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Write the dihedral group as $D_6=⟨r,s\mid r^6=s^2=1,srs=r^{-1}⟩$, where $r$ rotates through $\pi /3$. There are six rotations and six reflections, so the order is 12. Conjugation by $s$ reverses a rotation, and conjugation by $r$ sends $r^{j}s$ to $r^{j+2}s$. Consequently the conjugacy classes and their sizes areThe last two classes distinguish the reflection axes through opposite vertices and through opposite edge midpoints. Their sizes sum to 12.

The Burnside lemma states that the number of orbits of a finite group action isFor $m=2n-1\ge 3$, the dihedral group consists of $m$ rotations $r^j$ by $2\pi j/m$, and $m$ reflections $r^{j}s$, each across the line through one vertex and the midpoint of the opposite edge. Thus $|G|=2m=4n-2$.

For the pentagon, the identity fixes $v^5{e}^5$ colourings. Each of the four nonidentity rotations has a single cycle on vertices and a single cycle on edges, fixing $ve$ colourings. Each of the five reflections has one fixed vertex and two vertex pairs, and one fixed edge and two edge pairs, fixing $v^3{e}^3$ colourings. Applying the Burnside lemma to simultaneous vertex and edge colourings givesVertex colours and edge colours may be chosen independently; each symmetry must preserve both assignments.

Put $N=HK=⟨r⟩$, the normal subgroup of order $35$. By Cauchy's theorem for finite groups, $G$ contains an element $t$ of order two. Such an element cannot belong to the odd-order subgroup $N$. Thus $N\cap ⟨t⟩=\{1\}$ and $G=N⟨t⟩$, giving the semidirect productEvery automorphism of a finite cyclic group sends $r$ to $r^a$ for a unit $a$ modulo $35$. The relation $t^2=1$ forces the conjugation action of $t$ to square to the identity:The Chinese remainder theorem identifies a congruence class modulo $35$ with its pair of classes modulo $5$ and $7$. In each of these prime fields, the factorization $(a-1)(a+1)=0$ forces $a=1$ or $a=-1$. The choices are independent, giving exactly four homomorphisms from a finite cyclic group $C_2\to \mathrm{Aut}(C_{35})$. Explicitly,The factors $C_5$ and $C_7$ of $N$ are preserved by every automorphism, since each is its unique subgroup of that order. The four conjugation actions therefore give:

Here $D_n$ has group order $n$, as in the paper. The center of a group distinguishes these four groups of order seventy: its orders are $70,1,5,7$, respectively, so the four groups are pairwise nonisomorphic. Each construction is a semidirect product with $70$ elements, so every listed possibility occurs.

Let $|G|=p^2$. The allowed center of a group theorem gives $|Z(G)|>1$. By Lagrange's theorem its order is $p$ or $p^2$. In the first case $G/Z(G)$ has order $p$ and is a cyclic group. A cyclic quotient by the center forces $G$ to be Abelian: if $G/Z(G)=⟨gZ(G)⟩$, write arbitrary elements as $g^{a}z,g^{b}w$ with $z,w\in Z(G)$, and commute them directly. Hence in every caseFor two examples at prime-cube order, take the dihedral group of a square, $⟨r,s\mid r^4=s^2=1,srs=r^{-1}⟩$, and the quaternion group $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$. Both have order $8=2^3$. They are nonabelian because $sr\ne rs$ and $ij=k\ne -k=ji$, respectively. They are distinct examples even up to group isomorphism: the dihedral group has five elements of order two and the quaternion group has only one. The request asks for two examples; it does not require the prime to be arbitrary.

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.

Let $A=\mathrm{diag}(1,-1)$ and $B=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. They generate the dihedral group of a square, realized by $\{\pm I,\pm A,\pm B,\pm AB\}$. Its image in the projective linear group has four elements, since opposite matrices become equal. The projective classes of $A,B$ commute because $AB=-BA$, and each squares to the identity. Sending $[A^a{B}^b]$ to $(a,b)$ gives the Klein four-group. This is an example in which a nontrivial scalar commutator disappears after passing to a projective representation.

The six-element dihedral group generated by the displayed rotations already forces every invariant second-rank tensor to have the same form as an axially invariant second-rank tensor. The axial threefold rotation eliminates mixed entries and leaves only a planar scalar plus a planar antisymmetric part; the transverse half-turn eliminates the latter. No smaller finite subgroup suffices: cyclic rotation groups retain an axial antisymmetric tensor, while a four-element noncyclic rotation group retains arbitrary diagonal tensors along its three mutually perpendicular half-turn axes.

Independent sign changes and interchange of two coordinates in the natural dihedral group action force the displayed cubic normal form. Apart from the origin, its nondegenerate small equilibria lie on coordinate axes or on the diagonals. An axial branch has squared amplitude $\mu /a$ and eigenvalues $-2\mu ,\mu (1-b/a)$. A diagonal branch has component squared amplitude $\mu /(a+b)$ and eigenvalues $-2\mu ,2\mu (b-a)/(a+b)$. Subtracting the two nonzero equilibrium equations gives $(a-b)(A^2-B^2)=0$, excluding other cubic branches when $a\ne b$.

Put $\alpha =\sqrt[4]{7}$ and $L=\mathbb{Q}(\alpha ,i)$. The dihedral group $D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$ acts byThe ten subgroups and their fixed fields areThe normal subgroups are $D_8$, $⟨r⟩$, the two displayed Klein four-groups, $⟨r^2⟩$, and $\{1\}$.

A subgroup $H\le G$ is a subset of a group $G$ that is itself a group under the same group operation. It must contain the identity element of $G$ and be closed under products and inverse elements. Associativity is inherited from $G$, so it needs no new proof. The whole group and the trivial group $\{e\}$ are always subgroups. A proper subgroup satisfies $H\ne G$.

A useful single-condition test isTo see why it works, choose $a\in H$; then $a{a}^{-1}=e\in H$. Taking $a=e$ gives every inverse element $b^{-1}\in H$. Applying the test to $a$ and $b^{-1}$ then gives $ab\in H$. Conversely, every subgroup satisfies the test by closure under products and inverse elements. This also makes the additive version memorable: a nonempty subset of an additive group is a subgroup exactly when it is closed under differences. See McKernan's MIT lecture 2, Proposition 2.3, for the equivalent closure criterion.

For the square dihedral group $D_4$ above, $R=\{e,r,r^2,r^3\}$ is a cyclic subgroup: products add exponents modulo four, and $(r^k{)}^{-1}=r^{-k}$. The smaller collections $\{e,r^2\}$ and $\{e,s\}$ are subgroups of group order two. But $\{e,r\}$ is not a subgroup, since $r\cdot r=r^2$ lies outside it. A plausible-looking list of symmetries therefore need not constitute a complete collection of reversible operations.

Inside $(\mathbb{Z},+)$, the even numbers $2\mathbb{Z}$ form a subgroup: $2m-2n=2(m-n)$ is an even number. Its two cosets consist of the even numbers and the odd numbers, so its index of a subgroup is two. The parity group homomorphism $k\mapsto k\mathrm{mod}2$ has $2\mathbb{Z}$ as its kernel of a group homomorphism. A subgroup can thus identify exactly which information a composition-preserving map forgets.

Subgroups also describe residual symmetry. In the group action of $D_4$ on the four vertices, the stabilizer subgroup of $(1,1)$ is $\{e,sr\}$: the diagonal reflection $sr$ fixes that vertex, and the other six motions move it. Requiring a feature to stay fixed still leaves a group, because both composition and reversal preserve that requirement. This is why subgroups are useful for studying a group action one feature at a time.

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.

Take $r$ as a counterclockwise quarter-turn and $s$ as reflection across the horizontal axis. The eight symmetries of a square are $e,r,r^2,r^3,s,rs,r^{2}s,r^{3}s$. The last four are reflections across the horizontal axis, the diagonal $y=x$, the vertical axis and the diagonal $y=-x$, respectively. This is a dihedral group of order eight; the subscript convention $D_4$ counts polygon vertices, while some authors use $D_8$ for the same group.

A group generated by two involutions is dihedral in the extended convention allowing $C_2$ and the infinite dihedral group. If $a\ne b$, put $x=ab$, $y=b$; then $yx{y}^{-1}=x^{-1}$. Conversely, from the inversion relation, $y$ and $xy$ square to the identity and generate. If $xy=e$, the group is $C_2$ and the same nonidentity element is used twice. Exact order two, rather than merely a square equal to the identity, requires this exceptional case. This convention extends the usual finite dihedral group definition.

The two simple reflections generate a dihedral group. For $A_1\times A_1$, $A_2$, $B_2$, and $G_2$, the rotation has order $2,3,4,6$, so the Weyl groups have orders $4,6,8,12$ respectively.