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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.

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.

Conjugation by $P$ partitions $H$ into orbits. Each non-singleton orbit has size divisible by $p$, by the orbit-stabilizer theorem. The singleton orbits are exactly $H\cap Z(P)$. Since $|H|$ is divisible by $p$, $|H\cap Z(P)|\equiv 0(\mathrm{mod}p)$. This set contains $1$, so $\boxed{|H\cap Z(P)|\ge p}$.

For the deduction we also give the representation-theoretic ingredient behind Burnside's theorem: a nonabelian finite simple group has no nontrivial conjugacy class of prime power size. Here is a proof, so no stronger theorem is being assumed without justification. Suppose $g\ne 1$ has class size $p^b$. For an irreducible character $\chi$ of degree $d$ coprime to $p$, the class sum acts as the scalar $p^b\chi (g)/d$, which is an algebraic integer: the class sum has an integer-entry matrix in the regular representation, which contains every irreducible representation, so this scalar is an eigenvalue of an integer-entry matrix. Also $\chi (g)$ is an algebraic integer. Bezout's identity then makes $\chi (g)/d$ an algebraic integer. All its algebraic conjugates have modulus at most $1$, because character values of a representation are sums of $d$ roots of unity. An algebraic integer with this property is zero or a root of unity: the monic polynomials of all its powers have uniformly bounded integer coefficients, so two powers must coincide unless it is zero.

Every nontrivial irreducible representation of a simple group is faithful. If $\chi (g)/d$ were a root of unity, equality in the triangle inequality would force its representing matrix to be scalar, hence $g$ central, impossible. Thus $\chi (g)=0$ whenever $p\nmid d$, except for the trivial character of a representation. Orthogonality of the identity column and the $g$ column in the character table now givesThis would make $1/p$ an algebraic integer, contradicting the fact that rational algebraic integers are integers. The same proof covers class size $1$ directly through triviality of the center.

If an abelian subgroup $A$ had index $p^a$, every $g\in A\setminus \{1\}$ would have $A\subseteq C_G(g)$, making its class size a divisor of $p^a$, a contradiction. Such a $g$ exists: if $A=1$, then $G$ is a nontrivial p-group, whose center is nontrivial by the first argument with $H=P$. Therefore a nonabelian simple group has no abelian subgroup of prime power index.

Conjugating a cycle merely relabels its entries:Thus conjugate permutations have the same cycle type. Conversely, if $\sigma$ and $\tau$ have the same cycle type, match the entries of each cycle of $\sigma$ bijectively and in cyclic order with those of a cycle of $\tau$ of the same length. Extending these matches to a permutation $g$ gives $g\sigma g^{-1}=\tau$.

Let $C_{S_n}(\sigma )$ be the centralizer of $\sigma$. Its $S_n$-conjugacy class remains one $A_n$-conjugacy class exactly when $C_{S_n}(\sigma )$ contains an odd permutation. Indeed, in that case $C_{A_n}(\sigma )$ has index two in $C_{S_n}(\sigma )$, so the orbit-stabilizer theorem givesIf the centralizer contains only even permutations, the $A_n$ class has half the size and the $S_n$ class splits into two $A_n$ classes.

The even cycle types in $S_5$ areThe classes of the identity, a three-cycle, and a double transposition do not split: their centralizers contain an odd permutation. The centralizer of a five-cycle is its cyclic subgroup of order five, which lies in $A_5$, so that class splits in two. Therefore

The decomposition matrix separates into two connected components. The first contains ${\varphi }_1,{\varphi }_2$ and ${\chi }_1,{\chi }_{3A},{\chi }_{3B},{\chi }_4$; it is the principal block $B_0$. The second contains only ${\varphi }_3$ and ${\chi }_5$; since ${\varphi }_3(1)=5$ contains the full 5-part of $|A_5|=60$, this is a defect-zero representation and its block $B_1$ has defect group $1$.

For $D=1$, the normalizer is $N_G(1)=G$, so the Brauer correspondence is the identity and $B_1$ corresponds to itself.

The defect group of the principal block is a Sylow 5-subgroup $P\cong C_5$. There are six Sylow 5-subgroups in $A_5$, so the orbit-stabilizer theorem givesThe centralizer of a 5-cycle in $A_5$ is $P$, and an involution in the normalizer acts on $P$ by inversion. HenceIn characteristic five the simple $kN$-modules are inflated from $N/P\cong C_2$: their Brauer characters are ${\psi }_{+}=(1,1)$ and ${\psi }_{-}=(1,-1)$ on the identity and involution classes. If $1,\epsilon ,{\rho }_1,{\rho }_2$ are the two one-dimensional and two two-dimensional ordinary characters of $D_{10}$, their reductions areThe resulting decomposition matrix is connected, so these characters form the unique 5-block $b_0$ of $N$, with defect group $P$. By the Brauer first main theorem,This is the complete 5-modular blocks of A5 correspondence.

The singleton orbits are exactly $X^G$. Every other orbit represented by $x$ has size $|G|/|G_x|$ by the orbit-stabilizer theorem. Partitioning $X$ into its orbits of a group action therefore gives

For the second identity, countin two ways. Holding $g$ fixed givesHolding $x$ fixed gives $|Z|={\sum }_{x\in X}|G_x|$. Each orbit contributes $|G|/|G_x|$ points, each with stabilizer size $|G_x|$, and hence contributes $|G|$. Thus $|Z|=|G||X/G|$, proving Burnside lemma:

Burnside lemma says that the number of orbits of a finite group $G$ on a finite set $X$ isIndeed, double-count the set $\{(g,x):gx=x\}$. Counting first by $g$ gives the numerator. Counting first by $x$ gives ${\sum }_x|G_x|$; each orbit $O$ contributes $|O||G_x|=|G|$ by the orbit-stabilizer theorem, proving the formula.

Let $\pi (g)=|X^g|$ be the character of the permutation representation. Thenis, by Burnside's lemma, the number of orbits on $X\times X$. A two-transitive group action has exactly two such orbits: the diagonal and the ordered pairs of distinct points. Hence $⟨\pi ,\pi ⟩=2$. Transitivity also gives $⟨\pi ,1_G⟩=1$. If $\pi =1_G+\sum m_i{\chi }_i$ is its irreducible character decomposition, then $2=1+\sum m_i^2$, so exactly one nontrivial irreducible character occurs, with multiplicity one. Thus the permutation character has precisely two distinct irreducible summands.

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.

For a centered static Skyrmion $U_0$, the effective rotation-isorotation group is $G=SO(3{)}_{\mathrm{space}}\times SO(3{)}_{\mathrm{iso}}$. Its stabilizer subgroup $H$ consists of pairs obeying $A{U}_0(R^{-1}x)A^{†}=U_0(x)$. Thus distinct rigid orientations form the homogeneous space $G/H$, by the orbit-stabilizer theorem. The unit hedgehog has diagonal $SO(3)$ stabilizer and three orientation coordinates; the toroidal two-Skyrmion has a one-dimensional continuous stabilizer and five; the tetrahedral and cubic cases have discrete stabilizers and six. Adding three translations gives zero-mode orbit dimensions 6, 8, 9, 9. These are symmetry-generated low-energy coordinates, not a claim that non-Bogomolny multi-Skyrmion solutions possess an exact flat moduli space of arbitrary separations.

A matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts on the projective line by the Möbius transformation $x\mapsto (ax+b)/(cx+d)$, with the pole sent to infinity and infinity sent to $a/c$ when $c\ne 0$. The action is transitive because $\left(\begin{array}{cc}t& -1\\ 1& 0\end{array}\right)$ sends infinity to any finite $t$. The stabilizer subgroup of infinity consists of $\left(\begin{array}{cc}a& b\\ 0& a^{-1}\end{array}\right)$ with $a\ne 0$. The orbit-stabilizer theorem gives $|{\mathrm{SL}}_2({\mathbb{F}}_q)|=q(q^2-1)$. Its kernel is the scalar matrices of determinant one: a transformation fixing infinity has $c=0$, fixing zero then forces $b=0$, and fixing one gives $a=d$. Thus the kernel is $\{I,-I\}$, with these matrices coinciding in characteristic two.