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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.

Choose $x_0\in X$ with stabilizer subgroup $H$. Transitivity means that every point is $g{x}_0$ for some $g$. Its stabilizer subgroup issince $a(g{x}_0)=g{x}_0$ exactly when $g^{-1}ag\in H$. The kernel of a group homomorphism arising from the group action is the set of elements fixing every point, henceThis is the subgroup core. A group homomorphism is injective precisely when its kernel is trivial. The desired equivalence is therefore exactly the equivalence between a trivial subgroup core and a faithful group action. No finiteness assumption is required in this part.

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.

Write a member of the special linear group as $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with $ad-bc=1$. Its Möbius transformation is $z\mapsto (az+b)/(cz+d)$, with the usual values at a pole and at infinity. For nonreal $z$,This follows by multiplying numerator and denominator by $c\bar{z}+d$. Thus the sign of the imaginary part is preserved, while the extended real line is preserved as a set.

Fixing zero requires $b=0$. Therefore its stabilizer subgroup and group orbit areTranslations send zero to each finite real point, and $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ sends it to infinity.

For $gi=i$, comparing real and imaginary parts gives $a=d$ and $b=-c$, with $a^2+c^2=1$. Exactly the same conditions follow from $g(-i)=-i$. HenceFor any $x\in \mathbb{R}$ and $y>0$, the upper triangular matrixsends $i$ to $x+iy$ and $-i$ to $x-iy$. Therefore the two group orbits are the open complex upper half-plane and the open lower half-plane respectively. Along with the extended real line they exhaust the Riemann sphere, so there are exactly three orbits, the real determinant-one Möbius orbits.

Every $h\in H$ has the form $\left(\begin{array}{cc}a& b\\ 0& a^{-1}\end{array}\right)$ and acts by $z\mapsto a^{2}z+ab$. Thus $Hi$ is the entire complex upper half-plane, by the same explicit matrices $h_{x,y}$. Given $g\in G$, choose $h\in H$ such that $hi=gi$. Then $k=h^{-1}g$ fixes $i$, so $k\in SO(2)$ and $g=hk$. This proves the triangular-rotation factorization of real determinant-one matrices.

For uniqueness, if $h_1{k}_1=h_2{k}_2$, then $h_2^{-1}{h}_1=k_2{k}_1^{-1}\in H\cap SO(2)$. An upper triangular rotation matrix must have $\sin \theta =0$, soThere are consequently exactly two matrix factorizations, $(h,k)$ and $(-h,-k)$. Restricting the first diagonal entry of $h$ to be positive makes the factorization unique. For example, if $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and $q=\sqrt{c^2+d^2}$, that unique branch isAngles are understood modulo $2\pi$ when counting matrices; allowing unrestricted real angle representatives gives infinitely many labels for those same two factorizations.

Let $m=\mathrm{ord}(A)$, $n=\mathrm{ord}(B)$ and $r=\mathrm{ord}(AB)$. Cut the Riemann sphere along arcs joining three marked points, say $0,1,\infty$, so the complement is simply connected. Take copies labeled by the elements of $T$ and glue the slit banks with monodromies given by left multiplication by $A,B,(AB{)}^{-1}$. The product relation ensures consistency. The surface is connected because $A,B$ generate $T$. At a puncture with monodromy order $m$, fill each cycle using the local complex coordinate $z=w^m$; use the analogous charts at the other two punctures. This constructs the three-branch-point regular surface cover $N\to {\mathbb{P}}^1$.

Right multiplication by inverses defines a left action of $T$, commutes with the monodromy gluing and extends holomorphically over the filled points. Average a compatible metric over this finite group to obtain a metric for which $T$ acts by Riemannian isometries. When the genus is at least two, the unique compatible curvature-minus-one metric from the uniformization theorem is an invariant choice. Away from the three marked fibers the action is free. The nontrivial stabilizer subgroups are the conjugates of $⟨A⟩$, $⟨B⟩$ and $⟨AB⟩$, of orders $m,n,r$, omitting order-one groups.

There are $|T|/m$, $|T|/n$, $|T|/r$ points over the corresponding branch points. The Riemann-Hurwitz formula givesThis construction also covers the spherical and toroidal cases. If one monodromy has order one it is simply an unbranched marked fiber, with no nontrivial stabilizer.

For the supplied coset actions, faithfulness gives $m=n=3$. Compose permutations with the rightmost factor acting first. The two products $AB$ have cycle decompositionsThus $r=6$. The index is twelve and each subgroup has order eight, so $|T|=96$. ConsequentlyAll cycles of $A,B,AB$ on each coset set have the full orders $3,3,6$. No nontrivial power of any of these branch monodromies fixes a coset. The fixed-coset criterion means that each $U_i$ has trivial intersection with every conjugate of each cyclic branch stabilizer. Since these are all stabilizers on $N$, both $U_i$ act freely. Their quotient surfaces are therefore smooth closed surfaces, not cone-point orbifolds. Sunada theorem supplies isospectrality, and the unbranched Euler characteristic formula givesChoose the hyperbolic metric for these surfaces and their common genus-nine cover.

In this natural construction the two genus-two surfaces are isometric, by an orientation-reversing map. Define the sheet permutationDirect composition verifiesComplex conjugation of the base sphere fixes its three real branch points and reverses the two chosen generating loops, so their monodromies become the inverses. These identities supply a lift, after relabeling sheets by $P$, to an anticonformal map between the two branched covers. It extends across the branch points by the local power charts. An anticonformal map between closed hyperbolic Riemann surfaces is a Riemannian isometry, by uniqueness of the compatible hyperbolic metric.

Equivalently, take the hyperbolic triangle group of signature $(3,3,6)$. Reflection in the side joining the two order-three vertices conjugates its two vertex rotations to their inverses. The displayed sheet permutation identifies the reflected subgroup for the first covering with the subgroup for the second, up to an irrelevant change of base sheet. This is the orientation-reversing equivalence of branched-cover monodromy. The exhibited isometry is anticonformal, so the conclusion is about Riemannian isometry, without asserting that this map is holomorphic. Merely knowing the subgroups were almost conjugate would not have settled isometry; the explicit inverse-monodromy symmetry does. Nor does the data force this symmetry for every arbitrarily chosen invariant deformation of the cover's metric: it is present for the constructed hyperbolic metric.

Place the cube's vertices at $(\pm 1,\pm 1,\pm 1)$. Its full symmetry group of a cube consists of signed permutation matrices. To see completeness, a cube symmetry fixes its center of a group and permutes the six face normals, so it sends coordinate axes to signed coordinate axes; conversely every such matrix preserves the cube. Coordinate permutations and sign changes take any edge to any other edge, so this is a transitive group action on edges.

Choose the edge $e=\{(1,1,z):-1\le z\le 1\}$. A symmetry stabilizing it must preserve its axis direction and its midpoint $(1,1,0)$. It can swap the first two coordinates and independently reverse the third coordinate, but cannot change the signs of the two fixed coordinates. HenceThere are twelve edges and four elements of the stabilizer subgroup. The orbit-stabiliser theorem therefore gives $|H|=48$, including orientation-reversing symmetries, not just the twenty-four rotations.

The action defines a group homomorphism $H\to S_{12}$. If an element fixes every edge as a set, it fixes every vertex, since each vertex is the unique common point of its three incident edges. A cube isometry fixing all vertices is the identity. Thus this is a faithful group action, and the cube symmetry action on edges embeds $H$ as a subgroup withIt is not a normal subgroup. Central inversion $\mathbf{x}\mapsto -\mathbf{x}$ belongs to $H$ and swaps the twelve edges in six opposite pairs. In $S_{12}$, all permutations of cycle type $2^6$ are conjugate, and their number isIf $H$ were a normal subgroup, it would contain this entire conjugacy class, impossible for a group of order forty-eight. This avoids relying on a classification of normal subgroups of the symmetric group.

For $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in the special linear group over a finite field, define the Möbius transformation $M\cdot x=(ax+b)/(cx+d)$ when the denominator is nonzero, and set $M\cdot x=\infty$ when it is zero. At the extra point, $M\cdot \infty =a/c$ for $c\ne 0$, and $M\cdot \infty =\infty$ for $c=0$. Determinant one prevents simultaneous zero numerator and denominator.

The orbit-stabiliser theorem states $|G\cdot x|=[G:G_x]$, and for a finite group $|G|=|G\cdot x||G_x|$. The matrix $\left(\begin{array}{cc}t& -1\\ 1& 0\end{array}\right)$ sends infinity to any chosen $t\in {\mathbb{F}}_p$, so the orbit of infinity is the whole projective line, of size $p+1$. Its stabilizer subgroup iswhich has $p(p-1)$ elements. ConsequentlyThe SL2 action on a finite projective line need not be faithful: scalar matrices can fix every projective point. Faithfulness is not needed for this orbit calculation.

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 $|G^{+}|=(n+1)|G|$.

Here is the double-coset criterion for a one-point extension. Let $H=G_{\alpha }$, and suppose $x$ swaps $\omega$ and $\alpha$. Then $⟨G,x⟩$ is a one-point extension if and only ifFor necessity, in an extension $H$ fixes both $\omega$ and $\alpha$, so $x$ normalizes it and $x^2$ lies in it. Since $G$ is transitive on $\Omega$, the extension has exactly two double cosets relative to $G$: $G$ and $GxG$. For $g\notin H$, $xgx$ moves $\omega$ into $\Omega$ and is in the latter double coset.

For sufficiency, the displayed conditions make $G\cup GxG$ closed under multiplication. Products with middle element in $H$ reduce using $xhx=(xh{x}^{-1})x^2\in G$; those with middle element outside $H$ remain in $GxG$. A finite nonempty multiplication-closed set of permutations containing the identity is a group. It contains $G$ and $x$, hence equals $⟨G,x⟩$. Every element in $GxG$ moves $\omega$, while $G$ fixes it, giving the required stabilizer subgroup. The group is transitive because $G$ is transitive on $\Omega$ and $x$ moves the additional point.

For $1\le t\le |\Omega |$, a group is sharply t-transitive when any two ordered $t$-tuples of distinct points are related by exactly one group element. Equivalently its action on the set of such tuples is regular. In the finite caseand the stabilizer subgroup of an ordered $t$-tuple is trivial. The condition includes both existence and uniqueness, not just transitivity.

Put $n=|\Omega |\ge 2$, and let $H=G_{\alpha }$. Sharp two-transitivity gives $|G|=n(n-1)$ and $|H|=n-1$. A nonidentity element fixes at most one point. Counting the nonidentity elements in the $n$ stabilizer subgroups shows that there are $n-1$ fixed-point-free elements. Let $N$ be this set together with the identity. We first prove it is a normal subgroup, rather than presuming that fixed-point-free elements are closed under multiplication.

For $n=2$, $G=C_2$ and $N=G$ directly. Otherwise use complex characters of a finite group. Let $\pi$ be the permutation character and $\psi =\pi -1$, the character of the permutation representation with its constant line removed. For every nontrivial irreducible character $\theta$ of $H$, form the virtual characterIts values are $\theta (1)$ at the identity and at every fixed-point-free element. At an element with one fixed point, conjugate it to $h\in H\setminus \{1\}$; induction gives ${\phi }_{\theta }(g)=\theta (h)$, because there is exactly one fixed coset.

The identity and the $n-1$ fixed-point-free elements together contribute $n\theta (1)\overline{\eta (1)}$ to the inner product. The remaining elements are partitioned into the nonidentity parts of the $n$ stabilizer subgroups. Hence character orthogonality givesA virtual character of norm one is plus or minus an irreducible character: its coefficients in the irreducible-character basis are integers whose squares sum to one. Its positive degree $\theta (1)$ selects the plus sign. Thus each ${\phi }_{\theta }$ is an actual irreducible character.

For a group representation, $\chi (g)=\chi (1)$ holds exactly on its kernel: make the representation unitary and compare the sum of its unit-modulus eigenvalues with its dimension. All of $N$ therefore lies in the intersection of the kernels of the ${\phi }_{\theta }$. Conversely, a nonidentity element fixing a point gives $h\ne 1$ in $H$. Some nontrivial irreducible character of $H$ has $\theta (h)\ne \theta (1)$, since otherwise the regular representation of $H$ would not vanish at $h$. ConsequentlyThis proves normality and subgroup closure. It has order $n$ and no nonidentity element fixing a point, so it is a regular permutation subgroup.

Now $H$ acts transitively by conjugation on $N\setminus \{1\}$: identify an element of $N$ with its image of $\alpha$ and use transitivity of $H$ on the remaining points. Thus all nonidentity elements of $N$ have the same order. Taking a suitable power of one element shows this common order is a prime $p$. By Cauchy's theorem no other prime divides $|N|$, so $N$ is a $p$-group. Its nontrivial center is $H$-invariant, so transitivity forces the center to be all of $N$. Therefore $N$ is elementary abelian of order $p^d=n$.

Since $|H|=p^d-1$ is prime to $p$, $N$ is the unique Sylow $p$-subgroup of $G$. Uniqueness makes it characteristic under every group automorphism. We have provedThe character argument supplies the regular kernel of a finite sharply two-transitive group; the final Sylow argument establishes the stronger characteristic assertion.

Iwasawa's simplicity lemma states the following. Suppose $G$ acts faithfully and primitively on a set, and a stabilizer subgroup $H$ has an abelian normal subgroup $A$ whose $G$-conjugates generate $G$. Then every nontrivial normal subgroup $N$ contains $G^{\prime }$. In particular, if $G$ is nontrivial and perfect, then $G$ is simple.

Indeed a nontrivial normal subgroup in a faithful primitive action is transitive, so $G=NH$. Since $H$ normalizes $A$, all conjugates of $A$ have the same image in $G/N$. Those images generate the quotient, which is therefore abelian. This gives $G^{\prime }\le N$ and proves the stated conclusion.

Use the projective-line realization from part (ii). The subgroup $K$ of translations is abelian and normal in the stabilizer subgroup $G$ of infinity. We verify both remaining conditions of Iwasawa's simplicity lemma, instead of concluding simplicity from transitivity alone.

Let $R$ be generated by all conjugates of $K$ in $G^{+}$. It contains the matricesHere matrices act by fractional linear transformations. For $t\ne 0$,In particular $w(1)=x\in R$, and $w(t)w(1)=\mathrm{diag}(t,t^{-1})$ acts as multiplication by $t^2$. Squaring is a bijection of $F^{\ast }$, so every $M_s$ belongs to $R$. Hence $R$ contains $G$ and $x$, and $\boxed{R=G^{+}}$.

Choose $s\ne 1$. The commutator, with convention $[u,v]=uv{u}^{-1}{v}^{-1}$, isAs $t$ varies, this gives all translations. Thus $(G^{+}{)}^{\prime }$ contains $K$, and normality makes it contain every conjugate of $K$. Since those generate $G^{+}$, the group is a perfect group.

A sharply three-transitive action on nine points is a primitive group action, and this permutation action is a faithful group action. All Iwasawa hypotheses now hold, soIt may be identified with ${\mathrm{PSL}}_2(8)={\mathrm{PGL}}_2(8)$: the generators are fractional linear transformations and $|{\mathrm{PGL}}_2(8)|=8(8^2-1)=504$, while all nonzero field elements are squares. The simplicity proof above does not rely on assuming simplicity of that named family.

Adjoin the label $9=\infty$. Inversion on the projective line, with zero and infinity interchanged, isThe label $2=1\in F$ is fixed. Set $H=G_1=\{M_s:z\mapsto sz\}$, where the subscript one denotes the original point label, namely field zero. Then $x^2=1\in H$ and $x{M}_{s}x=M_{s^{-1}}$, so $x$ normalizes $H$.

For $g(z)=sz+t$ outside $H$, we have $t\ne 0$. On the projective line,The equality uses characteristic two and is valid as an equality of fractional linear transformations, including poles and infinity. ThusAll conditions of the double-coset criterion for a one-point extension hold. Therefore $G^{+}=⟨G,x⟩$ is a one-point extension, with $|G^{+}|=9\cdot 56=504$. Its stabilizer subgroup is sharply two-transitive, so its action on nine points is sharply three-transitive.

The symplectic group consists of the invertible linear maps preserving the alternating bilinear form:We use row-vector action in this question, which is the convention compatible with its printed upper-triangular flag stabilizer subgroup. Thus matrices preserve a form matrix $J$ by $gJ{g}^T=J$.

Count ordered symplectic bases. There are $q^{2m}-1$ choices for the first nonzero vector $e_1$. Nondegeneracy makes the equation $(e_1,f_1)=1$ a nonzero linear-functional equation, with $q^{2m-1}$ solutions. Their span is a nondegenerate plane; its orthogonal complement is symplectic of dimension $2m-2$. Repeating there givesEach symplectic basis is the image of a fixed one under exactly one form-preserving map, justifying the count as a group order. If $q=p^a$, every factor $q^{2i}-1$ is prime to $p$, so the exact $p$-part is $q^{m^2}$.

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 SU(2) group consists of complex $2\times 2$ matrices $U$ with $U^{†}U=I$ and $\det U=1$. The SO(3) group consists of real $3\times 3$ matrices $R$ with $R^{T}R=I$ and $\det R=1$. Differentiate these equations along a path through the identity. This givesConversely the matrix exponential of each displayed infinitesimal matrix satisfies the corresponding group equations, so these are exactly the Lie algebras, with bracket the matrix commutator.

Using the Pauli matrices, put $t_a=-i{\sigma }_a/2$. They form a real basis of the SU(2) Lie algebra. The Pauli matrix commutator identity gives $[t_a,t_b]={ϵ}_{abc}{t}_c$. Define $J_a$ on ${\mathbb{R}}^3$ by $J_{a}v=e_a\times v$. These are a basis of the SO(3) Lie algebra; the vector triple-product identity gives $[J_a,J_b]={ϵ}_{abc}{J}_c$. Thusis a real linear bijection preserving the bracket. This is the SU(2)-SO(3) Lie algebra isomorphism. It is a Lie-algebra isomorphism, not a group isomorphism: the Adjoint double cover from SU(2) to SO(3) has kernel $\{I,-I\}$.

The SU(3) group is defined similarly by $U^{†}U=I$ and $\det U=1$ on complex $3\times 3$ matrices. One SU(2) subgroup is $\{\mathrm{diag}(V,1):V\in SU(2)\}$. The real orthogonal matrices with determinant one form an SO(3) group subgroup, since real orthogonality is also complex unitarity.

Under the defining group action, the group orbit of $e_1$ is exactly the unit sphere in ${\mathbb{C}}^3$:Unitarity proves containment. Conversely, extend a unit vector $z$ to an orthonormal basis and use it as the first column of a unitary matrix. Multiplying the last column by the inverse of its determinant makes that determinant one without changing the first column. The isotropy group of $e_1$ must also preserve its orthogonal complement, and therefore isThis proves the unit sphere orbit of the defining special unitary action, including $S^5\cong SU(3)/SU(2)$.

For the complex quadric, write $z=x+iy$. Its defining relation separates intoThe Hermitian norm is then $z^{†}z=1+2|y{|}^2$, which is not constant on this set. For example, $e_1$ and $\sqrt{2}{e}_1+i{e}_2$ both satisfy the complex bilinear relation but have Hermitian norms one and three. Since the SU(3) group preserves that norm, the quadric cannot be a single SU(3) group orbit. In fact it is not even invariant under the whole group: $\mathrm{diag}(e^{it},e^{-it},1)e_1$ fails the bilinear relation when $e^{2it}\ne 1$.

The real SO(3) group does preserve the quadric, acting simultaneously on $x$ and $y$. Its group orbits are classified completely by $r=|y|\ge 0$. For $r=0$, $y=0$ and $x$ is a real unit vector, giving the group orbit $S^2$ and stabilizer subgroup $SO(2)$. For $r>0$, the vectors $x/\sqrt{1+r^2}$ and $y/r$ are orthonormal. Adjoining their cross product gives an oriented orthonormal frame. There is a unique rotation taking the standard frame to this one; hence the action is transitive at fixed $r$ and the stabilizer subgroup is trivial. Thus the real rotation orbits on a complex unit quadric areIn particular the real subgroup does not act transitively on the whole quadric. The parametrization $x=\sqrt{1+|y{|}^2}n$ with $n\in S^2$ and $y\perp n$ also identifies the quadric, as a real manifold, with the tangent bundle of $S^2$.