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Fix $\alpha \in \Omega$. If $\beta =g\alpha$, then normality of $H$ givesThus the transitive group action of $G$ permutes the orbits of a group action of $H$, so all $H$-orbits have the same cardinality $d$. They partition the prime-sized set $\Omega$, hence $d$ divides $|\Omega |$. The only possibilities are $d=1$ and $d=|\Omega |$. The first would make every element of $H$ fix every point, contrary to the assumption that $H$ acts nontrivially. Therefore $d=|\Omega |$ and

Now suppose $H\cap G_{\alpha }=\{1\}$. Define the orbit mapIt is surjective by transitivity of $H$. If $\theta (h_1)=\theta (h_2)$, then $h_2^{-1}{h}_1\in H\cap G_{\alpha }$, so $h_1=h_2$; hence $\theta$ is bijective. For $k\in G_{\alpha }$,Thus $\theta$ is an equivariant map: conjugation by $G_{\alpha }$ on $H$ corresponds exactly to the given action of $G_{\alpha }$ on $\Omega$.

Let $G$ act by left multiplication on the set $G/H$ of $n$ left cosets. This gives a group action homomorphismIts kernelis a normal subgroup of $G$, and the first isomorphism theorem gives $G/K\cong \mathrm{im}\phi$. By Lagrange's theorem, $|G/K|$ divides $n!$. The image acts transitively on the $n$ cosets, so the orbit-stabilizer theorem shows that $n$ divides $|G/K|$; in particular,

Now suppose $G$ is nonabelian and simple. Since $H$ is proper, the coset action is nontrivial, so $K\ne G$. Simplicity gives $K=\{e\}$, and $\phi$ embeds $G$ into $S_n$. Composing with the sign homomorphism givesIts kernel is normal. A nontrivial map would embed the simple group $G$ into the abelian group of order two, which is impossible because $G$ is nonabelian. The sign is therefore always $+1$, so

A group action of $G$ on $X$ is a map $(g,x)\mapsto gx$ such that $ex=x$ and $(gh)x=g(hx)$. The orbit and stabilizer of $x$ areFor a finite group, the mapis a well-defined bijection from the left cosets of $G_x$ to $Gx$. Each coset has $|G_x|$ elements, so the orbit-stabilizer theorem is

The Cauchy theorem for groups says that if a prime number $p$ divides $|G|$, then $G$ has an element of order $p$. To prove it, letThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$, which is divisible by $p$. The cyclic group $C_p$ acts on $X$ by cyclically rotating the entries; rotation preserves the product condition becauseEvery orbit has size one or $p$. The fixed points are exactly the tuples $(g,\dots ,g)$ with $g^p=e$. Their number is therefore divisible by $p$. Since the identity gives one fixed point, there is another, and its entry has order $p$.

Now let $|G|=33$. Cauchy's theorem gives a subgroup $H$ of order $11$. Let $H$ act by the conjugation action on the set $\mathcal{X}$ of subgroups of order $11$. It fixes $H$. If it also fixed $K\ne H$, then $H$ would normalize $K$; since $H\cap K=\{e\}$, the product $HK$ would be a subgroup of order $121$, which is impossible. Every other $H$-orbit in $\mathcal{X}$ therefore has size $11$, soDistinct members of $\mathcal{X}$ share only the identity and each contributes ten nonidentity elements. Hence $1+10|\mathcal{X}|\le 33$, forcing $|\mathcal{X}|=1$. Thus $H$ is a normal subgroup.

Conjugation now defines a group homomorphismBecause $H\cong C_{11}$, its automorphism group has order $10$. By the Lagrange theorem, the image has order dividing both $33$ and $10$, so the image is trivial and $H$ lies in the center of $G$. Cauchy's theorem also supplies $x$ of order $3$. If $h$ generates $H$, then $h$ and $x$ commute and $hx$ has order $\mathrm{lcm}(11,3)=33$. Therefore

Fordefine its action on the projective line $F\cup \{\infty \}$ by the Möbius transformationwith the usual conventions when the denominator vanishes or $x=\infty$. Matrix multiplication agrees with composition, so this is a group action. If $A$ fixes every projective point, then it fixes $0$ and $\infty$, forcing $b=c=0$, and fixing $1$ then gives $a=d$. Thus the kernel consists exactly of the nonzero scalar matrices $Z$. The induced group homomorphismis therefore injective. This is the projective general linear group action on the projective line.

For $F={\mathbb{F}}_4$,The scalar subgroup has order $|Z|=3$, soThe four classes represented byform a subgroup $P$ of order four, hence a Sylow subgroup for the prime two. Its action is translation $x\mapsto x+b$ on ${\mathbb{F}}_4$ and fixes $\infty$. For $b\ne 0$, characteristic two makes this permutation a product of two disjoint transpositions on the four finite points. It is therefore even, andThis realizes the Sylow 2-subgroup of PGL2 over F4.

Let $H={\mathrm{Stab}}_F(x)$. The identity belongs to $H$ because $F(e,x)=x$. If $g,h\in H$, the group action law givesso $gh\in H$. Finally, if $g\in H$, thenso $g^{-1}\in H$. The subgroup criterion therefore proves that ${\mathrm{Stab}}_F(x)$ is the stabilizer subgroup of $x$.

The group action law givesThus ${\varphi }_g$ is a bijection with inverse ${\varphi }_{g^{-1}}$, hence a permutation of $S$. Also, for every $s$,Therefore the assignment $\Phi (g)={\varphi }_g$ takes values in the symmetric group $\mathrm{Sym}(S)$ and satisfiesIt is a group homomorphism, with the group product on the right interpreted as composition of permutations.

A left group action of $G$ on a set $S$ is a function $\varphi :G\times S\to S$ satisfyingfor all $s\in S$ and $g,h\in G$. Writing $g\cdot s=\varphi (g,s)$, these are $e\cdot s=s$ and $(gh)\cdot s=g\cdot (h\cdot s)$. No commutativity is assumed, so the order of composition matters.

By Lagrange's theorem, the index of a subgroup is $[G:H]=75/25=3$. Thus the set of left cosets has three elements, and the preceding group action yields a group homomorphism $\Phi :G\to S_3$ after labelling them.

Its image is a subgroup of $S_3$, so its order divides $6$. By the first isomorphism theorem, $|\mathrm{im}\Phi |=|G|/|\ker \Phi |$ also divides $75$. Hence its order divides $\gcd (6,75)=3$, giving possibilities $1$ and $3$. The action on cosets is transitive: $g_2{g}_1^{-1}$ sends $g_{1}H$ to $g_{2}H$. It cannot be the trivial action on a three-element set, so the image does not have order $1$. ThereforeIt follows that $|\ker \Phi |=75/3=25$. Part (b)(i) gives $\ker \Phi \le H$, and $|H|=25$; equal finite orders force equality. A kernel of a group homomorphism is a normal subgroup, soThis is an instance of the fact that a subgroup of index the smallest prime divisor is normal.

Define $g\cdot (g_{1}H)=(g{g}_1)H$ on the set of left cosets. It is well-defined: if $g_{1}H=g_{2}H$, then $g_2=g_{1}h$ for some $h\in H$, so $(g{g}_2)H=(g{g}_{1}h)H=(g{g}_1)H$. The identity fixes each coset, and associativity in $G$ givesHence this is a left group action, and part (a) supplies its group homomorphism $\Phi :G\to \mathrm{Sym}(S)$.

An element of the kernel of a group homomorphism $\Phi$ fixes every coset, in particular $H$ itself. Thus $g\in \ker \Phi$ implies $gH=H$, which implies $g\in H$. ThereforeThe subgroup symbol is justified because a kernel of a group homomorphism is a subgroup, and the argument proves containment.

A left group action of $G$ on a set $S$ is a function $G\times S\to S$, written $(g,s)\mapsto g\cdot s$, satisfyingfor all $g,h\in G$, $s\in S$. The orbit of a group action through $s$ and the stabilizer subgroup of $s$ are, respectively,The identity element belongs to $G_s$. If $g,h\in G_s$, then $(gh)\cdot s=g\cdot (h\cdot s)=g\cdot s=s$, so $gh\in G_s$. If $g\in G_s$, applying $g^{-1}$ to $g\cdot s=s$ gives $s=g^{-1}\cdot s$, so $g^{-1}\in G_s$. These facts prove that the stabilizer subgroup is a subgroup of $G$.

For $h\in H$, define $\pi (h)(n)=hn{h}^{-1}$. Since $N$ is a normal subgroup, this sends $N$ into itself. It preserves multiplication, and its inverse is conjugation by $h^{-1}$, so it is an group automorphism of $N$. MoreoverThus $\pi :H\to \mathrm{Aut}(N)$ is a group homomorphism, induced by the group action of conjugation.

If $\pi$ is trivial, every element of $H$ commutes with every element of $N$. Henceis a group homomorphism, since $(n_1{h}_1)(n_2{h}_2)=n_1{n}_2{h}_1{h}_2$. Its image of a group homomorphism is a subgroup containing both $N$ and $H$, and every product $nh$ lies in $⟨N,H⟩$, so its image is exactly $⟨N,H⟩$. If $nh=e$, then $n=h^{-1}\in N\cap H=\{e\}$, so $n=h=e$. Thus $F$ is injective and gives

The stabilizer subgroup of $H$ for this conjugation group action is its normalizerIf it is all of $G$, the equality $gH{g}^{-1}=H$ holds for every $g\in G$, which is precisely the definition of a normal subgroup.

If instead $\mathrm{Stab}(H)=H$, the orbit-stabilizer theorem says that the group orbit of $H$ has size $[G:H]$. Every conjugate $gH{g}^{-1}$ is isomorphic to $H$, with group isomorphism $h\mapsto gh{g}^{-1}$, so this group orbit is a subset of $\Omega$. ThereforeThe group action need not be transitive on all of $\Omega$; this one group orbit supplies the lower bound.

A left group action is a map $a:G\times X\to X$ satisfyingDefine ${\rho }_a(g):X\to X$ by ${\rho }_a(g)(x)=a(g,x)$. The action axioms giveHence each ${\rho }_a(g)$ is a bijection, with inverse ${\rho }_a(g^{-1})$, and belongs to the symmetric group $\mathrm{Sym}(X)$. Finally,so ${\rho }_a:G\to \mathrm{Sym}(X)$ is a group homomorphism.

Let $v\in S$. Choose a unit vector $u$ perpendicular to $v$, and put $w=v\times u$. The cross product identities show that $(v,u,w)$ is an orthonormal basis with positive orientation: $u\times (v\times u)=v$, soThus the matrix $A_v$ with these columns belongs to the special orthogonal group $G$ and satisfies $A_{v}N=v$. For any $v,z\in S$, the element $A_z{A}_v^{-1}$ sends $v$ to $z$. Therefore the group action on the unit sphere is transitive.

For $X\in {\mathrm{SO}}_3(\mathbb{R})$, the defining identity $X^{T}X=I$ givesThus an orthogonal transformation preserves the Euclidean norm and sends the unit sphere to itself. Restricting the given group action to this invariant set preserves both action axioms.

If $XN=N$, the first column of $X$ is $N=e_1$. The other columns are unit vectors perpendicular to it, so the first row is also $(1,0,0)$. Hence $X=\mathrm{diag}(1,B)$ with $B^{T}B=I_2$ and $\det B=1$. Conversely, every such matrix lies in $G$ and fixes $N$. Since every element of the special orthogonal group in two dimensions has the planar rotation matrix form, the stabilizer subgroup isGeometrically these are exactly the rotations in three dimensions about the first coordinate axis.

If $G=\{e\}$, choose ordinary addition on $\mathbb{Z}$, giving the additive group of integers. Every power function is the unique identity function on the singleton $G$, so the group action axioms hold:Thus ordinary addition supplies the required group operation on $\mathbb{Z}$. Together with ii, existence is equivalent to $G$ being the trivial group.

Suppose an arbitrary binary operation $\star$ makes $\mathbb{Z}$ a group and makes $\rho$ a left group action on $G$. For any group action, each acting element induces a bijection: the function induced by its inverse element is its inverse function.

In particular, the acting element whose underlying integer is $0$ must induce a bijection $G\to G$. But its power function isa constant function. This is injective only when all elements of $G$ equal $e$. ThereforeThis proves the power-map group action obstruction for every possible $\star$; it never assumes that $\star$ is ordinary integer multiplication, or that the integer $0$ is the acting identity element.

The orbit of $x$ under conjugation is its conjugacy classIt contains $x$ by using $g=e$, so the conjugacy classes cover $G$. If $z=gx{g}^{-1}$, then $x=g^{-1}zg$, and the group action identity shows that conjugating either element further gives the same orbit; hence $C(z)=C(x)$. If $C(x)$ and $C(y)$ intersect at $z$, this gives $C(x)=C(z)=C(y)$. Thus distinct conjugacy classes are disjoint and the conjugacy classes form a partition of a set $G$.

Write $g\cdot x=gx{g}^{-1}$. If $e$ is the identity element of $G$, then $e\cdot x=ex{e}^{-1}=x$. Also, for $g,h,x\in G$,The middle equality uses associativity and the inverse element rule $(gh{)}^{-1}=h^{-1}{g}^{-1}$. Thus this is a left group action, the conjugation action of $G$ on itself.

A group action is a map $G\times \Omega \to \Omega$, $(g,\omega )\mapsto g\omega$, satisfying $1\omega =\omega$ and $(gh)\omega =g(h\omega )$. Its group orbit and stabilizer areThe stabilizer contains the identity. If $g,h$ fix $\omega$, then $gh$ fixes it; and $g\omega =\omega$ implies $g^{-1}\omega =\omega$ after applying $g^{-1}$. Thus it is a subgroup.

The map from left cosets $g{G}_{\omega }$ to group-orbit points $g\omega$ is well-defined, surjective and injective: $g\omega =h\omega$ is equivalent to $h^{-1}g\in G_{\omega }$. Therefore $|\mathrm{Orb}(\omega )|=[G:G_{\omega }]=|G|/|G_{\omega }|$ for finite $G$.

For conjugation in $S_3$, the group orbits are exactly the conjugacy classes, determined by cycle type:Conjugation sends $(i_1\cdots i_k)$ to $(g(i_1)\cdots g(i_k))$, so these displayed sets are invariant and each is one group orbit. Their sizes $1,3,2$ also agree with centralizer orders $6,2,3$.