Suppose there exists a homomorphism from a finite group $G$ onto $G'$, and that $G'$ has an element of order $n$. How do we prove that $G$ has an element of order $n$?
(From Lagrange's Theorem, we know that the order of an element must divide the order of a group. And also homomorphisms have this property that the order of an element's image must divide the order of the element.)
Let $f:G\rightarrow G'$ the morphism. Let $y\in G'$ which has order $n$ and $x\in G$ with $p(x)=y$, the subgroup $H$ of $G$ generated by $x$ is a cyclic group whose order $pn$ in fact $|H|=|kerf_{\mid H}|n$, $x^p$ has order $n$.