In D₄ (the symmetry group of a square), r is a 90° counterclockwise rotation and s is a reflection. Using the defining relation srs = r⁻¹, what is the composition s ∘ r ∘ s equal to?
Ar (a 90° counterclockwise rotation)
Br⁻¹ (a 90° clockwise rotation, i.e., r³)
Cr² (a 180° rotation)
Ds (the same reflection)
The relation srs = r⁻¹ is definitional for dihedral groups. It captures the key non-commutativity: conjugating the rotation by the reflection reverses it. Physically: if you reflect, then rotate counterclockwise 90°, then reflect back, the net effect is a clockwise 90° rotation. This is why r and s don't commute — sr ≠ rs.
Question 2 Multiple Choice
A student argues that D₅ (the symmetry group of a regular pentagon) must be abelian because all five rotations commute with each other, forming a cyclic subgroup. What is wrong with this reasoning?
ANothing — D₅ is actually abelian since n = 5 is odd
BThe rotations do form an abelian subgroup, but D₅ also contains 5 reflections, and reflections do not commute with rotations — so D₅ as a whole is non-abelian
CD₅ is non-abelian because not all rotations commute with each other
DThe argument is correct only for n ≥ 6; D₅ is a special case
The cyclic subgroup of rotations ⟨r⟩ is indeed abelian within Dₙ. But Dₙ also contains n reflections, and the relation srs = r⁻¹ shows that a reflection s does NOT commute with the rotation r (unless r = e or n ≤ 2). D₅ has 10 elements total and is non-abelian for the same reason as all Dₙ with n ≥ 3: rotation and reflection fail to commute.
Question 3 True / False
In any dihedral group Dₙ with n ≥ 3, the subgroup of rotations ⟨r⟩ = {e, r, r², ..., rⁿ⁻¹} is a normal subgroup.
TTrue
FFalse
Answer: True
A subgroup H is normal if gHg⁻¹ = H for all g. For ⟨r⟩ in Dₙ: conjugating any rotation rᵏ by another rotation gives another rotation (rotations form a subgroup, so the subgroup is closed under conjugation). Conjugating rᵏ by a reflection s gives s·rᵏ·s⁻¹ = (srs⁻¹)ᵏ = (r⁻¹)ᵏ = r⁻ᵏ, which is still in ⟨r⟩. So ⟨r⟩ is normal.
Question 4 True / False
D₃ is abelian because it has mainly 6 elements and most groups of order 6 are abelian.
TTrue
FFalse
Answer: False
D₃ is the smallest non-abelian group — in fact, D₃ ≅ S₃ (the symmetric group on 3 elements), which is non-abelian. The relation srs = r⁻¹ applies in D₃ just as in all Dₙ for n ≥ 3: a rotation followed by a reflection is not the same as the reflection followed by the rotation. Not all groups of order 6 are abelian: ℤ₆ is abelian, but D₃ ≅ S₃ is not.
Question 5 Short Answer
The defining relation srs = r⁻¹ in the dihedral group Dₙ is the key to why it is non-abelian. Explain what this relation says geometrically, and why it implies that s and r do not commute.
Think about your answer, then reveal below.
Model answer: Conjugating r by s reverses the rotation; geometrically, flipping then rotating then flipping back gives the opposite rotation, which means s·r ≠ r·s
The relation srs = r⁻¹ says: perform s, then r, then s again — the result is r⁻¹ (the reverse rotation). Rearranging: sr = r⁻¹s. If r and s commuted, we would need sr = rs, but instead sr = r⁻¹s ≠ rs (for n ≥ 3, where r ≠ r⁻¹). Geometrically: on a regular polygon, flipping then rotating gives a different result than rotating then flipping — because the reflection axis moves with the polygon under rotation.