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For a subgroup $H$ of a group $G$ and $g\in G$, the conjugate subgroup is the conjugate subsetIt is a subgroup because conjugation is a group automorphism. Restricting that group automorphism gives a group isomorphism from $H$ to $gH{g}^{-1}$, so conjugate subgroups have the same group order. This describes the same operations after moving their reference configuration. Conjugacy of subgroups is an equivalence relation. ^[ref]

For an equilateral triangle with vertices of a polygon labeled $1,2,3$ counterclockwise, its symmetry group is the symmetric group $S_3$, also the dihedral group of group order six. Let $r=(123)$ be the counterclockwise rotation through $120$ degrees, and let $s=(12)$ be the reflection fixing vertex of a polygon $3$. Products apply the rightmost permutation first. The cyclic subgroup $K=⟨s⟩=\{e,(12)\}$ contains the identity element and one reflection. Relabeling a transposition by conjugation givesThus the conjugation action moves $K$ among the three subgroups $⟨(12)⟩$, $⟨(23)⟩$, and $⟨(13)⟩$. Geometrically, the rotation moves the axis of the reflection to the next axis. Each reflection fixes a different vertex of a polygon, although group isomorphisms relate all three subgroups.

The normalizer tells when this change of reference leaves the subgroup unchanged. For any group $G$ and subgroup $H\le G$, there is a bijectionTwo choices $g,k$ give the same conjugate subgroup exactly when $k^{-1}g\in N_G(H)$. Consequently the index of a subgroup $[G:N_G(H)]$ counts the distinct conjugate subgroups, even for an infinite group. For a finite group the count is $|G|/|N_G(H)|$. These are cosets; they need not form a quotient group, since $N_G(H)$ need not be a normal subgroup of $G$. In the triangle example there are $6/2=3$ conjugate subgroups, while the cyclic subgroup $⟨r⟩$ has only one. ^[ref]

A left group action of a group $G$ on a set $X$ is a function $(g,x)\mapsto g\cdot x$ satisfyingwhere $e$ is the identity element of $G$. The second rule means that $h$ moves the point first and $g$ moves it next. These rules let group multiplication describe consistent rearrangements of a set; see Ackerman's algebra notes, definition 0.4.1.1.

For each $g$, the function $x\mapsto g\cdot x$ is a bijection: its inverse is $x\mapsto g^{-1}\cdot x$, using the inverse element of $g$. Equivalently, a group action is a group homomorphism $\rho :G\to \mathrm{Sym}(X)$ into the symmetric group of all permutations of $X$, with $\rho (g)(x)=g\cdot x$ and $\rho (gh)=\rho (g)\circ \rho (h)$. See MIT's group-action notes, definition 23.1.

For a concrete group action, put the vertices of a polygon of a square at $A=(1,1)$, $B=(-1,1)$, $C=(-1,-1)$ and $D=(1,-1)$. Let $r(x,y)=(-y,x)$ be a counterclockwise quarter-turn rotation and let $s(x,y)=(y,x)$ be a reflection. Their eight permutations of the vertices of a polygon form the dihedral groupHere $D_4$ means the symmetry group of a four-sided polygon, with group order eight; the alternative convention calls this same group $D_8$. Products apply the rightmost operation first. For example, $s\cdot B=D$ and $(rs)\cdot B=r\cdot D=A$. The same group also has separate group actions on the sides of a polygon, the diagonals of a polygon, and the singleton set containing the centroid of the square. Specifying $X$ matters: changing what is moved changes the orbits of a group action and the stabilizer subgroups.

A group action need not be a faithful group action. For example, the cyclic group $C_4=\mathbb{Z}/4\mathbb{Z}$ acts on $X=\{0,1\}$ by $k\cdot x=x+k(\mathrm{mod}2)$, using modular arithmetic. Only the parity of $k$ matters: $0,2$ fix both points and $1,3$ exchange them. The associated group homomorphism $C_4\to \mathrm{Sym}(X)$ has group kernel $\{0,2\}$, of group order two. Either point has a group orbit of size two and a stabilizer subgroup of group order two, so the orbit-stabilizer theorem gives $4=2\cdot 2$.

The useful questions are which positions can be exchanged, and which transformations survive after imposing a constraint such as keeping one marked point fixed. An orbit of a group action answers the first question; a stabilizer subgroup answers the second. This turns a list of transformations into a count of distinct configurations, and explains why several transformations can produce the same configuration.

A group records reversible operations that can be combined consistently. Group theory studies their subgroups, group homomorphisms, group actions, and quotient groups. A symmetry group makes the abstraction visible: transformations may change individual points while preserving a specified object or its structure. An orbit of a group action records which positions can be exchanged, while a stabilizer subgroup records which operations leave a selected feature fixed.

A polygon with six sides and six vertices. A regular hexagon has equal sides and angles, with each interior angle equal to $2\pi /3$. Its symmetry group consists of six rotations and six reflections, forming a dihedral group of order twelve.

The symmetry group of a hyperelastic material consists of invertible reference transformations $H$ preserving its strain energy density: $W(FH)=W(F)$ for every admissible deformation gradient $F$. A convention restricting reference transformations to determinant one restricts $H$ to the special linear group. Material isotropy requires all proper rotations in this group. This right action differs from the left action in material frame indifference.

Use a real physical field, so the three positive wavevectors are accompanied by their complex-conjugate modes. Modulo lattice translations, the effective symmetry group is $G={\mathbb{T}}^2⋊C_6$. Its translation group action isA $120^{\circ }$ rotation cyclically permutes the amplitudes; a half-turn $\kappa$ acts by simultaneous complex conjugation. Together these generate the sixfold rotation group. These formulas specify the group action independently of the choice of orientation of the three wavevectors.

For a nonzero real hexagon $(h,h,h)$, its stabilizer subgroup is the rotation group $C_6$. A translation fixes it only when all three phase factors equal one. Cyclic permutation forces a fixed vector to have equal components, and the half-turn forces those components to be real. Therefore $\mathrm{Fix}(C_6)=\{(a,a,a):a\in \mathbb{R}\}$. For a roll $(r,0,0)$ with nonzero real $r$, the stabilizer subgroup consists of the translations $T_{0,\theta }$ and the half-turn, with $\kappa T_{0,\theta }\kappa =T_{0,-\theta }$. It is $S^1⋊C_2$, isomorphic to $O(2)$ as an abstract group even though spatial reflections are absent. The circle eliminates $B,C$ from its fixed-point subspace of a group action, and the half-turn makes $A$ real. ThusIf unreduced physical translations are used, the full lattice kernel is included in both stabilizer subgroups. The real representation on the six critical components is absolutely irreducible: translations distinguish the three Fourier pairs, rotations permute them, and the half-turn rules out a complex scalar in the commutant. The equivariant branching lemma consequently supplies generic steady branches of both axial types at a transverse crossing of the common linear eigenvalue. It guarantees existence, not stability or exhaustiveness of the possible branches.

Translation equivariance severely restricts the monomials in $\dot{A}$. Through cubic order they are $A$, $\overline{B}\overline{C}$, $|A{|}^{2}A$, $|B{|}^{2}A$ and $|C{|}^{2}A$: each has the same translation weight as $A$, using ${\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3=0$. The half-turn forces their coefficients to be real; the $120^{\circ }$ rotation gives the two cyclic counterpart equations. A spatial reflection would interchange $B,C$ and require equal cross-couplings, but it is not a symmetry here. After a real amplitude rescaling makes the generic nonzero quadratic coefficient one, the rotational hexagon amplitude equations arewith the other two equations obtained by cyclic permutation. All four coefficients are real; $\delta$ records the absence of spatial reflection. The normalization assumes the resonant quadratic coefficient is nonzero; its vanishing would be an additional degeneracy.

For the roll, $r^2=\mu /{\nu }_1$. The radial eigenvalue is $-2\mu$ and the phase of the active mode is a neutral translation. The transverse variables $(b,\overline{c})$ have matrixIts two real eigenvalues, each occurring twice in the full real linearization, are $-({\nu }_2-{\nu }_1)r^2\pm \sqrt{{\delta }^2{r}^4+r^2}$. Define $d={\nu }_2-{\nu }_1$ and $D_r=d^2-{\delta }^2$. Strict attraction modulo translation is therefore equivalent toIndeed the larger transverse eigenvalue is negative precisely when $d{r}^2>\sqrt{{\delta }^2{r}^4+r^2}$, and squaring is legitimate only after requiring $d>0$. If ${\delta }^2>d^2$, that inequality is impossible. Equality ${\delta }^2=d^2$ is also insufficient because of the additional positive term $r^2$. Any nonzero roll that exists with ${\nu }_1<0$ already has a positive radial eigenvalue; the degenerate case ${\nu }_1=0$ has no ordinary isolated nonzero roll branch.

For hexagons put $N={\nu }_1+2{\nu }_2$. A nonzero real amplitude $h$ satisfies $\mu =N{h}^2-h$, or $h=(1\pm \sqrt{1+4N\mu })/(2N)$ when $N\ne 0$. Separate real and imaginary disturbances. The real Jacobian matrix is cyclic, with diagonal $-h-2{\nu }_1{h}^2$ and off-diagonal entries $h-2({\nu }_2\pm \delta )h^2$. Its eigenvalues areThe imaginary Jacobian matrix has every entry equal to $-h$, giving eigenvalues $-3h,0,0$. The two zero modes translate the hexagonal pattern. Thus the general strict stability conditions for any nonzero root $h$ areThese conditions state orbital stability transverse to translations, not asymptotic convergence to one prescribed spatial phase. Equalities require a nonlinear bifurcation calculation.

When ${\nu }_2>{\nu }_1>0$, both $d,N$ are positive. The upper positive hexagon branch is radially stable after its saddle-node bifurcation at $\mu =-1/(4N)$, $h=1/(2N)$; it remains stable until $h=1/d$. At this second threshold the pair ${\lambda }_{\pm }$ crosses the imaginary axis with nonzero frequency if $\delta \ne 0$. The crossing is transverse, since $dh/d\mu =1/(2Nh-1)$ and the derivative of $-2h+2d{h}^2$ with respect to $h$ is two there. Hence, subject to the usual nonzero cubic coefficient, this is a Hopf bifurcation:For $\delta =0$ the same threshold has two zero real eigenvalues, so it is not a Hopf bifurcation.

A possible phase portrait when neither steady branch is stable has an attracting periodic oscillation surrounding the unstable hexagon, with the three amplitudes waxing and waning cyclically. This is realized, for example, by ${\nu }_1=1$, ${\nu }_2=2$, $\delta =2$ just above ${\mu }_H=4$. Rolls are unstable for every positive $\mu$ because ${\delta }^2=4>d^2=1$. At the hexagon Hopf point $h=1$, the real spectrum is $-9,\pm 4i\sqrt{3}$. Evaluation of the cubic Hopf normal form with a unit-norm critical eigenvector gives $\mathrm{Re}{G}_{21}=-64/27<0$, so this example has a supercritical Hopf bifurcation and an attracting small limit cycle in the real invariant subspace. The figure shows such a cycle at $\mu =4.2$ and an approaching trajectory; the positive octant is forward invariant because an amplitude's derivative at zero is the product of the other two. This is a possible portrait, rather than a claim that every parameter set with unstable rolls and hexagons has the same attractor. Stability asserted for the periodic orbit here is within the stipulated real subspace.

Put the cube's vertices at $(\pm 1,\pm 1,\pm 1)$. Squared distances between distinct vertices are $4,8,12$ according as one, two or three coordinates differ. Four mutually equidistant vertices cannot all differ in one coordinate, since the cube has no triangles, or in three coordinates, since each vertex has only one antipode. They must differ in two coordinates. The two regular tetrahedra are therefore precisely the sets $xyz=1$ and $xyz=-1$.

The full symmetry group $G$ consists of signed coordinate permutations, so $|G|=2^3\cdot 3!=48$. These maps permute the two tetrahedra. The stabilizer $H$ of the first has order $24$ and acts faithfully on its four vertices: a linear map fixing the tetrahedron fixes a spanning set. Thus $H\cong S_4$. Central inversion $-I$ swaps the tetrahedra and commutes with every element of $H$, giving the direct-product decomposition of the cube symmetry groupHere $H$ preserves one tetrahedron; it is not the subgroup consisting only of rotations.

The relation $Rm=m{R}^{-1}$ implies $m{R}^j=R^{-j}m$. Move each reflection generator to the right in any word, cancel pairs using $m^2=I$, and reduce exponents modulo $N$. Every element then has a normal form $R^j$ or $R^{j}m$, $0\le j<N$.

For the actual symmetry group of a regular polygon these are $N$ distinct rotations and $N$ distinct reflections. The determinant distinguishes rotations from reflections in the planar action. Hence the source's dihedral group $D_N$ has order $2N$, with its subscript counting polygon vertices. The calculations below use that convention.

The table describes the symmetry group of a regular hexagon. Choose $R$ as a counterclockwise rotation through $\pi /3$ and $m_1$ as reflection in one symmetry axis. With $s=m_1$, the relations areThe rotation entries already are $I,R,\dots ,R^5$, and the table gives $R{m}_j=m_{j+1}$ with indices modulo six. ThereforeThe source writes $D_6$ with the subscript counting polygon vertices; this dihedral group has order 12.

A symmetry is a transformation of a physical system that preserves its laws: it maps admissible solutions and consistently transformed data to admissible solutions. The central symmetry of special relativity is Lorentz invariance, expressing equivalence of inertial frames and preservation of the Minkowski interval and the universal speed of light. With spacetime translations included, the spacetime symmetry group is the Poincaré group.

A practical sufficient check is to express an equation as an equality between Lorentz tensors of the same type, with every repeated upper/lower pair correctly contracted and the same free indices on both sides. Every ingredient, including sources and background fields, must have its stated Lorentz transformation rule. For example, $p^{\mu }{p}_{\mu }=-m^2{c}^2$ is a Lorentz scalar equality, and $d{p}^{\mu }/d\tau =q{F}^{\mu }{}_{\nu }{U}^{\nu }$ equates four-vectors. Such manifest Lorentz covariance makes the form of the equation the same in every inertial frame.

This is a sufficient test when the transformation rules are genuine physical ones, not a guarantee from writing index symbols. A fixed preferred four-vector could make an equation formally covariant if it too were transformed, while its chosen physical background breaks Lorentz invariance. Conversely, an equation written in three-dimensional components can be fully consistent with special relativity, as the Lorentz force is, even when the covariance is not manifest.

A polygon with five sides and five vertices. A regular pentagon has equal sides and angles; its symmetry group is the order-ten dihedral group, consisting of five rotations and five reflections. Each reflection axis passes through a vertex and the midpoint of the opposite edge.

The full symmetry group of a regular tetrahedron is the symmetric group $S_4$ acting on its vertices. Its subgroups with at most six elements have eight types up to conjugation, with the following representative subgroups and counts:A transposition is a reflection through the plane containing the two fixed vertices and the midpoint of the exchanged edge. A double transposition is a half-turn about the axis joining the midpoints of the two exchanged edges. A three-cycle is a one-third-turn about a vertex-to-opposite-face axis. A four-cycle is a quarter-turn composed with a reflection perpendicular to the opposite-edge midpoint axis; it is an improper orthogonal transformation.

The unique Klein four-group containing only double transpositions consists of the three opposite-edge half-turns and the identity element. The other three Klein four-groups consist of two perpendicular mirror reflections, their half-turn product, and the identity element. Each $S_3$ is a stabilizer subgroup of a vertex, including the three rotations and three mirror reflections preserving that vertex.

For completeness, a Klein four-group containing $(12)$ must contain the commuting involution $(34)$ or $(12)(34)$, yielding the displayed subgroup; one containing no transpositions must contain all three double transpositions. An $S_3$ subgroup has a unique cyclic subgroup of cardinality three, whose unique fixed vertex must therefore be preserved by the entire $S_3$. These observations and the disjoint cycle decomposition exhaust the types allowed by groups of order at most six. The counts sum to $25$.

Take the four-point setthe vertices of a polygon of a rectangle whose side lengths are four and two. Three choices of structure give three different symmetry groups on these same points.

As an unstructured set, $R$ has all $4!=24$ permutations as symmetries, forming the symmetric group $S_4$. If Euclidean distance is part of the data, the allowed maps are only the identity, a half-turn rotation, and the reflections across the two coordinate axes. They form a Klein four-group of group order four. To see that nothing is missing, fix the image of one vertex of a polygon: its three other distances, $2$, $4$, and $\sqrt{20}$, are distinct, so they uniquely determine the images of the remaining vertices.

Preserving affine structure permits more. The affine mapcycles all four points of $R$. It sends a horizontal side of length four to a vertical side of length two, so it is not an isometry. Yet it preserves the rectangle's vertex set and all affine combinations among those points.

Let $A=\mathrm{diag}(2,1)$ and let $r(x,y)=(-y,x)$ be the square's quarter-turn rotation. ThenConjugating every square symmetry by $A$ gives eight affine maps preserving $R$. This group isomorphism makes the affine symmetry group a dihedral group, although some of its maps do not preserve the original Euclidean distances. There are no further affine symmetries: an affine map preserves the common midpoint of the diagonals, so it must preserve opposite pairs. The image of one vertex has four choices and the image of an adjacent vertex has two; those choices determine the map through three noncollinear points.

ThusAn arbitrary permutation need not be an affine map. For example, exchanging the two top vertices while fixing the two bottom vertices sends the common diagonal midpoint to two different points, which no single affine map can do. The phrase "the symmetry of these points" becomes unambiguous only after stating which structure matters.

A symmetry group collects bijections of an object to itself that preserve a specified structure in both directions, with function composition as the operation. For a geometric object one may require Euclidean distances to be preserved, giving an isometry group; for a graph, one preserves adjacency, giving a graph automorphism group. For an unstructured finite set, every permutation is allowed. The structure to be preserved is part of the definition.

For a square with its usual geometric structure, the symmetry group consists of four rotations and four reflections. These form the dihedral group of group order eight. The existing group action example shows how its eight elements move the vertices of a polygon; changing the object to its edges or a marked configuration changes the group action being studied.^[ref]

Specifying the preserved data makes it possible to test a proposed symmetry exactly. After the preserved data are declared, function composition, the identity map, and inverse maps automatically preserve them too. Symmetry transformations therefore form a group, and questions about equivalent configurations become questions about group orbits, stabilizer subgroups, and group homomorphisms. This supplies reusable mathematics for comparing many different kinds of objects.

A symmetry transformation is a bijection preserving the data that define the object's symmetry group. Its individual group elements are symmetry transformations; the group operation combines them by function composition.