TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Classification of groups of order $p q$ for primes $p < q$

Statement

Let $p < q$ be primes.

If $p ∤ (q - 1)$ , every group of order $p q$ is cyclic.
If $p ∣ (q - 1)$ , there are exactly two isomorphism classes of groups of order $p q$ : the cyclic group $C_{p q}$ and one nonabelian semidirect product $C_{q} ⋊ C_{p}$ .

Facts & Assumptions

Given: Primes $p < q$ and a group $G$ of order $p q$ .

[L1]

The group $G$ has a normal subgroup $Q$ of order $q$ (If $p < q$ are primes and $∣ G ∣ = p q$ , then $G$ has a normal subgroup of order $q$ ).

[L2]

Cauchy's theorem supplies a subgroup $P$ of order $p$ , and every prime-order group is cyclic (Cauchy's theorem: if a prime $p$ divides $∣ G ∣$ , then $G$ has an element of order $p$ , A finite group of prime order is cyclic and every nonidentity element generates it).

[L3]

A normal factor and a complement with trivial intersection realise an external semidirect product ( Recognition theorem: $G = N H$ with $N ⊴ G$ , $N \cap H = 1$ exactly realises an external semidirect product).

[L4]

Nontrivial actions of $C_{p}$ on $C_{q}$ exist exactly when $p ∣ (q - 1)$ and give a unique semidirect-product type ( For primes $p < q$ , nontrivial actions of $C_{p}$ on $C_{q}$ exist exactly when $p ∣ (q - 1)$ and are unique up to automorphisms).

[L5]

A product with a normal subgroup is a subgroup; group powers obey the addition law; and $x^{m} = 1$ exactly when the order of $x$ divides $m$ (If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ , Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute, If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L6]

If two coprime integers divide an integer, then their product divides it (If $\gcd (a, b) = 1$ and $a ∣ b c$ then $a ∣ c$ ; and if $a ∣ c$ , $b ∣ c$ and $\gcd (a, b) = 1$ then $a b ∣ c$ ).

[L7]

The order of a subgroup divides the order of a finite group (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L8]

Finite cyclic groups are determined up to isomorphism by their order (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

Proof

technique · cases

1.1L1L2L5L7algebra

Choose $Q$ and $P$ as in [L1] and [L2]. By [L7], the order of $Q \cap P$ divides both primes, so the intersection is trivial. Since $Q$ is normal, $Q P$ is a subgroup by [L5]. Its cosets $x P$ for $x \in Q$ are distinct because $x P = x^{'} P$ would give $x^{' - 1} x \in Q \cap P$ , and each has $p$ elements. Thus $∣ Q P ∣ = p q = ∣ G ∣$ , so $G = Q P$ .

2.1step 1.1L2L3

Both $Q$ and $P$ are cyclic by [L2], and [L3] gives $G ≅ C_{q} ⋊ C_{p}$ .

3.1step 2.1L5L6

[assume-case first] Suppose the action is trivial. Let $x, y$ generate the commuting factors of orders $q, p$ . If $(x y)^{m} = 1$ , then $x^{m} = y^{- m}$ belongs to their trivial intersection, so $q ∣ m$ and $p ∣ m$ by [L5]. Since $p, q$ are coprime, [L6] gives $p q ∣ m$ . Hence $x y$ has order $p q$ , and $G$ is cyclic.

3.2step 2.1L4

[assume-case second] Suppose the action is nontrivial. Then [L4] says that this is possible exactly when $p ∣ (q - 1)$ and that all such products are isomorphic. The product is nonabelian because some element of $P$ acts nontrivially on $Q$ .

4.1step 3.1step 3.2L4L8∎

[cases-exhaustive] If $p ∤ (q - 1)$ only step 3.1 occurs. If $p ∣ (q - 1)$ , steps 3.1 and 3.2 give two types, distinguished by commutativity; [L8] gives uniqueness of the cyclic type and [L4] gives uniqueness of the nonabelian type.

Depends on

Used by

Dependency tree · two levels

85 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Group Theory (standard reference, not scraped)

Classification of groups of order pqpq for primes p<qp<q