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

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

Statement

Let $p < q$ be primes. A nontrivial action $C_{p} \to Aut (C_{q})$ exists if and only if $p ∣ (q - 1)$ . When it exists, all nontrivial such actions give isomorphic semidirect products $C_{q} ⋊ C_{p}$ .

Facts & Assumptions

Given: Primes $p < q$ .

[L1]

$Aut (C_{q}) ≅ (Z / q)^{\times}$ ( $Aut (C_{n}) ≅ (Z / n Z)^{\times}$ ).

[L2]

For prime $q$ , $φ (q) = q - 1$ ( $φ (1) = 1$ , and $φ (p) = p - 1$ for every prime $p$ ), and $φ (n) = ∣ (Z / n)^{\times} ∣$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ); together these give $∣ (Z / q)^{\times} ∣ = q - 1$ .

[L3]

A prime divisor of a finite group order occurs as the order of an element (Cauchy's theorem: if a prime $p$ divides $∣ G ∣$ , then $G$ has an element of order $p$ ).

[L4]

The congruence $x^{p} \equiv 1 (m o d q)$ has at most $p$ solution classes ( For prime $q$ and $d \geq 1$ , the congruence $x^{d} \equiv 1 (m o d q)$ has at most $d$ residue-class solutions).

[L5]

A homomorphism is injective exactly when its kernel is trivial (A group homomorphism is injective if and only if its kernel is trivial).

[L6]

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 $G$ ).

[L7]

Compatible changes by automorphisms of the two factors give isomorphic semidirect products (Actions changed by automorphisms of the kernel and complement give isomorphic semidirect products).

[L8]

Every nonidentity element of a group of prime order generates that group (A finite group of prime order is cyclic and every nonidentity element generates it).

Proof

technique · iff

1.1L5L6algebra

[forward] A nontrivial homomorphism $C_{p} \to Aut (C_{q})$ has a proper kernel. By [L6], its kernel has order dividing $p$ , so it is trivial. The homomorphism is therefore injective by [L5] and its image has order $p$ .

1.2L1L2L3

[reverse] If $p ∣ (q - 1)$ , [L2] and [L3] give an element of order $p$ in $(Z / q)^{\times}$ . Sending a generator of $C_{p}$ to this element defines a nontrivial action.

1.3L4L8algebra

Suppose $A$ and $B$ are two distinct subgroups of order $p$ in $(Z / q)^{\times}$ . Their intersection is trivial, because [L8] says a nonidentity element in the intersection would generate each prime-order group. Thus $A \cup B$ contains $2 p - 1 > p$ distinct roots of $x^{p} = 1$ , contradicting [L4]. There is consequently a unique subgroup of order $p$ .

2.1step 1.1L1L2L6

By [L1], the target is $(Z / q)^{\times}$ , of order $q - 1$ by [L2]. Its image has order $p$ by step 1.1, so [L6] gives $p ∣ (q - 1)$ .

3.1step 1.1step 1.3L7algebra∎

Any two nontrivial actions $α, β$ are isomorphisms from $C_{p}$ onto this unique subgroup. Hence $v = β^{- 1} α$ is an automorphism of $C_{p}$ and $β_{v (h)} = α_{h}$ for every $h$ . Taking the automorphism of $C_{q}$ to be the identity, the compatibility in [L7] gives isomorphic semidirect products.

Depends on

Used by

Classification of groups of order pq for primes p<q Theorem

Dependency tree · two levels

51 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)

For primes p<qp<q, nontrivial actions of CpCp​ on CqCq​ exist exactly when p∣(q−1)p∣(q−1) and are unique up to automorphisms