order of elements in finite groups

This article proves two elementary results regarding the orders of group elements in finite groups.

Theorem 1

Let $G$ be a finite group, and let $a\mathrm{\in }G$ and $b\mathrm{\in }G$ be elements of $G$ that commute with each other. Let $m\mathrm{=}\mathrm{|}a\mathrm{|}$, $n\mathrm{=}\mathrm{|}b\mathrm{|}$. If $\gcd \mathit{}\mathrm{(}m\mathrm{,}n\mathrm{)}\mathrm{=}\mathrm{1}$, then $m\mathit{}n\mathrm{=}\mathrm{|}a\mathit{}b\mathrm{|}$.

Proof. Note first that

$${(ab)}^{mn}=a^{mn}{b}^{mn}={(a^m)}^n{(b^n)}^m=e_G$$

since $a$ and $b$ commute with each other. Thus $|ab|\le mn$. Now suppose $|ab|=k$. Then

$$e_G={(ab)}^k={(ab)}^{km}=a^{km}{b}^{km}=b^{km}$$

and thus $n|km$. But $\gcd (m,n)=1$, so $n|k$. Similarly, $m|k$ and thus $mn|k=|ab|$. These two results together imply that $mn=k$.

Theorem 2

Let $G$ be a finite abelian group. If $G$ contains elements of orders $m$ and $n$, then it contains an element of order $\mathrm{lcm}\mathit{}\mathrm{(}m\mathrm{,}n\mathrm{)}$.

Proof. Choose $a$ and $b$ of orders $m$ and $n$ respectively, and write

$$\mathrm{lcm}(m,n)=\prod p_i^{k_i}$$

where the $p_i$ are distinct primes. Thus for each $i$, either $p_i^{k_i}\mid m$ or $p_i^{k_i}\mid n$. Thus either $a^{m/p_i^{k_i}}$ or $b^{n/p_i^{k_i}}$ has order $p_i^{k_i}$. Let this element be $c_i$. Now, the orders of the $c_i$ are pairwise coprime by construction, so

$$\left|\prod c_i\right|=\prod \left|c_i\right|=\mathrm{lcm}(m,n)$$

and thus $\prod c_i$ is the required element.

Title	order of elements in finite groups
Canonical name	OrderOfElementsInFiniteGroups
Date of creation	2013-03-22 16:34:02
Last modified on	2013-03-22 16:34:02
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	5
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 20A05