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First supply the unlettered preliminaries. Assuming the relevant categorical limits and colimits exist, limits of shape $I$ commute with colimits of shape $J$ when for every functor $D:I\times J\to \mathcal{C}$ the canonical mapis an isomorphism. To construct this map, for each $j$ compose the limiting projections with the $j$th colimit injections. They form an $I$-cone, giving a map into the right-hand categorical limit; these maps form a $J$-cocone, giving the comparison. This specifies the actual comparison required for commutation of limits and colimits.

A group regarded as a one-object category has its group elements as arrows, with composition given by multiplication. A functor from it to the Category of sets is a group action. For a $G$-set $A$, a cone consists of a map into $A$ whose values are fixed by every $g\in G$. Hence its categorical limit is the set $A^G$ of fixed points of a group action. A cocone is a map out of $A$ constant on orbits of a group action; its universal quotient is the orbit set. Thus

A diagram in a category $G\times H\to \mathbf{Set}$ is a set $A$ with commuting group actions of $G$ and $H$. Its limit–colimit comparison isCommutation makes $A^G$ stable under $H$. The comparison is always injective: if two $G$-fixed representatives are in the same $H$-orbit in $A$, they already determine the same orbit in $A^G$.

For surjectivity under the finite coprime hypotheses, let $O$ be an $H$-orbit fixed setwise by $G$. Its size divides $|H|$ by the orbit-stabilizer theorem. Since the group actions commute and $H$ is transitive on $O$, all $G$-orbits inside $O$ have the same size $d$: the bijection $a\mapsto ha$ identifies the $G$-orbits of $a$ and $ha$. Thus $d$ divides $|O|$, while the orbit-stabilizer theorem also gives $d\mid |G|$. The orders are coprime, so $d=1$. Every point in $O$ is therefore $G$-fixed, and $O$ lies in the image of $\theta$. We have proved commutation of fixed points and orbit quotients for coprime groups:The bijection is the canonical comparison and is natural in the set with commuting group actions, so it proves the requested commutation of limits and colimits in the Category of sets.

Let $q:G\twoheadrightarrow K$ and $r:H\twoheadrightarrow K$ be the two quotient homomorphisms. On the underlying set $A=K$, define the commuting group actionsThe inverse is essential for the right multiplication formula to define a left group action: $h_1\cdot (h_2\cdot k)=kr(h_2{)}^{-1}r(h_1{)}^{-1}=(h_1{h}_2)\cdot k$. Left and right multiplication commute.

Because $q$ is onto and $K$ is nontrivial, there is no fixed point of a group action for $G$: if every $q(g)k=k$, cancellation would make every element of $K$ the identity. Thus $A^G=\varnothing$. Because $r$ is onto, the $H$-action is transitive and $A/H$ is a singleton. The induced $G$-action on that singleton is trivial. Consequently the comparison iswhich is not an isomorphism. This proves the common quotient obstruction to commutation of fixed points and orbits, so limits of shape $G$ do not commute with colimits of shape $H$ in the Category of sets.

Restrict the quotient projection $\pi :G\to G/N$ to $H$. This is a group homomorphism withIts image is a nontrivial subgroup of the prime-order quotient group, because $H$ is not contained in $N$. By Lagrange's theorem for finite groups, the image has order $p$ and is all of $G/N$. The first isomorphism theorem for groups therefore gives

For the second assertion, partition $C$ into conjugacy classes for $N$. Because $N$ is normal, conjugation by $G$ permutes these smaller classes: $g{\mathrm{Cl}}_N(x)g^{-1}={\mathrm{Cl}}_N(gx{g}^{-1})$. This group action is transitive since any two elements of $C$ are conjugate in $G$. Elements of $N$ fix every $N$-class, so the action factors through $G/N$.

If $m$ is the number of smaller classes, the orbit-stabilizer theorem for this transitive action gives $m\mid |G/N|=p$. ThusDistinct conjugacy classes are disjoint because they are orbits of a group action. This proves conjugacy class splitting in a prime-index normal subgroup without assuming all $G$-conjugations are already conjugations by $N$.

For a group action, the orbit of a group action of $x$ is $Gx=\{gx:g\in G\}$, and its stabilizer subgroup is $G_x=\{g:gx=x\}$. The orbit-stabilizer theorem gives $|Gx||G_x|=|G|$: the map $g{G}_x\mapsto gx$ is a well-defined bijection from stabilizer cosets to the orbit.

Within one orbit, the stabilizer subgroups are conjugate, since $G_{hx}=h{G}_x{h}^{-1}$. Their sizes are therefore equal. Summing stabilizer sizes over a single orbit gives $|Gx||G_x|=|G|$, and summing over all $n$ orbits yieldsNow count the pairs $(g,x)$ fixed by the action in two ways. Holding $x$ fixed gives $|G_x|$ choices of $g$, whereas holding $g$ fixed gives $|\mathrm{Fix}(g)|$ choices of $x$. ThusCombining the two identities proves Burnside lemma:The finiteness hypotheses justify these cardinality sums. If $X$ is empty, every sum and the orbit count are zero, so the formula still applies.

A finite nonabelian simple group cannot be a p-group: a nontrivial finite p-group has nontrivial centre of a group, since its class equation makes every noncentral conjugacy class size divisible by $p$, and hence makes the centre size a positive multiple of $p$. That centre is a normal subgroup, so must be the whole group if $G$ is a simple group, making it abelian; an abelian simple group has prime order. Therefore every Sylow subgroup here is nontrivial and proper. It cannot be normal, so $n_p>1$.

To obtain a simple group embedding from Sylow conjugation, the conjugation action on Sylow subgroups gives a group homomorphism $\rho :G\to S_{n_p}$. This group action is transitive by the Sylow theorems and is nontrivial since $n_p>1$. Its kernel of a group homomorphism is a normal subgroup, so the defining property of a simple group makes the kernel of a group homomorphism trivial: $\rho$ is an embedding. Now compose with the sign of a permutation $S_{n_p}\to \{\pm 1\}$. A nontrivial composite would again be injective because $G$ is a simple group, embedding $G$ in a group of order two, impossible for a nonabelian group. Thus $\rho (G)$ lies in the alternating group $A_{n_p}$. By Lagrange's theorem,The argument only invokes the factorial formula once $n_p\ge 2$; the impossible case $n_p=2$ itself is also excluded by the embedding.

First use the nonzero loci, as required by the standard SL2R action on differentials. A nonzero holomorphic one-form has, away from its zeros, flat coordinateswhose changes of coordinate are translations. A nonzero holomorphic quadratic differential similarly has local flat coordinates $w=\int \sqrt{q}$, with $q=d{w}^2$ and changes of coordinate $w_j=\pm w_i+c_{ij}$. These are respectively translation surfaces and half-translation surfaces.

Identify a flat coordinate with a vector in ${\mathbb{R}}^2$. For $A\in {\mathrm{SL}}_2(\mathbb{R})$, replace every flat coordinate by $W=Aw$. Since $A$ preserves orientation and commutes with multiplication by $-1$, the new changes of coordinate areThey are holomorphic in the new coordinates, and so define a new complex structure. Define ${\omega }_A=dW$ or $q_A=d{W}^2$ in that structure. The forms glue because translations preserve $dW$, and the extra signs preserve $d{W}^2$.

The zeros also extend. A zero of order $m$ of a holomorphic one-form has cone angle $2\pi (m+1)$; a zero of order $m$ of a holomorphic quadratic differential has cone angle $(m+2)\pi$. The real-linear deformation preserves the corresponding winding multiplicity. Filling the cone in a local coordinate $\zeta$ gives $W={\zeta }^{m+1}$ in the first case, or a local branch of $W={\zeta }^{(m+2)/2}$ in the second. Thus the resulting forms are constant multiples of ${\zeta }^{m}d\zeta$ or ${\zeta }^{m}d{\zeta }^2$ and have the same zero orders. This verifies extension across the missing points, rather than merely producing an atlas on the punctured surface.

An isomorphism preserving the original differential identifies its flat coordinates up to the permitted translations or signs; applying $A$ identifies the deformed atlases too. Hence the construction descends to the corresponding moduli spaces. Applying $B$ after $A$ replaces $w$ by $BAw$, soThe area of a quadratic differential, and the analogous area of a holomorphic one-form, are preserved because $\det A=1$.

For $q={\omega }^2$, the flat coordinates obtained from $\omega$ already give the required half-translation surface atlas for $q$. The same replacement $w\mapsto Aw$ therefore constructs both deformations, and

The printed sets include identically zero differentials. They have no flat coordinates, so the customary geometric group action is defined on the nonzero loci. One can obtain a set-theoretic action on the displayed entire sets by declaring $A\cdot (X,0)=(X,0)$; the same equivariance identity then holds at zero. This extension is generally not continuous: as $t\omega \to 0$, the deformed underlying surface is the same $X_A$ for every real $t>0$, and can differ from $X$. Thus a claim about the standard continuous geometric group action requires the nonzero convention.

The analogous flat-coordinate construction does not define a full ${\mathrm{SL}}_2(\mathbb{R})$ action on arbitrary cubic differentials. Away from zeros, a nonzero holomorphic cubic differential has flat coordinatesTheir changes of coordinate are $w_j=\zeta w_i+b_{ij}$ with ${\zeta }^3=1$. A branch change is therefore a rotation through $2\pi /3$, not merely a sign. Applying a real-linear map $A$ changes its linear part to $AR{A}^{-1}$, where $R$ is that rotation. In general this is not a complex-linear map, so the proposed changes of coordinate are not holomorphic and cannot define the required deformed complex structure.

For example, take $A=\mathrm{diag}(2,1/2)$ and $R=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$ with $\theta =2\pi /3$. Thenwhose off-diagonal entries fail the condition for a complex-linear map. This dependence on the choice of cube-root coordinate is the obstruction even when considering descent from the locus $c={\omega }^3$: the three possible roots need not lead to the same deformation of the cubic pair.

The real matrices preserving orientation that normalize the order-three rotations are precisely the matrices of complex-linear maps; intersecting with ${\mathrm{SL}}_2(\mathbb{R})$ leaves $\mathrm{SO}(2)$. Indeed a nonreal rotation determines its complex structure, and conjugation to its inverse would reverse that structure's orientation. Thus there is a natural rotation action,The conclusion concerns the geometric group action analogous to that for translation surfaces and half-translation surfaces; it does not rule out artificial group actions unrelated to these atlases.

Give $V_n$ the action $(g\cdot P)(x,y)=P((x,y)g)$, where $(x,y)$ is a row vector. Substitution preserves homogeneous degree, and composition agrees with matrix multiplication, so this is a complex representation of SU(2) group. The diagonal torus $\mathrm{diag}(z,z^{-1})$ acts on $x^{n-j}{y}^j$ with weight $z^{n-2j}$; these weights are distinct.

A nonzero invariant subspace is invariant under the torus, so finite-dimensional Fourier projection onto its weight spaces supplies a nonzero monomial. Differentiating the group action and complexifying supplies the operators $E=x{\partial }_y$ and $F=y{\partial }_x$. Their repeated application moves between every successive monomial, with nonzero coefficients until an endpoint is reached:The invariant subspace therefore contains the whole monomial basis. ThusFor $n=0$ this is the one-dimensional trivial representation. Torus weight projections are legitimate because an invariant complex subspace is closed in finite dimension.

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 $x$.

An action is properly discontinuous when every compact set meets only finitely many of its translates. For an isometric action on a proper metric space, this gives the finiteness needed to extract a constant group element from a sequence with bounded source and target.

The orientation-preserving symmetries of a cube are the $3\times 3$ signed permutation matrices with determinant one. There are $3!2^2=24$ of them. Their group action on the four body diagonals identifies this group with the symmetric group $S_4$. Indeed choose diagonal directions $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$, whose sum is zero and whose only linear relation has all coefficients equal. A rotation fixing every diagonal sends each direction to itself or its negative; the relation forces all four signs equal. All negative would give $-I$, which has determinant $-1$, so the kernel of a group homomorphism is trivial. Both groups have order 24, giving the identification. Its face permutations have cycle types $1^6$, $1^{2}4$, $1^2{2}^2$, $2^3$ and $3^2$, occurring $1,6,3,6,8$ times respectively.

For nonzero holomorphic one-forms and holomorphic quadratic differentials, apply $A\in {\mathrm{SL}}_2(\mathbb{R})$ to every flat coordinate. Translation and sign transition maps remain of the same type, defining a new complex structure and differential. Composition gives a group action, zero orders persist and area is preserved. The zero section has no such atlas; fixing it gives only a set-theoretic extension that is generally discontinuous.