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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.

Burnside's lemma counts the orbits of a finite group action:Identify each face with its opposite vertex. Every symmetry induces the same permutation on these two four-element sets, so a permutation with $c$ cycles fixes $3^c{2}^c=6^c$ colorings. The rotational symmetry group is $A_4$, with one identity, eight three-cycles, and three products of two disjoint transpositions. Their cycle counts are four, two and two. Thus the rotation count isAllowing reflections gives the full tetrahedral group $S_4$. In addition there are six transpositions with three cycles and six four-cycles with one cycle. The count becomesEquivalently, 30 of the 141 rotational classes form 15 mirror pairs, while 111 rotational classes are their own mirror class.

The product of fixed-coloring counts is an instance of the orbit count for simultaneous colorings.

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

A left group action of $G$ on a set $S$ is a map $G\times S\to S$, written $(g,s)\mapsto g\cdot s$, satisfyingfor every $s\in S$ and $g,h\in G$, where $e$ is the identity element of $G$. In particular each $g$ acts by a bijection, whose inverse is the action of $g^{-1}$.

Given a group action, define $\rho (g)(j)=g\cdot j$. Each $\rho (g)$ is a permutation, since $\rho (g^{-1})$ is its inverse. The action law gives, for every $j$,Thus $\boxed{\rho :G\to S_n\text{ is a group homomorphism}}$.

Conversely, given a group homomorphism $\rho :G\to S_n$, put $g\cdot j=\rho (g)(j)$. Since a homomorphism sends the identity to the identity permutation, $e\cdot j=j$. AlsoThese are exactly the two group action axioms, so the construction defines the required action.

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.

A group action assigns to each $g\in G$ and $x\in \Omega$ a point $g\cdot x$ so that $1\cdot x=x$ and $(gh)\cdot x=g\cdot (h\cdot x)$. The orbit of $x$ is $Gx=\{g\cdot x:g\in G\}$. The action is transitive if there is exactly one orbit, equivalently any point can be carried to any other by some group element.

The identity element satisfies $e\ast x=ex{e}^{-1}=x$. For $g,h\in G$,Both action axioms hold, so conjugation is a group action on $G$.

The orbit of $s$ is $O_s=\{gs:g\in G\}$, and its stabilizer is $G_s=\{g\in G:gs=s\}$. The identity fixes $s$; if $g,h$ fix it then $(gh)s=g(hs)=s$; and $gs=s$ implies $g^{-1}s=s$. Thus $G_s$ is a subgroup.

A group action is transitive when it has just one orbit: for every $s,t\in S$ there is $g\in G$ with $gs=t$.

Two group actions on $S_1,S_2$ are equivalent if there is a bijection $\varphi :S_1\to S_2$ such that $\varphi (gs)=g\varphi (s)$ for all $g,s$. Such a bijection is equivariant.

The orbit-stabilizer theorem in this form says that the action on $O_s$ is equivalent to the left coset action on $G/G_s$, by $g{G}_s\mapsto gs$. This map is well-defined and injective because $gs=hs$ exactly when $h^{-1}g\in G_s$, and it is onto by the definition of the orbit. It is equivariant by multiplication.

Thus every transitive action is a coset action. Two such actions $G/H$ and $G/K$ are equivalent exactly when $H,K$ are conjugate: stabilizers at different points of one orbit are conjugate, and a conjugacy explicitly supplies a coset-action equivalence. This last classification is what makes counting by orbit types possible.

A cyclic group of order six has exactly one subgroup of each order $1,2,3,6$, and no others. Since it is abelian, these are also its four conjugacy classes of subgroups. Thus there is one transitive group action of each size $1,2,3,6$.

An action on six points is a disjoint union of its orbits. Its equivalence type is therefore determined by nonnegative integers satisfyingIf $n_6=1$ there is one possibility. If $n_6=0$, the possibilities $n_3=0,1,2$ allow respectively $4,2,1$ choices of $n_2$, with $n_1$ then fixed. HenceThis is counting finite group actions by transitive orbit types. Equivalently, a generator acts as a permutation whose cycles have lengths dividing six, and these are its eight possible cycle types on six points.

The orbit of $s$ is $O_s=\{gs:g\in G\}$, and its stabilizer is $G_s=\{g\in G:gs=s\}$. The identity fixes $s$; if $g,h$ fix it then $(gh)s=g(hs)=s$; and $gs=s$ implies $g^{-1}s=s$. Thus $G_s$ is a subgroup.

A group action is transitive when it has just one orbit: for every $s,t\in S$ there is $g\in G$ with $gs=t$.

Two group actions on $S_1,S_2$ are equivalent if there is a bijection $\varphi :S_1\to S_2$ such that $\varphi (gs)=g\varphi (s)$ for all $g,s$. Such a bijection is equivariant.

The orbit-stabilizer theorem in this form says that the action on $O_s$ is equivalent to the left coset action on $G/G_s$, by $g{G}_s\mapsto gs$. This map is well-defined and injective because $gs=hs$ exactly when $h^{-1}g\in G_s$, and it is onto by the definition of the orbit. It is equivariant by multiplication.

Thus every transitive action is a coset action. Two such actions $G/H$ and $G/K$ are equivalent exactly when $H,K$ are conjugate: stabilizers at different points of one orbit are conjugate, and a conjugacy explicitly supplies a coset-action equivalence. This last classification is what makes counting by orbit types possible.

A cyclic group of order six has exactly one subgroup of each order $1,2,3,6$, and no others. Since it is abelian, these are also its four conjugacy classes of subgroups. Thus there is one transitive group action of each size $1,2,3,6$.

An action on six points is a disjoint union of its orbits. Its equivalence type is therefore determined by nonnegative integers satisfyingIf $n_6=1$ there is one possibility. If $n_6=0$, the possibilities $n_3=0,1,2$ allow respectively $4,2,1$ choices of $n_2$, with $n_1$ then fixed. HenceThis is counting finite group actions by transitive orbit types. Equivalently, a generator acts as a permutation whose cycles have lengths dividing six, and these are its eight possible cycle types on six points.

The Burnside lemma states that the number of orbits of a finite group action isFor $m=2n-1\ge 3$, the dihedral group consists of $m$ rotations $r^j$ by $2\pi j/m$, and $m$ reflections $r^{j}s$, each across the line through one vertex and the midpoint of the opposite edge. Thus $|G|=2m=4n-2$.

For the pentagon, the identity fixes $v^5{e}^5$ colourings. Each of the four nonidentity rotations has a single cycle on vertices and a single cycle on edges, fixing $ve$ colourings. Each of the five reflections has one fixed vertex and two vertex pairs, and one fixed edge and two edge pairs, fixing $v^3{e}^3$ colourings. Applying the Burnside lemma to simultaneous vertex and edge colourings givesVertex colours and edge colours may be chosen independently; each symmetry must preserve both assignments.

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

Any nonzero vector in ${\mathbb{F}}_2^3$ extends to a basis. Sending one such basis to another gives an element of the general linear group, so its group action on the seven nonzero vectors is a transitive group action. Counting successive independent columns givesBy the orbit-stabilizer theorem, the stabilizer $H$ of a chosen nonzero vector $v$ has order $168/7=24$.

There are three two-dimensional linear subspaces containing $v$, corresponding to the three one-dimensional linear subspaces of ${\mathbb{F}}_2^3/⟨v⟩$. The stabilizer $H$ acts transitively on them: extend a basis of either plane beginning with $v$ to a basis of the whole vector space. Thus the stabilizer of a flag $⟨v⟩<P$ has order $24/3=8$.

If $Q\le G$ has order eight, each orbit of a group action of $Q$ on the seven nonzero vectors has size a power of two. Since seven is odd, at least one orbit of a group action has size one. Thus $Q$ fixes a nonzero vector $v$. Its group action on the three planes containing $v$ likewise has a fixed plane $P$. Therefore $Q$ is contained in, and by order equals, the stabilizer of that flag. The group action of $G$ is transitive on these flags, so all such subgroups are conjugate subgroups.

Distinct flags give distinct stabilizers. In a basis adapted to a flag, its order-eight stabilizer consists ofIts only common fixed nonzero vector is $e_1$. Its only invariant plane is $⟨e_1,e_2⟩$: an invariant plane is the kernel of a linear map of a nonzero linear functional, and over ${\mathbb{F}}_2$ that linear functional must be fixed; the displayed matrices have only the fixed nonzero linear functional $e_3^{\ast }$. Hence each subgroup specifies its flag uniquely. There are seven choices of $v$ and three choices of $P$, giving

The permutation module $M_F^{\mu }$ is the $F$-vector space with the $\mu$-tabloids as basis, with $S_n$ acting by permuting their entries. For a tableau $t$ with column stabilizer $C_t$, its polytabloid isThe Specht module $S_F^{\mu }$ is the submodule of $M_F^{\mu }$ spanned by these polytabloids, equivalently generated by any one $e_t$ under the group action.

The standard basis theorem states thatis an $F$-basis of $S_F^{\mu }$ in every characteristic. Its dimension is the number of standard tableaux of shape $\mu$; in the hook shape used below the basis is exactly the displayed $e_{1ij}$ with $2\le i<j\le n$.

Apply the Abelian group criterion for a product of two prime squares with the smaller and larger primes in their correct order. For $p<11$, all such groups are Abelian exactly when $p\nmid 11^2-1=120$. Of the primes $2,3,5,7$, only $7$ qualifies. For $p>11$, the criterion is $11\nmid p^2-1$, equivalently $p\not\equiv \pm 1(\mathrm{mod}11)$. Therefore the exact answer isEquivalently, among the stipulated primes $p\ne 11$, exclude $2,3,5$ and the residue classes $\pm 1$ modulo $11$.

For completeness, the necessity in that criterion supplies actual counterexamples at every excluded prime. If $\ell <q$ and $\ell \mid q^2-1$, the multiplicative group of a finite field is cyclic, so choose $\alpha \in {\mathbb{F}}_{q^2}^{\times }$ of order $\ell$. Let a generator of $C_{{\ell }^2}$ act on the additive group of ${\mathbb{F}}_{q^2}$ by multiplication by $\alpha$. This defines a nontrivial group action, hence a nonabelian semidirect product of order ${\ell }^2{q}^2$. Use $(\ell ,q)=(p,11)$ for $p=2,3,5$, and $(\ell ,q)=(11,p)$ for $p\equiv \pm 1(\mathrm{mod}11)$ with $p>11$. Thus no excluded prime can be restored to the answer.

This is the sum of content monomials of colorings up to the group action. It equals the cycle index specialized at $t_r=p_r(x)$. One may first use finitely many colors; stabilization defines the corresponding homogeneous symmetric function. Calling the specialized cycle index a cycle indicator makes $Z_G=F_G$ an equality in the same variables.