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For $x\in X$,The stabilizer contains the identity; if $g,h$ fix $x$, thenso it is a subgroup.

The orbit-stabilizer theorem statesDefineThe map is well defined and bijective: two elements give the same image exactly when they differ by an element of the stabilizer. Lagrange's theorem then proves the formula.

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

For subgroups $H,P\le G$, define $x\sim y$ when $y\in HxP$. The identity elements show reflexivity, inverses show symmetry, and multiplication shows transitivity. Thus the double cosets $HxP$ partition $G$.

Let $H$ act by left multiplication on the set of left cosets $G/P$. The orbit of $xP$ consists of the cosets contained in $HxP$, so its size is $|HxP|/|P|$. Its stabilizer isThe orbit-stabilizer theorem therefore gives

Suppose $P$ is a Sylow subgroup of $G$, with $|P|=p^a$, and write the largest power of $p$ dividing $|H|$ as $p^b$. If every $H\cap xP{x}^{-1}$ had order at most $p^{b-1}$, the displayed formula would make every double-coset size divisible by $p^{a+1}$. Their sum $|G|$ would then also be divisible by $p^{a+1}$, contradicting the choice of $P$. Hence some $H\cap xP{x}^{-1}$ has order $p^b$ and is a Sylow $p$-subgroup of $H$. This is the Sylow subgroup of a subgroup from double cosets argument.

To count the general linear group over a finite field $G{L}_n({\mathbb{F}}_p)$, choose its columns successively. There are $p^n-1$ choices for the first, $p^n-p$ for the second, and $p^n-p^k$ for column $k+1$. ConsequentlyNone of the factors $p^j-1$ is divisible by $p$, so the upper unitriangular group is a Sylow $p$-subgroup: it has one arbitrary field entry in each of the $n(n-1)/2$ positions above the diagonal and hence order $p^{n(n-1)/2}$. The permutation matrices form a subgroup isomorphic to the symmetric group $S_n$.

By Cayley theorem, every finite group $G$ embeds in $S_{|G|}$, and permutation matrices embed this symmetric group in $G{L}_{|G|}({\mathbb{F}}_p)$. The latter has the explicit Sylow $p$-subgroup just described, so the result proved in the first part, applied to the embedded copy of $G$, proves that every finite group has a Sylow $p$-subgroup.

The counting part of the Sylow theorems says that if $p^a$ is the largest power of $p$ dividing $|G|$, then the number $n_p$ of Sylow $p$-subgroups satisfies

Finally, let $|G|=pq$ with prime numbers $p>q$. The Sylow counts giveso $n_p=1$ and the Sylow $p$-subgroup is a normal subgroup. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal; their elements commute, so $G$ is their direct product and is abelian. In the nonabelian case one must therefore have $n_q=p$, whenceor equivalently $q\mid p-1$, as recorded by nonabelian group of order pq.

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

By transitivity and the orbit-stabilizer theorem,Inside $G_x\cong S_4$, there are three Sylow $2$-subgroups, each of order eight, and four Sylow $3$-subgroups, each of order three. These are also Sylow subgroups of $G$.

The orbit-stabilizer theorem states that for a finite group $G$ acting on a set $X$ and $x\in X$,Define $G/G_x\to Gx$ by $g{G}_x\mapsto gx$. This is well-defined because $g{G}_x=h{G}_x$ exactly when $h^{-1}g\in G_x$, equivalently $gx=hx$. The same equivalence proves injectivity, and the definition of the orbit proves surjectivity. Thus $|Gx|=[G:G_x]$, giving the formula.

The orbit-stabilizer theorem identifies the orbit of a group action of $s$ with the set of left cosets of $G_s$. Explicitly, there is a bijectionHere $G/G_s$ denotes a set of cosets; it is not asserted to be a quotient group. To prove the identification, observe thatThis equivalence proves both well-definedness and injectivity. Every element of the orbit of a group action has the form $g\cdot s$, proving surjectivity.

In particular, for a finite group, the cardinalities satisfyThe last equality follows from Lagrange's theorem applied to the stabilizer subgroup.

For a free group action, every stabilizer subgroup is $\{e\}$. By the orbit-stabilizer theorem, every orbit of a group action has cardinality $|G|$. Hence the two free group actions have the same numberof orbits of a group action. Choose one representative $s_i$ from each $\cdot$ orbit of a group action and one representative $t_i$ from each $\ast$ orbit of a group action, and match them using $1\le i\le r$.

Every $s\in S$ has a unique expression $s=g\cdot s_i$. The orbit of a group action determines $i$, and the free group action property determines $g$: if $g\cdot s_i=h\cdot s_i$, then $(h^{-1}g)\cdot s_i=s_i$, forcing $g=h$. Therefore the ruledefines a function $f:S\to S$ without ambiguity. For $h\in G$,which proves the required identity for every $s$.

Indeed, $f$ is a bijection: its inverse is $g\ast t_i\mapsto g\cdot s_i$, well defined because the second is also a free group action. If $S$ is empty, the unique empty function already satisfies the identity. Thus the result holds in every case, and proves the stronger statement that free finite actions with equal cardinality are isomorphic.

The orbits of a group action form a partition of a set $S$: every element lies in its own orbit of a group action, and two intersecting orbits of a group action coincide because each point can be moved back and then forward using inverse elements. By the orbit-stabilizer theorem, each orbit of a group action has cardinality dividing $8$, so its size is $1,2,4$ or $8$.

If every orbit of a group action had size $2,4$ or $8$, their sizes would add to an even number, contrary to $|S|=17$. Therefore there is a one-element orbit of a group action, say $\{s^{\prime }\}$. Every $g\in G$ sends $s^{\prime }$ into this same singleton, soThus $s^{\prime }$ is a fixed point of a group action. In fact the number of fixed points of a group action is odd, since all other orbits of a group action have even cardinality.

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.

The polynomial is irreducible by Eisenstein's criterion at $2$. Its Galois group therefore acts transitively on five roots, so its order is divisible by $5$ by the orbit-stabilizer theorem.

To count the real roots, differentiate:It is positive on $(-\infty ,0)$, negative on $(0,\sqrt[3]{4})$, and positive thereafter. The local maximum is $f(0)=2>0$ and the local minimum is $2-6(\sqrt[3]{4}{)}^2<0$. The limits at infinity have opposite signs, so there are exactly three real roots and one conjugate pair of nonreal roots. Complex conjugation fixes the three real roots and interchanges the other two; its permutation is a transposition. The prime-degree subgroup result supplied in the paper now gives

Write the group action as $(\omega ,g)\mapsto \omega g$. A right group action satisfies $\omega 1=\omega$ and $(\omega g)h=\omega (gh)$. Its orbit and stabilizer areThe identity belongs to $H$. If $g,h\in H$, then $\omega gh=(\omega g)h=\omega$, so $gh\in H$. Also $\omega g=\omega$ implies $\omega =(\omega g)g^{-1}=\omega g^{-1}$, so $g^{-1}\in H$. Thus $H$ is a subgroup.

For the orbit-stabilizer theorem, the map $G\to \mathrm{Orb}(\omega )$, $g\mapsto \omega g$, is onto, andIts fibers are precisely the right cosets $Hh$, each containing $|H|$ elements. Counting these disjoint fibers givesFor the symmetric group $S_3$, the six subgroups areThe group action $H\mapsto g^{-1}Hg$ has four orbits: each of $\{1\}$, $A_3$, and $S_3$ is fixed, while the three order-two subgroups form one orbit. Here an orbit is a collection of subgroups, rather than a collection of individual permutations.

Fix ${\omega }_1$. For any $u,v\ne {\omega }_1$, apply the given two-point transitivity property to $({\omega }_1,u)$ and $({\omega }_1,v)$. The resulting element fixes ${\omega }_1$ and sends $u$ to $v$, so the stabilizer is transitive on the complement.

The whole group is transitive on $\Omega$: extend any desired first-coordinate pair by choosing a different second point on each side. By the orbit-stabilizer theorem, $|G|=n|{\mathrm{Stab}}_G({\omega }_1)|$. Applying the same theorem inside the stabilizer to its orbit of size $n-1$ shows that $n-1$ divides the stabilizer order. HenceIn the action on ordered pairs there are exactly two orbits. One is the diagonal, transitive because the action on $\Omega$ is transitive. The other comprises distinct-coordinate pairs and is transitive by hypothesis. Equality of the two coordinates is preserved by every group element, so the two sets cannot merge into one orbit. Faithfulness of the action is not required.

For a group action, the orbit and stabilizer areThe orbit-stabilizer theorem for a finite group states

For a regular tetrahedron, rotations fixing a vertex rotate about the line from that vertex through the opposite face centre. Their angles are $0,2\pi /3,4\pi /3$, so the stabilizer is a cyclic group of order three. Rotations act transitively on the four vertices, giving $|G_1|=4\cdot 3=12$ by the orbit-stabilizer theorem.

For a cube, the corresponding axis is the body diagonal through the fixed vertex and its opposite. The three incident edges can be cyclically permuted by the same three angles, so again the stabilizer is cyclic of order three. The orbit has eight vertices. Hence

The permutation inverse is ${\rho }_{g^{-1}}$: applying the two maps successively sends $\omega$ to $(\omega g)g^{-1}=\omega$. With the paper's right-action convention, permutation products also act on the right, andThus $\boxed{\rho (gh)=\rho (g)\rho (h)}$. If function composition is written on the left instead, the same right action is represented by $g\mapsto {\rho }_{g^{-1}}$; keeping these conventions consistent avoids an artificial reversal. The kernel consists precisely of elements fixing every point, so

The rotational tetrahedral group acts transitively on its four vertices. A rotation fixing a vertex must preserve its opposite face and rotate its three vertices cyclically; its stabilizer has order three. The orbit-stabilizer theorem gives $|G|=4\cdot 3=12$. The rotations are the identity, two nontrivial rotations about each of the four vertex-to-opposite-face axes, and a half-turn about each of the three opposite-edge-midpoint axes. Therefore the counts by element order areA rotation fixing all vertices fixes three linearly independent vertex vectors and is the identity. The vertex permutation action is therefore faithful. The listed rotations induce the identity, 3-cycles and double transpositions, all even permutations. Its injective image has order 12, the full order of $A_4$. Hence $\boxed{G\cong A_4}$.

Count the pairs $(g,\omega )$ for which $g$ fixes $\omega$. Counting first by group elements gives ${\sum }_g\mathrm{Fix}(g)$, while counting first by points gives ${\sum }_{\omega }|\mathrm{Stab}(\omega )|$. On a group orbit of size $r$, orbit-stabilizer theorem makes each stabilizer have size $|G|/r$, so that group orbit contributes $|G|$ to the second sum. Dividing by $|G|$ proves Burnside's lemma:A tetrahedron has four vertices. A rotational symmetry is determined by the image of one vertex, with four possibilities, and the cyclic orientation of the three vertices around it, with three possibilities. All these choices are realizable, so its rotational symmetry group of a tetrahedron has order $12$. More explicitly, its elements are the identity, the two nontrivial rotational symmetries around each of four vertex-to-opposite-face axes, and the half-turn about each of three axes through opposite edge midpoints. Thus there is one element of order 1, eight of order 3, and three of order 2. The induced permutations of vertices are the even permutations, identifying the group with $A_4$.

Apply the group action to the set of edge colourings. A colouring is fixed by a rotational symmetry precisely when its colour is constant on each cycle of the induced edge permutation. The identity has six edge cycles and fixes $4^6$ colourings. A vertex-axis rotational symmetry has two cycles of length three, fixing $4^2$. An opposite-edge half-turn fixes the two edges on its axis as sets and exchanges the other four in two pairs, giving four cycles and $4^4$ fixed colourings. ThereforeReflections are excluded because the equivalence in the source uses rotational symmetries.

The stabilizer of the indicated flag consists of block upper triangular matrices with diagonal block sizes $1,2,2$. The diagonal blocks can be any elements of $\mathrm{GL}(1,q)$, $\mathrm{GL}(2,q)$ and $\mathrm{GL}(2,q)$. The upper off-diagonal blocks have $2+2+4=8$ arbitrary entries. Thus the stabilizer order is $(q-1)|\mathrm{GL}(2,q){|}^2{q}^8$. The orbit-stabilizer theorem gives

The stabilizer of the indicated flag consists of block upper triangular matrices with diagonal block sizes $1,2,2$. The diagonal blocks can be any elements of $\mathrm{GL}(1,q)$, $\mathrm{GL}(2,q)$ and $\mathrm{GL}(2,q)$. The upper off-diagonal blocks have $2+2+4=8$ arbitrary entries. Thus the stabilizer order is $(q-1)|\mathrm{GL}(2,q){|}^2{q}^8$. The orbit-stabilizer theorem gives

Let $H=Z(x)$. The map $G/H\to C(x)$ sending $gH$ to $gx{g}^{-1}$ is well defined: multiplying $g$ on the right by an element commuting with $x$ leaves its conjugate unchanged. It is onto by the orbit definition. If $gx{g}^{-1}=hx{h}^{-1}$, then $h^{-1}g$ commutes with $x$, so $gH=hH$, proving injectivity.

Every coset has $|H|$ elements, so $|C(x)|=|G|/|H|$. ThereforeThis is the orbit-stabilizer theorem specialized to conjugation.