Algebra I: Section 6. The structure of groups. 6.1 Direct products of ...

Figure 6.3. Action of σρθσ on basis vectors in R 2 .
To prove this, we show that the set S of elements listed in (19) is already a group. Then, since
Dn = 〈ρθ, σ〉 ⊇ S and S is a subgroup containing the generators, the two sets must be equal;
uniqueness of the factorization follows. Clearly I ∈ S, and S · S ⊆ S because
(ρ k θ σℓ )(ρ r θ σs ) = ρ k θ (σℓ ρ r θ σ−ℓ )σ ℓ+s
= ρ k θ(σ ℓ ρθσ −ℓ ) r σ ℓ+s
� �r σ ℓ+s
= ρ k θ
ρ (−1)ℓ
θ
= ρ k+(−1)ℓr θ σ ℓ+s ∈ S
Finally S −1 = S, and S is a subgroup, because
(ρ k θσ ℓ ) −1 = σ −ℓ ρ −k
θ
= σ ℓ ρ −k
θ σℓ σ −ℓ
= (σ ℓ ρθσ ℓ ) −k σ −ℓ
=
�
ρ (−1)ℓ
θ
(conjugation by σ ℓ is an automorphism)
(repeated use of (18B))
� −k
σ −ℓ
= ρ −(−1)ℓk θ σ −ℓ ∈ S
(because σ = σ −1 )
From (19) we see that Dn is a finite group of operators on the plane, with |Dn| = 2n.
As for uniqueness, the identity ρk θσℓ = ρr θσs implies ρ k−r
θ = σs−ℓ . Unless both exponents
are congruent to zero, the transformation on the left has determinant +1 while the one on the
right has determinant −1, which is impossible.
The unique decomposition (19) shows that the subgroups N = 〈ρθ〉 and H = 〈σ〉 have the
properties (i) N ∩ H = (e), (ii) NH = Dn. Furthermore N is a normal subgroup, a fact that
is apparent once we compute the action of H on N by conjugation:
(20) φσ(ρ k θ ) = σρk θ σ−1 = (σρθσ −1 ) k = ρ −k
θ
Thus φσ is the inversion automorphism J : n → n −1 . Then conjugation by an arbitrary element
g = ρ r θ σs ∈ Dn has the following effect
∈ N
gρ k θg−1 = (ρ r θσs )ρ k θ (ρrθ σs ) −1 = ρ r θ (σsρ k θσ−s )ρ −r
θ = ρrθ (ρ(−1)s k
θ )ρ −r
θ = ρ(−1)s k
θ
because N is abelian. This proves Dn ∼ = Zn ×φ Z2 under the action H × N → N in (20). �
6.2.7 Corollary (Multiplication Law in Dn). If two elements of Dn are written in factored
form (19), the group operations take the form
(a) (ρk θσℓ ) · (ρr θσs ) = ρ k+(−1)ℓr θ σℓ+s (b) (ρk θσℓ ) −1 = ρ −(−1)ℓk θ σ−ℓ 16