TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03

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If $g$ and $h$ have finite orders $m$ and $n$ , then $ι (ord (g, h)) = lcm (ι (m), ι (n))$ in $G \times H$

Statement

Let $ι : N \to Z$ be the canonical embedding. If $g \in G$ and $h \in H$ have finite orders $m, n \geq 1$ , then in the external direct product

$ι (ord (g, h)) = lcm (ι (m), ι (n)) .$

Facts & Assumptions

Given: Groups $G, H$ , elements $g \in G, h \in H$ , and positive natural numbers $m, n$ with $ord (g) = m$ and $ord (h) = n$ .

[L1]

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

[L2]

If an element has finite order $r$ , then for every natural $k$ its $k$ th power is the identity exactly when $ι (r) ∣ ι (k)$ ; equivalently, $r$ is the least positive natural exponent taking it to the identity (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, 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]

For positive $m, n$ , the integer $L = lcm (ι (m), ι (n))$ is a positive common multiple of $ι (m)$ and $ι (n)$ , and it divides every common multiple. Thus $L = ι (ℓ)$ for a unique natural $ℓ \geq 1$ (Common multiple, and the least common multiple $lcm (a, b)$ , taken to be $0$ when $a = 0$ or $b = 0$ , Every common multiple of $a$ and $b$ is a multiple of $lcm (a, b)$ , and $\gcd (a, b) \cdot lcm (a, b) = ∣ a b ∣$ , The naturals embed in the integers).

[L4]

Induction is valid for natural-number powers (The principle of mathematical induction).

Proof

technique · direct

1.1

For every natural $k$ , $(g, h)^{k} = (g^{k}, h^{k})$ : it holds at $k = 0$ , and the successor step follows by componentwise multiplication.

L1L4given

2.1

Let $ℓ$ be the natural from [L3]. Since $ι (m) ∣ ι (ℓ)$ and $ι (n) ∣ ι (ℓ)$ , [L2] and step 1.1 give $(g, h)^{ℓ} = (e_{G}, e_{H})$ .

step 1.1L2L3given

2.2

If $(g, h)^{k} = (e_{G}, e_{H})$ for a positive natural $k$ , then step 1.1 gives $g^{k} = e_{G}$ and $h^{k} = e_{H}$ . Hence $ι (m) ∣ ι (k)$ and $ι (n) ∣ ι (k)$ .

step 1.1L2given

3.1

By [L3], the two divisibilities of step 2.2 imply $ι (ℓ) ∣ ι (k)$ . As $ℓ, k \geq 1$ , this forces $ℓ \leq k$ : an integer quotient $q$ with $ι (k) = q ι (ℓ)$ is positive and hence at least $1$ . Thus $ℓ$ is the least positive exponent sending $(g, h)$ to the identity.

step 2.1step 2.2L3algebra

4.1

The definition of element order gives $ord (g, h) = ℓ$ . Applying $ι$ and using [L3] gives the displayed equality.

step 3.1L2L3∎

Depends on

Used by

Every abelian group of order n is cyclic if and only if n is squarefree Corollary
Invariant factors determine the order and exponent of a finite abelian group Corollary
There are exactly two isomorphism classes of groups of order 105 Corollary
The cyclic group ℤ/4 is not isomorphic to ℤ/2×ℤ/2 Counterexample
Elementary divisors regroup uniquely into invariant factors Lemma
Carmichael's λ(n) is the maximum order of a unit modulo n Proposition

Dependency tree · two levels

46 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

Sharifi, Abstract Algebra, direct products (standard reference, not scraped)

If gg and hh have finite orders mm and nn, then ι(ord⁡(g,h))=lcm⁡(ι(m),ι(n))ι(ord(g,h))=lcm(ι(m),ι(n)) in G×HG×H