Try
A triangle has rotations r=(1 2 3) and a reflection s=(1 2). Permutations act right to left. Every symmetry can be written uniquely as rⁱsʲ, with i taken modulo 3 and j modulo 2. The reflection reverses rotation: srs⁻¹=r⁻¹. Thus
(rⁱsʲ)(rᵏsˡ)=r^(i+(−1)ʲk)s^(j+l).
The first coordinate does not simply add i+k: it depends on the second coordinate j. For example (r s)r=s, whereas r(r s)=r²s. Verify each result by the displayed rule or by following the three labels.
This interaction is the reason S₃ is not the direct product C₃×C₂. A semidirect product records the interaction as an action by automorphisms. We will construct its operation, prove every group axiom, and explain when a group already in front of us has this form.
Predict
An automorphism of a group N is a bijective homomorphism N→N. Automorphisms form a group Aut(N) under composition. Their composition and inverse preserve multiplication, and the identity map is their identity element.
For Cₙ generated by a, predict which choices a↦aᵘ can define an automorphism. Sending a generator to the identity cannot work unless the group is trivial. Sending it to another generator should work; we will prove precisely that u must be coprime to n.
An action of H by automorphisms of N is a homomorphism α:H→Aut(N). Write αh for the automorphism corresponding to h; the convention is α(hk)=αh∘αk. Choosing one automorphism for each element is not enough. The choices must respect the multiplication of H. Predict what goes wrong if a generator of order three is assigned an automorphism whose cube is not the identity.
Experiment
Represent S₃ by pairs (i,j) in C₃×C₂, but use the interacting operation (i,j)(k,l)=(i+(−1)ʲk,j+l). Reduce coordinates modulo 3 and 2. Fill in the last row before checking it.
| First pair | Second pair | Action on k: (−1)ʲk modulo 3 | Add i to that result | Product |
|---|
| (1,0) | (1,1) | 1 | 1+1 | (2,1) |
| (1,1) | (1,0) | −1≡2 | 1+2≡0 | (0,1) |
| (2,1) | (1,1) | −1≡2 | 2+2≡1 | (1,0) |
| (1,1) | (1,1) | −1≡2 | 1+2≡0 | (0,0) |
Now deliberately use an incompatible rule. On pairs in C₅×C₃, choose representatives j=0,1,2 and propose (i,j)(k,l)=(i+2ʲk,j+l), reducing the result modulo 5 and 3. Multiplication by 2 is an automorphism of C₅, but its cube is multiplication by 8≡3, not the identity. Test associativity on x=(0,2), y=(0,1), z=(1,0).
Check the failed action
The product xy is (0,0), so (xy)z=(1,0). But yz=(2,1), and x(yz)=(4·2,0)=(3,0). The results differ. The rule is a well-defined operation after choosing residue representatives, but it is not associative. The problem is the incompatible wrap from exponent 3 to exponent 0, not a failure of multiplication by 2 to be invertible on C₅.
Understand
Automorphisms of a cyclic group
Let n≥2 and Cₙ=⟨a⟩. A homomorphism from Cₙ to itself is determined by a’s image. Any choice a↦aᵘ defines a homomorphism φᵤ(aᵏ)=aᵘᵏ: changing k by a multiple of n does not change the result, and products add exponents. Its image is generated by aᵘ, which has order n/gcd(n,u). Hence φᵤ is surjective, and therefore bijective on the finite set, exactly when gcd(n,u)=1.
The same criterion characterizes the units in ℤ/nℤ. If gcd(n,u)=1, a Bézout identity uv+nw=1 provides an inverse residue v. Conversely uv≡1 modulo n forces every common divisor of u and n to divide 1. Composition satisfies φᵤ∘φᵥ=φᵤᵥ, and equal homomorphisms give the same residue by evaluating at a. Thus Aut(Cₙ) is isomorphic to the multiplicative group of units modulo n. For n=1 the group and its automorphism group are both trivial; we can handle that case directly without a convention about units in the zero ring.
For example Aut(C₅) has multipliers 1,2,3,4. Inversion, multiplier −1, has square equal to 1 and defines an action of C₂. But multiplier 2 does not define an action of C₃. In general a proposed action of Cₘ with generator sent to φ exists exactly when φᵐ is the identity: define α(j)=φʲ and check independence of the exponent representative.
The external construction, with all axioms
Let N,H be groups and let α:H→Aut(N) be a homomorphism. No finiteness or commutativity is assumed here. Define N⋊αH to have underlying set N×H and multiplication
(n,h)(m,k)=(n αh(m),hk).
Closure follows because αh(m)∈N. To check associativity, introduce a third pair (q,l). The first coordinate of ((n,h)(m,k))(q,l) is n αh(m) α(hk)(q). The first coordinate of (n,h)((m,k)(q,l)) is n αh(m αk(q)), which equals n αh(m) αh(αk(q)) because αh is a homomorphism. The action law makes these equal. The second coordinates agree by associativity in H. We have used both the automorphism property and the homomorphism H→Aut(N).
Write eN and eH for the two identities. The identity pair is (eN,eH), since α(eH) is the identity map and automorphisms preserve eN. The inverse is
(n,h)⁻¹=(α(h⁻¹)(n⁻¹),h⁻¹).
Multiplying in the given order gives first coordinate n αh(α(h⁻¹)(n⁻¹))=eN. Multiplying in the reverse order gives α(h⁻¹)(n⁻¹) α(h⁻¹)(n)=eN. Both second coordinates are eH. This proves the two-sided inverse and finishes the group axioms.
The two subgroups and the quotient
The maps n↦(n,eH) and h↦(eN,h) are injective homomorphisms. Denote their images by N₀ and H₀. Their intersection is the identity pair, and each pair uniquely factors as (n,eH)(eN,h). The projection π(n,h)=h is an onto homomorphism with kernel N₀, so N₀ is normal and the quotient (N⋊αH)/N₀ is isomorphic to H. The map h↦(eN,h) is a homomorphic section of π.
Conjugation by (eN,h) sends (n,eH) to (αh(n),eH), recovering the chosen action. More generally, conjugation by (m,h) sends it to (m αh(n)m⁻¹,eH). The subgroup H₀ need not be normal. If it were normal, the commutator of an element of N₀ with one of H₀ would lie in both normal subgroups and therefore be the identity. The displayed conjugation would then force αh(n)=n for all h,n. Conversely, a trivial action gives the ordinary direct product and H₀ is normal. Thus this specified complementary subgroup is normal exactly when the action is trivial.
If N or H is trivial the formulas still work and give the other group. When both are finite, the order is |N||H| because the underlying set is their Cartesian product. This counting statement is not a proof of associativity or an isomorphism.
Recover the construction inside a group
Suppose N and H are subgroups of G, N is normal in G, N∩H is trivial, and NH=G. Define αh(n)=hnh⁻¹. Normality keeps this in N; conjugation and its inverse are homomorphisms, and conjugation by hk is conjugation by h after conjugation by k. Thus α:H→Aut(N) is a valid action.
The map Φ:N⋊αH→G given by Φ(n,h)=nh is a homomorphism because
(nh)(mk)=n(hmh⁻¹)hk.
It is onto by NH=G. If nh=mk, then m⁻¹n=kh⁻¹ belongs to N∩H, so n=m and h=k. This proves injectivity and unique factorization. Hence Φ is an isomorphism. Every hypothesis has a role: normality defines the action, NH=G gives existence, and the intersection condition gives uniqueness.
In S₃, N=⟨(1 2 3)⟩ is normal of index two and H=⟨(1 2)⟩ has trivial intersection with it. Their six distinct products exhaust S₃, and conjugation is inversion. Thus the pair construction in the table really is S₃. For the dihedral group Dₙ of order 2n, with n≥3, the same proof uses rotations r and a reflection s to identify Dₙ≅Cₙ⋊C₂ with inversion action. This reuses the geometric existence and normal forms in dihedral symmetry. The pair construction also makes sense at n=2, when inversion is trivial and it gives C₂×C₂; no claim about a nondegenerate two-sided polygon is needed.
Which changes of labels preserve the action?
Suppose β:N→N′ and γ:H→H′ are isomorphisms, with actions α and α′. The coordinate map (n,h)↦(β(n),γ(h)) is an isomorphism of the semidirect products provided
β(αh(n))=α′(γ(h))(β(n)) for every h,n.
Indeed, this equation makes the first coordinates of the two product calculations agree, while γ preserves the second coordinates. The map is bijective because β and γ are. Conversely, if this coordinate map is a homomorphism, apply it to conjugation by (eN,h) to recover exactly the displayed compatibility. This characterizes isomorphisms of this prescribed coordinate form. It does not assert that every abstract isomorphism must preserve the particular factors we selected.
Apply
Change the invalid C₅ example to C₇×C₃ with the same multiplier 2. Check that it now defines a group. Compute (3,1)(5,2), the reverse product, and the inverse of (3,1). State precisely where the earlier associativity objection has disappeared.
Check the changed action and calculations
Multiplication by 2 is invertible modulo 7, and 2³=8≡1, so it defines a C₃-action. The products are (3+2·5,0)=(6,0) and (5+4·3,0)=(3,0), proving noncommutativity. The inverse formula gives (−2⁻¹·3,−1)=(2,2), since 2⁻¹=4 modulo 7. Check both products: 3+2·2≡0 and 2+4·3≡0. Exponents differing by three now give the same automorphism, so the action law survives reduction of the second coordinate.
Finally, replace inversion on a cyclic group by inversion on an arbitrary group N. Is it always an automorphism? Compute (nm)⁻¹=m⁻¹n⁻¹ and compare it with n⁻¹m⁻¹. Inversion is a homomorphism exactly when N is abelian. A familiar action formula cannot be transferred to a new group without checking its hypotheses.
Continue
Return to cyclic groups and orders for the order-of-a-power formula, direct products for the commuting special case, and normal subgroups and quotients for the quotient construction.
Continue to split extensions and groups of order pq. A known normal subgroup and quotient do not automatically supply a complementary subgroup. The next lesson proves the splitting criterion and applies it to a complete finite-order classification.
