$S_3$ as a Frobenius group

Given: The symmetric group $S_3=\mathrm{Sym}(\{0,1,2\})$, the transposition $(01)$, and $H=⟨(01)⟩$.

Elements and conjugacy classes 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 $\mathrm{\mid }{S}_3\mathrm{\mid }=3\cdot 2\cdot 1=6$; in cycle notation the six elements are $\mathrm{id}=[0,1,2]$, $(01)=[1,0,2]$, $(02)=[2,1,0]$, $(12)=[0,2,1]$, $(012)=[1,2,0]$ and $(021)=[2,0,1]$, that is $S_3=\{\mathrm{id},(01),(02),(12),(012),(021)\}$. Conjugating a cycle relabels its entries: $g(ab)g^{-1}=(g(a)g(b))$ and $g(abc)g^{-1}=(g(a)g(b)g(c))$ for $g\in S_3$. Hence every conjugate of the transposition $(01)$ is again a transposition, and as $g$ runs through $S_3$ the ordered pair $(g(0),g(1))$ runs through all six ordered pairs of distinct entries, so the conjugate $(g(0)g(1))$ takes each of the three values $(01),(02),(12)$ and the conjugacy class of $(01)$ is $\{(01),(02),(12)\}$; every conjugate of the $3$-cycle $(012)$ is again a $3$-cycle, and $g=\mathrm{id}$ and $g=(01)$ give the conjugates $(012)$ and $(102)=(021)$, so the conjugacy class of $(012)$ is $\{(012),(021)\}$; and the identity is conjugate only to itself. These three classes partition the six elements of $S_3$, so they are its conjugacy classes (The finite symmetric group $S_n$, one-line notation, and cycle notation, Conjugating a cycle relabels each entry: $g(a_1\dots a_k)g^{-1}=(g(a_1)\dots g(a_k))$).

Orders in $S_3$: a transposition $(ab)$ satisfies $(ab{)}^2=\mathrm{id}$ and $(ab)\mathrm{\ne }\mathrm{id}$, so has order $2$; a $3$-cycle $\sigma$ satisfies ${\sigma }^3=\mathrm{id}$ and $\sigma \mathrm{\ne }\mathrm{id}$, so has order $3$ (The finite symmetric group $S_n$, one-line notation, and cycle notation, The order $\mathrm{\mid }G\mathrm{\mid }$ of a finite group and the order $\mathrm{ord}(g)$ of an element, with $\mathrm{ord}(g)=\mathrm{\infty }$ when no positive power of $g$ is the identity, $⟨g⟩=\{g^n:n\in \mathbb{Z}\}$, and every cyclic group is abelian).

$A_3\le S_3$ is a normal subgroup of order $3$, and $A_3\mathrm{\ne }{S}_3$; a group of order $3$ is cyclic, hence contains an element of order $3$, so $A_3$ consists of the identity together with the two elements of order $3$ of $S_3$, namely $A_3=\{\mathrm{id},(012),(021)\}$ by [F1] and [F2] (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 $n=0,1$, A finite group of prime order is cyclic and every nonidentity element generates it, Normal subgroup: invariance under conjugation, Subgroup).

Orders and subgroups: $\mathrm{\mid }{S}_3\mathrm{\mid }=6$ by [F1]; a subgroup's order divides the group order; the intersection of two subgroups is a subgroup; and two distinct subgroups of order $2$ have trivial intersection, because their intersection is a subgroup of each of them, so has order dividing $2$, and has order $2$ only if it equals both, which would make them equal (Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$, The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$, One-step subgroup test: a nonempty $H\subseteq G$ is a subgroup iff $g{h}^{-1}\in H$ for all $g,h\in H$; the identity and the inverses of $H$ are then those of $G$, Subgroup).

Conjugation of cycle symbols: for $g\in S_3$ and a cycle $(a_1\dots a_k)$ of $S_3$, $g(a_1\dots a_k)g^{-1}=(g(a_1)\dots g(a_k))$; in particular conjugation sends a transposition to a transposition and a $3$-cycle to a $3$-cycle, and for the $3$-cycle $(012)$ and $h=(01)$ it gives $h(012)h^{-1}=(h(0)h(1)h(2))=(102)=(021)$ (Conjugating a cycle relabels each entry: $g(a_1\dots a_k)g^{-1}=(g(a_1)\dots g(a_k))$, The finite symmetric group $S_n$, one-line notation, and cycle notation).

Free-action criterion and uniqueness: if $N,H\le G$ with $G=N⋊H$, $1<N$, $1<H$ and every $1\mathrm{\ne }h\in H$ fixes only the identity of $N$ under conjugation, then $H$ is a Frobenius complement of $G$; and if $G$ is a Frobenius group with complement $H$ and kernel set $N$, then $N⊴G$, $G=NH$, $N\cap H=\{1\}$, and $N$ is the unique normal subgroup $M⊴G$ with $MH=G$ and $M\cap H=\{1\}$ (Frobenius groups and fixed point free actions, Frobenius semidirect product decomposition, An internal semidirect product and a complement to a normal subgroup, Left group actions, transitive actions, and faithful actions, Normal subgroup: invariance under conjugation).

Conjugation invariance of the Frobenius-complement property: if $H$ is a Frobenius complement of $G$ and $g\in G$, then $H^g=gH{g}^{-1}$ is a subgroup of $G$ isomorphic to $H$, and it is again a Frobenius complement, because $H^g\cap x{H}^g{x}^{-1}=(H\cap g^{-1}xH{x}^{-1}g{)}^g$ (Conjugation $x\mapsto gx{g}^{-1}$ is an automorphism, The conjugacy class ${\mathrm{Cl}}_G(x)$ and centralizer $C_G(x)$ of an element, In a group $e^{-1}=e$, $(g^{-1}{)}^{-1}=g$ and $(gh{)}^{-1}=h^{-1}{g}^{-1}$, the order of the last product being essential, Subgroup).