# 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 $\mathrm{gcd}\mathit{}\mathrm{(}m\mathrm{,}n\mathrm{)}\mathrm{=}\mathrm{1}$, 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 $\mathrm{gcd}(m,n)=1$, 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 |