Prove that $G$ has a element whose order is least common multiple of $m$ and $n$.

Question

Let $G$ be an abelian group and suppose that $G$ has elements of order $m$ and $n$ respectively. Prove that $G$ has an element of order $\mathrm{lcm}[m,n]$

Answer 1

If $m\mid n$ or $n\mid m$, there is nothing to be shown. Let $a,b$ have orders $m,n$ and let $d=\gcd(m,n)$. Then consider $x=a+b$. Then $mx=mb\ne 0$, $nx=na\ne0$ and $\frac{nm}dx = \frac nd mx =\frac nd mb=\frac md nb=0$ we conclude that the order of $x$ divides $\operatorname{lcm}(m,n)$, but neither $m$ nor $n$.