TITLE: Order of element equal to least common multiple
QUESTION [13 upvotes]: Let $G$ be a group, and $a,b\in G$. Suppose $\operatorname{ord}(a)=m, \operatorname{ord}(b)=n$, and that $ab=ba$. Prove that there is an element $c\in G$ such that $\operatorname{ord}(c)=\operatorname{lcm}(m,n)$.

Let $A=\operatorname{lcm}(m,n)$. I consider $(ab)^A=a^Ab^A=1$, so the order of $ab$ divides $A$. What can we do to find an element with order exactly $A$?

REPLY [4 votes]: Let $m=\prod_{i=1}^t p_i^{m_i}$ and $n=\prod_{i=1}^t p_i^{n_i}$ where $p_1,p_2,\cdots,p_t$ are distinct primes and $m_i,n_i \geq 0$.
Furthermore, may assume that $m_i < n_i$ if $1 \leq i \leq l$ and $m_i \geq n_i$ if $l+1 \leq i \leq t$.
Let $c=\prod_{i=1}^l p_i^{m_i},d=\prod_{i=l+1}^t p_i^{n_i}$.
We can conclude that $o(a^c)=\prod_{i=l+1}^t p_i^{m_i}$ and $o(b^d)=\prod_{i=1}^l p_i^{n_i}$, clearly that $(o(a^c),o(b^d))=1$ and $a^cb^d=b^da^c$. Hence the order of $a^cb^d$ is  $\prod_{i=1}^l p_i^{n_i} \cdot \prod_{i=l+1}^t p_i^{m_i}=[m,n]$, which completes the proof.
Remark: The method above can also be used to prove the existence of an element in $G$ with order $(m,n)$.