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

1.1L1L4given

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

2.1step 1.1L2L3given

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

2.2step 1.1L2given

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 $ι (n) ∣ ι (k)$ .

3.1step 2.1step 2.2L3algebra

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.

4.1step 3.1L2L3∎

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

Depends on

Dependency tree · 2 levels

Sources

Sharifi, Abstract Algebra, direct products

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