Sylow II: in a finite group every $p$-subgroup lies in a conjugate of any Sylow $p$-subgroup, and the Sylow $p$-subgroups form a single conjugacy class

Given: The hypotheses and objects in the Statement.

Let $G$ be finite, let $p$ be prime, and write $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$. Then $G$ has a subgroup of order $p^a$, hence a Sylow $p$-subgroup (def-sylow-p-subgroup). (Sylow I: every finite group has a Sylow $p$-subgroup).

If a finite $p$-group $P$ acts on a finite set $X$, then $\mathrm{\mid }X\mathrm{\mid }\equiv \mathrm{\mid }{X}^P\mathrm{\mid }(\mathrm{m}\mathrm{o}\mathrm{d}p)\mathrm{.}$. (If a finite $p$-group $P$ acts on a finite set $X$, then $\mathrm{\mid }X\mathrm{\mid }\equiv \mathrm{\mid }{X}^P\mathrm{\mid }(\mathrm{m}\mathrm{o}\mathrm{d}p)$).

For a finite group $G$ and a prime $p$, let ${\mathrm{Syl}}_p(G)$ be the set of Sylow $p$-subgroups (def-sylow-p-subgroup). Define $n_p(G):=\mathrm{\mid }{\mathrm{Syl}}_p(G)\mathrm{\mid }\mathrm{.}$ This cardinal is defined even before existence is proved because ${\mathrm{Syl}}_p(G)$ is a subset of the finite power set of $G$; thm-sylow-first-theorem later shows it is nonzero. (The number $n_p(G)$ of Sylow $p$-subgroups).

Let $H\le G$ be a subgroup (def-subgroup). The normalizer of $H$ in $G$ is $N_G(H):=\{g\in G:gH{g}^{-1}=H\}\mathrm{.}$ Thus $g\in N_G(H)$ exactly when the conjugation automorphism $c_g$ preserves $H$ setwise (thm-conjugation-is-an-automorphism). The subgroup property is proved in lem-centralizers-and-normalizers-are-subgroups. (The normalizer $N_G(H)=\{g\in G:gH{g}^{-1}=H\}$ of a subgroup).

Let $G$ be a finite group and $H\le G$. Then $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }\mathrm{.}$ Consequently, under the canonical embedding $\iota :\mathbb{N}\to \mathbb{Z}$, $\mathrm{\mid }H\mathrm{\mid }$ divides $\mathrm{\mid }G\mathrm{\mid }$. (Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$).