Decide when an extension splits

Try

Reduction modulo two gives a surjective homomorphism π:C₄→C₂, where Cₙ denotes the additive integers modulo n. Can we choose a representative for each output so that the representative choice itself preserves addition? Sending [0] to [0] and [1] to [1] chooses representatives, but fails: [1]+[1]=[0] in C₂ while [1]+[1]=[2] in C₄.

Try the other possible lift of [1], namely [3]. It also has order four, so doubling it is [2], not zero. Thus neither possible choice gives a homomorphism. A choice of representatives is not automatically a section in the algebraic sense we need.

Predict

Let N be a normal subgroup of any group G, let Q=G/N, and let π:G→Q be the quotient map. A homomorphic section is a homomorphism s:Q→G with π∘s equal to the identity on Q. We say the extension splits when such a section exists. A complement to N is a subgroup H≤G with N∩H=❴e❵ and NH=G. Normality of H is not required.

Predict how these definitions connect. A section chooses one lift of every coset compatibly with multiplication; a complement contains exactly one such lift per coset. Review semidirect products for the action construction. Splitting will produce an action, but does not force that action to be trivial or the result to be a direct product.

Experiment

Construct a group of order 21 with elements (a,i), where a is a residue modulo 7 and i a residue modulo 3. Let the second coordinate act on the first by multiplication by 2ⁱ modulo 7. This is well-defined because 2³≡1 modulo 7. The semidirect multiplication is

(a,i)(b,j) = (a + 2ⁱb, i+j),

with each coordinate reduced in its own modulus. Predict the products in the table; the first two demonstrate that order matters.

The subgroup N consists of all (a,0); the subgroup H consists of (0,i). Projection onto i has the homomorphic section i↦(0,i). Every element is uniquely (a,0)(0,i), and the two subgroups intersect only at identity. This extension splits but is nonabelian, as the first two products show. In contrast, the reduction C₄→C₂ in Try has no section and therefore will have no complement.

Understand

Sections and complements are equivalent

Suppose s is a homomorphic section. It is injective, because s(q)=s(q′) implies q=πs(q)=πs(q′)=q′. Its image H is a subgroup. If h=s(q) lies in N, then q=π(h)=e, so h=e. For any g∈G, put h=s(π(g)). Then π(gh⁻¹)=e, hence gh⁻¹∈N, so g∈NH. This constructs a complement.

Conversely, suppose H is a complement. The restriction π|H has kernel N∩H=❴e❵. It is onto Q because every g=nh has π(g)=π(h). Its inverse s:Q→H⊆G is a homomorphism: the unique lifts in H of q and q′ multiply to the unique lift of qq′. It is a section. This proof does not use finiteness.

A complement determines the action h·n=hnh⁻¹ on N. Normality keeps this inside N, and conjugation preserves products and composes correctly. The map N⋊H→G, (n,h)↦nh, is a homomorphism because nhn′h′=n(hn′h⁻¹)hh′. It is onto by NH=G. If nh=n′h′, then n′⁻¹n=h′h⁻¹ lies in N∩H, giving n=n′ and h=h′. Thus it is an isomorphism. This is the splitting mechanism, not a cardinality-only argument.

For C₄→C₂, the kernel is ❴[0],[2]❵. The Try calculation rules out a section. Equivalently the only subgroup of order two in C₄ is this kernel itself, so it cannot be a complement. Although every fiber has two elements, equal fiber sizes do not create a splitting.

Classify every group of order pq

Now p and q are primes with p<q, and |G|=pq. By Sylow theory, the number of q-Sylow subgroups divides p and is 1 modulo q. Its only possible value is 1: the other divisor p is strictly between 1 and q. Thus there is a normal subgroup N of order q, cyclic because its order is prime.

A p-Sylow subgroup H exists and is cyclic of order p. Its intersection with N is trivial by Lagrange. The products nh are distinct for distinct pairs, since equality puts a difference in the intersection. Thus NH has pq elements and is G. Every group of order pq is therefore a split extension C_q⋊C_p. This conclusion used the prime-order and distinct-prime hypotheses; it does not contradict the C₄ example.

Choose generators x of N and y of H. Conjugation has the form yxy⁻¹=xᵘ, with u a nonzero residue modulo q, and yᵖ=e forces uᵖ≡1 modulo q. Conversely any such u defines the action i↦(a↦uⁱa) and hence a group on F_q×C_p. The semidirect-product theorem proves the group axioms. The map (a,i)↦xᵃyⁱ from this constructed group to G preserves the multiplication and is bijective by the unique normal forms above.

Every automorphism of C_q sends x to xᵘ for a unique nonzero u, and every such image defines an automorphism. Composition multiplies the residues, so Aut(C_q)≅F_q×, the multiplicative group of the field F_q. This group is cyclic of order q−1 by the proved finite-field multiplicative cyclicity theorem. We need that theorem, not merely the cardinality of the automorphism group.

If u=1, the factors commute, giving C_q×C_p≅C_pq; for example the pair of generators has order pq. A nontrivial action has u of order p, because its order divides the prime p. In a cyclic group of order q−1, such an element exists exactly when p divides q−1. If p does not divide q−1, only the cyclic group occurs.

Why all nontrivial actions give one isomorphism type

Assume p divides q−1. The cyclic group F_q× has a unique subgroup of order p. Any two elements u,v of order p generate this same subgroup, so u=vᵏ for some k not divisible by p. Compare the groups with product parameters u and v. The map

(a,i) ↦ (a,ki)

from the u group to the v group is bijective, since multiplication by k permutes C_p. Its compatibility with products is exactly v^(ki)=uⁱ. Thus every nontrivial action yields the same isomorphism type. Each such group is nonabelian, since (0,1)(1,0)=(u,1) differs from (1,0)(0,1)=(1,1). It cannot be isomorphic to the cyclic group. We have proved existence, exhaustion, uniqueness within each case, and a distinction between the cases.

Consequently there is one group of order pq up to isomorphism when p does not divide q−1, and exactly two when it does. This is a classification statement for distinct primes p<q, not a general rule for all composite orders.

Apply

For order 21, the nontrivial parameters modulo 7 are 2 and 4. Find an explicit isomorphism between the two resulting pair groups. Then change to order 55: use p=5,q=11, and check that u=3 has order five modulo 11. Compare it with v=4. Do not infer uniqueness merely because each action is nontrivial.

Check both changed-generator identifications

For order 21, 4²≡2 modulo 7, so (a,i)↦(a,2i) maps the parameter-2 group to the parameter-4 group. The target generator squared acts by 2, as required. These give the single nonabelian type, alongside C₂₁.

Modulo 11 the powers of 3 are 3,9,5,4,1, showing exact order five. Since 4⁴≡3 modulo 11, (a,i)↦(a,4i) identifies the parameter-3 group with the parameter-4 group. The nontrivial elements of the unique order-five subgroup are 3,9,5,4; every nontrivial action uses one of them and is isomorphic by the same generator-change argument. Therefore the two types of order 55 are C₅₅ and this nonabelian semidirect product. These conclusions include actual multiplication-preserving bijections, not just equal cardinalities.

As a further check, explain why order 35 has only the cyclic type, whereas order 39 has a nonabelian type. For 35, 5 does not divide 6. For 39, 3 divides 12, and multiplication by 3 modulo 13 has order three. In each case the normal q subgroup and a p complement must be established before applying the action argument.

Continue

Return to small-order classification to compare the earlier order-four and order-six arguments with this general proof. Keep four obligations visible: construct an example, prove every possible group fits the construction, identify equivalent actions, and distinguish the remaining isomorphism types.

A section is additional structure, not something supplied by a quotient map alone. A split extension can be nonabelian, and an abelian extension can fail to split. The saved tasks ask for the maps and hypotheses that separate these possibilities; a list of orders alone is not a proof.