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  "body": "\n## Proof\n\n1. The set $N_G(P) H = \\{xh : x \\in N_G(P),\\ h \\in H\\}$ is a subset of $G$. So it remains to show that each $g \\in G$ lies in it.\n2. Let $g \\in G$. Since $H$ is [[normal|0W77CVBCHQ]] in $G$ and $P \\subseteq H$, the [[conjugate subgroup|0D6WHTTH7T]] $gPg^{-1}$ is contained in $H$.\n3. The map $x \\mapsto gxg^{-1}$ is a bijection from $P$ to $gPg^{-1}$. So $gPg^{-1}$ is a subgroup of $H$ with $|gPg^{-1}| = |P|$.\n4. Since $P$ is a [[Sylow $p$-subgroup|0JWZX6FKSH]] of $H$, its order is the largest power of $p$ dividing $|H|$. By step 3, $gPg^{-1}$ is also a Sylow $p$-subgroup of $H$.\n5. By the [[second Sylow theorem|09GCQ5STVJ]] in the group $H$, there is $h \\in H$ with $gPg^{-1} = hPh^{-1}$.\n6. Let $x = h^{-1} g$. Then $xPx^{-1} = h^{-1}(gPg^{-1})h = h^{-1}(hPh^{-1})h = P$. So $x$ lies in the [[normaliser|0AT1ZSNM25]] $N_G(P)$.\n7. Then $g = hx = x \\,(x^{-1} h x)$.\n8. Since $H$ is normal in $G$, the element $x^{-1} h x$ lies in $H$.\n9. By steps 6 to 8, $g$ lies in the [[product|0NS1YTGM19]] $N_G(P) H$.\n10. Since $g \\in G$ was arbitrary, $G = N_G(P) H$.\n\n## Idea\n\nConjugating $P$ by any $g \\in G$ gives another Sylow subgroup of $H$, because $H$ is normal. Inside $H$ all Sylow subgroups are conjugate, so an element $h \\in H$ undoes the conjugation by $g$. Then $h^{-1} g$ normalises $P$, which splits $g$ into a normalising part and a part in $H$.\n",
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