjugation acts in each coordinate. Also $L_0 = \\{e\\}$ and $L_c = X \\times Y$.\n24. For $(x, y) \\in L_{i+1}$ and $(g, h) \\in X \\times Y$, the element $(x, y)^{-1}(g, h)^{-1}(x, y)(g, h)$ equals $(x^{-1}g^{-1}xg,\\ y^{-1}h^{-1}yh)$. It lies in $L_i$ by step 1. By step 1 again, $(L_i)$ is a central series, so $X \\times Y$ is nilpotent.\n25. Let $\\alpha \\colon X \\to X'$ be an isomorphism and let $(N_i)$ be a central series of $X$. Then $(\\alpha(N_i))$ is a central series of $X'$, because $\\alpha$ preserves the condition of step 1. So a group isomorphic to a nilpotent group is nilpotent.\n26. The [[order of each element|08V7JHS3QZ]] of $P_k$ divides $p_k^{a_k}$. So $P_k$ is a finite $p_k$-group, and it is nilpotent because [[finite $p$-groups are nilpotent|081WTKYX6C]].\n27. We show by induction on $k$ that $Q_k$ is nilpotent. The group $Q_0 = \\{e\\}$ is nilpotent. For $k \\geq 1$, steps 4 and 21 give $Q_k \\cong Q_{k-1} \\times P_k$. By steps 24 to 26 and the induction hypothesis, $Q_k$ is nilpotent.\n28. So $G = Q_r$ is nilpotent. This proves 1 and closes the cycle of implications.\n\n## Idea\n\nIn a nilpotent group, every proper subgroup grows strictly when one passes to its normaliser. The Frattini argument shows that the normaliser of a Sylow normaliser does not grow, so the Sylow normaliser is all of $G$. Once each Sylow subgroup is normal, coprime orders give trivial intersections, and $G$ splits as their direct product. Products of $p$-groups are nilpotent, which closes the circle.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "20F18",
      "20D15"
    ],
    "name": "Proof that a finite group is nilpotent iff it is the product of its Sylow subgroups",
    "proves": "00Y9N80VSS",
    "sources": [
      {
        "locator": "Properties",
        "url": "https://en.wikipedia.org/wiki/Nilpotent_group"
      }
    ],
    "tag": "N2W6PECZAE",
    "uses": [
      "0DZPD2MG33",
      "0S7Y9Q990Q",
      "0W77CVBCHQ",
      "0YYWCNTPWR",
      "0JWZX6FKSH",
      "0AT1ZSNM25",
      "0QTZDKJEDV",
      "09GCQ5STVJ",
      "0V98WY94B8",
      "0Q2MVQ7JMT",
      "0PPP2J2DJV",
      "08PY4GNAPK",
      "3C53TD",
      "0YP7WYXF9V",
      "085PND4KNT",
      "08V7JHS3QZ",
      "081WTKYX6C"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/N2W6PECZAE/",
  "links": [
    {
      "field": "proves",
      "target": "00Y9N80VSS"
    },
    {
      "field": "uses",
      "target": "0DZPD2MG33"
    },
    {
      "field": "uses",
      "target": "0S7Y9Q990Q"
    },
    {
      "field": "uses",
      "target": "0W77CVBCHQ"
    },
    {
      "field": "uses",
      "target": "0YYWCNTPWR"
    },
    {
      "field": "uses",
      "target": "0JWZX6FKSH"
    },
    {
      "field": "uses",
      "target": "0AT1ZSNM25"
    },
    {
      "field": "uses",
      "target": "0QTZDKJEDV"
    },
    {
      "field": "uses",
      "target": "09GCQ5STVJ"
    },
    {
      "field": "uses",
      "target": "0V98WY94B8"
    },
    {
      "field": "uses",
      "target": "0Q2MVQ7JMT"
    },
    {
      "field": "uses",
      "target": "0PPP2J2DJV"
    },
    {
      "field": "uses",
      "target": "08PY4GNAPK"
    },
    {
      "field": "uses",
      "target": "3C53TD"
    },
    {
      "field": "uses",
      "target": "0YP7WYXF9V"
    },
    {
      "field": "uses",
      "target": "085PND4KNT"
    },
    {
      "field": "uses",
      "target": "08V7JHS3QZ"
    },
    {
      "field": "uses",
      "target": "081WTKYX6C"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/N2W6PECZAE.md",
  "msc": [
    "20F18",
    "20D15"
  ],
  "name": "Proof that a finite group is nilpotent iff it is the product of its Sylow subgroups",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/N2W/N2W6PECZAE.md",
  "status": null,
  "tag": "N2W6PECZAE",
  "uri": "m2://N2W6PECZAE",
  "url": "https://math2.tamnd.com/wiki/proof-that-a-finite-group-is-nilpotent-iff-it-is-the-product-of-its-sylow-subgroups/",
  "writing": {
    "tag": "N2W6PECZAE",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}