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A mathematician associated with foundational work in analysis, complex analysis, permutation groups and elasticity.^[ref]

For a finite group $G$, choose one representative $x_i$ from each conjugacy class with more than one element. The class equation iswhere $Z(G)$ is the center of a group and $C_G(x_i)$ is the centralizer of $x_i$. Each central element contributes its own one-element conjugacy class. Every remaining conjugacy class contributes its size, given by the orbit-stabilizer theorem. The sum counts each element exactly once. See McKernan, Group actions and automorphisms, lemma 23.6.

For conjugacy classes of the symmetric group on three letters, take $x_1=(12)$ and $x_2=(123)$. Their centralizers have sizes two and three, and the center of a group has size one, givingFor an abelian group, every centralizer is the whole group, every conjugacy class has size one, and the sum over noncentral representatives is empty.

The power of this counting identity is that the sizes of conjugacy classes constrain group structure. For a nontrivial finite p-group with group order $p^n$, every noncentral conjugacy class has size a positive power of $p$, by Lagrange's theorem. The class equation therefore gives $|Z(G)|\equiv 0(\mathrm{mod}p)$. Since the identity element lies in the center of a group, the center of a group must have at least $p$ elements. This obtains an element commuting with the entire group without constructing one. See McKernan, lemma 23.7.

Historically, these ideas belong to the nineteenth-century development of permutation groups and normal subgroups. Milne's historical opening records Galois's introduction of normal subgroups in 1832 and Jordan's systematic treatment of permutation groups in 1870. This is context for the conjugation action and the class equation, rather than a claim that either author first stated the modern formula.^[ref]

A group acts on the left cosets of a subgroup by left multiplication; its kernel is the core of that subgroup.

Let $H\le G$ be a subgroup. The notation $G/H$ here means the set of left cosets $\{aH:a\in G\}$, even when $H$ is not a normal subgroup. DefineThis gives a group action without needing a quotient group structure on $G/H$.^[ref]

The representative check is essential: if $aH=bH$, then $a=bh$ for some $h\in H$, and $(ga)H=(gbh)H=(gb)H$. Thus equal input cosets give equal output cosets. The identity element fixes every coset, and associativity gives $g_1\cdot (g_2\cdot aH)=(g_1{g}_2)\cdot aH$. Each induced map is a permutation, with inverse induced by $g^{-1}$. This is a transitive group action because $b{a}^{-1}$ sends $aH$ to $bH$.

The stabilizer subgroup of $H$ is $H$ itself. At a general coset, the stabilizer subgroup and the group kernel areConsequently the coset action is a faithful group action exactly when the subgroup core is trivial.^[ref]

Indeed, $g\cdot (aH)=aH$ exactly when $a^{-1}ga\in H$, or $g\in aH{a}^{-1}$. Intersecting these membership conditions over all cosets gives the group kernel. Conjugation permutes the subgroups in this intersection, making it a normal subgroup of $G$ contained in $H$. Conversely, any normal subgroup of $G$ contained in $H$ lies in every $aH{a}^{-1}$. This explains why the subgroup core is the largest such normal subgroup: it is exactly the portion of $H$ invisible at every label, rather than merely at the label $H$.

For a normal subgroup $H$, every $aH{a}^{-1}$ equals $H$, so $\ker {\rho }_H=H$. For $H=\{e\}$, the construction is the regular group action behind Cayley's theorem. Between these extremes, a nontrivial subgroup with trivial subgroup core can give a faithful group action on fewer labels. If the index of a subgroup is finite, $[G:H]=n$, the first isomorphism theorem givesThus a subgroup of small index of a subgroup exposes a small permutation group associated with $G$. The two examples below and in nonfaithful coset action of the symmetric group on three letters show exactly which information can survive.

For a permutation group, $c_r(g)$ is the number of length-$r$ cycles of $g$. The cycle index records the average cycle structure. On coloring a finite underlying set, a coloring fixed by $g$ is constant on each cycle. Replacing $t_r$ by the power-sum symmetric polynomial $p_r(x)$ therefore gives the pattern inventory by the weighted Burnside lemma.

An ordered permutation group is order n-transitive if it can send any increasing $n$-tuple to any other increasing $n$-tuple. This compares tuples with the same relative order, unlike unrestricted n-transitive group actions. In the usual transitive chain setting, order two-transitivity implies every finite order-transitivity for a lattice-ordered permutation group.

Always true for a nonempty finite constraint domain, as conventionally assumed. Write $q=|D|$ and identify its values with $\{0,\dots ,q-1\}$. For each satisfying assignment $(x_1,\dots ,x_n)$, simultaneously replace every value byThis permutation of the common constraint domain preserves every equality and inequality, hence every all-different constraint. The resulting $q$ assignments are distinct: equality of even one coordinate before and after a shift forces $a=0(\mathrm{mod}q)$. There is at least one variable, since $m>0$ and $k\ge 2$.

The cyclic group of shifts therefore partitions the solution set into disjoint orbits of size $q$. It follows thatThe conclusion includes an unsatisfiable instance, whose solution count is zero. Full permutation group orbits need not have size $q!$, because unused values can be permuted without changing an assignment; the cyclic-shift argument avoids that mistake.

If an empty constraint domain is permitted, there are zero assignments here, but the wording “divisible by $|D|$” then asks about divisibility by zero, which is ordinarily left undefined. The asserted divisibility uses the standard nonempty-domain convention, and the no-solution conclusion still holds for that degenerate input.

Heinrich Weber's 1882 paper concerns prime numbers represented by quadratic forms. Its definition abstracts a finite group from the nature of its elements: what matters is a closed binary operation, associativity, and left and right cancellation laws. Here degree $h$ means the number of elements, now called the group order; it is not the number of letters acted on by a permutation group. This matters when relating the passage to earlier work on substitutions. The definition fits the use of groups of classes of quadratic forms, where the elements are already mathematical objects more elaborate than individual permutations.^[ref][ref]

The apparent omission of an identity element and inverse elements is not fatal. Every nonempty finite cancellative semigroup is a group. Here is a proof identifying exactly where the assumption of a finite set enters. For fixed $a\in G$, the maps $L_a(x)=ax$ and $R_a(x)=xa$ are injective by cancellation. An injective function from a finite set to itself is surjective, so choose $e$ with $ae=a$. For every $x$,and left cancellation gives $ex=x$. To show $e$ is also a right identity element, useRight cancellation gives $xe=x$. Thus the same $e$ works on both sides for every $x$; it has not merely been chosen as an element fixing $a$.

Surjectivity now gives $b,c$ with $ab=e$ and $ca=e$. By associativity,So $b$ is a two-sided inverse element of $a$. Since $a$ was arbitrary, the usual group axioms follow. The nonempty hypothesis is tacit in a system of $h$ elements intended to have positive $h$: the empty system would satisfy the displayed universal conditions vacuously but would have no identity element.

The infinite case exposes the force of the qualification. The positive integers under addition are closed, associative and cancellative, but have no additive identity element. Even the nonnegative integers have an identity element without additive inverse elements for positive elements. The failure is precisely that injective functions of an infinite set to itself need not be surjective. Heinrich Weber's conditions are therefore not an unrestricted replacement for the modern group axioms.

Historically the abstraction is important because it allows arguments to be applied across different kinds of elements. An assumption of commutativity is absent: class-group applications do not make commutativity part of the general definition. For example, the symmetric group $S_3$ meets the conditions but is noncommutative. The definition also does not require a prior presentation as permutations, although Cayley's theorem provides such a representation. The passage documents the separation of an algebraic structure from any one representation of it, with finiteness carrying logical work that later axiom systems usually assign to explicit identity element and inverse element axioms.

Augustin-Louis Cauchy's importance lies in the interaction of foundations, methods and applications. He made the justification of limits, series and integrals a programme for teaching and research; developed complex analysis into a source of powerful general methods; and contributed to permutation groups, geometry and continuum mechanics. His work is best assessed through particular achievements and their limitations, rather than by treating every theorem now bearing his name as a modern statement he already possessed. His engineering training and subsequent mathematical career also make a rigid separation of pure and applied work historically unhelpful.^[ref]

The Cours d'analyse by Augustin-Louis Cauchy (1821) makes convergence a condition requiring argument. A series is studied through its partial sums; a limit describes a variable approaching a fixed value; an infinitesimal is a variable tending to zero, rather than simply a mysterious fixed quantity smaller than every positive number. The introduction explicitly challenges unrestricted transfers of rules from finite expressions to infinite ones. This is not an abandonment of calculation: it seeks to identify when calculations are valid.^[ref]

For example, the Cauchy criterion for series asks whether tails can all be made small:This is a modern formulation of the convergence requirement, with the completeness of the real numbers understood. Merely requiring $a_n\to 0$ is weaker. In the harmonic series, grouping the terms from $2^j+1$ to $2^{j+1}$ gives a block of at least $1/2$, so its tails cannot satisfy the criterion. In contrast, the root test compares a tail with a geometric series: if $\mathrm{limsup}|a_n{|}^{1/n}<q<1$, then eventually $|a_n|\le q^n$, which proves absolute convergence. Such arguments explain the usefulness of convergence tests instead of reducing Augustin-Louis Cauchy's contribution to a vocabulary of definitions.

His Résumé des leçons sur le calcul infinitésimal by Augustin-Louis Cauchy (1823) extends the programme to differentiation and integration. For a continuous function on a compact interval, an integral can be approached through sums of values times subinterval lengths. A modern justification uses uniform continuity: for a partition of mesh at most $\delta$, two choices of sample points give sums differing by at most $(b-a){\omega }_f(\delta )$, where the modulus of continuity ${\omega }_f(\delta )$ tends to zero. Refinement then establishes the common limit. This reconstruction makes clear why control of error, rather than a geometrical picture of area alone, is decisive. It does not attribute the later full Riemann integral theory for discontinuous functions to the 1823 text.^[ref]

The achievement has limits. The 1821 Sum theorem of Cauchy asserts continuity of sums of convergent series of continuous functions in terms that, under a pointwise reading, are insufficient. The original proof treats the tail as negligible without establishing the uniform control needed near the point concerned. A modern diagnostic is $f_n(x)=x^n$ on $[0,1]$: every $f_n$ is continuous, but its pointwise limit is zero for $x<1$ and one at $1$. Thus a proof must distinguish pointwise convergence from uniform convergence. Augustin-Louis Cauchy's use of variable and infinitesimal language deserves historical interpretation, but it does not license silently importing modern hypotheses into his printed theorem.^[ref]

In complex analysis, the Mémoire sur les intégrales définies, prises entre des limites imaginaires by Augustin-Louis Cauchy (1825) exemplifies a new productive setting for integration. A contour integral depends on a path in the complex plane; deforming that path and accounting for isolated singularities can evaluate otherwise difficult definite integrals. In modern form, Cauchy's integral theorem applies to a holomorphic function on a suitable domain and Cauchy integral formula states, for a circle enclosing $z$ and lying with its interior in that domain,The importance is structural: local values and derivatives are determined by data on a surrounding path. Historical proofs and hypotheses should be distinguished from later general forms of the theorem.^[ref]

A short modern consequence measures the power of this method. For an entire function bounded by $M$, the derivative version of Cauchy integral formula on a circle of radius $R$ gives $|f^{\prime }(z)|\le M/R$. Since $R$ is arbitrary, $f^{\prime }=0$, proving Liouville's theorem. If a nonconstant polynomial $P$ had no root, then $1/P$ would be an entire function, bounded outside a large disc by the leading term and inside the disc by compactness. Liouville's theorem would force it to be constant, a contradiction. This proves the fundamental theorem of algebra. The chain is an assessment of the reach of the methods, not a claim that Augustin-Louis Cauchy alone originated every named result.

Augustin-Louis Cauchy's algebra concerns concrete permutations as well as analytical formulas. His 1845 investigations developed the theory of substitutions; Cauchy theorem for groups says that a prime number dividing a finite group order occurs as an element order. One can appreciate its generality from a modern argument: rotate the coordinates of the tuples $(g_1,\dots ,g_p)$ satisfying $g_1\cdots g_p=e$. Rotation preserves this condition, since $g_2\cdots g_p{g}_1=g_1^{-1}(g_1\cdots g_p)g_1=e$. There are $|G{|}^{p-1}$ such tuples; nonfixed rotation orbits have $p$ elements, and fixed tuples correspond to $g^p=e$. When $p\mid |G|$, the number of fixed tuples is divisible by $p$, so besides $e$ there is an element of element order $p$. This is explanatory mathematics, not Augustin-Louis Cauchy's original proof. The dating itself needs care: Peter Michael Neumann's archival argument places the major burst of work in 1845 despite an 1844 date traditionally attached to a volume.^[ref]

Finally, Augustin-Louis Cauchy's work on elasticity shows the strength of making a local mathematical object out of physical interactions. Instead of describing only a force perpendicular to a surface, one studies the traction on a plane through a point, with its direction dependent on the plane's unit normal. In modern notation the relation is $\mathbf{t}(\mathbf{n})=\mathbf{\sigma }\mathbf{n}$, where the Cauchy stress tensor contains the necessary directional information. Together with balance of linear momentum, this yields equations for the internal motion of a body. The modern tensor notation is later explanatory language; the conceptual contribution is the reduction of the forces on arbitrary internal planes to a finite collection of local components.^[ref]

Augustin-Louis Cauchy's lasting contribution was a powerful research programme, not a finished foundation. His insistence on justified passage to a limit reshaped analysis, while complex integration, substitutions and local stresses opened methods whose value extended beyond their first applications. The inadequacy of a pointwise convergence hypothesis, the need to state domains in complex analysis, and the later construction of the real numbers show that nineteenth-century rigour remained a collective and revisable achievement. These qualifications strengthen an assessment of his influence by identifying what later mathematicians could extend and repair.

The independence statements are relative consistency statements. We construct models from an ambient model of ZFC; the construction is an interpretation and does not assume that the ambient model is transitive. Foundation is already true in the ordinary ambient membership structure, so its positive consistency direction is immediate.

For the negative direction, construct an extensional cumulative universe over Quine atoms. Let $A$ be a countably infinite set of formal symbols $q_a$. Give each $q_a$ the extension $\{q_a\}$. For a set $X$ of already constructed objects, let $S(X)$ be its unique representative: set $S(\{q_a\})=q_a$, and otherwise use a fresh tagged code $(1,X)$. Atom codes have a different tag. Begin with $U_0=A$ and putSet $W={\bigcup }_{\alpha }{U}_{\alpha }$. The interpreted membership relation $E$ gives $S(X)$ exactly the members $X$, and gives $q_a$ exactly the member $q_a$. This convention matters: creating a second representative for $\{q_a\}$ would violate extensionality. Every object is uniquely determined by its extension, so $E$ satisfies full extensionality. Every set-sized collection $X$ of objects has bounded construction rank, by ambient replacement, and hence has its representative $S(X)$ in $W$.

Here are the remaining axiom verifications. Empty set and pairing use $S(\varnothing )$ and $S(\{x,y\})$. If $E(y)=X$, its interpreted union is $S({\bigcup }_{x\in X}E(x))$. Its interpreted power set issince every subset extension has a unique representative. The pure, atom-free finite hierarchy supplies a copy of $\omega$ and an inductive set. For separation, translate an $E$-formula into the ambient language, with quantifiers restricted to the definable class $W$; ambient separation gives the appropriate subset of $X$, and $S$ represents it. For replacement, uniqueness in the interpreted functional relation gives an ambient definable functional image of the set $X$. Ambient replacement makes that image set-sized, and $S$ again represents it. These arguments also establish that all the operations used have bounded construction rank.

Ambient choice gives choice in $W$: choose a member from each nonempty extension in a set-sized family and represent the graph, using interpreted ordered pairs. But $q_{a}E{q}_a$, and the nonempty set with extension $\{q_a\}$ has no $E$-minimal member. ThusTogether with the ordinary model, this proves independence of foundation from the other ZF axioms, even when choice is retained.

To extend the construction to failure of choice, let the full permutation group of $A$ act on $W$ byThe singleton convention is equivariant, so this is an automorphism of $(W,E)$. An object has finite support if some finite $F\subseteq A$ has the property that every permutation fixing $F$ pointwise fixes the object. Retain objects all of whose membership descendants have finite support, stopping the hereditary recursion at the atomic self-loops. This gives the hereditarily finite-supported Quine-atom model $\mathcal{H}$; each $q_a$ belongs to it, with support $\{q_a\}$.

Extensionality persists because every retained object's members are retained. Empty set, pairing, infinity and union preserve hereditary finite support. For separation on a retained set $x$, the union of finite supports for $x$ and the formula's parameters supports the separated extension: permutations fixing those supports preserve satisfaction in $\mathcal{H}$. All its members are retained, so its representative is retained. For replacement, the same support fixes the functional range, because an automorphism must take the unique output at an input to the unique output at its transported input. Ambient replacement bounds that range; all its members are in $\mathcal{H}$, so its representative belongs to $\mathcal{H}$. Finally, the collection of all retained subset representatives of $x$ is an ambient set, contained in the full interpreted power set of $x$. A support of $x$ permutes this collection onto itself, and its members are retained. Its representative is therefore the internal power set in $\mathcal{H}$. This verifies every ZF axiom except foundation; foundation still fails at each $q_a$.

The atom set $A$ and the family $[A{]}^2$ of its two-element subsets have empty support and are retained. Suppose a choice function $f$ on $[A{]}^2$ were retained, with finite support $F$. Choose distinct $q_a,q_b$ outside $F$ and transpose them. The transposition fixes $f$ and fixes the unordered pair $\{q_a,q_b\}$, but moves whichever of its two members $f$ selects. Equivariance would require that selected member to be fixed, a contradiction. HenceThe full universe $W$ gives the corresponding positive-choice model. Choice is therefore independent of ZF without foundation. Unlike a permutation model with ordinary urelements, this model retains full pure-set extensionality: the atomic objects are self-membered sets, not empty objects exempted from extensionality.

The constructible-universe interpretation gives $\mathrm{Con}(\mathrm{ZF})\Rightarrow \mathrm{Con}(\mathrm{ZFC})$. Thus the consistency assumptions in both displayed conclusions can equivalently be taken to be $\mathrm{Con}(\mathrm{ZF})$; no additional consistency strength is being used.

The isomorphic pair is $A_8\cong PS{L}_4(2)$; $PS{L}_3(4)$ is different. All three have order $20160$, so order alone does not settle the question; explicitly $|PS{L}_3(4)|=(64-1)(64-4)(64-16)/(3\cdot 3)=20160$. We give a concrete realization and an element-order distinction.

On $W={\bigwedge }^2{\mathbb{F}}_2^4$, the Pfaffian isIt is a nonsingular plus-type quadratic form: the three displayed pairs give a hyperbolic decomposition. Change of basis multiplies the Pfaffian by the determinant, which is always one in $G{L}_4(2)$. Hence the exterior-square realization of PSL4 over F2 embeds $G{L}_4(2)$ into $O_6^{+}(2)$. The kernel of a group homomorphism is trivial: an element acting trivially on the exterior square fixes every decomposable two-space, then every one-space as an intersection of two-spaces, and hence is a scalar; over ${\mathbb{F}}_2$ this scalar is one.

For a second model let $E$ be the even-weight vector subspace of ${\mathbb{F}}_2^8$ and quotient by the all-one vector. The dot product descends to a nondegenerate alternating bilinear form on the six-dimensional quotient, andis well defined, since complementing an even subset changes half its weight from $w/2$ to $4-w/2$. It has $(1+70+1)/2=36$ zero vectors, so is plus type. Coordinate permutations give a faithful $S_8$ action: fixing the quotient fixes each weight-two subset, because its complement has weight six, and fixing all two-subsets forces the identity permutation.

From the independently derived symplectic order and quadratic orbit count below,Thus $O_6^{+}(2)\cong S_8$ and the image of $G{L}_4(2)=S{L}_4(2)=PS{L}_4(2)$ is its index-two subgroup $A_8$. The index-two subgroup is unique: a nontrivial group homomorphism $S_8\to C_2$ sends every transposition to the same nonidentity element and is the sign map.

To separate $PS{L}_3(4)$, observe that $A_8$ contains $(12345)(678)$ of order 15. An odd-order projective element of $PS{L}_3(4)$ lifts to a semisimple linear operator, since its order upstairs divides three times an odd order in characteristic two. The possible eigenspace degree patterns are $1+1+1$, $2+1$ and $3$. In the split case all eigenvalues have order dividing three, so the projective element has no order 15. In the $2+1$ case the eigenvalues are $\lambda ,{\lambda }^4,{\lambda }^{-5}$ with $\lambda \in {\mathbb{F}}_{16}^{\times }$; its fifth power is scalar because ${\lambda }^{15}=1$, so its projective order divides five. In the irreducible case the determinant-one torus has order $4^2+4+1=21$ and quotienting its scalar subgroup of order three gives projective order dividing seven. There is therefore no element of order 15 in $PS{L}_3(4)$.

Several small-parameter coincidences connect the finite classical groups to permutation groups. The projective-line action immediately gives $PS{L}_2(2)\cong S_3$, $PS{L}_2(3)\cong A_4$, $PG{L}_2(3)\cong S_4$ and $PS{L}_2(4)\cong A_5$ by faithful actions of degrees three, four and five and their orders. Also $PS{L}_2(5)\cong A_5$: its involutions have Klein-four centralizers, giving five Sylow subgroups; their conjugation action is faithful and embeds its order-60 simple group into $S_5$. Simplicity here can be checked from class sizes $1,15,20,12,12$: no proper nontrivial normal class union has order dividing 60. The same five-subgroup action extends to $PG{L}_2(5)$, with trivial kernel of a group homomorphism and order 120, giving $S_5$. Indeed its kernel intersects $PS{L}_2(5)$ trivially, so centralizes that normal subgroup. Commuting projectively with both upper and lower elementary unipotents forces a representing matrix to be scalar, so this kernel is trivial.

The less immediate coincidence $PS{L}_2(9)\cong A_6$ can be exhibited on the ten unordered triple partitions used in Question 4. Let $a=(123)$, $b=(456)$, $w=(12)(36)$, assign $\infty$ to $123\mid 456$ and $0$ to $126\mid 345$, and label the other partitions by applying $a^x{b}^y$ to the zero partition, with $t=x+iy\in {\mathbb{F}}_9$ and $i^2=-1$. Inspection of these nine partitions givesTranslations and inversion generate $PS{L}_2(9)$: conjugating the upper unitriangular matrices by inversion gives the lower ones, and elementary elimination generates $S{L}_2(9)$. Their projective action has order $9(9^2-1)/2=360$. These partition permutations come from even permutations of six letters, whose partition action is faithful as proved below; hence they fill $A_6$.

Finally, on the even-weight module of ${\mathbb{F}}_2^6$ modulo its all-one vector, coordinate permutations preserve a nondegenerate alternating bilinear form in dimension four and act faithfully, again by the weight-two-subset argument. Both $S_6$ and $S{p}_4(2)$ have order 720, proving $S{p}_4(2)\cong S_6$, with commutator subgroup $A_6$. These examples illustrate both the need to quotient group centers and the low-rank exceptions to general simplicity. They do not make equal orders sufficient for group isomorphism, as the two order-20160 examples already demonstrate.

Using the printed convention that the projective points represent lines of binary quartics, the locus of GIT stable points isThe root at infinity is included. We determine stability by calculating every relevant algebraic one-parameter subgroup weight.

Use the natural action $(g\cdot f)(X,Y)=f(g^{-1}(X,Y))$. Every nontrivial algebraic one-parameter subgroup of $S{L}_2$ is conjugate to ${\lambda }_m(t)=\mathrm{diag}(t^m,t^{-m})$ for some integer $m>0$: its two integral weights of a representation sum to zero, and a basis of weight vectors can be rescaled to have determinant one. WriteIf $r=\min \{j:a_j\ne 0\}$, then $r$ is the multiplicity of a root at the projective point $[1:0]$: in the local coordinate $Y/X$, the polynomial begins with a nonzero multiple of $(Y/X{)}^r$.

The Hilbert-Mumford criterion for projective stability, with $\mu =-\min$(active weights), now givesBeing a GIT stable point requires this number to be strictly positive for every nontrivial subgroup. Conjugating the subgroup lets $[1:0]$ range over the entire projective line. Thus the root-multiplicity criterion for stable binary forms becomes $r_p(f)<2$ at every point $p$. A nonzero degree-four form over $\mathbb{C}$ has four projective roots counted with multiplicity, so this means precisely four distinct roots. This is equivalently nonvanishing of its discriminant.

The boundary cases show why a weak inequality would give the wrong answer. A triple or quadruple root can be moved to $[1:0]$, leaving only strictly positive weights; its affine representative tends to zero under that subgroup, so it is unstable. If all root multiplicities are at most two, all Hilbert-Mumford values are nonnegative and the point is a GIT semistable point. A double root gives a zero value for a suitable subgroup, so never represents a GIT stable point. More concretely, after putting that double root at $[1:0]$ one hasand ${\lambda }_1(t)\cdot f\to a{X}^2{Y}^2$. When there are three distinct roots, this limiting form, representing a GIT semistable point with two distinct roots, is outside the original group orbit, so that group orbit is not closed in the semistable locus. When there are two double roots, a change of coordinates gives $X^2{Y}^2$, which has the positive-dimensional diagonal torus as stabilizer. Neither case represents a GIT stable point.

For a form with four distinct roots, the projective stabilizer is finite: it permutes the four roots, and a Möbius transformation fixing three distinct points is the identity. Thus its image in $PG{L}_2$ injects into the finite permutation group on the roots. The kernel of a group homomorphism for $S{L}_2\to PG{L}_2$ is $\{\pm I\}$, also finite. This agrees with the finite-stabilizer condition in geometric invariant theory. The weight calculation already supplies the required strict stability condition for every conjugate subgroup.

Subgroups $U,V$ of a finite group $T$ have Gassmann equivalence whenSumming over the conjugacy classes gives $|U|=|V|$. Equivalently, their coset permutation representations have the same character of a representation. In particular, almost conjugate subgroups need not be conjugate.

Put $p=\mathrm{ord}A$, $q=\mathrm{ord}B$, and $r=\mathrm{ord}(AB)$. Begin with the sphere punctured at three points. Its fundamental group is $⟨a,b,c\mid abc=1⟩$, and the epimorphism sending $a,b,c$ to $A,B,(AB{)}^{-1}$ defines a connected regular $|T|$-sheeted covering space. Around the punctures the monodromies have cycles of lengths $p,q,r$. Fill each lifted puncture by a disk, using the local branched covering models $z\mapsto z^p$, $z\mapsto z^q$ and $z\mapsto z^r$. This produces a closed oriented surface $M$, with $T$ acting by homeomorphisms that are smooth in these local models. This construction works even if one of $p,q,r$ is one.

Triangulate the base sphere with the three marked points as vertices and two faces. Lifting this triangulation givesso the Euler characteristic isIn the hyperbolic case $1/p+1/q+1/r<1$, this is equivalently the triangle cover construction for Sunada surfaces. Take the hyperbolic triangle groupand its epimorphism $\theta :\Delta \to T$ with the same images. All finite-order elements of $\Delta$ are conjugate to powers of its three elliptic generators. Because their images have the same exact orders, $K=\ker \theta$ is torsion-free and $M=K\backslash {\mathbb{H}}^2$. It can be assembled from $2|T|$ hyperbolic triangles of angles $\pi /p,\pi /q,\pi /r$. Both examples below belong to this hyperbolic case.

For $M_1=U\backslash M$ and $M_2=V\backslash M$ to be smooth surfaces normally covered by $M$, the subgroup actions must be free. The precise condition is that neither subgroup meet any conjugate in $T$ of a nonidentity power of $A$, $B$ or $AB$. These are exactly the possible point stabilizers. Then $M\to M_1$ and $M\to M_2$ are regular coverings, with deck transformation groups $U,V$ respectively. No normality of $U,V$ in $T$ is needed: normality would concern the further covering of the base orbifold, not these coverings from $M$. For a free subgroup $W$,

Choose any $T$-invariant smooth Riemannian metric on $M$, for example by averaging an arbitrary metric. It descends to both quotients. On functions let $R_t$ be the unitary action of $t\in T$, and let $E=e^{-s{\Delta }_M}$ for $s>0$. The averaging operator $P_W=|W{|}^{-1}{\sum }_{w\in W}{R}_w$ projects orthogonally onto the $W$-invariant functions, which identify with functions on $W\backslash M$. The trace formula underlying the Sunada theorem isThe expression $\mathrm{Tr}(R_w{e}^{-s{\Delta }_M})$ is a class function, because ${\Delta }_M$ commutes with $T$. Thus Gassmann equivalence makes the two sums identical. Alternatively, apply the same averaging formula to each finite-dimensional Laplacian eigenspace: its invariant dimension is $|W{|}^{-1}{\sum }_{w\in W}{\chi }_{\lambda }(w)$. This proves equality of every eigenvalue multiplicity and hence isospectrality of the two quotient metrics.

If $V=tU{t}^{-1}$ for some $t\in T$, the action of $t$ gives an isometry between the quotients for every such metric. Nonconjugacy is therefore necessary. To make it sufficient, choose the quotient orbifold metric generically so that its regular locus has no nonidentity local isometries; this is the equivariant, or orbifold, version of the Sunada local isometry lemma. Smooth perturbations on regular coordinate balls provide this genericity while their lifts remain smooth on $M$. Any isometry between $M_1$ and $M_2$, when composed locally with their projections, would then identify the same points of the base regular locus. By continuation it is an isomorphism of their coverings of that locus, and the classification of connected covering spaces forces $U,V$ to be conjugate in $T$. Thus nonconjugate subgroups give non-isometric isospectral quotients for a suitable generic invariant metric. The genericity belongs to the base regular locus; the covering metric on $M$ itself necessarily has the prescribed deck symmetries.

For genus two, an explicit choice is the following permutation group on twelve labels, with permutation products composed from right to left:These parameters are those in Kang's genus-two construction. They admit a direct finite verification. Starting with $T_0=\{1\}$ and iterating $T_{j+1}=T_j\cup T_{j}A\cup T_{j}B$, the successive sizes are $1,3,7,13,23,39,61,81,93,96$, after which the set is unchanged. Thus $|T|=96$, and direct multiplication gives $(p,q,r)=(3,3,6)$ and $|U|=|V|=8$. The conjugacy-class triples $(|C|,|C\cap U|,|C\cap V|)$, obtained as $C=\{tx{t}^{-1}:t\in T\}$, areThis proves Gassmann equivalence. The subgroup $V$ fixes the labels $5,11$, whereas $U$ has no common fixed label, so they cannot be conjugate even in the full symmetric group. Both are groups of order eight, so they meet no nonidentity order-three element. The only order-two power of $AB$ iswhich is central in $T$ and lies in neither subgroup. This also excludes its conjugates and proves freeness. The quotient Euler characteristics and genera are thereforeThe covering surface $M$ has genus nine.

For genus three, use $T={\mathrm{GL}}_3({\mathbb{F}}_2)$, of order $(8-1)(8-2)(8-4)=168$. Let $U$ stabilize a nonzero vector and let $V$ stabilize a two-dimensional subspace. Both have index seven and order twenty-four. An element $t$ fixes $2^{\dim \ker (t-I)}-1$ nonzero vectors. Hyperplanes correspond to nonzero dual vectors over ${\mathbb{F}}_2$, so the number fixed in the dual action is $2^{\dim \ker (t^T-I)}-1$, the same number. Equality of these permutation characters proves that the point and hyperplane stabilizers are almost conjugate.

They are not conjugate. In coordinates where $U$ fixes $e_1$, its elements include all matrices $\left(\begin{array}{cc}1& \alpha \\ 0& C\end{array}\right)$ with arbitrary row $\alpha$ and $C\in {\mathrm{GL}}_2({\mathbb{F}}_2)$. An invariant plane containing $e_1$ would project to a line invariant under every $C$, which is impossible. A plane not containing $e_1$ would be the graph of a linear functional on the quotient, and arbitrary $\alpha$ would move that graph. Thus $U$ stabilizes no plane, whereas every conjugate of $V$ does.

Choose the concrete generatorsOver ${\mathbb{F}}_2$, each of $A,B,AB$ has order seven. The same closure iteration has sizes $1,3,7,15,31,59,103,144,168$, verifying that $A,B$ generate $T$. Since seven does not divide $24$, neither stabilizer contains a nonidentity power of these generators, or any conjugate of such a power. We may therefore use $(p,q,r)=(7,7,7)$ in the triangle cover construction for Sunada surfaces. The quotient calculation isHere the covering surface $M$ has genus forty-nine. Choosing the generic invariant metrics described above completes both requested pairs.

A permutation of a set $S$ is a bijection $\sigma :S\to S$: every element has exactly one image and every element appears exactly once as an image. For a finite set, a permutation can also be recorded as an ordering of all its elements, once a reference order has been chosen. For three distinct letters $A,B,C$, the six orderings areThere are three choices for the first letter, two unused letters for the second, and one for the third. Every path in the tree gives one complete ordering, and every ordering has one such path, so the rule of product gives $3\cdot 2\cdot 1=3!=6$, a factorial.

Ordering versus mapping. Label the reference positions $1,\dots ,n$. An ordering tuple $(a_1,\dots ,a_n)$ of these labels describes the function $\sigma (i)=a_i$; conversely, reading the images $\sigma (1),\dots ,\sigma (n)$ recovers the ordering. For example,means $1\mapsto 3$, $2\mapsto 1$, $3\mapsto 4$, $4\mapsto 2$. Its ordered image list is $(3,1,4,2)$, whereas its cycle decomposition of a permutation is $(1342)$: starting at $1$, follow its successive images back to $1$. Parentheses in the latter expression encode arrows around a permutation cycle, rather than an image list in reference order. ^[ref]

The two viewpoints answer different useful questions about the same object. An ordering tells us which object occupies each position; a bijection tells us where each label is sent and allows successive rearrangements through function composition. All permutations of $n$ labels form the symmetric group, whose subgroups are permutation groups. The counting viewpoint supplies $n!$ possibilities, a factorial; the mapping viewpoint reveals how rearrangements combine.

Composition convention. Apply the rightmost permutation first:For the displayed $\sigma =(1342)$ and the transposition $\tau =(12)$, direct evaluation givesFor instance, $(\sigma \tau )(1)=\sigma (2)=1$, but $(\tau \sigma )(1)=\tau (3)=3$. Thus function composition of permutations need not be commutative; the omitted labels in these permutation cycles are fixed points.

The permutation group of a three-element set has six elements: the identity, three transpositions, and two three-cycles. A three-cycle and any transposition generate it, with relations $x^3=h^2=e$, $hx{h}^{-1}=x^{-1}$. Its alternating group is the normal cyclic subgroup of order three. It is the unique nonabelian isomorphism type of a group of order six.

For a transitive finite permutation group, suppose each point stabilizer has orbits of a common size $r$ on the other points. If a nontrivial block system exists, all two-point stabilizers are trivial. Indeed for block size $b$, $r\mid n-1$ gives $\gcd (r,b)=1$. An orbit of a block not containing $\alpha$ has size at most $br$ and divisible by $br$, so its b point-orbits are disjoint. A stabilizer of $\alpha$ and a point of that block fixes the whole block. Interchanging the two points also fixes the block of $\alpha$. Equal two-point-stabilizer orders then show that each two-point stabilizer fixes every point, and faithfulness makes it trivial.