Try
Consider the quarter-turn R with rows (0,−1),(1,0), and the reflection S with rows (1,0),(0,−1). Both preserve lengths. Predict whether applying S after R is the same as applying R after S. On the vector (1,0), the two results are (0,−1) and (0,1). The order matters.
The eight maps Rʲ and RʲS, for j=0,1,2,3, are the symmetries of a square centered at the origin. We will prove why every finite group of plane isometries reduces to either rotations alone or rotations together with a reflection—even when its original formulas contain translations.
An isometry preserves all pairwise Euclidean distances. The existing affine orthogonal classification proves that every such map of the plane has the form x↦Qx+b, where QᵀQ=I. Here we use that theorem, including its conclusion that the map is bijective. We do not assume that preserving distances from just one point is sufficient.
Predict
Define O(2) to be the group of real 2-by-2 matrices Q satisfying QᵀQ=I, and SO(2) its determinant-one subgroup. Determinants in O(2) are ±1 because det(Q)²=1. Predict the second column once the first column is the unit vector (cos θ,sin θ).
Now compare two order-two groups: the half-turn group ❴I,−I❵ and the reflection group ❴I,S❵. Both are abstractly C₂. Are they conjugate as groups of plane isometries? Record a property of their nonidentity maps that survives a change of coordinates.
Experiment
Write Rθ for the rotation with rows (cos θ,−sin θ),(sin θ,cos θ). Use S=diag(1,−1). Products act on column vectors from right to left. Fill the final column before checking it.
| Map | First column | Second column | Image of (1,0) |
|---|
| Rθ | (cos θ,sin θ) | (−sin θ,cos θ) | (cos θ,sin θ) |
| RθS | (cos θ,sin θ) | (sin θ,−cos θ) | (cos θ,sin θ) |
| SRπ/2 | (0,−1) | (−1,0) | (0,−1) |
| Rπ/2S | (0,1) | (1,0) | (0,1) |
The first two maps agree on (1,0) but differ on (0,1). One sample vector cannot certify equality of linear maps. The two product rows express SRθ=R−θS: reflecting reverses the rotation angle.
Change the angle to 2π/3. Predict the number of distinct maps generated by R₂π/₃ and S, their multiplication rule, and how many reverse orientation.
Check the changed group
There are six maps R₂πj/₃Sᵉ with j=0,1,2 and e=0,1. The rotation exponents distinguish maps with the same e, while determinant distinguishes the two values of e. With exponents reduced modulo 3, multiplication is (j,e)(k,d)=(j+(−1)ᵉk,e+d), where the second coordinate is reduced modulo 2. Exactly the three maps with e=1 reverse orientation. This is the dihedral group of order six, not the cyclic group of order six.
Understand
All orthogonal plane matrices
If the first column of Q∈O(2) is (a,b), then a²+b²=1. A unit vector perpendicular to it is either (−b,a) or (b,−a); these are the only two possibilities because its perpendicular space is one-dimensional. Write a=cos θ,b=sin θ. The first choice gives Rθ with determinant 1, and the second gives RθS with determinant −1. Thus these are all the matrices, not merely examples.
Identify (x,y) with the complex number z=x+iy. Rθ acts by multiplication by e^(iθ), while S acts by complex conjugation. Multiplication of rotations corresponds to multiplication of unit complex numbers; the map from the unit circle to SO(2) is injective because the image of 1 determines the multiplier, and surjective by the column classification. Therefore SO(2) is isomorphic to the unit complex numbers under multiplication.
Every Q has a unique form RθSᵉ with θ taken modulo 2π and e∈❴0,1❵. Since S Rθ S=R−θ, the product is
(RθSᵉ)(RφSᵈ)=R(θ+(−1)ᵉφ) S^(e+d).
This is precisely the external semidirect product SO(2)⋊C₂, where the nonidentity element of C₂ acts by inversion. The normal rotation subgroup is the kernel of determinant, the copy of C₂ is generated by S, and their intersection is the identity. These observations agree with the proved internal and external semidirect criteria.
Each RφS is a reflection, not a glide: it fixes the line at angle φ/2 and negates its perpendicular line. This follows by conjugating S by Rφ/₂. The classification here concerns linear maps fixing the origin; affine glide reflections occur outside O(2).
A finite rotation subgroup is cyclic
Let H≤SO(2) be finite. If H consists only of the identity, it is C₁. Otherwise represent each element by its angle in [0,2π), and let θ be the least strictly positive angle present. Such a minimum exists because the set is finite and nonempty.
For any angle α in H, real division gives α=qθ+r with integer q≥0 and 0≤r<θ. The matrix RαRθ^(−q)=Rr belongs to H. Minimality forces r=0. Thus every element is a power of Rθ.
It remains to prove the angle and exact order, not merely generation. Divide 2π by θ: write 2π=nθ+r with n a positive integer and 0≤r<θ. If r>0, the group element Rθ^(−n) has representative angle r, contradicting minimality. Hence r=0 and θ=2π/n. The n powers with exponents 0 through n−1 are distinct, and every group element is one of them. Therefore |H|=n, and the subgroup is exactly the standard rotations by multiples of 2π/n. Finiteness is essential: rotations by all rational multiples of 2π have no least positive angle.
All finite subgroups of O(2)
Let G≤O(2) be finite and H=G∩SO(2). If every element preserves orientation, G=H is the cyclic group just found. Otherwise choose a reflection T=RφS in G. Any other orientation-reversing element g has gT⁻¹∈H, so G is the disjoint union H and HT. Write H=❴R₂πj/ₙ:0≤j<n❵, including n=1. Then
G = ❴R₂πj/ₙ, R₂πj/ₙT : 0≤j<n❵,
T²=I, and T R₂π/ₙ T=R−₂π/ₙ. Consequently G has exactly 2n elements and is isomorphic to Cₙ⋊C₂ with inversion action. This is the dihedral group of order 2n, using the convention that Dₙ has order 2n. For n=1 it is just a single reflection and the identity. For n=2 inversion on C₂ is trivial, and the four-element group is C₂×C₂. Noncommutativity begins at n≥3.
Conjugating by the rotation R−φ/₂ makes T into S and leaves H unchanged. Hence all such reflection-containing groups with the same n are conjugate to the standard model. Rotational cyclic groups and reflection-containing groups cannot be conjugate: determinant is preserved by conjugation and the former contain only determinant-one matrices. In particular the rotation group C₂ and reflection group D₁ are abstractly isomorphic but not conjugate geometric actions. Their nonidentity fixed sets are a point versus a line, another invariant distinction.
Finite affine groups have a common fixed point
Let G be a finite group of plane isometries, now allowing translations in their formulas. Fix any point p and define
c = (1/|G|) Σ over g∈G of g(p).
For h(x)=Qx+b in G, affine linearity gives h(c)=(1/|G|)Σ h(g(p)). Left multiplication g↦hg permutes G, so this average equals c. Thus every member fixes c. The same argument may be written as the average of the distinct orbit points: all orbit points appear with equal stabilizer multiplicity in the group-indexed sum.
Let τ(x)=x+c. Then τ⁻¹gτ fixes the origin, so its affine formula has zero translation part and it belongs to O(2). Translation conjugacy reduces every finite plane-isometry group to the linear classification above. The proof uses a finite sum divided by the positive integer |G| in the real numbers; it does not extend by an undefined average over an infinite group.
The common fixed point need not be unique. A group generated by one reflection fixes its entire axis, while a nontrivial finite rotation group fixes just its center. The identity group fixes the whole plane. A nonzero translation and a nontrivial glide reflection each generate an infinite group, consistent with their absence from this finite classification.
Apply
Move the square example away from the origin. Let
F(x,y)=(1−y,x+3), T(x,y)=(x,4−y).
Determine a common fixed point, conjugate both maps to linear matrices, and prove that the generated group has exactly eight elements. Then average the orbit of (0,0) under F alone. Does that average certify a fixed point for every member of the group generated by F and T without any further check?
Check the affine group and the averaging claim
The fixed-point equations for F give c=(−1,2), and direct substitution shows T(c)=c. With τ(u)=u+c, the conjugates are Rπ/₂ and S. The eight distinct normal forms therefore prove the full group has order eight. The F-orbit of (0,0) is (0,0),(1,3),(−2,4),(−3,1); its average is (−1,2).
Averaging under the subgroup generated by F proves only that F and its powers fix the average. In this example direct checking of T(c)=c completes the argument. Alternatively average over the full eight-element group, whose invariance proof covers both generators at once. A subgroup average cannot generally certify invariance under a larger group without a separate reason.
Continue
Return to plane reflections and affine classification for fixed lines and glide reflections, or dihedral normal forms for symbolic multiplication. Keep three statements separate: an abstract group isomorphism, a matrix realization, and conjugacy of two geometric actions.
Continue to spatial rotations and pole counts. In three dimensions, rotation axes and their stabilizers supply useful finite-group counts, but a list of possible counts alone will not establish a complete geometric classification.
