Frobenius Groups and the Normal Complement Theorem — Examples

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).

Given: A finite field $F$ with $\mathrm{\mid }F\mathrm{\mid }=q>2$, its additive group $F^{+}$ and multiplicative group $F^{\times }$, and the set $G$ of affine maps $f_{a,b}$.

Field arithmetic (Field): $0\mathrm{\ne }1$; $(F,+)$ is an abelian group with identity $0$, so $F^{+}=(F,+)$ is a group with $\mathrm{\mid }{F}^{+}\mathrm{\mid }=q$; $(F\setminus \{0\},\cdot )$ is an abelian group with identity $1$, so $F^{\times }=(F\setminus \{0\},\cdot )$ is a group, where $a\in F^{\times }$ means $a\mathrm{\ne }0$ and then $a$ has a multiplicative inverse $a^{-1}$ with $a{a}^{-1}=1$; multiplication distributes over addition, so $a(x+y)=ax+ay$ and $(a+c)x=ax+cx$; from $a\mathrm{\ne }1$ and $(a-1)b=0$ it follows that $b=0$, because $a-1\mathrm{\ne }0$ is invertible; and $\mathrm{\mid }{F}^{\times }\mathrm{\mid }=q-1$ because $F^{\times }=F\setminus \{0\}$ has the $q$ elements of $F$ except $0$.

Composition and inversion of affine maps: for $a,c\in F^{\times }$ and $b,d\in F$, $f_{a,b}\circ f_{c,d}=f_{ac,ad+b}$, and $f_{a,b}$ is a bijection with two-sided inverse $f_{a^{-1},-a^{-1}b}$; in particular $f_{1,0}={\mathrm{id}}_F$ (The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition, Field).

Subgroup criterion: a nonempty subset of a group closed under products and inverses is a subgroup (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, 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).

Cardinalities: $\mathrm{\mid }F\mathrm{\mid }=q$, $\mathrm{\mid }{F}^{\times }\mathrm{\mid }=q-1$, $\mathrm{\mid }{F}^{\times }\times F\mathrm{\mid }=\mathrm{\mid }{F}^{\times }\mathrm{\mid }\cdot \mathrm{\mid }F\mathrm{\mid }=(q-1)q$, and $\mathrm{\mid }{F}^{+}\mathrm{\mid }=q$ (The cardinality $\mid A\mid$ of a finite set, The product rule: $\mid A\times B\mid =\mid A\mid \mid B\mid$, and $\mid {\prod }_{i<m}{A}_i\mid ={\prod }_{i<m}\mid A_i\mid$, [F1]).

Normal subgroups and internal semidirect products: $K⊴G$ means $gK{g}^{-1}=K$ for all $g\in G$; if $K,H\le G$ with $K⊴G$, $G=KH$ and $K\cap H=\{1\}$, then $G=K⋊H$ is an internal semidirect product (Normal subgroup: invariance under conjugation, An internal semidirect product and a complement to a normal subgroup, Subgroup).

Conjugation is an automorphism and $c\in C_G(R)$ centralizes every element of a subgroup $R$; $f\in G$ fixes an element $k\in K$ under conjugation exactly when $fk{f}^{-1}=k$ (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).

Free-action criterion and kernel 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 for a Frobenius group $G$ with complement $H$ and kernel $N$ one has $N⊴G$, $G=NH$, $N\cap H=\{1\}$ and $N$ is the unique normal subgroup $M$ with $MH=G$ and $M\cap H=\{1\}$ (Frobenius groups and fixed point free actions, Frobenius semidirect product decomposition, Normal subgroup: invariance under conjugation).

Given: The symmetric group $S_4=\mathrm{Sym}(\{0,1,2,3\})$, its natural action on $\{0,1,2,3\}$, and $H=\{\sigma \in S_4:\sigma (2)=2\}$.

Cycle notation: a transposition $(ab)$ exchanges $a$ and $b$ and fixes every other letter; $(ab{)}^2=\mathrm{id}$; and a permutation of the four letters fixing two of them is determined by what it does to the remaining two, so the permutations fixing both $2$ and $3$ are exactly $\mathrm{id}$ and $(01)$ (The finite symmetric group $S_n$, one-line notation, and cycle notation).

Stabilizers, cosets and conjugation: for a group $G$ acting on a set $X$ and $x\in X$, the stabilizer $G_x=\{g:g\cdot x=x\}$ is a subgroup; point stabilizers of points in one orbit are conjugate: for $g\in G$ one has $g{G}_x{g}^{-1}=G_{g\cdot x}$, because $g\sigma g^{-1}$ fixes $g\cdot x$ exactly when $\sigma$ fixes $x$ (The orbit $G\cdot x$ and stabilizer $G_x$ of a point in a group action, Orbit-stabiliser: $G\mathrm{/}{G}_x\to G\cdot x$, $g{G}_x\mapsto g\cdot x$, is a well-defined bijection, Left group actions, transitive actions, and faithful actions, Subgroup, 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).

The transitivity of the natural action is the statement that for all $i,j\in \{0,1,2,3\}$ there is $\sigma \in S_4$ with $\sigma (i)=j$: if $i=j$ take $\sigma =\mathrm{id}$, and if $i\mathrm{\ne }j$ take the transposition $\sigma =(ij)$, which sends $i$ to $j$ by [F1]; the action is nonregular because the nonidentity element $(01)$ fixes the point $2$ by [F1] (Left group actions, transitive actions, and faithful actions).

Characterization of Frobenius complements by the coset action: for a finite group $G$ and a subgroup $\{1\}<H<G$, the subgroup $H$ is a Frobenius complement exactly when the left action of $G$ on $G\mathrm{/}H$ is transitive, nonregular, and every nonidentity element of $G$ fixes at most one coset (Frobenius permutation action characterization, Left and right cosets $gH$ and $Hg$ of a subgroup, $x\in aH$ iff $a^{-1}x\in H$, and $aH=bH$ iff $a^{-1}b\in H$).

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).