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

Aut(Z3, +), which in turn are fully determined by where they send the nontrivial element a �= e
in Z2. There are just two possible semidirect products:
Group G (1) : in which Φ(a) = [1] 3 , the identity element in U3. We have Z2 = {e, a}
and the corresponding automorphisms of Z3 are
φ (1)
e
= φ(1)
a = τ [1] = idZ3
All elements in Z2 go to the identity map so we have the trivial action of Z2 on
Z3. The resulting group is the abelian direct product G (1) = Z3 × Z2 which, by the
Chinese Remainder Theorem, is isomorphic to the cyclic group Z6.
Group G (2) : in which Φ(a) = [2] 3 = [−1] 3 . Then the corresponding automorphisms
of Z3 are
φ (2)
e = idZ3
φ (2)
a = τ [−1]
Now φa is the inversion map J : [ℓ] → [−1][ℓ] = −[ℓ] on the additive group Z3.
The semidirect product in which Z2 acts by inversion of Zn is precisely the dihedral
group Dn, so G (2) ∼ = D3. [ This can be seen by observing that the elements
satisfy the identities
ρ = ([1] n , [0] 2 ) and σ = ([0] n , [1] 2 )
o(ρ) = n o(σ) = 2 σρσ −1 = ρ −1
characteristic of the dihedral groups.] There are no other possibilities for groups of
order 6. In particular, up to isomorphism the only abelian group of order 6 is Z6.
Incidentally, we have indirectly proved that the permutation group S3 is isomorphic to D3
because both are noncommutative and of order 6. �
6.2.18 Exercise. Verify that the multiplication law in G (2) = Zn ×φ Z2 is given by
([i], [j]) · ([k], [ℓ]) = ([i] + τ j
[−1] ([k]), [j + ℓ])
= ([i + (−1) j k] , [j + ℓ])
Use this to show that the elements ρ, σ above satisfy the “dihedral identities.” �
6.2.19 Exercise. Can you devise a bijective map ψ : S3 → D3 that effects the isomorphism
mentioned above?
Hint: What subgroups in S3 might play the roles of N = {e, ρθ, ρ 2 θ } and H = {e, σ} in D3? �
In the preceding discussion we ended up having to determine all possible homomorphisms
Φ : (Z2, +) → (U3, · ) ∼ = Aut(Z3, +)
That was easy because the groups were quite small, and cyclic. More generally, to determine the
possible semidirect products N ×φ H we must determine all homomorphisms Φ : H → Aut(N).
In attacking this question you should remember that a homomorphism is completely determined
once you know where it sends the generators of H. Another useful constraint is the fact that
• The order of the image group Φ(H) must divide the order of Aut(G).
• The order of the image group Φ(H) must also divide the order of H.
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