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Every nontrivial subgroup of the quaternion group contains $-1$. Indeed, it either contains $-1$ directly, or contains one of $\pm i,\pm j,\pm k$, whose square is $-1$.

Suppose $Q_8$ acts on an $n$-point set with $n<8$. Every group orbit has size at most $n<8$. By the orbit-stabilizer theorem, each point's stabilizer subgroup has order $8/|Q_{8}x|>1$, so it contains $-1$. Consequently $-1$ fixes every point, and the group action is not faithful. An embedding in $S_n$ would give a faithful group action on $n$ letters, which is impossible. ThusThe left-regular group action on eight points is faithful by Cayley theorem, so the minimal faithful permutation degree of the quaternion group is exactly eight. The proof allows multiple group orbits; restricting only to transitive actions would not suffice.

For a group action of $G$ on $X$, let $G_x=\{g:gx=x\}$ be the stabilizer subgroup of $x$, and $Gx$ its group orbit. The orbit-stabilizer theorem is the bijectionIt is well defined because elements in the same left coset differ by an element fixing $x$. Conversely, $gx=hx$ implies $h^{-1}g\in G_x$, so the cosets coincide. Surjectivity is the definition of the group orbit. For a finite group, counting the cosets givesTo obtain Cayley theorem, let $G$ act on its own underlying set by left multiplication. The map ${\lambda }_g:x\mapsto gx$ is a permutation, with inverse ${\lambda }_{g^{-1}}$, and ${\lambda }_g{\lambda }_h={\lambda }_{gh}$. If ${\lambda }_g$ is the identity permutation, evaluating at the identity element gives $g=e$. Thus $g\mapsto {\lambda }_g$ is an injective group homomorphism. When $|G|=n$, labeling this set by $n$ letters identifies its image with a subgroup of $S_n$. Every finite group embeds in the symmetric group on its own elements.

Use the standard implicit assumption that $p$ is prime, so ${\mathbb{F}}_p$ is a finite field. The matrices in question form $G{L}_2({\mathbb{F}}_p)$, the general linear group over a finite field. Products stay in the set because determinants multiply, the identity has determinant one, andhas entries in the finite field and nonzero determinant. Together with associativity this verifies the group axioms. The multiplication of matrices with column vectors is a group action because $Iv=v$ and $(gh)v=g(hv)$.

Let $C=⟨g⟩$ be the order-$p$ cyclic subgroup. By the orbit-stabilizer theorem, its group orbits on ${\mathbb{F}}_p^2$ have size either one or $p$. If $f$ is the number of fixed vectors, counting all $p^2$ vectors gives $f\equiv p^2\equiv 0(\mathrm{mod}p)$. Since the zero vector is fixed, $f\ge 1$ and hence $f\ge p$. There is a nonzero fixed vector $v$.

Extend $v$ to a basis $(v,w)$. In this basis $g$ has matrix $\left(\begin{array}{cc}1& b\\ 0& d\end{array}\right)$. From $g^p=I$ we get $d^p=1$; in the prime finite field, $d^p=d$, so $d=1$. Since $g$ has order $p>1$, it is not the identity, and $b\ne 0$. Replace $w$ by $b^{-1}w$. Then $g(b^{-1}w)=v+b^{-1}w$, and in the new basisThus all the order-p matrices in GL2 over the prime field form one conjugacy class. This proof also works at $p=2$. Primality is essential to the first assertion: for modulus four, $\mathrm{diag}(2,1)$ has nonzero determinant but no inverse, so the set defined using merely nonzero determinants would not be a group.

The conjugacy class of $x$ is its orbit under the conjugation action, and its stabilizer is the centralizer $C_G(x)$. The orbit-stabilizer theorem givesBy Lagrange's theorem, the order of the centralizer divides $p^n=|G|$, so the index is a power of $p$. A conjugacy class has size one exactly when its element belongs to the center of a group.

Separate these singleton classes in the class equation:where the sum takes one representative of each nonsingleton conjugacy class. Every summand is divisible by $p$. Thus $p\mid |Z(G)|$, and the identity ensures $|Z(G)|>0$. Consequently $|Z(G)|\ge p$. This is the nontrivial center of a finite p-group property.

Extend each element of $G$ to fix $\omega$. A one-point extension of a permutation group is a transitive permutation group $G^{+}\le S_{{\Omega }^{+}}$ such thatwith the induced action on $\Omega$ equal to the prescribed action. The stabilizer subgroup equality is the substantive condition: merely adjoining an element that moves $\omega$ does not suffice. If $|\Omega |=n$, the orbit-stabilizer theorem gives $|G^{+}|=(n+1)|G|$.

Taking diagonal blocks is a homomorphism $P\to L$. Its kernel is $Q$, and block-diagonal inclusion is a section. For any $g\in P$, remove its diagonal element $\ell \in L$ by $g{\ell }^{-1}\in Q$. Also $Q\cap L=1$. ThusUsing $|{\mathrm{GL}}_k(q)|=q^{k(k-1)/2}{\prod }_{i=1}^k(q^i-1)$ and part (a), the product simplifies toThe exponent simplification is $2k(m-k)+k(k+1)/2+k(k-1)/2+(m-k{)}^2=m^2$.

For an independent orbit-stabilizer theorem count, choose an ordered independent isotropic tuple $f_1,\dots ,f_k$. After $r$ choices, their span has $q^r$ elements and its orthogonal complement has dimension $2m-r$. The next choice therefore has $q^{2m-r}-q^r$ possibilities. The number of tuples isEvery totally isotropic $k$-space has $|{\mathrm{GL}}_k(q)|$ ordered bases, so the number of such spaces isEach tuple extends to a symplectic basis by successively choosing paired partners and taking orthogonal complements. Consequently the symplectic group is transitive on these spaces. The stabilizer subgroup of $W$ also preserves $W^{\perp }$, and so is exactly $P$. Dividing $|{\mathrm{Sp}}_{2m}(q)|$ by $N_k$ reproduces the boxed answer. Dividing by $T_k$ instead would count the pointwise stabilizer subgroup of the ordered tuple, a different subgroup.

The Möbius transformations are bijections, so they preserve which coordinates of a triple are equal. Conversely, part (i) implies transitivity within each equality pattern. For a pattern with fewer than three distinct entries, extend the chosen distinct points to a triple before applying part (i); this works also for $p=2$, when the projective line has exactly three points. There are therefore five orbits of a group action.

We first count $|G{L}_2({\mathbb{F}}_p)|$. Its first column can be any nonzero vector, giving $p^2-1$ choices. The second must be outside the first column's one-dimensional span, giving $p^2-p$ choices. HenceFor the all-equal pattern take $(\infty ,\infty ,\infty )$. Its stabilizer subgroup consists of upper triangular matricesof order $p(p-1{)}^2$. Its group orbit has order $p+1$.

The three exactly-two-equal patterns have representatives $(\infty ,\infty ,0)$, $(\infty ,0,\infty )$ and $(0,\infty ,\infty )$. Each stabilizer subgroup fixes $0$ and $\infty$ individually, so it is the diagonal subgroup of order $(p-1{)}^2$. Each group orbit has order $p(p+1)$; their repeated-coordinate positions keep these three orbits distinct.

For the all-distinct pattern use $(0,1,\infty )$. Its stabilizer subgroup is the scalar subgroup of order $p-1$, and the group orbit has order $(p+1)p(p-1)$. Stabilizers of arbitrary triples are conjugates of the displayed representative stabilizers. In summary the equality-pattern orbits of projective triples haveThe sizes add to $(p+1{)}^3$, and each orbit size times its stabilizer order is $|G|$, as in the orbit-stabilizer theorem.

The identity permutation fixes every multivariate polynomial. With permutation composition $(\sigma \tau )(i)=\sigma (\tau (i))$, substitution givesThus this is a left variable-permutation action on polynomials. The substitution acts on variable names; using the displayed definition avoids reversing the composition order.

For the given multivariate polynomial, the group orbit corresponds to the three partitions of four indices into two unordered pairs:These three polynomials are distinct, and every pairing occurs under a permutation. The orbit-stabilizer theorem givesThe eight elements of the stabiliser subgroup independently interchange the members of each pair, and may also interchange the two pairs.

Applying a permutation to the elements of an $r$-element subset preserves its cardinality. The identity fixes each subset andso this is the symmetric group action on subsets. For two $r$-element subsets $U,W$, choose a bijection $U\to W$ and a bijection between their complements; together they define a permutation of the full $n$-element set. Hence the action is transitive.

A member of the stabiliser subgroup must permute $U$ internally and its complement internally, with independent choices. ThusThis also agrees with the orbit-stabilizer theorem; $0!=1$ handles $r=n$.

A left group action is a map $(g,x)\mapsto g\cdot x$ with $e\cdot x=x$ and $(gh)\cdot x=g\cdot (h\cdot x)$. The group orbit and stabiliser subgroup areThe identity belongs to $G_x$. If $g,h\in G_x$, then $(gh)\cdot x=g\cdot x=x$; also $g^{-1}\cdot x=g^{-1}\cdot (g\cdot x)=x$. Thus $G_x$ is a subgroup.

For the orbit-stabilizer theorem, the map $g{G}_x\mapsto g\cdot x$ from left cosets to the group orbit is a bijection: $g\cdot x=h\cdot x$ if and only if $h^{-1}g\in G_x$, which is precisely $g{G}_x=h{G}_x$. Each left coset has $|G_x|$ elements, since multiplication by $g$ is a bijection from $G_x$ to $g{G}_x$, and the left cosets partition $G$. For finite $G$ this proves

Every $g\in G$ lies in $HgK$, since both subgroups contain the identity. If two double cosets intersect, chooseThen $g^{\prime }=h_2^{-1}{h}_{1}g{k}_1{k}_2^{-1}\in HgK$, so every element of $H{g}^{\prime }K$ lies in $HgK$. Reversing the argument gives equality. Thus the double cosets partition the group.

Unlike single cosets, double cosets can have different cardinalities. For finite $G$, decompose $HgK$ into left cosets of $K$, or apply the orbit-stabilizer theorem to the action of $H$ on $gK$. Its stabiliser subgroup is $H\cap gK{g}^{-1}$, givingThis formula explains the counterexamples in the following parts.

The identity belongs to the center of a group. If $z,w$ commute with every $g$, then $zw$ does too; rearranging $gz=zg$ also shows that $z^{-1}$ commutes with $g$. Hence $Z(G)$ is a subgroup. Moreover $gz{g}^{-1}=z$ for $z\in Z(G)$, so it is a normal subgroup.

Suppose $|G|=p^n>1$. The conjugation action partitions $G$ into conjugacy classes. The orbit-stabilizer theorem gives class size $[G:C_G(x)]$, where $C_G(x)$ is the centralizer of $x$. By Lagrange's theorem, this size is a power of $p$; it is one exactly when $x$ is central. Thus the class equation has the formIt follows that $p\mid |Z(G)|$. Since the identity is central, $|Z(G)|\ge 1$, and thereforeThis is the nontrivial-centre property of a finite p-group.

For a group action of a finite group $G$ on a set and a point $x$, the orbit-stabilizer theorem iswhere $G_x=\{g:gx=x\}$ is the stabilizer subgroup. The map $g{G}_x\mapsto gx$ is well defined because elements of the same coset act identically on $x$. If $gx=hx$, then $h^{-1}g\in G_x$, so their cosets agree; every orbit point has the form $gx$, giving surjectivity. This is a bijection. Lagrange's theorem, $|G|=[G:H]|H|$ for any subgroup $H$, yields the counting formula.

Apply the conjugation action of the finite p-group $G$ to its nontrivial normal subgroup $N$. Normality ensures the action stays inside $N$. The singleton orbits of a group action are exactly the elements of $N\cap Z(G)$. Every larger orbit of a group action has size a positive power of $p$, by the orbit-stabilizer theorem and Lagrange's theorem. ThereforeSince $N$ is nontrivial, its order is divisible by $p$. The intersection contains the identity and has cardinality a positive multiple of $p$, so it contains a nonidentity element. This proves the central intersection property of normal subgroups of finite p-groups.

For the proper subgroup $H$, let $H$ act by left multiplication on the left cosets $G/H$. A coset $gH$ is fixed by every $h\in H$ exactly when $g^{-1}Hg\subseteq H$. Both groups have the same finite size, so this inclusion is equality; the fixed cosets are precisely those with $g\in N_G(H)$. Hence their number is $[N_G(H):H]$. All non-singleton orbits of a group action have size divisible by $p$, soSince $H$ is proper, $[G:H]$ is a positive power of $p$ greater than one. The fixed-coset count is nonzero, because $H$ itself is fixed, and is divisible by $p$. Therefore the normalizer strictly contains $H$:This is the normalizer condition for finite p-groups. The proof also covers $H=\{1\}$, whose action fixes all cosets.

Write $|G|=p^{r}m$ with $p$ prime and $p\nmid m$. The Sylow theorems assert that a subgroup of order $p^r$ exists; every p-group subgroup is contained in a conjugate of a Sylow subgroup; all Sylow subgroups are conjugate; and their number $n_p$ satisfies $n_p\equiv 1(\mathrm{mod}p)$ and $n_p\mid m$.

For the containment and conjugacy assertions, fix a Sylow subgroup $P$ and let any p-group subgroup $Q$ act by left multiplication on the $m$ left cosets $G/P$. By the orbit-stabilizer theorem, every orbit has size a power of $p$. Since $p\nmid m$, at least one orbit has size one. If $gP$ is such a fixed point, then $qgP=gP$ for all $q\in Q$, equivalently $g^{-1}Qg\subseteq P$. Thus $Q\subseteq gP{g}^{-1}$. If $Q$ too has order $p^r$, this inclusion is equality, proving conjugacy. This is Sylow containment from a coset fixed point.

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.