A Sylow $p$-subgroup is normal if and only if it is unique

Given: The hypotheses and objects in the Statement.

Let $G$ be finite, let $P$ be a Sylow $p$-subgroup, and let $H\le G$ be a $p$-subgroup. There is $g\in G$ with $H\le gP{g}^{-1}$. In particular, for every Sylow $p$-subgroup $Q$ there is $g\in G$ with $Q=gP{g}^{-1}$, so the Sylow $p$-subgroups form one conjugacy class. (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).

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 $G$ be a group and let $N\le G$ be a subgroup (def-subgroup). For $g\in G$, write $gN{g}^{-1}:=\{gn{g}^{-1}:n\in N\}\mathrm{.}$ The subgroup $N$ is normal in $G$ when $gN{g}^{-1}=N\text{for every }g\in G\mathrm{.}$ In that case write $N⊴G$. Equivalently, every inner conjugation of $G$ maps $N$ onto itself. The connection with equality of the left and right cosets of def-coset is proved in thm-normal-subgroup-characterisations. (Normal subgroup: invariance under conjugation).