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

Statement

Let $\iota:\mathbb N\to\mathbb Z$ be the canonical embedding. If $g\in G$ and $h\in H$ have finite orders $m,n\ge1$ , then in the external direct product

$\iota(\operatorname{ord}(g,h))=\operatorname{lcm}(\iota(m),\iota(n)).$

Facts & Assumptions

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

[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 $\iota(r)\mid\iota(k)$ ; equivalently, $r$ is the least positive natural exponent taking it to the identity (The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity, If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ 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 $L=\operatorname{lcm}(\iota(m),\iota(n))$ is a positive common multiple of $\iota(m)$ and $\iota(n)$ , and it divides every common multiple. Thus $L=\iota(\ell)$ for a unique natural $\ell\ge1$ (Common multiple, and the least common multiple $\operatorname{lcm}(a,b)$ , taken to be $0$ when $a = 0$ or $b = 0$ , Every common multiple of $a$ and $b$ is a multiple of $\operatorname{lcm}(a,b)$ , and $\gcd(a,b) \cdot \operatorname{lcm}(a,b) = |ab|$ , 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 $\ell$ be the natural from [L3]. Since $\iota(m)\mid\iota(\ell)$ and $\iota(n)\mid\iota(\ell)$ , [L2] and step 1.1 give $(g,h)^\ell=(e_G,e_H)$ .

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 $\iota(m)\mid\iota(k)$ and $\iota(n)\mid\iota(k)$ .

step 1.1L2given

3.1

By [L3], the two divisibilities of step 2.2 imply $\iota(\ell)\mid\iota(k)$ . As $\ell,k\ge1$ , this forces $\ell\le k$ : an integer quotient $q$ with $\iota(k)=q\iota(\ell)$ is positive and hence at least $1$ . Thus $\ell$ is the least positive exponent sending $(g,h)$ to the identity.

step 2.1step 2.2L3algebra

4.1

The definition of element order gives $\operatorname{ord}(g,h)=\ell$ . Applying $\iota$ and using [L3] gives the displayed equality.

step 3.1L2L3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 87 results over 24 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

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

If ggg and hhh have finite orders mmm and nnn, then ι(ord⁡(g,h))=lcm⁡(ι(m),ι(n))\iota(\operatorname{ord}(g,h))=\operatorname{lcm}(\iota(m),\iota(n))ι(ord(g,h))=lcm(ι(m),ι(n)) in G×HG\times HG×H