If $G$ is a finite group, $H$ is a subgroup of $G$, and there is an element of $G/H$ of order $n$, then there is an element of $G$ of order $n$ 
$G$ is a finite group, $H$ is a normal subgroup. There is an element in $G/H$ of order $n$. Prove that there is an element in $G$ of order $n$.

Note: I have proved that there exists an element in $G$, say $b$, such that $b^n=e$. I'm stuck in proving that $n$ is the smallest element, i.e. $n$ is the order of $b$.
Please provide a new proof if possible!

If $gH$ has order $n$ we get $g^n\in H$. Let $t$ be the order of $g^n$. Then the order of $g$ is $nt$ (why?). In the cyclic subgroup generated by $g$ there exists an element of order $n$ (why?).
