# order of elements in finite groups

###### Theorem 1

Let $G$ be a finite group, and let $a\in G$ and $b\in G$ be elements of $G$ that commute with each other. Let $m=\lvert a\rvert$, $n=\lvert b\rvert$. If $\gcd(m,n)=1$, then $mn=\lvert ab\rvert$.

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 $\lvert ab\rvert\leq mn$. Now suppose $\lvert ab\rvert=k$. 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 $mn|k=\lvert ab\rvert$. 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 $\mathrm{lcm}(m,n)$.

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

 $\left\lvert\prod c_{i}\right\rvert=\prod\left\lvert c_{i}\right\rvert=\mathrm{% lcm}(m,n)$

and thus $\prod c_{i}$ is the required element.

Title order of elements in finite groups OrderOfElementsInFiniteGroups 2013-03-22 16:34:02 2013-03-22 16:34:02 rm50 (10146) rm50 (10146) 5 rm50 (10146) Theorem msc 20A05