Frobenius normal two complement for S_3

Given: The symmetric group $S_3=\mathrm{Sym}(\{0,1,2\})$ and the prime $p=2$.

Elements, conjugacy classes and order-$2$ subgroups of $S_3$: one-line notation identifies the permutations of $\{0,1,2\}$ with the lists $[b_0,b_1,b_2]$ whose entries are $0,1,2$ each occurring once, so $\mathrm{\mid }{S}_3\mathrm{\mid }=3\cdot 2\cdot 1=6$ and $S_3=\{\mathrm{id},(01),(02),(12),(012),(021)\}$; conjugation relabels the entries of a cycle, $g(ab)g^{-1}=(g(a)g(b))$ and $g(abc)g^{-1}=(g(a)g(b)g(c))$, so the conjugacy classes are $\{\mathrm{id}\}$, $\{(01),(02),(12)\}$ and $\{(012),(021)\}$; and the conjugates $⟨(01){⟩}^g=⟨(g(0)g(1))⟩$ of $⟨(01)⟩$ are exactly the three subgroups $\{\mathrm{id},(01)\},\{\mathrm{id},(02)\},\{\mathrm{id},(12)\}$ of order $2$ (The finite symmetric group $S_n$, one-line notation, and cycle notation, Conjugating a cycle relabels each entry: $g(a_1\dots a_k)g^{-1}=(g(a_1)\dots g(a_k))$, $S_3$ as a Frobenius group).

$A_3⊴S_3$ has order $3$ and $A_3=\{\mathrm{id},(012),(021)\}$ (The alternating group $A_n=\ker (\mathrm{sgn})$ of even permutations, $A_n$ is normal in $S_n$; for $n\ge 2$, $2\mathrm{\mid }{A}_n\mathrm{\mid }=n!$, while $A_n=S_n$ for $n=0,1$, $S_3$ as a Frobenius group, Normal subgroup: invariance under conjugation).

Sylow and subgroup order facts: $\mathrm{\mid }{S}_3\mathrm{\mid }=6=2\cdot 3$, so the exact power of $2$ dividing $\mathrm{\mid }{S}_3\mathrm{\mid }$ is $2$, and every subgroup of order $2$ is a Sylow $2$-subgroup; a subgroup of $S_3$ has order dividing $6$, so every $2$-subgroup of $S_3$ has order $1$ or $2$; a group of order $2$ is a finite $2$-group (Sylow $p$-subgroups of a finite group, Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$, A finite $p$-group has order $p^n$ for a prime $p$ and some $n\in \mathbb{N}$, Every subgroup of a finite $p$-group has order a power of $p$).

Normal $p$-complement: a normal subgroup $K⊴G$ with $p\nmid \mathrm{\mid }K\mathrm{\mid }$ and $[G:K]$ a power of $p$ is a normal $p$-complement of $G$; the trivial subgroup has order $1$, which is prime to every $p$, and if $G$ is a finite $p$-group then $K=\{1\}$ is a normal $p$-complement because $[G:\{1\}]=\mathrm{\mid }G\mathrm{\mid }$ is a power of $p$ (Normal p complement and p nilpotent group, Sylow $p$-subgroups of a finite group, A finite $p$-group has order $p^n$ for a prime $p$ and some $n\in \mathbb{N}$, Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$).

Local normalizers and fusion: for $1\mathrm{\ne }Q\le P$ with $P\in {\mathrm{Syl}}_p(G)$ the normalizer $N_G(Q)$ is the local subgroup of the theory; and $P$ controls fusion in $P$ with respect to $G$ when every $G$-conjugacy between two elements of $P$ is realized by an element of $P$ (P local normalizer for normal complement theory, Control of fusion in a sylow p subgroup, The conjugacy class ${\mathrm{Cl}}_G(x)$ and centralizer $C_G(x)$ of an element, Conjugation $x\mapsto gx{g}^{-1}$ is an automorphism).

Frobenius normal $p$-complement theorem: for a finite group $G$, a prime $p$ and $P\in {\mathrm{Syl}}_p(G)$, the conditions (a) $G$ has a normal $p$-complement, (b) every $N_G(Q)$ with $1\mathrm{\ne }Q\le P$ has a normal $p$-complement, and (c) $P$ controls fusion in $P$, are equivalent (Frobenius normal p complement theorem).