Use relations without assuming rotations commute with reflections

Try

Label the vertices of a regular n-gon by residues zero through n−1. Let r send i to i+1, a one-step rotation, and let s send i to −i, a reflection. We write Dₙ for this symmetry group of order 2n, with n≥3. Some books use a different subscript convention, so always state the order explicitly.

Compute sr and rs on vertex zero and vertex one. A reflection reverses the direction in which a subsequent rotation moves around the polygon.

Predict

Predict the result of reflecting, rotating one step, then reflecting again. Under right-to-left composition this is srs. Tracking a general label i should produce i−1, not i+1. This is the relation srs=r⁻¹, the source of noncommutativity.

Experiment

Use a square, so exponents of r are reduced modulo four. The symmetries have forms rᵃ or rᵃs.

Product	Move a rotation past s	Normal form
sr	r⁻¹s	r³s
(rs)(r²)	r r⁻²s	r³s
(r²s)(rs)	r²r⁻¹s²	r

Verify the last product by applying both maps to every vertex. It is a rotation even though each factor is a reflection. Reversing the two reflection factors produces the inverse rotation, illustrating why the order of composition matters.

Understand

The defining relations rⁿ=e, s²=e and sr=r⁻¹s reduce every word to rᵃsᵇ, with a modulo n and b modulo two. Multiplication becomes (rᵃsᵇ)(rᶜsᵈ)=r^(a+(−1)ᵇc)s^(b+d). The sign change records whether a rotation passes through a reflection.

These 2n forms are distinct for n≥3. Within either family, the image of zero identifies a. If rᵃ=rᶜs, evaluating zero gives a=c. Evaluating one would then give a+1=a−1 modulo n, requiring n to divide two, impossible. The normal forms therefore describe exactly 2n different permutations, not merely an upper bound from a rewriting rule.

Every rᵃs is an involution: its square is rᵃr⁻ᵃ=e, and it is not identity by distinctness. Rotations have order n/gcd(n,a). In a square, the half-turn r² is also an involution, while the quarter-turns have order four. Element order alone does not distinguish every geometric type, but it supplies useful invariants.

The group is nonabelian for n≥3. If r and s commuted, the relation srs=r⁻¹ would also equal r, forcing r²=e, contrary to r having order n. At n=2 the abstract presentation instead gives an abelian four-element group. This special case does not satisfy the same ordinary polygon argument, so do not silently apply the distinctness proof there.

Apply

Construct a product table for the six symmetries of a triangle using exponent pairs (a,b). Check associativity either by identifying the entries as actual permutations or by verifying the multiplication rule algebraically. A table that merely looks regular is not a proof of associativity.

Use a noncommuting pair to distinguish D₃ from the cyclic group of order six. Both have six elements, but an isomorphism must preserve multiplication and therefore commutativity. Matching cardinality alone never supplies an isomorphism.

Continue

Next count subgroups through their cosets. The rotation subgroup of a dihedral group will supply a recurring example of an index-two subgroup and, later, a quotient group.