20A05-OrderOfElementsInFiniteGroups.tex

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\pmtitle{order of elements in finite groups}
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\begin{document}
This article proves two elementary results regarding the orders of group elements in finite groups.

\begin{thm} 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 $mn=\lvert ab\rvert$.
\end{thm}
\textbf{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 $mn=k$.

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

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