LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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A direct product of two finite cyclic groups is cyclic if and only if their orders are coprime

Statement

If $C_{m}$ and $C_{n}$ are finite cyclic groups of orders $m, n \geq 1$ , then $C_{m} \times C_{n}$ is cyclic if and only if $\gcd (m, n) = 1$ .

Facts & Assumptions

Given: Cyclic groups $C_{m} = ⟨ g ⟩$ and $C_{n} = ⟨ h ⟩$ .

[L1]

The external direct product uses componentwise multiplication (The external direct product $G \times H$ with componentwise multiplication) and is a group ( $G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections).

[L2]

A power of an element of order $r$ is the identity exactly when its exponent is divisible by $r$ (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 $j = k$ ).

[L3]

$lcm (r, s)$ is the least positive common multiple of $r$ and $s$ (Common multiple, and the least common multiple $lcm (a, b)$ , taken to be $0$ when $a = 0$ or $b = 0$ ).

[L4]

For positive integers $m, n$ , $\gcd (m, n) lcm (m, n) = m n$ (Every common multiple of $a$ and $b$ is a multiple of $lcm (a, b)$ , and $\gcd (a, b) \cdot lcm (a, b) = ∣ a b ∣$ ).

Proof

technique · direct

1.1

For $(x, y) \in C_{m} \times C_{n}$ with coordinate orders $r$ and $s$ , [L1] and [L2] show that $(x, y)^{t} = (1, 1)$ exactly when both $r ∣ t$ and $s ∣ t$ ; its order is therefore $lcm (r, s)$ by [L3].

L1L2L3

2.1

If $\gcd (m, n) = 1$ , [L4] and step 1.1 give $ord (g, h) = lcm (m, n) = m n$ , so $(g, h)$ generates the product of order $m n$ .

step 1.1L4algebra

3.1

Conversely, if the product is cyclic, a generator $(x, y)$ has order $m n$ . Its coordinate orders divide $m$ and $n$ , so step 1.1 gives $m n = lcm (r, s) \leq lcm (m, n) \leq m n$ ; hence $lcm (m, n) = m n$ , and [L4] gives $\gcd (m, n) = 1$ .

step 1.1L2L3L4algebra∎

Depends on

Used by

A positive integer admits a primitive root exactly when it is 1, 2, 4, pᵏ, or 2pᵏ for an odd prime p Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 83 results over 22 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

William Stein, Elementary Number Theory, Lemma 2.5.7 (standard reference, not scraped)