Questions: Semidirect Products

Question 1

In D_n = ℤ_n ⋊ ℤ_2, how does the generator of ℤ_2 act on ℤ_n?

• A. Trivially: k ↦ k
• B. By inversion: k ↦ -k
• C. By doubling: k ↦ 2k
• D. By adding 1: k ↦ k + 1

Answer: (B)

A reflection s satisfies s r s⁻¹ = r⁻¹ for each rotation r, so conjugation by s inverts rotations. In additive notation on ℤ_n this is k ↦ -k, which is an automorphism. Adding 1 is not a homomorphism, and the trivial action would give the abelian group ℤ_n × ℤ_2.

Question 2

If N is normal in G, H is a subgroup, N ∩ H = {e} and NH = G, then G is isomorphic to the direct product N × H.

• T. True
• F. False

Answer: False

These conditions make G an internal semidirect product. G is the direct product only if H is also normal, which is equivalent to the conjugation action of H on N being trivial. S_3 has normal subgroup A_3 ≅ ℤ_3 and complement ⟨(1 2)⟩ ≅ ℤ_2, but S_3 is nonabelian, so it is not ℤ_3 × ℤ_2.

Question 3

Explain why there is a nonabelian group of order 21 but no nonabelian group of the form ℤ_5 ⋊ ℤ_3.

Model answer: Aut(ℤ_7) ≅ ℤ_6 contains elements of order 3, so a nontrivial homomorphism ℤ_3 → Aut(ℤ_7) exists. Aut(ℤ_5) ≅ ℤ_4 has no element of order 3, so every homomorphism ℤ_3 → Aut(ℤ_5) is trivial and the product is ℤ_15.

A nontrivial action requires the order of a generator's image to divide both 3 and |Aut(N)|. Since 3 divides 6 but not 4, only ℤ_7 admits a twisted product with ℤ_3. This matches the general fact that a nonabelian group of order pq with p < q exists exactly when p divides q - 1.

Question 4

Using the rule (n₁, h₁)(n₂, h₂) = (n₁·φ_{h₁}(n₂), h₁h₂), what is the inverse of (n, h)?

Model answer: (φ_{h⁻¹}(n⁻¹), h⁻¹)

Check: (n, h)(φ_{h⁻¹}(n⁻¹), h⁻¹) = (n·φ_h(φ_{h⁻¹}(n⁻¹)), e) = (n·n⁻¹, e) = (e, e), using φ_h ∘ φ_{h⁻¹} = φ_e = identity.